[i]Hey I cannot find out how to do this:[/i]
I have many constants in this paper ($C_{1}, C_{2}, C_{3},\ldots$).  I was wondering how can one have Latex number these and reference each one.  So, this way when I have to add a new constant say between $C_{4}$ and $C_{5}$, then I don't have to change the old $C_{5}$, $C_{6}$, and all the others afterwards.  Thanks for all the help!

That's an interesting idea and can be achieved with [quote]\newcounter{ccounter}
\newcommand{\C}{\addtocounter{ccounter}{1}C_\arabic{ccounter}}
The set of constants is $\$$\{\C,\C,\C,\ldots\}$\$$[/quote]You may want to use something other than \C if that conflicts with anything else.

Now, I am running into a problem.  I am not able to label and reference each constant.  I tried using

\newcounter{ccounter}
\newcommand{\C}{\refstepcounter{ccounter}C_\arabic{ccounter}}

but still cannot label individual constants
for example:

...\C\label{one}...

....C_\ref{one}....

does not work.  Please help and thank you.

What is going wrong? Both of the following work for me:
[code]\newcounter{ccounter}
\newcommand{\C}{\refstepcounter{ccounter}C_\arabic{ccounter}}
$\C\label{one}$\\
$\C\label{two}$\\
$C_\ref{one}$ \\
$C_\ref{two}$[/code]
[code]\newcounter{ccounter}
\newcommand{\C}{\addtocounter{ccounter}{1}C_\arabic{ccounter}}
$\C\label{one}$\\
$\C\label{two}$\\
$C_\ref{one}$ \\
$C_\ref{two}$[/code]

Yes, I tried and your example does work (although I needed extra curly brackets when referencing as a subscript).

But, some how in my tex file the numbering is off when I reference certain constants; i.e. the counter is messed up.  For example, I have something like what follows:
\begin{theorem}
\begin{equation}
...\C\label{one}...
\end{equation}
where C_\ref{one}=....
\end{theorem}
(dollar signs omitted)

The problem is I get C_4 in the equation and then the part after the equation gives C_5=...(instead of C_4).

The counter is off for some reason.

In order to see what is going on (eg how you've defined the theorem environment) post a complete but very short document that shows the problem. If you use [code ]...[/code ] tag then you can leave the dollar signs in and they won't be misinterpreted.

OK, please bare with me.  I think the problem is that
\begin{equation}
\end{equation}
commands seems to be adding to the numbering in the referencing of the counter.

They shouldn't affect things, unless you've messed with the equation counter. This silly document works fine:
[code]\documentclass[12pt]{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}

\begin{document}
\newtheorem{theorem}{Theorem}
\newcounter{cccounter}
\newcommand{\C}{\refstepcounter{cccounter}C_\arabic{cccounter}}
\begin{theorem}
\begin{equation}
...\C\label{one}...
\end{equation}
where $C_\ref{one}=$....
\end{theorem}

\begin{theorem}
\begin{equation}
...\C\label{t}...
\end{equation}
where $C_\ref{t}=$....
\end{theorem}

\begin{theorem}
\begin{equation}
...\C\label{h}...
\end{equation}
where $C_\ref{h}=$....
\end{theorem}

\begin{theorem}
\begin{equation}
...\C\label{u}...
\end{equation}
where $C_\ref{u}=$....
\end{theorem}

\end{document}[/code]

[code]
\documentclass[12pt]{article}
\pagestyle{plain}

\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{amscd}

\newcommand{\abs}[1]{\left| #1 \right|}

\newtheorem{definition}{Definition}[section]
\newtheorem{theorem}[definition]{Theorem}
\newtheorem{axiom}[definition]{Axiom}
\newtheorem{corollary}[definition]{Corollary}
\newtheorem{conjecture}[definition]{Conjecture}
\newtheorem{counterexample}[definition]{Counterexample}
\newtheorem{lemma}[definition]{Lemma}
\newtheorem{sublemma}[definition]{Sublemma}
\newtheorem{proposition}[definition]{Proposition}
\newtheorem{question}[definition]{Question}
\newtheorem{remark}[definition]{Remark}
\newtheorem{application}[definition]{Application}
\newtheorem{idea}[definition]{Idea}

