inequalities
induction
logarithms
calculus
derivative
inequalities proposed

If $a_{1}$, $a_{2}$, $\ldots$, $a_{n}\geq 0$ are such that \[a_{1}^{2}+\cdots+a_{n}^{2}=1,\]
 then find the maximum value of the product $(1-a_{1})\cdots (1-a_{n})$.

[quote="N.T.TUAN"]If $n=1$ then it is $0$. Now if $n>1$.

I assume that we solved following: If $a_{1},a_{2}\geq 0$ and $a_{1}^{2}+a_{2}^{2}\leq 1$. Find maximum of $(1-a_{1})(1-a_{2})$.

I think it is easy and I called value that is $M$.

By induction on $n$ we can prove : If $n>1$ and $a_{1},...,a_{n}\geq 0$ such that $a_{1}^{2}+...+a_{n}^{2}\leq 1$ then $(1-a_{1})...(1-a_{n})\leq M$.

Answer : $M$ if $n>1$, $0$ if $n=1$.

Am I true?  :maybe:[/quote]

I think you're wrong because the value of $M$ you will get is 1 that holds when you take every $a_{i}=0$. So the sum of the squares satisfy your condition, but clearly $M<1$ so that's not the solution.

[quote="Megus"]Alternatively we can play around with Karamata's and $f(x)=\ln (1-\sqrt{x})$ which is concave on $[0,\frac{1}{4}]$ and convex on $[\frac{1}{4},1)$.[/quote]

Can you show in detail?

the final and correct answer is that the maximum realizes for x3=x4=...=xn=0 and x1=x1=sqrt2 /2

Just want to ask who is the author of this very nice problem? Is he in this forum?

I have one, but it's very, very ugly. It took me six pages in the contest.
I think that a solution with Karamata may be more acceptable :)

[b][color=blue]Lemma.[/color][/b] Let $x,y \in [0,1]$ such that $x^{2}+y^{2}= c \in [0,1]$. If $c \in \left[ 0, \frac12 \right]$, then the maximum of $(1-x)(1-y)$ is reached when one of the variables is $0$. Else, the maximum is reached either when $x=y$ or when $xy = 0$.

[i]Proof.[/i] It's straightforward using Lagrange Multipliers, but I don't think that anyone is too eager to see such a proof.

Now, can you complete it, silouan? The method I used in the last part is quite common on Mathlinks.

[quote="hungkhtn"][quote="silouan"]Could someone post a solution?[/quote]

I will show it for you, silouan.

First, we realize that this problem is totally equivalent to the following one (by contradiction method)

If $x_{1},x_{2},...,x_{n}\in [0,1]$ such that $x_{1}x_{2}...x_{n}=(1-\frac{1}{\sqrt{2}})^{2}$ then
\[f(x_{1},x_{2},...,x_{n}) = (1-x_{1})^{2}+(1-x_{2})^{2}+...+(1-x_{n})^{2}\le 1 (1) \]
I choose this renewed form because it it easier to reason. Now, let's start with (1). Denote $k=1-\frac{1}{\sqrt{2}}$.

Remark. If $x,y\in [0,1]$ such that $x+y+xy \ge 1$ then
\[(1-x)^{2}+(1-y)^{2}\le (1-xy)^{2}. \]
Indeed, just notice that
\[(1-x)^{2}+(1-y)^{2}-(1-xy)^{2}=(x+y-1)^{2}-x^{2}y^{2}\]

\[=(x-1)(y-1)(x+y+xy-1) \le 0. \]
Returning to the initial inequality. Just notice that if $x_{1}\le x_{2}\le ...\le x_{n}$ then $x_{1}x_{2}x_{3}\ge k^{2}$ so $x_{2}x_{3}\ge k^{4/3}$, so
\[x_{2}+x_{3}+x_{2}+x_{3}\ge 2k^{2/3}+k^{4/3}= 1.07 \ge 1 \]
(oh... it is so approximating, greatly luck). According to the remake, we refer
\[f(x_{1},x_{2},...,x_{n}) \le f(x_{1},x_{2}x_{3},1,x_{4},...,x_{n}) \le f(x_{1},x_{2}x_{3}x_{4},1,1,x_{5},...,x_{n}) \le ... \le f(x_{1},x_{2}x_{3}...x_{n},1,1,...,1), \]
By this step, the problem returns to initial case $n=2$ (just for $2$ numbers). We have done.

Remark. It is such a very nice inequality, I must add it to my book, thank its author :lol:[/quote]
Very nice  :wink:  ! And very special , too !
My solution :
$n=2$ we need to find the maximum of :
$g(x,y)=(1-\frac{x}{\sqrt{x^{2}+y^{2}}})(1-\frac{y}{\sqrt{x^{2}+y^{2}}})$  $\forall x,y \in [0,+\infty)$

Put $y=1$ we have $g'(x,1)=\frac{(x-1)(x^{2}+1)^{2}-(x-1)(x+1)(x^{2}+1)^{\frac{3}{2}}}{(x^{2}+1)^{\frac{7}{2}}}$

Easy to show that $max g(x,1)=(1-\frac{1}{\sqrt{2}})^{2}$ with the equality if $x=1$ or $max g(x,y)=(1-\frac{1}{\sqrt{2}})^{2}$ with the equality if $x=y$

If $n\geq\3$ put $f(x_{1},x_{2},.....,x_{n})=(1-x_{1})(1-x_{2})..(1-x_{n})$
Notice that $x_{i}\geq\ 0 \leftrightarrow 1-x_{i}\leq\ 1$ we have :
$f(x_{1},x_{2},.....,x_{n}) \leq\ f(x_{1},x_{2},.....,0) \leq\ .... \leq\ f(x_{1},x_{2},0,0,.......,0)$ We return to $n=2$ which is easy !  :D

Thank you,

Your way is short, too, Tan. I never believe that you are only at 11th grade.

It is not elememtary, VASC (why you affirm that the maximum exists at $(x_{1},x_{2},...,x_{n})$). However, it is very natural and short.

[quote="Vasc"]I think you are wrong because $x_{1}^{2}+x_{2}^{2}\le 1$ and the inequality
$f(x_{1},x_{2},0,0,.......,0) \le (1-\frac 1{\sqrt{2}})^{2}$
doesn't hold (see $x_{1}=x_{2}=0$). :(
[/quote]
You are right , Vasc ! But in my solution , I think $f(x_{1},x_{2},0,0,0,.....,0)$ with $x_{i}$ satisfy $\sum x_{i}^{2}=1$ mean $x_{3}=x_{4}=.....=x_{n}=0$ ( so that we have $x_{1}^{2}+x_{2}^{2}=1$ ) and we can reutrn to case $n=2$ ! Am I wrong ?

[quote="hungkhtn"]
\]
 Just notice that if $x_{1}\le x_{2}\le ...\le x_{n}$ then $x_{1}x_{2}x_{3}\ge k^{2}$ so $x_{2}x_{3}\ge k^{4/3}$, so
\[
[/quote]

Why is this?

All right. I see.

But it is not very convenient to prove in the exam that $2k^{2/3}+k^{4/3}\geq 1$!

I understand. But it is not hard to check. Let see other solutions when you must count the derivative, so complicated! The fact that $2k^{2/3}+k^{4/3}~1.07~1$ also shows that this problem is very nice and sharp.

[quote="radio"]Hmm... actually how can it be proved?[/quote]

.. I think this is not a hard and interesting question. Calculating in inequalities is indispensable, and by a calculator, you can check it easily (I am not sure if Romania Exam allows calculator). However, if you still ask for this question, I will answer

\[2k^{2/3}+k^{4/3}\ge 1 \Leftrightarrow k^{2/3}\ge \sqrt{2}-1\]
\[\Leftrightarrow (\sqrt{2}-1)^{2}\ge \sqrt{2}(\sqrt{2}-1)^{3}\Leftrightarrow 1 \ge \sqrt{2}(\sqrt{2}-1) \Leftrightarrow \sqrt{2}\ge 1\]

It is obvious. But it is boring,

[quote="hungkhtn"][quote="silouan"]Could someone post a solution?[/quote]

I will show it for you, silouan.

First, we realize that this problem is totally equivalent to the following one (by contradiction method)

If $x_{1},x_{2},...,x_{n}\in [0,1]$ such that $x_{1}x_{2}...x_{n}=(1-\frac{1}{\sqrt{2}})^{2}$ then
\[f(x_{1},x_{2},...,x_{n}) = (1-x_{1})^{2}+(1-x_{2})^{2}+...+(1-x_{n})^{2}\le 1 (1) \]
I choose this renewed form because it it easier to reason. Now, let's start with (1). Denote $k=1-\frac{1}{\sqrt{2}}$.

Remark. If $x,y\in [0,1]$ such that $x+y+xy \ge 1$ then
\[(1-x)^{2}+(1-y)^{2}\le (1-xy)^{2}. \]
Indeed, just notice that
\[(1-x)^{2}+(1-y)^{2}-(1-xy)^{2}=(x+y-1)^{2}-x^{2}y^{2}\]

\[=(x-1)(y-1)(x+y+xy-1) \le 0. \]
Returning to the initial inequality. Just notice that if $x_{1}\le x_{2}\le ...\le x_{n}$ then $x_{1}x_{2}x_{3}\ge k^{2}$ so $x_{2}x_{3}\ge k^{4/3}$, so
\[x_{2}+x_{3}+x_{2}+x_{3}\ge 2k^{2/3}+k^{4/3}= 1.07 \ge 1 \]
(oh... it is so approximating, greatly luck). According to the remake, we refer
\[f(x_{1},x_{2},...,x_{n}) \le f(x_{1},x_{2}x_{3},1,x_{4},...,x_{n}) \le f(x_{1},x_{2}x_{3}x_{4},1,1,x_{5},...,x_{n}) \le ... \le f(x_{1},x_{2}x_{3}...x_{n},1,1,...,1), \]
By this step, the problem returns to initial case $n=2$ (just for $2$ numbers). We have done.

Remark. It is such a very nice inequality, I must add it to my book, thank its author :lol:[/quote]

can someone clarify why is this an equivalent problem, please?

It is because he applied lagrange multipliers incorrectly. Since $ x_i\ge 0$, we have to check the border of that inequality, or when $ x_i\equal{}0$, etc. This gives the correct answer.

I also believe that he solved the system incorrectly. I got that the max occurs when M variables are a, N variables are (1-a) and P variables are 0 where M+N+P=n


inequalities
algorithm
inequalities proposed

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1.$ Prove that
$ \frac {36}{a^2b \plus{} b^2c \plus{} c^2a} \plus{} \frac {1}{abc} \ge 343.$
When does equality hold?[/quote]

 :( I believe you have find spcial method to solve out this type.Could you give some detail here.

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1.$ Prove that
$ \frac {36}{a^2b \plus{} b^2c \plus{} c^2a} \plus{} \frac {1}{abc} \ge 343.$
When does equality hold?[/quote]
See also here: http://canhang2007.wordpress.com/2009/10/28/inequality-10-v-q-b-can/ to know in which cases equality occurs.

[quote="can_hang2007"]I am sorry for asking this. But it is just a little curiosity. Why do so many Chinese people use computer to solve inequalities? Is it an interesting work?[/quote]
A good question. can_hang2007,
People interested in computer based reasoning would be happy with the [i][b]algorithms[/b][/i] they create for verifying thousands inequalities by a single program, while others interested in pen-and-paper based deduction would be happy with the [i][b]flexible skills [/b][/i]to prove inequalities individually and heuristically.

Thank you very much for your answer, Guest. I hope you will create a good program to check inequalities. :-)

Don't see a clear proof yet. You all believe its well done?

The stronger inequality:
\begin{align*} a^2b+b^2c+c^2a \le  \frac{9261 r^2 -862r+73}{484} \end{align*}
still holds, equality occurs also at $a=b=c.$
The following is my idea:
\begin{align*} 2\left(a^2b+b^2c+c^2a \right) &= pq-3r+ \left(a-b \right) \left(a-c \right) \left(b-c \right) \\ & \le q-3r +  \left[\frac{77}{2} \,\frac{ \left(a-b \right)^2 \left(a-c \right)^2 \left(b-c \right)^2}{ 26 -99 q+189 r } + \frac{26 -99 q+189 r}{154} \right] \\ &\le  \frac{9261 r^2 -862r+73}{242} .\end{align*}
The last inequality is equivalent to: $$ \left(539q+3087r+234 \right) \left(22q-63r-5 \right)^2 \ge 0,$$ here, we used the identity $ \left(a-b \right)^2 \left(a-c \right)^2 \left(b-c \right)^2= q^2-4q^3+18qr-27r^2-4r .$

[quote=can_hang2007]Let $ a,b,c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1.$ Prove that
$ \frac {36}{a^2b \plus{} b^2c \plus{} c^2a} \plus{} \frac {1}{abc} \ge 343.$
When does equality hold?[/quote]
Does there exist a nice Cauchy-Schwarz/AM-GM proof ?


Let a sequence: $ a_1\equal{}a_2\equal{}12\sqrt2,a_{n\plus{}1}\equal{}\minus{}a_n\sqrt{a_{n\minus{}1}^2\plus{}1}\plus{}a_{n\minus{}1}\sqrt{a_n^2\plus{}1}\ \forall n\geq2$.
Prove that: $ \sqrt{2\plus{}2\sqrt{a_n^2\plus{}1}}\in N \ \forall n\geq1$.

[quote="Moubinool"]try argsh(x)[/quote]
More detail, please!

[quote="tdl"]Let a sequence: $ a_1 \equal{} a_2 \equal{} 12\sqrt2,a_{n \plus{} 1} \equal{} \minus{} a_n\sqrt {a_{n \minus{} 1}^2 \plus{} 1} \plus{} a_{n \minus{} 1}\sqrt {a_n^2 \plus{} 1}\ \forall n\geq2$.
Prove that: $ \sqrt {2 \plus{} 2\sqrt {a_n^2 \plus{} 1}}\in N \ \forall n\geq1$.[/quote]
Following Moubinool's hint :

Let $ u_n \equal{} \text{argsh}(a_n)$ such that $ a_n \equal{} \sinh(u_n)$

The equation becomes $ \sinh(u_{n \plus{} 1}) \equal{} \sinh(u_{n \minus{} 1} \minus{} u_n)$ and so : $ u_{n \plus{} 1} \equal{} u_{n \minus{} 1} \minus{} u_n$ with $ u_1 \equal{} u_2 \equal{} \text{argsh}(12\sqrt 2)$

The quantity $ b_n \equal{} \sqrt {2 \plus{} 2\sqrt {a_n^2 \plus{} 1}}$ becomes $ b_n \equal{} 2\cosh(\frac {u_n}2)$

Since $ a_1 \equal{} a_2 \equal{} \text{argsh}(12\sqrt 2)$, we get $ \cosh(a_1) \equal{} \cosh(a_2) \equal{} 17$ and so $ b_1 \equal{} b_2 \equal{} 6$

It's immediate to see that $ \frac {u_n}2 \equal{} k_n\frac {u_1}2$ for some $ k\in\mathbb Z$
And it's easy to see that $ 2\cosh(x)\in\mathbb N$ $ \implies$ $ 2\cosh(nx)\in\mathbb N$ (just raise $ e^x \plus{} e^{ \minus{} x}$ to power $ n$)

Hence the result


water passes through and between the endodermal cells into the vascular cylinder in the root

i think this statement is false what do you think?

it is not false. Water moves through endodermal cells in symplastic transport, and between cells in aopoplactic transport.
 :lol:

Wait a second...I believe water crosses the endodermis and enters the vascular cylinder through the symplastic route...It can't go through that apoplast because of the Casparian Strip. After is crosses the endodermis, though, I believe some water goes again into the apoplastic route.


What resources would people on this forum recommend for studying for the SAT II Physics? My background is Physics C, both parts. However, I have zero exposure to anything that is not Mech or E&M, and my school offers nothing on anything besides that...

Since you already have Physics C knowledge, learning topics other than Mechanics and E/M shouldn't be that hard for you. Get a good Physics B book (maybe Cliffs or Princeton Review) and a SAT II Physics book, since Physics B basically covers everything on the SAT II.

When are you taking it? I'm taking mines in June.

http://www.sparknotes.com/testprep/books/sat2/physics/

Everything you need is right there, for free. That's all I used (I hadn't taken a physics class of any kind), and I managed to get an 800.

[quote="Eugevir"]http://www.sparknotes.com/testprep/books/sat2/physics/[/quote]

What...the...heck...

You mean I just made a $15 donation to Barnes and Noble for no reason?

The material looks almost exactly the same as AP Physics B... I should probably take it before I forget everything ;)


$X,Y$  commute and are diagonalisable there exist a common basis in wich
there are similar to diagonals matrices.
Eingenvalues of
$X$ are $exp(2i \pi k/a)$
$Y$ are $exp(2i \pi q/b)$

Eingenvalues of $X+Y$ are $exp(2i \pi k/a)+exp(2i \pi q/b) \neq 0 $ since $a; \;b$ are

numere intregi impar, so $X+Y$ is inversabila

[quote="silviu"]Fie  n un numar natural ,n>1 ,a,b numere intregi impare si matricile :\[
X,Y \in M_n (C),a.i.:XY = YX,si:X^a  = Y^b  = I_n .
\]Demonstrati ca matricea:\[
X + Yinversabila
\][/quote]
X + Y divides X^ab + Y^ab = 2 I_n.


[quote="dieuhuynh"]Let $m,n$ be positive integers and $p$ be a prime number .Show that if $\frac{7^m+p.2^n}{7^m-p.2^n}$ is a integer then it is a prime number.[/quote]

$\frac{7^m+p.2^n}{7^m-p.2^n}=\frac{7^m-p.2^n+p.2^{n+1}}{7^m-p.2^n}=1+\frac{p.2^{n+1}}{7^m-p.2^n}$

Let $\frac{p.2^{n+1}}{7^m-p.2^n}=a$, $a \in N$

$p.2^{n+1}=a.7^m-ap.2^n$
$p.2^n(2+a)=a.7^m$

So we have case 1: $p=7$ or case 2: $a=py$.

Case 1:
$2^n(2+a)=a.7^{m-1}$

$a=\frac{2^n(2+a)}{7^{m-1}}$.
Since $(2^n,7^{m-1})=1$, so $\frac{2+a}{7^{m-1}}=x$, $x \in N$.

And also $x=\frac{a}{2^n}$.

From this we get:

$x.2^n=x.7^{m-1}-2$
$x(7^{m-1}-2^n)=2$

If $x=1$, checking $\mod 2$ gives us that equation has no solutions.
If $x=2$, checking $\mod 7$ also gives us that equation has no solutions.

Case 2:
$p.2^n(2+py)=py.7^m$
$2^n(2+py)=y.7^m$

This implies $y=t.2^n$, $t \in N$, so we have:

$2^n(2+pt.2^n)=t.2^n.7^m$
$2+pt.2^n=t.7^m$
$t(7^m-p.2^n)=2$

If $t=1$, checking $\mod 2$ gives us that equation has no solutions.
If $t=2$ then $7^m-p.2^n=1$, so denominator of given fraction must be $1$.

If we have $p>3$ then it is of the form $6k+1$ or $6k-1$. Put these into $7^m-p.2^n=1$ and check $\mod 6$. We see that equation has no solutions. Therefore $p \leq 3$.
If $p=2$ then checking $\mod 4$ gives us that equation has no solutions.
So the only possible value of $p$ is $3$.

Now we have the equation:
$7^m-3.2^n=1$

If $n=1$, then $m=1$ which gives the solution $(m,n,p)=(1,1,3)$.
If $n=2$, then $m$ is not integer.
If $n\geq3$ then checking $\mod 8$ gives us that $m=2b$, $b \in N$.

$7^{2b}-3.2^n=1$
$(7^b-1)(7^b+1)=3.2^n$

We have case 1:
$7^b-1=3.2^s$
$7^b+1=2^r$
where $s,r \in N$, and $r+s=n$.

$3.2^s+1=2^r-1$
$2^s(2^{r-s}-3)=2$
$2^{s-1}(2^{r-s}-3)=1$

From this we get $s=1, r=3$, so $n=4$. Then $m=2$ and we have another solution $(m,n,p)=(2,4,3)$.

In case 2:
$7^b-1=2^s$
$7^b+1=3.2^r$
where $s,r \in N$, and $r+s=n$.

$2^s+1=3.2^r-1$
$2^r(3-2^{s-r})=2$
$2^{r-1}(3-2^{s-r})=1$

We get that $r=1, s=2$, so $n=3$. But in this case $m$ is not an integer.

So we have two triples $(m,n,p)=(1,1,3), (2,4,3)$ when given fraction is integer.

With first triple fraction is equal $13$ and with second it is $97$. We see that they are both primes.


inequalities
number theory unsolved
number theory

ehh ... this is a stupid question I am sure :D  , but it came when I was solving a problem.
find all the triplets $(a,b,c)$ of positive integers such that $\frac{1}{a} +\frac{1}{b}+ \frac{1}{c}$ is a positive integer.

well, it must have been posted before...

a standard method is to use inequalities. the problem reduces in finding triplets of (a,b,c) such that abc divides ab+ac+bc which means that ab+ac+bc >= abc. Now you have that min(a,b,c) <= 3, otherwise abc >= 3bc, abc >= 3ac, abc >= 3ab. Adding the last three inequlities you take abc >= ab+ac+bc, a contradiction. Note that (a,b,c)=(3,3,3) is the only solution when min(a,b,c)=3. So, wlog suppose that a >= b >= c and c <= 2. Then 1/a+1/b+1/c <= 3/c(1). Now taking possibilities on the value of c(=1,2) and if needed on the value of 1/a+1/b+1/c using (1) we can easily find all such triplets.


We oll knaw thot same peaple canfuse o certoin poir af letters. Which poir wauld couse the funniest ond mast dromotic effect?
We zll know thzt some people confuse z certzin pzir of letters. Which pzir would czuse the funniest znd most drzmztic effect?
We all know that some people confuse a certain pair of letters. Which pair would cause the funniest and most dramatic effect?
We all know that some people focuse a fertain pair of letters. Which pair would fause the cunniest and most dramatic ecceft?
We all know that some people confuse a cedtain paid of letteds. Which paid woulr cause the funniest and most rdamatic effect?
[color=green]We all know that fome people consufe a certain pair of letterf. Which pair would caufe the sunnieft and moft dramatic essect?[/color] WINNAH!
We arr know that some peopre confuse a certain pail of rettels. Which pail would cause the funniest and most dlamatic effect?
We all know that some people coyfuse a certaiy pair of letters. Which pair would cause the fuyyiest ayd most dramatic effect?

We [b]oll [/b][b]knaw [/b][b]thot [/b][b]same [/b][b]peaple [/b][b]canfuse [/b][b]o [/b][b]certoin [/b][b]poir [/b][b]af [/b]letters. Which [b]poir [/b][b]wauld [/b][b]couse [/b]the funniest [b]ond [/b][b]mast [/b][b]dromotic [/b]effect? :)
We zll know thzt some people confuse z certzin pzir of letters. Which pzir would czuse the funniest znd most drzmztic effect?
We all know that some people confuse a certain pair of letters. Which pair would cause the funniest and most dramatic effect?
We all know that some people focuse a fertain pair of letters. Which pair would fause the cunniest and most dramatic ecceft?
We all know that some people confuse a cedtain paid of letteds. Which paid woulr cause the funniest and most rdamatic effect?
We all know that fome people consufe a certain pair of letterf. Which pair would caufe the sunnieft and moft dramatic essect?
We arr know that some peopre confuse a certain pail of rettels. Which pail would cause the funniest and most dlamatic effect?
We all know that some people coyfuse a certaiy pair of letters. Which pair would cause the fuyyiest ayd most dramatic effect?

One time I was going up in an airplane. My mom is Japanese, so she can confuse the letters L and R sometimes, so she said, "oh, _________, you're gonna fry!" She really meant to say fly...lol

F and S are funniest...example: this s*cks! replace s... :P


topology
real analysis
real analysis unsolved

assume that $X$ is metric space with no isolated point
a)prove that if $X$ is seprable then we can decompose $X$ to countable dense subset.
b)if we dont know that $X$ is sep or not is it true?(i do not know this part)

What does "decompose $X$ to countable dense subset" mean? :? Could you please clarify? I understand it to mean parittioning $X$ into countable dense subsets, but in this case the answer to the second question would trivially be "no", because $X$ may have no countable dense subsets.

i mean partitioning $X$ to countable,      dense subsets :)

[quote="sam-n"]b)if we dont know that $X$ is sep or not is it true?(i do not know this part)[/quote]
If $X$ is not separable how can we extract a countable dense subsets? :?

Unfortunately, this nice problem will be already solved when I wake up next morning :(

I'm probably just tired right now, but I think we can give counterexamples to the first question. :?

What if our space is uncountable, but there is a countable ball around a point, say $x_0$? Call this ball $B(x_0,r)$. Since the space is partitioned in countable subsets, we must have uncountably many such subsets. Since in $B$ we must have a point from each subset, it means that we must have uncountably many points in $B$, which is countable. This is a contradiction.

To give a concrete example, we can take our space to be the real set consisting of $[0,1]$ and the rationals in $[1,2]$.

What am I doing wrong here? :?

Indeed, you are wrong. You forgot that Sam-n likes to not mention some "unnecessary" conditions. I think, he omitted the fact that $X$ is a complete space.

Am I right, Sam-n?

But if you're right and sam-n did forget to mention a condition, then, after all reasonable definitions of being wrong, I was not, was I? :)

Anyway, let's assume this is indeed the case. Then, if $c$ is the continuum cardinality, every ball in $X$ has $c$ points. For each point $x\in X$, we can choose a decreasing sequence of balls $B_n(x,r_n(x)),\ \forall x, r_n(x)\to 0$ s.t. for each $x$ and each $n$ there is at least a point of $X$ in $B_n(x,r_n(x))\setminus B_{n+1}(x,r_{n+1}(x))$. Then $B_n(x,r_n(x))\setminus B_{n+1}(x,r_{n+1}(x))$ actually contains $c$ points. For each $n$, we can extract countably many $B_n$'s which cover $X$. For each of these, label the points in $B_n\setminus B_{n+1}$ with the real numbers. We do this for each $n$, and we get the desired construction, I believe (unless I'm wrong, of course ;)).

Excuse me disturbing you again, Your Highness.
Please, explain how to find countably many $B_n$, which cover $X$.

The $B_n$ cover $X$ (there is one centered at each point of $X$). Our metric space is separable, so it's Lindelof (from each open covering we can extract a countable subcovering).

Myth, are you mad at me? :?

Edit: I've also edited my previous post: the $r_n$ used there actually depend on $x$, so I replaced them with $r_n(x)$.

Excuse me again, Your Highness.

What should we do with collisions, when two sets $B_n(x)\setminus B_{n+1}(x)$ and $B_n(y)\setminus B_{n+1}(y)$ intersects? Each of these sets has its own labeling. How to choose a number for points in this intersection?
And second question. We have different labelings for each $n$. It is possible that the same point will finally has infinitely many different labelings. What should we do here?

Sorry for my dumb questions.