\newcounter{ccounter}
\newcommand{\C}{\refstepcounter{ccounter}C_\arabic{ccounter}}

\begin{document}

Let
\begin{equation}\label{lfl}
h(x)=x^2.
\end{equation}
This seems to mess up the constants.

\begin{corollary}
Given $4 \leq a < b$ and the prime divisors of $a$ and $b$ come from
the first $k$ primes, then
\begin{equation}
b-a>a/(\log a)^{\C\label{what}}
\end{equation}
where
$C_{\ref{what}}=C_{\ref{what}}(k)=\min\{8e\cdot30^{k+3}k^{4.5},16
\cdot 2^{6k+20}\} k\log\log k$.  Moreover, if $k$ is sufficiently
large, then
\begin{equation}
b-a>a/(\log a)^{\C\label{next}}
\end{equation}
where $C_{\ref{next}}=C_{\ref{next}}(k)=30^{k+4}$.
\end{corollary}

\end{document}
[/code]

That works fine! I get $b-a>a/(\log a)^{C_{1}}$ and $C_{2}=C_{2}(k)$ and $C_{3}=C_{3}(k)$ in the document. Isn't that OK?

If you are getting something else, get rid of the .aux file and then run latex twice. It could be you have an old .aux file around (it contains the references).

This is weird.  I just tried the same code but changed the latter two equations:

[code]
\documentclass[12pt]{article}
\pagestyle{plain}

\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{amscd}

\newcommand{\abs}[1]{\left| #1 \right|}

\newtheorem{definition}{Definition}[section]
\newtheorem{theorem}[definition]{Theorem}
\newtheorem{axiom}[definition]{Axiom}
\newtheorem{corollary}[definition]{Corollary}
\newtheorem{conjecture}[definition]{Conjecture}
\newtheorem{counterexample}[definition]{Counterexample}
\newtheorem{lemma}[definition]{Lemma}
\newtheorem{sublemma}[definition]{Sublemma}
\newtheorem{proposition}[definition]{Proposition}
\newtheorem{question}[definition]{Question}
\newtheorem{remark}[definition]{Remark}
\newtheorem{application}[definition]{Application}
\newtheorem{idea}[definition]{Idea}

\newcounter{ccounter}
\newcommand{\C}{\refstepcounter{ccounter}C_\arabic{ccounter}}

\begin{document}

Let
\begin{equation}\label{lfl}
h(x)=x^2.
\end{equation}
This seems to mess up the constants.

\begin{corollary}
Given $4 \leq a < b$ and the prime divisors of $a$ and $b$ come from
the first $k$ primes, then
\begin{equation}
\C\label{what}
\end{equation}
where
$C_{\ref{what}}=C_{\ref{what}}(k)=\min\{8e\cdot30^{k+3}k^{4.5},16
\cdot 2^{6k+20}\} k\log\log k$.  Moreover, if $k$ is sufficiently
large, then
\begin{equation}
\C\label{next}
\end{equation}
where $C_{\ref{next}}=C_{\ref{next}}(k)=30^{k+4}$.
\end{corollary}

\end{document}
[/code]

and this worked fine for me.  But, the previous post did not.

Ah. I see the problem. I'll look at it now :)

The problem is the way you are refencing the counter. If you use \ref you are referencing the label counter not the C counter. If you try \begin{equation*} you'll get strange things like $C_{0.1}$ :)

So instead of writing C_{\ref{what}} use C_{\theccounter} and of course you then don't need labels either. So this will work:
[code]Let
\begin{equation}\label{lfl}
h(x)=x^2.
\end{equation}
This seems to mess up the constants.
\begin{corollary}
Given $4 \leq a < b$ and the prime divisors of $a$ and $b$ come from
the first $k$ primes, then
\begin{equation}
b-a>a/(\log a)^{\C}
\end{equation}
where
$C_{\theccounter}=C_{\theccounter}(k)=\min\{8e\cdot30^{k+3}k^{4.5},16
\cdot 2^{6k+20}\} k\log\log k$.  Moreover, if $k$ is sufficiently
large, then
\begin{equation}
b-a>a/(\log a)^{\C}
\end{equation}
where $C_{\theccounter}=C_{\theccounter}(k)=30^{k+4}$.
\end{corollary} [/code]

Thank you very much for all your help.  I guess the only problem with using this method is that I cannot reference different constants later throughout the proof.  I will look into this a little more.  Thanks again!