P.S. I am really embarrassed  :blush:, but I never heard about Lindelof spaces before. I think it is from topology course. I have found it in the net.

Are you sure? Or you just don't want to continue this discussion because of me?  :? :(

I am sure, don't worry :). I'll have another try. I think I have a promising track, but I'd like to check it out.

However, I would really want to know what this "your Highness" nonsense is all about. Don't get me wrong: you can kepp calling me that if you like (you can call me Little Red Riding Hood if it pleases you :)), but you sound mad at me, and I have no idea why you would be mad at me :? :( .

And another try:

First choose a countable dense set $x_1,x_2,\ldots$. Now, around $x_i$, take the ball $B_n=B(x_n\frac 1n)$.

We now look at the collection $\mathcal C$ of sets of disjoint countable dense sets $\{A_i\}_{i\in I}$ with the property that all the $A_i$ are included in $\bigcup B_n$. We give it a partial order by making $\{A_i\}_{i\in I}\le \{U_j\}_{j\in J}$ iff for each $i$, there is a $j$ s.t. $A_i=U_j$. We can apply Zorn's Lemma on $\mathcal C$ (the hypotheses needed for Zorn's Lemma are easy to check) to get a maximal set of disjoint countable dense sets with that property, $\{A_i\}_{i\in I}$.

Assume now that $|I|<c$. In this case, the union of the $A_i$'s has less than $c$ elements. This means that from each $B_n$ we can extract a point which is not from any of the $A_i$'s (these points need not be distinct). The newly extracted set is countable and dense, and we can add it to our family $\{A_i\}$, contradicting its maximality.

The above shows that $|I|=c$, or, in other words, we can find $c$ disjoint countable dense subsets in our space. The points left outside $\bigcup A_i$ are at most $c$. This means that we can put each such point in a different $A_i$, thuis enlarging the $A_i$'s so that they stay countable (and dense, of course), and this time they partition $X$.

I hope to God this works now. My sincerest apologies if it still doesn't.


I want to know what people think about this bad math at Verizon:

The difference between .002 cents and .002 dollars (.2 cents).

http://media.putfile.com/Verizon-Bad-Math''

It really hurts to listen to the idiots at Verizon struggle with this.

The audio file is an hour long, 43 minutes, and 33 seconds. I'm not listening to all that for that simple of a question.

Reading a question, I don't find anything unusual, employees aren't usually very good especially at these big corporations.

You really only need to listen to the first minute, unless it intrigues you enough to keep going.  But the point was, it shouldn't take 43 minutes to resolve this.

If you don't wanna listen, [url=http://verizonmath.blogspot.com/]here[/url] is a written summary, as well as a conclusion (what happened to resolve the issue).

Wow, that's pretty sad.

However, I think the people at Verizon, have done this on purpose, to take advantage of the majority of Americans, who are like the representatives at Verizon.

That's hilarious/sad.. I wouldn't think it would be that much of a baffling topic.

Rep: "Hm...."

*5 minute silence*

Rep: "Well this is Canada..."

*5 minute silence*

Rep: "Hm..."

Are you guys kidding me?  You aren't really that gullible, are you?  This is obviously a joke.  Pretty funny though :D

Edit: lol, if you listen through to about 25 minutes (don't think there is anything past this?) the lady gives him her number.  I called it and got the voicemail of Andrea, which was the name in the clip. LOL, and her mailbox was full, wonder why :P

So it's either true or a pretty good hoax.

It's not a joke.

verizonmath.blogspot.com

Yep, not a joke.. I've learned to do a little research before replying on topics like these :P.

The guy could have had some better approaches to explaining the difference between cents and dollars, as it is not easy to conceptualize .002 cents (which was the problem), but he was very patient and it paid off, I suppose.

Pretty interesting, actually.

Well, yeah, I've seen that blog before.  But anyone can make a Blogger blog...

I mean, there's no way they can be that stupid.  So it's either a prank, all the employees at Verizon are told to go along with this error, or they indoctrinate each Verizon employee with this kind of math.  I guess any of those three is very possible at this point.

Edit: after a little more searching, I'm starting to believe it might actually have been legitimate.  I mean, I just didn't think people could be that utterly stupid.

It's really not [i]that[/i] hard to imagine.. many people like to visualize money  when they think about it, so visualizing .002 cents is just not possible for many and it leads to the confusion seen here.


I don't know if this is easy enough to put in HSB.

Prove that no right triangles exist such that the length of each side is the square of an integer.

[hide="uhhhh"]ok this means by pythagorean theorem $(x^{2})^{2}+(y^{2})^{2})=(z^{2})^{2}\rightarrow x^{4}+y^{4}=z^{4}$ By FLT theres no integer solutions to this equation[/hide]

[hide]

$(a^{2})^{2}+(b^{2})^{2}=(c^{2})^{2}$

$a^{2}=m^{2}-n^{2}$
$b^{2}=2mn$
$c^{2}=m^{2}+n^{2}$
for some integers m and n such that $m\ge n$

If m and n are not divisible by two, then $2mn$ cannot be a perfect square, since it only contains one two. If p is the rest of m and n besides the two's, then

$n=p2^{x}$
$m=p2^{x+y}$

for some integers x and y.

$c^{2}=n^{2}+m^{2}$
$c^{2}=(p2^{x})^{2}+(p2^{x+y})^{2}$
$c^{2}=p^{2}2^{2x}+p^{2}2^{2x+2y}$
$c^{2}=p^{2}2^{2x}(1+2^{2y})$

Since $c^{2}$ is a perfect square and so is $p^{2}2^{2x}$, so that means that $1+2^{2y}$ is one as well. $2^{2y}$ is a perfect square too, and the only perfect squares that differ by 1 are 0 and 1. That would make $2^{2y}=0$, which is impossible. A contradiction.

[/hide]


counting distinguishability AoPSwiki search

How many ways are there to put 4 distinguishable balls into 2 indistinguishable boxes?

I want to make sure this is a good way of describing it.
$ 2^4\equal{}16$. Since there are 2 indistinguable boxes, we have $ 2!$. $ \frac{16}{2}\equal{}8$
But how would you do it if they were both indistinguable?

I'm kind of confused.  On AoPSWiki, when I search [url=http://www.artofproblemsolving.com/Wiki/index.php/Balls_and_urns]Balls and Urns[/url], I get that, if we are distributing distinguishable to indistinguishable like in this case, we would do $ \binom{4 \plus{} 2 \minus{} 1}{2} \equal{} 10$ instead.  However, myyellowducky's solution makes sense to be as well.  Could someone explain please?  As to answer myyellowducky's question, here's my explanation.  When they are both indistinguishable, first write out all the cases possible.  We could have $ \{4,0\}, \{3,1\}, \{2,2\}, \{1,3\},$ and $ \{0,4\}$.  But since the boxes are also indistinguishable, we count $ \{4,0\}$ the same as $ \{0,4\}$, and same with $ \{3,1\}$ and $ \{1,3\}$.  So our only choices are actually $ \{4,0\}, \{3,1\},$ and $ \{2,2\}$, giving us $ 3$ choices.

Thanks jxl28 for answering my question. :)

I don't really know how AoPS Wiki said $ 4\plus{}2\minus{}1$...I know why it says $ 4\plus{}2$ (4 balls, 2 boxes), but I'm not so sure about the $ 1$. Maybe it's an overcounting thing?

There is only one way to place 4 balls in one box and none in another. Then, there are 4 ways to place 3 balls on one box and 1 in another. As for placing 2 balls in each box, let us call the balls A, B, C, and D. There are 3 possibilities in that case: A and B in one and C and D in the other, A and C in one and B and D in the other, and A and D in one and B and C in the other. Hence, there are a total of 1+4+3=8 possibilities.

[hide=Why is this wrong?]Let's do casework. There's $\binom{4}{4}$ ways, to put all 4 balls, in 1 box, since the boxes are indistinguishable. Next, there's $\binom{4}{1}$ ways, to get 1 ball, to put in 1 box. Finally, there's $\binom{4}{2}$ ways, to get 2 balls, to put in 1 box, and 2 to put, in another. Adding, we get $6 + 4 + 1$, or $11$.[/hide]

[hide = @pycops] when you do two balls you count things like (b_1,b_2),(b_3,b_4) different from (b_3,b_4),(b_1,b_2) what would you do to change that?[/hide]

what the heck, this is literally an EXAMPLE problem from chapter 5, why is it a Level 20 problem in alcumus???

Why doesn't dividing by 2 generalize to more than 2 boxes? Is it because of special cases for sums (what I mean is that something like $(2,3,3)$ would become undercounted in the division)?


ceiling function
floor function
combinatorics proposed
combinatorics

let $ n \in \mathbb{N}$ and k be such that $ 1 \le k \le n$. Find the number of ordered $ k\minus{}$tuples $ \{ a_1,a_2,a_3, \cdots , a_k \}$ of integers such that $ 1\le a_j \le n$, for $ 1 \le j \le k$ and [u]either[/u] there exist $ l,m \in \{ 1,2, \cdots , k \}$ such that $ l<m$ but $ a_l>a_m$ [u]or[/u] there exists $ l \in \{ 1, 2, \cdots , k \}$ such that $ a_l \minus{}l$ is an odd number.

Is the answer: $ n^k\minus{}\left(\left\lceil{\frac {n \minus{} k \plus{} 1}{2}} \right\rceil \plus{} p_{k \minus{} 1}(1)\left\lceil{\frac {n \minus{} k \minus{} 1}{2}} \right\rceil \plus{} p_{k \minus{} 1}(2)\left\lceil{\frac {n \minus{} k \minus{} 3}{2}} \right\rceil \plus{} ...\right)$? (where $ p_{k \minus{} 1}(l)$ is the number of the  representations of $ l$ as a sum of at most $ k \minus{} 1$ non-negative numbers (for example $ p_{3}(4) \equal{} 4$ because $ 4 \equal{} 4 \equal{} 3 \plus{} 1 \equal{} 2 \plus{} 2 \equal{} 2 \plus{} 1 \plus{} 1$))

We will count the number of tuples satisfying neither of the two conditions.

Let $ b_i$ equal $ a_i\plus{}i$.   The condition that $ a_i\minus{}i$ is never odd is equivalent to the condition that $ b_i$ is always even.  The condition that the $ a_i$ are sorted in non-decreasing order is equivalent to the condition that the $ b_i$ are sorted in increasing order, and the condition that the $ a_i$ are between $ 1$ and $ n$ is equivalent to the condition that the $ b_i$ are between $ 1$ and $ n\plus{}k$.  In other words, the $ b_i$ are increasing sequences of even integers between $ 1$ and $ n\plus{}k$, which are in one to one correspondence with $ k$ element subsets of a set of size $ \lfloor \frac{n\plus{}k}{2} \rfloor$.

Subtracting, we see that the number of tuples is $ n^k\minus{}\binom{\lfloor \frac{n\plus{}k}{2} \rfloor }{k}$.

[quote="kevinatcausa"]We will count the number of tuples satisfying neither of the two conditions.

Let $ b_i$ equal $ a_i \plus{} i$.   The condition that $ a_i \minus{} i$ is never odd is equivalent to the condition that $ b_i$ is always even.  The condition that the $ a_i$ are sorted in non-decreasing order is equivalent to the condition that the $ b_i$ are sorted in increasing order, and the condition that the $ a_i$ are between $ 1$ and $ n$ is equivalent to the condition that the $ b_i$ are between $ 1$ and $ n \plus{} k$.  In other words, the $ b_i$ are increasing sequences of even integers between $ 1$ and $ n \plus{} k$, which are in one to one correspondence with $ k$ element subsets of a set of size $ \lfloor \frac {n \plus{} k}{2} \rfloor$.

Subtracting, we see that the number of tuples is $ n^k \minus{} \binom{\lfloor \frac {n \plus{} k}{2} \rfloor }{k}$.[/quote]
But i think the problem requires only one of the two conditions to be satisfied. :noo:  I would like to have an answer in that case. :(  :(

What I showed was that there were $ \binom{\lfloor \frac{n\plus{}k}{2} \rfloor}{k}$ tuples satisfying [b]neither[/b] of the two conditions.  Since there are $ n^k$ tuples total, all you need to do is subtract to get the number of tuples satisfying [b]either[/b] condition.

[quote="kevinatcausa"]What I showed was that there were $ \binom{\lfloor \frac {n \plus{} k}{2} \rfloor}{k}$ tuples satisfying [b]neither[/b] of the two conditions.  Since there are $ n^k$ tuples total, all you need to do is subtract to get the number of tuples satisfying [b]either[/b] condition.[/quote]

i Agree that your solution is correct. But i would like to have an answer in the case the intersection is not counted. i.e.  you have given $ (Universal Set)\minus{}(A^{c} union B^{c})$.
i want $ (A union B)\minus{}(A intersection B)$. :lol:

The words to describe what you're looking for are "exclusive or."

Thanks. i will edit the question then.

the edited question:
[quote="ith_power"]let $ n \in \mathbb{N}$ and k be such that $ 1 \le k \le n$. Find the number of ordered $ k \minus{}$tuples $ \{ a_1,a_2,a_3, \cdots , a_k \}$ of integers such that $ 1\le a_j \le n$, for $ 1 \le j \le k$ and [u]either[/u] there exist $ l,m \in \{ 1,2, \cdots , k \}$ such that $ l < m$ but $ a_l > a_m$ [u][i](exclusive)[/i] or[/u] there exists $ l \in \{ 1, 2, \cdots , k \}$ such that $ a_l \minus{} l$ is an odd number.[/quote]


If a and b are two roots of x^4+x^3-1=0, prove that ab is a root of x^6 + x^4 + x^3 - x^2 - 1 = 0.

In a more readable form:
If $a$ and $b$ are two roots of $x^{4}+x^{3}-1=0,$ prove that $ab$ is a root of $x^{6}+x^{4}+x^{3}-x^{2}-1 = 0.$

[hide="Proof"]Let the roots be $a,b,c,d$.  Then we have \[\left(x-a\right)\left(x-b\right)\left(x-c\right)\left(x-d\right)=0.\] This means that \begin{eqnarray}a+b+c+d&=&-1 \\ abcd&=&-1 \\ abc+abd+acd+bcd&=&0 \\ ab+ac+ad+bc+bd+cd&=&0\end{eqnarray} Due to $\left(4\right)$, there is no $x^{5}$ term in the $6$th degree polynomial.  Also, since $abcd=-1$, the constant term is $-1$ as well because the product of the terms in $\left(4\right)$ is $\left(-1\right)^{3}=-1$.  We also find the $x^{4}$, $x^{3}$, and $-x^{2}$ terms as well due to the same reasoning that gave equations $\left(1\right)-\left(4\right)$ and the following equation: \[\left(x-ab\right)\left(x-ac\right)\left(x-ad\right)\left(x-bc\right)\left(x-bd\right)\left(x-cd\right)=0.\] [/hide]


Prove that
if $r$ is a rational number then$r=x^3+y^3+z^3$ with $x,y,z$ be rational numbers.
AND if $r$ is a positive rational number then $r=x^3+y^3-z^3$ with $x,y,z$ be positive rational numbers.

since $z$ does not have to be positive, if the first is true, then plug in $-z$ for $z$, and we have proved the second. to prove the first, we first let $x=\frac{x_1}{x_2}$, $y=\frac{y_1}{y_2}$, and $z=\frac{z_1}{z_2}$. Cubing and combining, we get...

$\displaystyle\frac{(x_1y_2z_2)^3+(x_2y_1z_2)^3+(x_2y_2z_1)^3}{x_2^3y_2^3z_2^3}$

not exactly sure where this gets us, but you can probably prove that for certain $x_1,x_2,y_1,y_2,z_1,z_2$, this yields all rational numbers.

maybe that'll help.


Ring Theory
superior algebra
superior algebra solved

Let $R$ be commutative ring with identity and $P$ be it's prime ideal. It is readily that the set $S = R-P$ is multiplicative. Prove that the localization ring $S^{-1}R$ is local.

Let $I$ be an ideal of $S^{-1}R$. If $I\cap R \not \subseteq P$, then an element from $S$ would be in $I$, which is invertible, thus we would have $I=S^{-1}R$.
So we know that $I\cap R \subseteq P$ for all proper ideals $I$. But $I \subseteq S^{-1}(I\cap R)$ (for every $i \in I$, we can find $s \in S$ with $si \in R$; in fact both sets are equal).
Thus $I \subset S^{-1}P$, the latter being the only maximal ideal by this.


Could you solve this by elementary method other than direct summation?

1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 + 1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16 + 1/17 =?

Not calculus

Since you are only summing 13 terms, why would you want to do it another way?

[quote="Arne"]Of course that's not possible.[/quote]

hmm... why not possible? just wondering...

Because there is no closed formula or something for the harmonic series.

If there would be a nice way, a closed formula would exist :P  :D


summer program
Mathcamp
AMC
AIME
AwesomeMath
Columbia

I'll warn you right now that this isn't so much a question as it is me complaining / making a suggestion.

Today I received a letter in the mail from Canada / USA Mathcamp, which was along the lines of, "Due to your good score on the AMC, blah blah, we think it would be a great idea for you to apply here." I was excited to receive the letter, as I have heard of the camp but didn't really think much of applying before. And suddenly, my heart sank when I saw that the application deadline was 10 days from today. Maybe enough time to solve the problems, maybe not, not to mention I have to get letters of recommendation too.

Did most people who received this letter get it earlier than now? The actual letter itself is dated March 20, almost a month ago. What happened? If most people did get it around this time, I hardly think that one week is enough time to prepare a good application.

Similar happened to me last year. Well, first I received an invitation letter that cited Canadian Open pretty early in March; I was able to hand out recommendation forms to appropriate teachers prior to spring break and solve the quiz while enjoying my break. I completed the online application by the deadline. Meanwhile, I found another of the same letter citing my AMC scores after the spring term started in early April (in my school mailbox -- I can't access it when school is closed). I am not sure about when exactly the letter arrived during break, but it was definitely quite late.

The same happened to me this year (posted my thread right after getting the letter, since there isn't all that much time to do it...)
The letter was very nice and all, and it really made me want to get into the camp, but it definetely was a bit to late to serve its purpose effectively, in my opinion. If you send them out that late and want people to apply, there are many good scorers that really don't even try the form (sure, those that are really interested in the camp will still apply, but there are a bunch of people who i can imagine would just drop it because they feel they have to liitle time). At my school there are 2 juniors who aren't doing it because they are confident that they got into awesome math. They would usually definetely still apply, and even if it were only to solve the problems, but the fact that it was so late discouraged them and made them a lot more apathetic about the matter.

So if there is a way to get them out faster (of course this may not be possible because of logistic matters)
I would definetely suggest sending them earlier.

my 2 cents  :)

Lulze: I got mine very late as well, just last week, but thankfully I had known about the camp beforehand. However, I think it'd be a great idea for mathcamp to send it out earlier so more people who dont know about it get a chance [more than 10 days] to apply/do the problems

It would definitely be better if we could get these out earlier, but logistics prevents it.  We have to wait for the AMC to get scored, and then for us to receive the scores.  By then we have the brochures, envelopes, and letters printed (which is why the date on the letter is so early).  We then mail it out first class, rather than bulk mail, in order to get it out as soon as possible, but that's still not early enough!

So, yes: if we could get it out any earlier we would.  It's a real pity that this is so hard!

[quote="Smartguy"]Do you only mail it out to people who made the AIME?[/quote]
I don't think they do, because I got a letter too, and I didn't make AIME (111).

[quote="Smartguy"]Do you only mail it out to people who made the AIME?[/quote]

We've found that timed tests are not perfect predictors of mathematical ability in the camp environment.  I'm actually not sure what the cutoff is --- I think it's around the top 6000 --- but it's intentionally broad because there are very different kinds of skills required for different kinds of mathematics.

I applied to Mathcamp after getting the letter. That is, I applied two years after first getting the letter, and got several months to work on the quiz.

The late letter certainly does keep a lot of people applying (when I mentioned MC to the mother of someone who went to AwesomeMath, she cited "the brochure came too late" as a reason for not applying). Though, sad as it is, I'm not sure sending it out earlier would do anything more than lowering the acceptance rate and making a few hundred more people cry about not getting in. :(

Those of you who are reading this thread who received a letter -- do you still happen to have the envelope?  If you do, can you let us know the date of the postmark on the envelope, and can you confirm that it was mailed first class?  (how much is the postage that's printed on the envelope?)

Our mailers claim that the letters were mailed on March 25th, so there's no reason for so many people to be getting the letter so late.

[quote="ztbb"][quote="shelly32494"]My envelope doesnt have a postmark. nor does it have a postage thing. all it says in teh upper right hand corner is "non-profit org. U.S. postage P-A-I-D Columbia, SC Permit #535" ://[/quote]

Aha.  There's the problem.  Our mailers are not following instructions.[/quote]

Interesting.  Anyone else who's received a letter: could you also post the postage/postmark date?  I've heard some differing reports here and it'd be nice for us to get a better idea so we can improve this in the future!

Thanks!

[quote="shelly32494"]My envelope doesnt have a postmark. nor does it have a postage thing. all it says in teh upper right hand corner is "non-profit org. U.S. postage P-A-I-D Columbia, SC Permit #535" ://[/quote]

Same thing for mines.

don't think i still have mine, but i'll check for it.


Expected Value Part 1

We learn about expected value by playing a pair of games involving a die.

Thats a rather nice series of lectures. I think my students will really enjoy it. Thanks for making it available to the world.

Question - Is it okay if I share this with the world?

Yes it's ok if you share it with the world..I assume that people from all around the world are already using this website..You should share the knowledge :).

Whenever you make intentional mistakes my soul screams :D


geometry
parallelogram
symmetry
combinatorics proposed
combinatorics

An equilateral triangle of side length $ n$ is divided into $ n^2$ unit equilateral triangles. Find the number $ f(n)$ of parallelograms bounded by sides in the so obtained grid.

I remember reading this problem in Andreescu's 'Counting Strategies' book.

[hide]By symmetry, we just need to count parallelograms whose sides are parallel to left and right sides of the triangle and then multiply by 3.

Add $ n\plus{}2$ points below the triangle to create one more row. For every parallelogram, move down along the line segment downwards along each of the sides and mark the bottom point among the newly constructed $ n\plus{}2$ points. Every parallelogram , marks a unique set of four points.

Hence the number of parallelograms is $ 3{n\plus{}2\choose 4}$[/hide]


Hi guys!
How I know a lot of medalists on IMO from not so development counterys after finishing theire schools went to stady in the best universitys of the foreign counterys like USA,Canada or UK.
I think some of you have this kind of experience too,and I hope you can tell me,what advantage does a medal(maybe sliver or gold medal) on IMO give someone to enter to a well-know university?
Thank you beforehand! :wink:

I know we have AoPSers/MathLinkers from various parts of the former Eastern Bloc and Soviet Union (which I guess all count as developing countries today) who are now studying at top United States universities like Harvard, Princeton, and MIT. I am less familiar with how many young people from countries like, for instance, Iran, Vietnam, and other strong IMO countries are doing in seeking or gaining admission to United States universities.

In general, all of the better United States universities like to admit students from all over the world, as that provides a better education to everyone who attends, through the "diversity" of the students on campus. By United States visa regulations, most foreign students must prove that they have full financial support for their studies, and only a few universities (mostly the most famous ones) fully fund scholarships for foreign students. So there is a general tendency for there to be more undergraduate students from countries that have government support for their students to study abroad (e.g., Singapore and Malaysia) or high levels of wealth so that families can pay their children's costs themselves (e.g., Canada, Taiwan, the Republic of Korea, and other developed countries). Some countries that formerly had government support programs for international undergraduate study and friendly diplomatic relations with the United States no longer do, so over the last few decades the list of the top ten countries sending students to the United States for undergraduate study has changed.

An IMO medal is certainly helpful, but it is NOT a guarantee of admission. The math departments at the top universities are not large enough to admit all IMO participants, and they also admit students who have no IMO experience at all. Doing your best in all your other courses (expecially English) will help, and of course you have to apply to get in at all.

Good luck. And thanks, by the way, for putting some Armenian script in your signature. That was interesting.

Thanks a lot for replay tokenadult,but I would like to hear from somebody who has passed through this,I mean I want to know more concretely,what examinations we have to give for enter to that universitys?,when and how those examinations usually takes place?and something like this :)  :roll:.
Thanks again!!! :wink:

Oh, examinations is an easy question. For admission to the United States universities, you have to take the SAT I college admission examination,

http://www.collegeboard.com/student/testing/sat/about/SATI.html

and for the most selective universities also some SAT II subject tests

http://www.collegeboard.com/student/testing/sat/about/SATII.html

Those are MUCH less influential in admission decisions in the United States than college admission tests are in most countries, but still very important.

P.S. I forgot to mention in my last reply that some of the top research universities in the United States have full scholarship support based on financial need for ALL admitted applicants, foreign or domestic, so you should definitely apply to those universities if you can't afford to pay full tuition yourself, and apply for financial aid.

Dear tokenadult,can you tell me a little more about the [b]college admission tests[/b]?
[quote]P.S. I forgot to mention in my last reply that some of the top research universities in the United States have full scholarship support based on financial need for ALL admitted applicants, foreign or domestic, so you should definitely apply to those universities if you can't afford to pay full tuition yourself, and apply for financial aid.[/quote]
How can I apply to them?
Also when I have to do that,befor enter to an university or after?
Thanks :) !

I'm not that knowledgeable about other universities, but for Princeton you can find all of the application forms on their website (including the one for financial aid). They claim to meet everyone's financial need, and they were pretty generous with me. If you'd like, I can give you contact info for a few of my friends who are international.

Guys I have another question,don't you know,is it possible to try to enter to two or tree universitys at once?
[quote="Xevarion"] If you'd like, I can give you contact info for a few of my friends who are international.[/quote]
If it's not difficult for you :) .

You can apply to as many universities as you like. However, most places require you to pay US$\$$50-100 in order to apply, so it might not be worthwhile. I think in the US most people apply to 3-6, though there are plenty of people who apply to more.

I'll pm you with contact info for some of my friends after I ask their permission.

Essays matter a lot for admissions. That is your opportunity to tell the university to which you apply why you are a good choice for them out of all of the international applicants. (Also it is your opportunity to show how well you know English!)

About essay,can anybody give me a copy of somebodys essay?
I think it will help me to write mine :) .

[quote="Soarer"]As a matter of fact, my friend, who got 2 IMO silver medals, was turned down by 4 US universities. I forgot which 4 he applied though.[/quote]
i heard about that kind of case quite often (i think..).
but i don't see how come those medalists turned down by colleges....(they are likely to do well in school and tests..)
is it because they devote most of their time in math?

[quote="kimby_102"]but i don't see how come those medalists turned down by colleges.[/quote]

Applicants who do well in math but poorly in written English will look less desirable than applicants who do well in both to English-language universities. An IMO silver medal is a rare achievement, worldwide, and an indication of one kind of math ability, but it is a commonplace achievement in the applicant pool of a few universities, and all of those universities will choose some prospective math majors with no IMO participation at all. So the IMO silver medalist still has to practice English, learn science, and develop personal qualities that contribute to a college community.