I thought you might want to reference the constants elsewhere so rethought the answer. Basically you need to use \refstepcounter in the equation environment rather than in the definition of the counter. You can only have one label per equation so it works well. You could even redefine the equation environment to include \refstepcounter.

Anyway, this document works. Note the references at the end.
[code]\documentclass[12pt]{article}
\pagestyle{plain}

\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{amscd}

\newcommand{\abs}[1]{\left| #1 \right|}

\newtheorem{definition}{Definition}[section]
\newtheorem{theorem}[definition]{Theorem}
\newtheorem{axiom}[definition]{Axiom}
\newtheorem{corollary}[definition]{Corollary}
\newtheorem{conjecture}[definition]{Conjecture}
\newtheorem{counterexample}[definition]{Counterexample}
\newtheorem{lemma}[definition]{Lemma}
\newtheorem{sublemma}[definition]{Sublemma}
\newtheorem{proposition}[definition]{Proposition}
\newtheorem{question}[definition]{Question}
\newtheorem{remark}[definition]{Remark}
\newtheorem{application}[definition]{Application}
\newtheorem{idea}[definition]{Idea}

\newcounter{ccounter}
\newcommand{\C}{C_\arabic{ccounter}}

\begin{document}

Let
\begin{equation}\label{lfl}
h(x)=x^2.
\end{equation}
This seems to mess up the constants.

\begin{corollary}
Given $4 \leq a < b$ and the prime divisors of $a$ and $b$ come from
the first $k$ primes, then
\begin{equation}
\refstepcounter{ccounter}
b-a>a/(\log a)^{\C\label{what}} \\
\end{equation}
where
$C_{\ref{what}}=C_{\ref{what}}(k)=\min\{8e\cdot30^{k+3}k^{4.5},16
\cdot 2^{6k+20}\} k\log\log k$.  Moreover, if $k$ is sufficiently
large, then
\begin{equation}
\refstepcounter{ccounter}
b-a>a/(\log a)^{\C\label{next}}
\end{equation}
where $C_{\ref{next}}=C_{\ref{next}}(k)=30^{k+4}$.
\end{corollary}
Now we can refer to $C_{\ref{what}}$ and $C_{\ref{next}}$

\end{document} [/code]

I like this a lot! Especially after using the code and definition below:
[code]%~~~CONSTANT COUNTER~~~
\newcounter{ccounter}
\newcommand{\C}{C_\arabic{ccounter}}
\def\rsc{\refstepcounter{ccounter}}
[/code]
It seems that this way I can number the constants as used in certain theorems, equations, etc.
Thank you very much Stevem!


topology
advanced fields
advanced fields unsolved

Is every Lindelof space a Hausdorff space? Is every Lindelof space a second-countable Hausdorff space? If not, what are some examples?

A rather non-illuminating example is that every set $ X$ with the indiscrete topology cannot be Hausdorff and is trivially Lindelof.

The uncountable excluded point topology is Lindelof and not second-countable. (Excluded point: Let $ U\subset X$ and $ p\in X$, then $ U$ is open iff $ p\notin U$).


设整数a,b的最大公约数为1,整数c,d同样.证明(ac+bd)与(ad-bc)或者(ad+bc)与(ac-bd)的最大公约数为1.

反例: $ a \equal{} \minus{} 4, b \equal{} 7, c \equal{} 8, d \equal{} 1$.

我验证了楼上的反例，果然如此，可见原命题为假。
楼上该不会是用电脑穷举的吧

该命题等价于证明$ (a \plus{} bi)(c \plus{} di)$或$ (a \plus{} bi)(c \minus{} di)$无实因子。令$ x, y$为两不共轭的虚高斯素数，再令$ a \plus{} bi \equal{} xy, c \plus{} di \equal{} x\overline{y}$，则$ y\overline{y}|(a \plus{} bi)(c \plus{} di)$，$ x\overline{x}|(a \plus{} bi)(c \minus{} di)$，与原命题矛盾。