If you look at the matter mathematically, it's easy to see why anyone is turned down: scarcity. There are too few spaces at the most desirable colleges to admit all who desire them.

[quote="Soarer"]As a matter of fact, my friend, who got 2 IMO silver medals, was turned down by 4 US universities. I forgot which 4 he applied though.[/quote]
Soarer don't  you know what was the risone of that,his bad English or something else?

i guess the reason is that his results in public exam were not so good. He was highly specialized in math, physics and programming, and his results in other subjects weren't good, I guess.

Top schools have very high standards for English skills for *undergraduate* international students.  The applicant pool is huge and diverse and the slots are few, so they can afford to be picky.  Consider the fact that being an *American* with an IMO medal is no guarantee of admission at certain schools.

Hi! I know a person that had no medals in IMO but plenty of them in other international competition (BMO) and our country's MO and he was invited by MIT. So I think that just an IMO medal is not the ideal by many of medals even in a "lower" competition. Anyway I am intresting of studying to MIT too. Any information would be welcome.

thatin where are you from?
 and what do you mean with "...invited by MIT..."

I imagine it just means accepted.  In any event, going to IMO is a far better credential than virtually any other at the high school level, and it is far more prestigious than making it to a national olympiad, but the usual 'finite number of spaces' and 'universities look at more than just mathematical prowess' caveats apply.

Hi all!!! :wink:

Hi Eagle! :wink:
Eagle I want to ask,are you going to be in Vietnam?

I think thatin's friend took TOEFL and SAT too(and I don't think that his results were bad),as must do everyone who wants to apply to a university in the USA!
Last days a lot of my friends from IMO was sent me information they have about admitting to the US universities ,and as I have understood everyone(peoples who finished a school in US,don't have to take TOEFL!) have to take that exams!
And also it's important to write some essays , they wants ,about yourself and about some other things ,because they want to see what other interests you have beside mathematic.
Thanks everyone for your repleys,for advises you gave me,you have too much helped me to realize what I have to do to admitte to US.
Now I have to work hard on my English because SAT is to hard ,also I am working on my Physics,as my Physics isn’t so good either:(.

Hi Tiks.

I still don't know if I'm going to be in vietnam cause they didn't decide yet if they will allow me to enter the TST.
but anyway, I am applying to some US universities this year and hope to get accepted somewhere.
this is the second year I apply, and I worked hard to improve my application.
If i dont get accepted nowhere in US I will most probably end up in IUB.
can someone tell me if applying for the second time [or first time but after I finished high-school] really destroys my chances to get accepted in a university.

Hi Eagle!
I hope I will meet you again in Vietnam or in US :) ,hwo knows maybe we will enter to the same university :) .
Guys I have another question :blush: .
What do you think about this
Princeton [color=red]VS[/color] MIT  :D
More seriously,I want to know your opinions about this universitys!
Thanks :wink:

[quote="Albanian Eagle"]thatin where are you from?
 [/quote]
Let me try to guess that $thatin$ is from Greece :) ,aren't you $thatin$ :lol: ?

Well of course since I am at Princeton I am hugely biased in favor of it. However since I don't go to MIT I can't compare the college experience there to that at Princeton.

I suggest that you apply to both and don't worry about choosing unless you get accepted by both.

You can find a lot of information about each university on their websites. You will get better answers from the people on this forum if you ask about specific things.

well, you said that your friend had prizes in your national olympiad and I wanted to know which national olympiad...
I don't understand your post above completely, though.

[quote="thatin"]he was invited to attend a summer course I do not know what it is. he didn't took TOEFL or SAT.
By the way, is TOEFL necessary or are any other certificates that are accepted?
how Tiks has so many smiles in his post?
and why did you asked that A.Eagle?
and why ( :rotfl: )[/quote]
As I know TOEFL is necessary for everyone hwo goes to US for studing,even if you going to study in a school.
Thatin don't you like my smiles :lol:  :P ?

[quote="thatin"]he was invited to attend a summer course I do not know what it is. he didn't took TOEFL or SAT.
By the way, is TOEFL necessary or are any other certificates that are accepted?
how Tiks has so many smiles in his post?
and why did you asked that A.Eagle?
and why ( :rotfl: )[/quote]

Will he pay the fees himself or has he received scholarship from MIT? [hide]BTW I also have 3 friends: one 2 times IPHO gold medalist and two IBO silver and bronze medalists. Unfortunately they got rejected to all universities they applied.The case of the physicist was the worst.He was the absolute(first of first as in IMO) theory part winner of IPHO. :( And also I know two 2 times ICHO gold medalists and one time bronze medalists who were also rejected. :( The universities that rejected these geniuses :lol:  were: Harvard,Princeton,MIT,Caltech and a few more.[/hide]

[quote="son of Skanderbeg"]Hey spider boy did you go to IUB?[/quote]

Hi Son of Skanderbeg!How are you? Currently I am waiting for visa(7 weeks :( ).But good news is that it seems they will give it tomorrow. :lol: But I am not sure yet.The embassy is always full with surprises.BTW nice avatar!!

My guess is that thatin means his friend was accepted in [i]RSI[/i]. RSI and MIT are completely different; don't mix them up (like many Greeks tend to do), you'll only confuse other people and mostly yourself.

To thatin: Εννοείς το Φραγγίσκο;

[quote="thatin"]he didn't pay as far as I know. What's IUB?[/quote]

IUB=International University Bremen located in Bremen, Germany.
I will probably study math.

[quote="thatin"]I like your smiles but they are 5 and the limit is 3! How did you make that?[/quote]
Limit isn't 3 :P ,if you write long posts you can put more than 3 smales   :P .

Hi all!
I just started to work on my application essay and I have some difficulties with its organization. I thought might some of you who had already entered to a university in the USA could send me his or her application essay, or gave some advice about writing it.
Thanks beforehand! :wink:
Tigran


[hide="solution"]We know that $(a+b-c)(a-b+c)(-a+b+c)\leq abc$ therefore $(1-2c)(1-2b)(1-2a)\leq abc$, so $4(ab+bc+ca)\leq 1+9abc$. Now we need only prove $\frac{1+9abc}{4}\leq \frac{9abc+2}{7}$ or $abc\leq \frac{1}{27}$, but by AM-GM we have $abc\leq (\frac{a+b+c}{3})^{3}=\frac{1}{27}$. Therefore we're done![/hide]

Nice solution :)
Can anyone find other solutions to this problems? :D

[quote="nayel"][hide="another solution but not very nice"]
we need to prove that
\[7(a+b+c)(ab+bc+ca)\le 9abc+2(a+b+c)^{3}\]
which is equivalent to
\[7((a+b+c)(ab+bc+ca)-abc)\le 2abc+2(a^{3}+b^{3}+c^{3})+6(a+b)(b+c)(c+a) \]
or
\[7(a+b)(b+c)(c+a)\le 2abc+2(a^{3}+b^{3}+c^{3})+6(a+b)(b+c)(c+a) \]

\[\iff (a+b)(b+c)(c+a)\le 2(a^{3}+b^{3}+c^{3}+abc) \]
it reduces to
\[2(a^{3}+b^{3}+c^{3})\ge a^{2}b+b^{2}c+c^{2}a+ab^{2}+bc^{2}+ca^{2}\]
which is just rearrangement. :) [/hide][/quote]
That's right. It's one of my solutions :D
But I denote: $a= \frac{x}{x+y+z}, b= \frac{y}{x+y+z}, c= \frac{z}{x+y+z}$ :)

A moment ago, I suggested using:
$(a+1)(b+1)(c+1) \ge 8(1-a)(1-b)(1-c)$
But I deleted it because I believed that it's useless. And now I find that I miscalculated and it's right :)
It's a nice solution, isn't it?

[quote="tunganh"][quote="tjhance"][quote="N.T.TUAN"]We know that $(a+b-c)(a-b+c)(-a+b+c)\leq abc$ [/quote]

Why is that?[/quote]
Denote $a+b-c=x, a-b+c=y,-a+b+c=z$
Then the inequality becomes $xyz \le \frac{(x+y)(y+z)(z+x)}{8}$ which is clearly true by AM-GM :D[/quote]
Your proof is wrong.  :( Why $x,$ $y$ and $z$ are non-negative numbers? :wink:

[quote="tunganh"][quote="tjhance"][quote="N.T.TUAN"]We know that $(a+b-c)(a-b+c)(-a+b+c)\leq abc$ [/quote]

Why is that?[/quote]
Denote $a+b-c=x, a-b+c=y,-a+b+c=z$
Then the inequality becomes $xyz \le \frac{(x+y)(y+z)(z+x)}{8}$ which is clearly true by AM-GM :D[/quote]

ack i'm sorry but I'm still not seeing it...

[quote="tunganh"]Let $a,b,c$ be positive reals such that $a+b+c=1$. Prove that:
$7(ab+bc+ca) \le 9abc+2$[/quote]

This is just Schur: $7(ab+bc+ca)(a+b+c) \leq 9abc+2(a+b+c)^{3}$.

[quote="arqady"][/quote][quote="tunganh"][quote="tjhance"][quote="N.T.TUAN"]We know that $(a+b-c)(a-b+c)(-a+b+c)\leq abc$ [/quote]

Why is that?[/quote]
Denote $a+b-c=x, a-b+c=y,-a+b+c=z$
Then the inequality becomes $xyz \le \frac{(x+y)(y+z)(z+x)}{8}$ which is clearly true by AM-GM :D[/quote][quote="arqady"]
Your proof is wrong.  :( Why $x,$ $y$ and $z$ are non-negative numbers? :wink:[/quote]
I'm sorry :)
But the inequality just follow Schur inequality :D

We easily have $(a+b-c)(a-b+c)=[a-(b-c)][a+(b-c)]=a^{2}-(b-c)^{2}\leq a^{2}$
Similarity we have three inequalities and when we multiply them, we can complete the proof.
This is the short form of Schur, I think

We also can use following assertion: Let $P(x,y,z)$ be a symmetric polynomial , then a) is equivalent to b), where
a)$P(x,y,z)\geq 0\forall x,y,z\geq 0$
b)$P(1,1,1)\geq 0,P(1,1,0)\geq 0, P(1,0,0)\geq 0$.

the above inequalities equivalent to
$f(ab)=ab(7-9c)+7c(1-c)-2 \leq 0$
because $0 \leq ab \leq \frac{(1-c)^{2}}{4}$ then all we need to do is to prove that $f(0);f(\frac{(1-c)^{2}}{4}\leq 0$
$f(0)=-7(c-\frac{1}{2})^{2}-\frac{1}{4}\leq 0$
and $f(\frac{(1-c)^{2}}{4}) =-\frac{(c+1)(3c-1)^{2}}{4}\leq 0$
so we are done  :)


function
inequalities proposed
inequalities

$ a,b,c\in[\frac{1}{4},4]$
$ abc \equal{} 1$
Find minimum off function
$ f \equal{} 8(a\plus{}b\plus{}c)\minus{}5(ab\plus{}bc\plus{}ca)$

Let $ f(a,b,c) \equal{} 8(a\plus{}b\plus{}c)\minus{}5(ab\plus{}bc\plus{}ca)$.

Suppose $ a \equal{}\min(a,b,c)$ then $ a\le 1$.

We have

$ f(a,b,c)\minus{}f(a,\sqrt{bc},\sqrt{bc}) \equal{} (8\minus{}5a)[\sqrt{b}\minus{}\sqrt{c}]^{2}\ge 0$.

We find the minimum value of

$ g(a,t) \equal{} 8(a\plus{}2t)\minus{}5(2at\plus{}t^{2})$(*) with $ at^{2}\equal{} 1,t\ge 1\ge a$.
Because $ a\ge\frac{1}{4}$ so $ t\le 2$

Put $ a \equal{}\frac{1}{t^{2}}$ in (*).

Consider

$ h(x) \equal{} 8(\frac{1}{x^{2}}\plus{}2t)\minus{}5(\frac{2}{x}\plus{}t^{2}),x\in[1,2]$.

$ h'(x) \equal{} 16\minus{}\frac{16}{x^{3}}\minus{}10t\plus{}\frac{10}{x^{2}}\equal{}\frac{(x^{3}\minus{}1)(16\minus{}10x)}{x^{3}}$

$ h'(x) \equal{} o\Leftrightarrow x \equal{} 1$ or $ x \equal{}\frac{8}{5}$.

Then $ h(x)\ge\min(h(1),h(2)) \equal{} 9$

Equality occurs only  for $ a \equal{} b \equal{} c \equal{} 1$ or $ a \equal{}\frac{1}{4},b \equal{} c \equal{} 2$ and cyclic . :)


Let $a,b,c,d$ be nonegative real numbers such that
\[a^{2}+b^{2}+c^{2}+d^{2}=1,\]
prove that
\[\frac1{1-ab}+\frac1{1-bc}+\frac1{1-cd}+\frac1{1-da}\leq\frac{16}3.\]

[quote="zaizai-hoang"]humh... Above inequality is not new and not hard ... toanhocmuonmau is author of it ... He send me in February, 2007  :wink:[/quote]

I created it in 2006 with a complicated two page solution.

Very similar one just posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=854641#854641]here[/url]
Does my method on the bottom post work?

Also where is the solution of this?

yes ... It is also my solution  :P I posted it in box Death Match of VI Forum :P


geometry
quadratics
LaTeX
rectangle
algebra
quadratic formula

[quote="joejia"]An edge of a square can be expressed as $4x-15$ meters or as $20-3x$ meters. What is its area in square meters.[/quote]
[hide]
$ 4x-15=20-3x$

$ 7x-15=20$

$ 7x=35$

$ x=5$

  And its area is $5^2$ or 25 meters squared.[/hide]

[quote="xdl615x"]Or you can use factoring:

[hide]

(4x-5)(-3x+20) = -12x^2+80x+15x-100 =

12x^2-95x+100.**<--Asteric

95+sqrt[9025-4800]/24

or

95+sqrt[9025-4800]/24

=

65+95/24 or 65-95/24 =

20/3 or -5/4 but can't be negative so

20/3? Or is it not possible to change the sign of the equation before using the quadratic equation? (Check asteric)
[/hide][/quote]

xdl615x:  [hide]There are several problems with your solution.  First of all, the notation could be clearer.  Using the standard rules of "order of operation", aka "PEMDAS", the division by 24 only applies to the rightmost term.  Since you are using the quadratic formula, I believe you intended the division to apply to both terms.  You absolutely must use parentheses to show your intent.  I would also strongly recommend that you use LaTeX, since it makes mathematical expressions much more readable.

Second, you miscopied a term from the original problem.  We were given that $4x-15$ was a side of the square.  You appear to have used $4x-5$ as a side of the square.

Finally, you multiplied the two side lengths, which would give you the area.  Then you set the area equal to zero and solved for $x$, using the quadratic equation.  Your final values of $x$ are simply the result of separately setting each of the height and width to zero:

$20-3x=0$
$20=3x$
$\frac{20}{3}=x$

But the problem did not ask for a long skinny rectangle whose area is zero.  The problem stated the figure was a square.  This is a case where the setup of the problem is crucial.[/hide]

G-UNIT:  [hide]To better assist beginning problem solvers, I would have welcomed a couple of words of english before diving into the equation.  But, your solution is concise and correct![/hide]

[hide]
The sides of a square are equal. Therefore. $ 4x-15= 20-3x$, $7x=35$, $x=5$. Thus, the square has side length of 5, and an area of $5^2=25$.[/hide]

[quote="joejia"]An edge of a square can be expressed as $4x-15$ meters or as $20-3x$ meters. What is its area in square meters.[/quote]

[hide]$4x-15=20-3x$

$7x=35$

$x=5$

So each side of the square is $4*5-15$ or $5$

So the area of the square is $5^2$ or $25$[/hide]

[hide]
The sides of a square are always equal!
$4x-15=20-3x\\7x=35\\x=5$
The side length of the square is $4*5-15=5$
The area is: $5^2=25$

DONE

[/hide]


Ok, so the question is:

A ball is dropped off a roof, at a height of 3 feet.

Every bounce it rises to $ \frac{2}{3}$ of its previous height.

On what bounce will the ball first be below $ \frac{1}{2}$ feet?

(A) 3 (B) 4 (C) 5 (D) 6 (E) 7.

Pretty straightforward right?

However:
ANSWERS
1. (B) 14
2. (A) 8671
3. (A) Sunday
4. (C) 5
5. (E) 22
6. (D) 3 : 5
7. (E) 127
8. (D) 80
9. (D) 2% gain
10. (D) 34
11. (A) 7
[b]12. (C) 5, (D) 6, (E) 7[/b]
...

So a typo?

 Table of Item Difficulty for the 2008
                      American Mathematics Contest  8
                 Tabulation of responses by    5693 Students
                             Percent of Responses

    ITEM       A         B            C           D            E       OMITS       ?
    ____    _____   _____     _____     _____     _____     _____     _____

     .......................
   [b]12   12.14     23.27     29.77*    18.81*    14.26*     1.56      0.18[/b]
     .......................

(*)  Correct Response

(This is NJ)

Hmm?

The question is not ambiguous, and even if it was, [i]3[/i] answers??

[quote="BOGTRO"]A ball is dropped off a roof, at a height of 3 feet.

Every bounce it rises to $ \frac {2}{3}$ of its previous height.

On what bounce will the ball first be below $ \frac {1}{2}$ feet?
[/quote]
This is not the wording of the AMC 8 problem.  In particular, the original problem never said "first be below".  Here is the original wording.

[quote="AMC 8"]A ball is dropped from a height of 3 meters.  On its first bounce it rises to a height of 2 meters.  It keeps falling and bouncing to $ \frac{2}{3}$ of the height it reached in the previous bounce.  On which bounce will it not rise to a height of 0.5 meters?[/quote]

I second Ravi.

The answer was initially (c) only. It was later changed to 5, 6, and 7 because it was ambiguous since for every bounce after the fifth, the ball will not reach a height of 1/2 feet. This is the case because my friend scored a 24 at first with the original answer key that came in the teacher's packet, but after they changed that answer he got a perfect score.

Maybe you have a rephrased version of the test?

I odn't have a test, I was contrusting that from memory.

Anyway, I got a 24 for some weird reason when my coach told me I got a 25, so I was wondering if it had something to do with this (I selected C)

EDIT: Not to mention my answers were the same as the official answers.

I don't know; maybe request what you got wrong?


A single one-ounce commemorative silver bar is 2'' long, $ 1\frac18$'' wide, and $ \frac3{32}$'' thick. How many of these one-ounce bars could be obtained from the silver contained in 3 blocks, given that each silver block is 6'' long, 4'' wide and 3'' thick? Assume there is no waste.

The volume, in cubic inches, of each bar, is $ 2 \cdot \frac{9}{8} \cdot \frac{3}{32} \equal{} \frac{27}{128}$.

The volume of all our blocks combined is $ 6 \cdot 4 \cdot 3 \cdot 3 \equal{} 216$.

We can get $ \frac{216}{\frac{27}{128}} \equal{} \frac{216 \cdot 128}{27} \equal{} 1024$ silver bars.


Consider a  $\triangle {ABC}$ and a point $E$ in $AC$ such that $AE=EC$ and a point $O$ in $BE$ such that $BO=OE$. If the straight line $AO$ cuts $BC$ at $D$, and $AO=12$, find $OD$.

[quote="ciprian"]Just Menelaos for the triangles BCE and ADC.[/quote]

But this problem is for little students, so please, no this kind of theory

Well, i'll post the answer, but please do not open if you don't want to

[hide]$\boxed {\boxed {4}}$[/hide]

Well, i solved it without using menelaos, can any of you post an elegant solution without that theory.........

Ok. Draw a paralel line to AC from O. Then it is easy to calculate the value of CB:DB. Then use area ratios to show that AD:OD=CB:DB.

Don't you think that this problem is too easy for Olympiad Section?

[quote="shobber"]Don't you think that this problem is too easy for Olympiad Section?[/quote]

I don't know actually


Bisectoarele $(AA', (BB', (CC'$ ale triunghiului $ABC$ intersecteaza cercul sau circumscris respectiv in $A'', B'', C''$.
Sa se arate ca: \[\frac{1}{\sqrt{A'A''}}+\frac{1}{\sqrt{B'B''}}+\frac{1}{\sqrt{C'C''}}\geq\sqrt{\frac{18}{R}}.\]

[quote="andyciup"]Este o problema foarte frumoasa.Am o solutie[pe care o s-o trimit la revista] in care folosesc inegalitatea NESBITT+$l_{a}\geq h_{a}$ si puterea punctului $A`$ fata de cercul circumscris triunghiului $ABC$.Poate postez maine solutia completa.[/quote]

Ma bucur. Am si eu o solutie frumoasa.

Asa..din moment ce am promis ceva,ma tin de cuvant :lol: Iata solutia mea:
Voi arata mai intai ca $\frac1{\sqrt{A'A''}}=\frac{AB+AC}{BC}\cdot\sqrt{\frac{AA'}{AB\cdot AC}}$.Iata cum:
Prin puterea punctului $A'$ fata de cercul circumscris triunghiului ABC,avem \[A'A''=\frac{A'B\cdot A'C}{AA'}\iff\]  \[\iff\frac1{A'A''}=\frac{AA'}{A'B\cdot A'C}=(\frac{AA'}{{A'C}^{2}}): (\frac{A'B}{A'C})=\]  \[=\frac{AA'}{{A'C}^{2}}\cdot\frac{AC}{AB}\] Acum mai avem de facut un singur pas:folosim urmatoarea
Lema:$A'C=\frac{BC\cdot AC}{AB+AC}$,de unde rezulta ca \[\frac1{\sqrt{A'A''}}=\frac{AB+AC}{BC}\cdot\sqrt{\frac{AA'}{AB\cdot AC}}(*)\] .Deci pasul a fost demonstrat.Acum ne "jucam" cu termenul drept:
$\frac{AB+AC}{BC}\cdot\sqrt{\frac{AA'}{AB\cdot AC}}\geq\frac{AB+AC}{BC}\cdot\sqrt{\frac{AD}{AB\cdot AC}}$[unde D=piciorul inaltimii din A]

Acum avem \[\frac{AD}{AB\cdot AC}=\frac{2S}{AB\cdot BC\cdot AC}=\frac1{2R}\] Din relatia asta din urma si din $(*)$ avem ca \[\frac1{\sqrt{A'A''}}\geq\frac1{\sqrt{2R}}\cdot\frac{AB+AC}{BC}\] .Adunand ceilalti doi termeni simetrici,rezulta ca \[\frac1{\sqrt{A'A''}}+\frac1{\sqrt{B'B''}}+\frac1{\sqrt{C'C''}}\geq\frac1{\sqrt{2R}}\cdot({\frac{AB+BC}{AC}+\frac{BC+BA}{CA}+\frac{AC+AB}{BC})}\]  \[\geq\sqrt{\frac{18}{R}}\] Si problema e demonstrata :lol:
[Nota:in primul post spuneam de inegalitatea NESBITT,ei bine,nu am folosit-o,ci am folosit celebra $\frac{a}{b}+\frac{b}{a}\geq 2$]
Asta e solutia mea :)

[quote="Marius Damian"]Frumoasa solutie![/quote]
Multumesc :) Care este solutia dumneavoastra?
Si inca un lucru:Daca in primul post tot vbeam de inegalitatea NESBITT,mi-am dat seama cum o pot include in solutie:Folosesc urmatoarea
 [b]Lema:[/b]Daca $a,b,c\geq 0$,atunci avem \[\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\geq 4\left( \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\] :lol:

Cu intarziere, iata si solutia mea:

Trebuie sa demonstram ca \[\sum{\frac{1}{\sqrt{A'A''}}}\geq \sqrt{\frac{18}{R}}.\] Pentru inceput, din puterea punctului $A'$ fata de cercul circumscris triunghiului $ABC$ si din inegalitatea $GM-AM$, avem \[AA'\cdot A'A''=BA'\cdot CA'\leq\frac{(BA'+CA')^{2}}{4}=\frac{a^{2}}{4}\] si analoagele.
De aici, tinand cont ca $AA'\geq h_{a}$, $h_{a}=\frac{2S}{a}$ si $S=\frac{abc}{4R}$, putem scrie \[\sum{\frac{1}{\sqrt{A'A''}}}\geq\sum{\frac{2\sqrt{AA'}}{a}}\geq \sum{\frac{2\sqrt{h_{a}}}{a}}=2\sqrt{2}\sum{\frac{\sqrt{S}}{a\sqrt{a}}}=\sqrt{2}\sum{\frac{\sqrt{bc}}{a\sqrt{R}}}.\] A mai ramas \[{\sum{\frac{\sqrt{bc}}{a}}\geq 3}\] inegalitate evidenta, conform $AM-GM$ si solutia se incheie.
[color=blue]Observatie.[/color] In enuntul problemei, in loc de bisectoare, pot fi ceviene oarecare concurente si concluzia ramane valabila.

[quote="perfect_radio"]Amandoua solutiile sunt frumoase.
[quote="Marius Damian"]
[color=blue]Observatie.[/color] In enuntul problemei, in loc de bisectoare, pot fi ceviene oarecare concurente si concluzia ramane valabila.[/quote]
Nu cred ca trebuie sa fie nici macar concurente dreptele alea. Ar trebui doar $A^\prime \in (BC)$ etc.[/quote]

Ar fi interesanta o extindere in spatiu.


function
number theory unsolved
number theory

$ \mu(n)\plus{}\mu(n\plus{}1)\plus{}\mu(n\plus{}2)\equal{}3$

where $ n \equal{} \text{natural number and }$
         $ \mu(n)     \equal{} \text{\ mobius function}\\
                    \text{\ \ \ \ \ \ } \equal{} \: 0 \text{\ if not square free}\\
                    \text{\ \ \ \ \ \ \ \ \ \ } (\minus{}1)^r \text{\ if } n\equal{}(p_1)(p_2)...(p_r)$

catch the n
(ex) 33(3x11), 34(2x17), 35(5x7)

(1) $ n$ is odd. (If $ n$ is even, then one of $ n$ and $ n \plus{} 2$ is not square free.)

(2) $ u(n) \plus{} u(n \plus{} 1) \plus{} u(n \plus{} 2) \equal{} 3$ follows $ u(n) \equal{} u(n \plus{} 1) \equal{} u(n \plus{} 2) \equal{} 1$ and so

$ n \equal{} \prod_{i \equal{} 1}^{2k_1}p_{1,i}$ where $ 2 < p_{1,1} < p_{1,2} < \ldots < p_{1,2k_1}$ are primes,

$ n \plus{} 1 \equal{} \prod_{i \equal{} 2}^{2k_2}p_{2,i}$ where $ 2 \equal{} p_{2,1} < p_{2,2} < \ldots < p_{2,2k_2}$ are primes and

$ n \plus{} 2 \equal{} \prod_{i \equal{} 1}^{2k_3}p_{3,i}$ where $ 2 < p_{3,1} < p_{3,2} < \ldots < p_{3,2k_3}$ are primes.