楼主的问题少了一个条件，好像是要求a，c互质；b，d互质
我记得在做一道关于2平方和的问题时会发现本质是楼主提出的这个问题。


let $ f: X\rightarrow Y$ be a function between topological spaces:

Show that $ f$ is continuous iff for all $ A  \subset X$, $ f(\overline{A})  \subset  \overline{f(A)}$

$ \Rightarrow$: Suppose $ f$ is continuous.  For any $ A$ we have $ A\subset f^{ \minus{} 1}\left(\overline{f(A)}\right)$, but by continuity $ f^{ \minus{} 1}\left(\overline{f(A)}\right)$ is closed, and hence $ \overline{A}\subset f^{ \minus{} 1}\left(\overline{f(A)}\right)\Rightarrow f\left(\overline{A}\right)\subset \overline{f(A)}$.
$ \Leftarrow$: Suppose $ f(\overline{A})\subset \overline{f(A)}$ for each $ A\subset X$.  Let $ B\subset Y$ be any closed set, and suppose $ f^{ \minus{} 1}(B)$ is not closed, then there is an $ x\in \overline{f^{ \minus{} 1}(B)}$ such that $ x\not \in f^{ \minus{} 1}(B)$.  But then $ f(x)\not \in B \equal{} \overline{B}$, contradiction.  Therefore, $ f^{ \minus{} 1}(B)$ is closed, so $ f$ is continuous.


Does anyone know where I can find some of the tutorials by Arkadii Slinko. The website doent work :maybe:

I think the website you're talking about is now here:
[url]http://www.math.auckland.ac.nz/olympiad/[/url]

I m talking about the stuff that was/is on the matholymp website


One side of an isosceles triangle has a length of 7. What are the possible lengths of the other two sides?

i think the other side can be anything in the range (0,14) if the two equal sides are 7.

[hide]YOu didn't say that that the side that has a length of seven is the side that is equal to another side. So, the equal sides can both be 4 and up. They can not be under 4 because then one side will be equal to more than the sum of the other 2 sides.

If seven is the side that is equal to another side then obviously one other side will be seven, but the other side can be any of these numbers 1,2,3,4,5,6,7,8,9,10,11,12,13. Since you're saying that isoceles is 2 or more of the same sides i guess 7 can be included as well.[/hide]

A isosceles is a triangle with [i]at least[/i] two sides of equal length.


Problem 2 (1995)
Study the convergence of the sequence defined by u(0)>=0
and
u(n+1) =sqrt(u(n)) + 1/(n+1)

for all nonnegative intergers n
-------------------------------------------------------------------
Problem 2 (1996)
Let a,b be positive intergers with a odd.Define the sequence u(n) as
follows

u(0) =b

u(n+1) =  1/2.u(n) if u(n) is even or u(n) +a otherwise

i)Show that u(n) <= a for some n in N

ii)Show that u(n) is periodic from some point onwards.

Problem 2 (1996) :

1)
if u(n) > a :
. if u(n) even u(n+1)=u(n)/2 < u(n)
. if u(n) odd u(n+2)=(u(n)+a)/2 < u(n)

So it's possible to construct a sub-sequence strictly decreasing of u(n) until u(n)<=a.

2)
as soon as u(n) reaches a previously reached value u(n) becomes periodic, and since there are only finite values <= a ....


[quote="mathematica"]Find all positive integers like $n$ so that:

$\tau (n)+\phi (n)=\sigma (n)$[/quote]
$\tau(n)$ is number of divisors or number of prime divisors?

[quote="Rust"]
$\tau(n)$ is number of divisors or number of prime divisors?[/quote]
$\tau (n)$ is the number of divisors.

[quote="Ilthigore"]I think that's right.

Suppose not, then $\phi(n)+\tau(n) \leq n<\sigma(n)$, I think.[/quote]

Well $\phi(4)+\tau(4)=5>4$, but that was a good hint. If you bump it up to $n+1$ then it works. I think this is a solution...

Suppose $n$ has $k$ divisors. Then $\phi(n) \le n-k+1$ and $\tau(n)=k$, so that adding gives $\phi(n)+\tau(n) \le n+1 \le \sigma(n)$ in which equality can happen iff $n$ is prime, since if $n$ is not prime then $\sigma(n)>n+1$ and if $n$ is prime then $\phi(n)+\tau(n)=p-1+2=p+1=\sigma(n)$.