Does the general solution exist ?   :?:

Maybe there is solution.
Actually my book has some solution of exercise problem,
but some problems don't have solution like this problem;;


Three vertices of a regular 2n+1 sided polygon are chosen at random. Find the probability that the center of the polygon lies inside the resulting triangle.

i had typed up a long solution, but the site was down and i lost it, so here's the condensed version [hide]it doesn't matter where the first vertex is. the second vertex can be anywhere on the right hand side (we only need to consider this by symmetry). if we number the vertices such that the first vertex is 0 and the last vertex is 2n, and the second vertex is i and the third vertex is j, 1:le:i:le:n, and n+1:le:j:le:n+i, in order for the center to be inside. The probability of that happening is i/(2n-1). We average this for i from 1 to n and get (n+1)/2(2n-1). Trying this on some small values, we find that n=1 (triangle) and n=2 (pentagon) give 1 and 1/2 like they should.[/hide]


linear algebra
matrix
algebra
polynomial
vector

Let A be non-zero square matrix with the property that A^3 = 0, where 0 is the zero matrix, but with A being otherwise arbitrary.

(a) Express (I−A)^−1 as a polynomial in A, where I is the identity matrix.

(b) Find a 3×3 matrix satisfying B^2 is not equal to  0, B^3 = 0.

I + A + A^2

and

the matrix with 1's along the superdiagonal and 0's elsewhere

i.e. if i, j, k are standard basis vectors, Bi = j, Bj = k, Bk = 0.


In Cartesian coordinate, let be given two points $A(0,6),\ B(2,5)$ and line $(d): x-2y+2=0$. Find point $M \in (d)$ such that $|MA-MB|$ takes the greatest value.

huh, this problem is not bad, why noone?


this is just karamata for ln(x) i think...so suppose $a_1\geq a_2\geq ... \geq a_n$. then $a_n\leq S - (n-1)a_n$, $a_n+a_{n-1}\leq 2S - (n-1)a_{n-1} - (n-1)a_n$, ..., but $a_1+...+a_n=nS - (n-1)(a_1+...+a_n)$, so we can use karamata for ln(x) which is konkave, so the inequality follows.

Peter

what is Karamata inequality? (sorry for my ignorance  :blush: )

Or: by the AM-GM: [hide][/hide], and multiply with the analogous ones.

karamata is the following: let $ a_1\geq a_2\geq ... \geq a_n$ and $ b_1\geq b_2\geq ... \geq b_n$ be finite sequences such that $ b$ majorizes $ a$. this means that $ a_n\leq b_n$, $ a_n \plus{} a_{n \minus{} 1}\leq b_n \plus{} b_{n \minus{} 1}$, ..., $ a_n \plus{} ... \plus{} a_2\leq b_n \plus{} ... \plus{} b_2$ but $ a_n \plus{} ... \plus{} a_1 \equal{} b_n \plus{} ... \plus{} b_1$. let f be a convex(respectively concave) function. then the following inequality holds: $ f(a_1) \plus{} ... \plus{} f(a_n)\geq f(b_1) \plus{} ... \plus{} f(b_n)$(respectively $ f(a_1) \plus{} ... \plus{} f(a_n)\leq f(b_1) \plus{} ... \plus{} f(b_n)$). hope i got no typo.

Peter

Um,I think we haven't had $ S-(n-1)a_i \geq 0 $yet.So we can't take ln or AM-GM.
I think this is really an interesting inequality.For $ n=3 $,it is Schur

What a pity,but this one is wrong in the case $ S-(n-1)a_i<0 $.For $n=4$,take $a_1=0.1;a_2=1;a_3=3;a_4=4$ :(


If $ a,b,c > 0$, then
$ \sum\sqrt{ab(a\plus{}b)}\geq\sqrt{\prod(a\plus{}b)}\plus{}\frac{8}{5}\sqrt{abc}\plus{}\frac{1}{5}\sqrt{\prod(a\plus{}b)\plus{}10abc}$

[quote="filomen"]If $ a,b,c > 0$, then
$ \sum\sqrt {ab(a \plus{} b)}\geq\sqrt {\prod(a \plus{} b)} \plus{} \frac {8}{5}\sqrt {abc} \plus{} \frac {1}{5}\sqrt {\prod(a \plus{} b) \plus{} 10abc}$[/quote]

not true...

try $ a\equal{}b\equal{}c\equal{}1$ .

[quote="filomen"]If $ a,b,c > 0$, then
$ \sum\sqrt {ab(a \plus{} b)}\geq\sqrt {\prod(a \plus{} b)} \plus{} \frac {8}{5}\sqrt {abc} \plus{} \frac {1}{5}\sqrt {\prod(a \plus{} b) \plus{} 10abc}$[/quote]

It is try!!!! Sorry  :(   :(   :(


induction
real analysis
real analysis unsolved

I am feeling very stupid as I am drawing a blank on how to even start on the following problem.

The the sum of of S,
      S = 1*1! + 2*2! + 3*3! + ....... + n*n!

Can someone help me get started?

Thanks!

The closed form is $ (n \plus{} 1)! \minus{} 1$.  Do it first by induction, that'll give you some idea for why it's true and you should see a direct proof.

Thank you!  I guess I was trying to make it harder than it was!

Or:
$ S\equal{}\sum_{k\equal{}1}^n k \cdot k! \equal{} \sum_{k\equal{}1}^n \left( (k\plus{}1) \cdot k! \minus{} k! \right) \equal{} \sum_{k\equal{}2}^{n\plus{}1} k! \minus{} \sum_{k\equal{}1}^n k! \equal{} (n\plus{}1)!\minus{}1$
 :wink:


quadratics
superior algebra
superior algebra unsolved

Prove that $ \mathbb{Q}\left(\sqrt{2},\sqrt[3]{2},\sqrt[4]{2},...\right)$ is an infinite, algebraic extension of Q.

Prove that every quadratic extension of $ \mathbb{Q}$ is of the form $ \mathbb{Q}\left(d\right)$ for some square-free integer d.

The extension is algebraic because it is generated by algebraic elements. Show that it contains a subextension of degree $ n$ for all $ n$, thus it must have infinite degree.

For the second one, let $ K$ be a quadratic extension of $ \mathbb Q$. Let $ \alpha \in K \backslash \mathbb Q$. Show that:
- $ K\equal{}\mathbb Q[\alpha]$.
- $ \alpha$ is a root of an equation $ ax^2\plus{}bx\plus{}c\equal{}0$ with rational $ a,b,c$ and $ a \neq 0$.
- $ \mathbb Q[\alpha] \equal{} \mathbb Q[\sqrt{b^2\minus{}4ac}]$.
- For any rational $ d$, there exists a rational $ e$ such that $ D\equal{}de^2$ is a squarefree integer.
- Take $ d\equal{}\sqrt{b^2\minus{}4ac}$ and conclude $ \mathbb Q[\sqrt{d}]\equal{}\mathbb Q[\sqrt{D}]$.


I don't know about you guys, but I am excited about the ARML contest this year!

I hope everyone is having a good end to their school year, and learning a good deal of math everyday! Do your best on the problems that we sent out. We will have solutions to them and discuss some of them at our upcoming practices.

Note that ARML has switched to 10 questions for the individual round this year. The explanation for why is given here: http://www.arml.com/arml_registration_2009/competition_rules.php
Also note that calculators will no longer allowed on the Team Round beginning this year.

I imagine now no.'s 8 and 10 will be the hardest. They'll probably add an easier one into the front half and a harder one into the back half of the problems. Adding even one more "hard one" should space the scores out a little more so there aren't so many ties at the top.

Hi, I'm Anurag Deeconda, and I'm a junior at Spring valley. I haven't actually received a letter confirming this, but from talking to Sean, I found out that I'm on the All State Math Team, so I thought I would introduce myself on this thread. I qualified for the AIME this year and got a 7 I think. I got 800 on SAT I Math, and 790 and 800 on SAT II Math I and II. I'm on the Spring Valley Math team, and have taken AP Stats and am taking AP Calculus AB. I'll be taking AP Calculus BC next year.

Does anyone have a good link to past ARML questions online? I couldn't make it to the practice yesterday, so I would like to work on them at home. Also, when is the next practice?

Thanks,
Anurag

Anurag, I sent you a private message. You can see the problems from last year's contest here: http://arml.com/2008contest.php

And you can find the problems of the 2007 contest here: http://arml.com/2007contest.php

The contest consists of 4 parts.
- A Team Round (20 minutes, 10 problems, worked on by all 15 people on a team)
- A Power Round (1 hour, proof based questions, worked on by all 15 people on a team)
- An Individual Round (10 minutes for 2 problems, 10 total problems (changed from 8 in past years), worked on individually)
- A Relay Round (groups of 3, each person gets a problem and must pass their answer to the next person in order for it to be used in their problem. 5 points given for groups whose third person turns in the correct answer at 3 minutes, 3 points for groups that turn in the correct answer at 6 minutes)

- Tiebreaker questions are used to decide overall winners
- Super Relay event is just for fun

No calculators allowed.


Try to see if you can make any sense of this
[url]http://stoa.usp.br/cesarabp/files/1729/9522/2009-Bravo-Small-Finite-Fields.pdf[/url]

jacquelinek, why do you keep editing your post to become nonsense after it is answered?
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=304517[/url]

This is a lot like spamming.


hi
new poster here so opportunity to say hay too

i have the following prob
[code]
\begin{equation}
\begin{split}
\left ( "big equation goes here" \\
       "the big equation continues" \right )
\end{split}
\end{equation}
[/code]

the problem is that i need to use "\left (" and "\right )" to adapt to the height of the equation, but need to split it cos its long.
what i did and seemed to solve the problem was doing a "false right" and "false left" and despite tex editors dont like it, the compliler behave as i wanted, i mean:

[code]
\begin{equation}
\begin{split}
\left ( "big equation goes here" \right \\
\left "the big equation continues" \right )
\end{split}
\end{equation}
[/code]
note the lack of parethesis in the second "\right" and in the second "\left" between the split (or new line as you want "\\")
there must be a more elegant way of doing this right?
any help much appreciated
regards
c

[quote="choza"]what i did and seemed to solve the problem was doing a "false right" and "false left" and despite tex editors dont like it, the compliler behave as i wanted, [/quote]
That's because the "false" left and right are missing dots so should be \left. and \right. with a dot after each command.
[code]\begin{equation}
 \begin{split}
  \left ( "big equation goes here" \right. \\
  \left. "the big equation continues" \right )
 \end{split}
\end{equation}[/code]
This is a common way of saying you only want one bracket. Another way, that doesn't involve pairing is to use \bigl(, \Bigl(, \biggl(, \Biggl( etc $\bigl( \Bigl( \biggl( \Biggl($ which, given that you have loaded the amsmath package, resize themselves according to the equations you've used.

thanks for the great tip... Much appreciated  :D


Post here if you competed in the Oregon/Washington science bowl which occurred at the University of Portland today :!:

I want to see if there are any mathlinkers who went there today.

I doubt there are more than a couple at most from Oregon, but I have high hopes for Washington :wink:

[quote="nat mc"]i competed in science bowl, but at terre haute.  We won, so we're going to nationals.[/quote]

see you there nat mc.

My friend is from Washington and he told me his high school won this year. (That's Skyview HS.)

In case anyone cares, my old high school won too (A&M Consolidated HS).

I'm trying to go as a volunteer, but they might not take me because I applied kinda late.


quadratics
function
Columbia
inequalities
inequalities unsolved

Let $a,b$ are  real numbers. Prove that:
$3(a+b+1)^{2}+1\ge\ 3ab$

Uh,
$3(a+b+1)^{2}+1 = (3a^{2}+3ab+3b^{2}+6a+6b+4)+3ab \ge 3ab$
Is there something missing? Or is the "very easy" title accurate for once? ;)

We can denote that $f(a)=3a^{2}+3x(y+2)+3y^{2}+6y+5$ by a quadratic function of a, by calculate $\Delta$ we have what we must do! :D
It is the Columbia 2001 TST?

Maybe it shoudl be for all real numbers?

Yes, of course, you can find the original of this form the collection of Hojoo Lee!

Yes, i have edited problem to correct one. Any way this one is easy. And my solution is using quaratic function that similar to with [b]NkMAsTeR[/b]

Another question(easy):

If $\prod a_{i}=1$, and $a_{1},a_{2}, ....,a_{n}$ are  positive numbers, prove that ineq. is true:

$\sqrt{a_{1}}+\sqrt{a_{2}}+...+\sqrt{a_{n}}\leq a_{1}+a_{2}+...+a_{n}$
 :wink:

[quote="hien"]Let $ a,b$ are  real numbers. Prove that $ 3(a \plus{} b \plus{} 1)^{2} \plus{} 1\ge\ 3ab.$[/quote]$ 3(a \plus{} b \plus{} 1)^{2} \plus{} 1\minus{}\ 3ab\equal{}\frac{1}{4}(3a\plus{}3b\plus{}4)^2\plus{}\frac{3}{4}(a\minus{}b)^2\geq 0.$

This problem is introduced as Exercise 1.19 on page 32 in new book :

Ji Chen, Chao-Cheng Ji, [url=http://product.dangdang.com/product.aspx?product_id=20670358]Algebraic Inequalities (Mathematical Olympiad Proposition people Lecture)[/url], Shanghai Scientific & Technological Educational Publishing House, 2009, 225p. (ISBN 978-7-5428-4848-2/O·613)


factorial
modular arithmetic
number theory proposed
number theory

$ x! \plus{}y!\plus{}z!\equal{}w!$ $ solve$

WLOG assume $ x\leqslant y\leqslant z(\leqslant w)$. If $ x \equal{} 0$ or $ x \equal{} 1$, then

$ 1 \plus{} y! \plus{} z! \equal{} w!$

The LHS is $ \equiv 1\pmod{y!}$ and the RHS is $ \equiv 0\pmod{y!}$, so $ y! \equal{} 1$, but $ 2 \plus{} z! \equal{} w!$ has no solution.

For $ x\geqslant 2$ we get

$ 1 \plus{} (x \plus{} 1)(x \plus{} 2)\cdots y \plus{} (x \plus{} 1)(x \plus{} 2)\cdots z \equal{} (x \plus{} 1)(x \plus{} 2)\cdots w$

Again we come to contradiction modulo $ x \plus{} 1$, hence $ x \equal{} y \equal{} z$ (since $ x\plus{}1$ is at least $ 3$), giving

$ 3x! \equal{} w!\implies 3 \equal{} (x \plus{} 1)(x \plus{} 2)\cdots w$.

Since $ 3$ is prime, the only possibility is $ 3 \equal{} x \plus{} 1 \equal{} w$, so $ (2,2,2,3)$ is the only solution.

A good solution but it had been post before.


parameterization
complex numbers
absolute value

Last problem on the complex numbers chapter (9) of vol 2.
The coefficients of x^n + a_(n-1) x^(n-1)...a_1x + a_0 = 0 are real and satisfy 0 < a_0 <_ a_1.... <_ a_(n-1) <_ 1.  Z is a complex root of the equation with abs(z) >_ 1. Show that z^(n+1) = 1.

Ok now that Ive got all that typed up, let me chalk up my case. This is what i did:

We have z is a (n+1)th root of 1, therefore
z^n + z^n-1 + ... z + 1 = 0.                (1.1)
Plugging z back into the initial equation, we get z^n + a_(n-1)z^(n-1) + ... a_1z + a_0 = line (1.1). If we relate the terms now, we get a_(n-1) through to a_0 are all equal to 1, which satisfies the parameters given.

Is this a valid solution? The one given in the book is both quite long and gets confusing (I'll talk about the confusing parts after this solution is checked).

[color=cyan]Unfortunately, you showed a converse of the problem given, which is not equivalent to what you needed to show.  You proved that you could pick the a(i) such that the (n+1)st root of 1 would be a root -- that is not the same as showing that for any selection of the a(i), if there is a root with absolute value >= 1, then it is an (n+1)st root of one.

Does that make sense?[/color]


if f(x)is greater that 0                                                                                f(x+y)=F(x)f(y)                                                                                f(x)=f(y) if x=y                                                                                proove that f(0)=1.,f(-x)=1/f(x),f(x-y)=f(x)/f(y)                                                                                find f(2-x)+f(x-6)=27

f(x+0)=f(0)f(x) so either f(x) = 0 or f(0)=1, but since f(x)>0, we must have f(0)=1

f(-x+x) = f(x)f(-x), so 1 = f(0) = f(x) f(-x) and thus f(-x) = 1/f(x)

f(x-y) = f(x+(-y)) = f(x)f(-y) = f(x)/f(y)

I don't understand your last question...


Hello,
Does anybody know a good place (online or off) that i can learn the amc 10 material that i absolutely need to know?
i have seen the aops amc 10 class and am contemplating enrolling in that. any comments on how that class went? did you learn a lot and such.

Thank you a lot in advance.

Buy AOPS vol. 1.
It has basically covers all the concepts of the AMC 10.
Also try old AMC 10 tests for practice, if you run out of those tests, try some of the old AHSMEs or Mathcounts state tests (which are about same level).

I guess if you have time, the class here will probably help you the most...

but anyways, personally, if you have good teachers in school and listen and pay attention, you could most likely pass the AMC 10 just by the concepts you learned in algebra 1/geometry in school

Get your hands on all of the AMC 10 problems you possibly can, and go over them. Mark all the ones you didn't understand the first time around or got wrong, and do them again. From the math I've seen in middle/high school curriculums, it probably won't help you too much on the AMC's.

AHSME's might help, but I've noticed that they have a very different "flavor" than the AMC 10's. Definitely, though, I wouldn't be practicing Mathcounts States to help for AMC10, since it has a completely different style.

[quote="jhangil"]Hi I am currently in AMC 10 prep course here and it's really fun. You learn in class by doing problems and solving them together and after class you get homework to solve together with other students and post solutions in the class forum. I would recommend it if you really care about AMC 10.
Also the AOPS V 1 is a great book I haven't finished it yet, but so far it has enriched me with great knowledge.[/quote]

I took that course before, and I think its REALLY good. (AMC10 score as a 7th grader was 138.0)

@ piggyinc and the future
You guys seem to have completely different viewpoints on this...
Why?
But overall, the class at aops seems pretty popular

One caveat I offer in general, if you find most of the material easy (i.e. you're already averaging above a 120), you may want to consider taking the AMC 12 problem series instead of AMC 10 problem series.

Personally, I took both in 10th grade, and I felt like I consistently mastered the AMC 10 material, but struggled and learned a lot with the AMC 12 material.


induction
arithmetic sequence
algebra solved
algebra

Prove that the serie of natural <>0 numbers (an)n>=1 is an arithmetic progression if and only if there exist a>0 so that a(n+1)=a(n)+[a(n)/(n+a)].

[quote="Lagrangia"]well this one I happen to know from a contest in which I competed! Dan Barbilian 10th grade this year ...

here is the solution which I gave at the contest and got a big round result! I think it is correct-- my solution :D!

if a0,a1,a2.. are natural and r=a(n+1)-a(n) which is also natural then a(n+1)=a(n)+r=a(n)+[r]. we just need to find a>0 so that r=a(n)/(n+a) which gives us a=a0/r because: [a(n)/(n+a)]=[(a0+nr)/(n+a0/r)]=[r]=r.
now reciprocally: if a(n+1)=a(n)+[a(n)/(n+a)] for any n from N then:
a1=a0+[a0/a], and so on. Lets denote now r=[a0/a]=> r<=a0/a<r+1 => ra<=a0<a(r+1)=>r(a+1)<a0+r<(a+1)(r+1) so then r<=a1/(a+1)<r+1 and we have r natural => [a1/(a+1)]=r.
 using induction we get r=[an/(a+n)] so a(n+1)=a(n)+r so (an) is an arithmetical progression!

:D[/quote]

I don't know what you write. Other users, please help to solve that problem


If $ a > 1$, then the sum of the real solutions of \[\sqrt{a \minus{} \sqrt{a \plus{} x}} \equal{} x\] is equal to

$ \textbf{(A)}\ \sqrt{a} \minus{} 1\qquad
\textbf{(B)}\ \frac{\sqrt{a} \minus{} 1}{2}\qquad
\textbf{(C)}\ \sqrt{a \minus{} 1}\qquad
\textbf{(D)}\ \frac{\sqrt{a \minus{} 1}}{2}\qquad
\textbf{(E)}\ \frac{\sqrt{4a \minus{} 3} \minus{} 1}{2}$

[hide]
Squaring twice, we have

$ x^4 \minus{} 2ax^2 \plus{} a^2 \equal{} a \plus{} x$

which factors as $ ( \minus{} x^2 \minus{} x \plus{} a \plus{} 1)( \minus{} x^2 \plus{} x \plus{} a)$

From this, we can see the possible roots are $ \frac {\pm \sqrt {4a \minus{} 3} \minus{} 1}2$ and $ \frac {\pm \sqrt {4a \plus{} 1} \plus{} 1}2$.

However, each squaring introduces a condition: in the first, we have $ x \ge 0$, from which we can conclude that the sign of the discriminant is positive.

The second squaring demands that $ x^2 \minus{} a \le 0$. Testing both solutions:

a)$ (\frac {\sqrt {4a \minus{} 3} \minus{} 1}2)^2 \minus{} a \equal{} \frac {4a \minus{} 4 \minus{} 2\sqrt {4a \minus{} 3} \minus{} 4a}4 \equal{} \frac { \minus{} 4 \minus{} 2\sqrt {4a \minus{} 3}}4$. This is definitely negative, so $ x \equal{} \frac {\sqrt {4a \minus{} 3} \minus{} 1}2$ passes the test.

b)$ (\frac {\sqrt {4a \plus{} 1} \plus{} 1}2)^2 \minus{} a \equal{} \frac {4a \plus{} 2 \plus{} 2\sqrt {4a \plus{} 1} \minus{} 4a}4 \equal{} \frac {2(1 \plus{} \sqrt {4a \plus{} 1})}{4}$. This is positive. This solution does not work.

Therefore, the only real solution is $ \frac {\sqrt {4a \minus{} 3} \minus{} 1}2$, or answer $ \boxed{\textbf{(E)}}$.
[/hide]

Is there a quick way to find that factorization?

Well, it's not a correct factorization; expanding gives a $ \minus{}2x^2$ term and others.

[quote="ffao"]
Squaring twice, we have

$ x^4 \minus{} 2ax^2 \plus{} a^2 \equal{} a \plus{} x$

which factors as $ ( \minus{} x^2 \minus{} x \plus{} a \plus{} 1)( \minus{} x^2 \plus{} x \plus{} a)$
[/quote]
I think this was a typo, should be $ ( \minus{} x^2 \minus{} x \plus{} a \minus{} 1)( \minus{} x^2 \plus{} x \plus{} a)$
To obtain the factorization rearange as $ a^2\minus{}(2x^2\plus{}1)a\plus{}x^4\minus{}x\equal{}0$ and solve as a quadratic in a.

Can't we just plug in a constant lol? Also, I think this has been posted recently in C&P


Let $ f: \mathbb{N}\rightarrow \mathbb{N}$ satisfy $ f(xf(y))\equal{}yf(x)$ for any $ x,y\in \mathbb{N}$. Find the least possible value of $ f(2007)$.

if we have to find $ X\equal{}min\{f(2007): \ f\ is\ solution\}$
$ \forall n\in\mathbb{N}: \ f(n)\equal{}0$ is solution
then $ X\equal{}0$  :!:  :!:

________
if you mean $ \mathbb{N}\equal{}\{1,2,....\}$

i)$ x\equal{}1$ gives $ fof(y)\equal{}f(1)f(y)$
ii)$ y\equal{}f(z)$ gives $ f(f(1)xz)\equal{}f(x)f(z)$
iii)$ y\equal{}1$ gives $ f(f(1)x)\equal{}f(x)$

iii),ii) gives $ f(x)\equal{}f(f(1).1.x)\equal{}f(1)f(x)$ so $ f(1)\equal{}1$

then $ fof(x)\equal{}x$ and $ f(xy)\equal{}f(x)f(y)$
and we have ($ \forall x,y\in\mathbb{N}: \ fof(x)\equal{}x,f(xy)\equal{}f(x)f(y)\iff \forall x,y\in\mathbb{N}: \ f(xf(y))\equal{}xf(y)$)

if $ x\equal{}\prod_{k\equal{}1}^{m}p_i^{a_i}$ where $ p_i$ are primes then $ f(x)\equal{}\prod_{k\equal{}1}^{m}g(p_i)^{a_i}$ where $ g: \mathbb{N}\to\mathbb{N}$ such that $ \forall x\ gog(x)\equal{}x,g(1)\equal{}1$
and we have $ f(2007)\equal{}g(3)^2g(223)\ge 2^2.3\equal{}12$

[quote="aviateurpilot"]
and we have ($ \forall x,y\in\mathbb{N}: \ fof(x) \equal{} x,f(xy) \equal{} f(x)f(y)\iff \forall x,y\in\mathbb{N}: \ f(xf(y)) \equal{} xf(y)$)

[/quote]

Why is this?

And how can you find minimum? Please explain in detail.

I mean, why if $ f(f(x))\equal{}x$ and $ f(xy)\equal{}f(x)f(y)$ and you know the initial condition is satisfied?

And for $ f(2007)\equal{}12$ I don't understand how you define that function.


I don't think there's enough info to solve this problem, because the perpendiculars can intersect the bisectors at any point along their lengths...

I can't get the answer yet. Im gonna keep trying

I think this is what the question means...not sure though.

[quote="neelnal"][quote="bfrock2000"]In triangle ABC, perpendiculars from A to the bisectors of angle and angle C meet the bisectors in D and E, respectively. The line through D and E intersects AC at X and AB at Y. what fractional part of the area of the triangle ABC is the area of region XYBC.[/quote]

Your first sentence makes no sense. "angle and angle"???

Also, is perpendiculars correct terminology? Do you mean Line is perpendicular to another line?[/quote]
That is correct terminology, and he meant "angle B and angle C." :D

ragnarok, that's correct, though $B$ and $C$ should be switched.

awww i dont think thats clear enuf but oh well I See it now


1.   The volume of a pyramid can be evaluated by using the equation V =Ah/3

where h is the height of the pyramid and A is the area of the base of the pyramid. The problem is
to design such a triangular pyramid where the volume is V = 72 cm3 and the height is
h = 12 cm. The base of the pyramid is a rectangular isosceles triangle.
a) Determine the area of the base of the pyramid. Pay attention to the unit of your
answer. (1 point)
b) Draw a picture of the base triangle of the pyramid in a proper scale. (2 points)

The area of the base is found by substitution $ 18cm^2$
It is isosceles right triangle, so each leg takes $ 2\sqrt 3 cm$ length

might i suggest that the next time ur topic title be a little more mathematical maybe.... u can post it in your concerned language in your own national community..

Title edited and topic moved to a more appropriate forum.


linear algebra
matrix
linear algebra unsolved

I do not have the answer of this one:
What are those endomorphisms $f: M_n(K) \to M_n(K)$ that map a matrix $M$ to an other matrix $f(M)$ such that:
$M$ is similar to $f(M)$ for every $M \in M_n(K)$

In particular $f(A)$ and $A$ have the same rank, thus we can use the problem from Ens written exam (yes, there it was over $C$, but easy remarks show that the proof is true over any field) to conclude that $f$ must be of the form $f(X)=AXB$ or $ f(X)=A^{t}XB$. From here it is not difficult, I think.