I wonder if 4 is the only composite integer for which this happens?


Four rectangular paper strips of length $10$ and width $1$ are overlapping perpendicularly in a square pattern such that each strip meets two other strips. How much area is covered?

\[ \text{(A)}\ 36 \qquad \text{(B)}\ 40 \qquad \text{(C)}\ 44 \qquad \text{(D)}\ 98 \qquad \text{(E)}\ 100  \]

[hide]
Since each rectangle has a width of 1, you have to remove 4 from the original area.
$40-4=36$
So, $A$[/hide]

pascal_1234 you are wrong, the answer is E which is 100 because draw a diagram and you will realize.

Bbbbeeeeeeeeewwwwwwaaaaaarrrrrrrreeeeeeeee of what lies ahead

[quote="blackscorpio1120"]actually i think pascal's way of solving the problem is incorrect
but his answer is correct

[hide]
each side is 10 units
making the square a total of 100 sq. units
if you set up the diagram you have to reduce each side by 2 since each side is covered with 1 unit
so now the inner square is 8 by 8 units and the outer is 10 by 10

$100 - 64 = 36$

so the asnwer is, $A$
[/hide][/quote]

yeah i'm gonna have to go with blackscorpio. Try diagramming it. You can make a square, or anything similar to the # sign....

you'll get 4 overlapped squares which are $1 unit^2$. The other areas of remaining rectangles are going to be 8 each. Since there are 4 "8-unit squared" areas...

4+8+8+8+8 = 36

Pascal's way is faster. You have 4 strips each of area 10 for a total of 40, and 4 little boxes with area of 1 where they overlap, so the answer is 40-4=36. Your way is longer--adding up each of the individual rectangles. ;)

How is pascal wrong? The four strips have an area of 10 units each and added together gives us 40 units.

Yeah, pascal is right. (And hey, my sister is actually right too! :P)


Anyone know where I can get a free audio converter to convert diff. types of audio files?

Specifically, I am looking for one that works with RealPlayer files.  All the ones I've found don't work with .rm's

Itunes can....but the best one I can think of is Cool Edit Pro 2.0 but you have to buy it

Is there a free MIDI to Wav or MIDI to MP3 converter?

dbpoweramp. google it. its free, but the mp3 conversion lasts only like 20 days or something. You can always download freee mp3 codecs though if you really need, use LAME or something like that...

Thats dbPowerAMp

real all in one media convertor.
you have to pay $\$12$ to get the full software though, but it works perfectly. i useit  myself.


[quote="MCrawford"]A parking structure contains 300 monthly spaces, 1500 daily spaces, and 100 executive spaces.  If the structure is to be enlarged proportionally to 2090 spaces, how many daily spaces should be added?[/quote]

[hide="solution"]
It currently has 1500 + 300 + 100, or 1900, spaces. Increasing it to 2090 means increasing it by 190 spaces, which means it is increased by 10%, There are 1500 daily spaces, and 10% of that means [b]150 daily spaces should be added[/b].[/hide]

[quote="MCrawford"]A parking structure contains 300 monthly spaces, 1500 daily spaces, and 100 executive spaces.  If the structure is to be enlarged proportionally to 2090 spaces, how many daily spaces should be added?[/quote]

[hide]there are 1900 spaces.
you need 190 more
that is 1/10
1500/10=150[/hide]

[color=indigo]Solution:[/color] [color=white]150 more daily spaces[/color]


It was inspired by a lot easier related problem on my classes today.

Determine if there $ \exists_{n\in \mathbb{N}}$ such that $ \forall_{k\in \mathbb{N} \ \wedge \ A \equal{} \{a_{1},...,a_{n}\} \equal{} \{k,k \plus{} 1,...,k \plus{} (n \minus{} 1)\}} \ \exists \ B\subset C \equal{} \{1,2,...,n\}$ such that $ \sum_{i\in B}a_{i} \in \mathbb{P}$

Any ideas?  :wink:

[quote="polskimisiek"]It was inspired by a lot easier related problem on my classes today.

Determine if there $ \exists_{n\in \mathbb{N}}$ such that $ \forall_{k\in \mathbb{N} \ \wedge \ A = \{a_{1},...,a_{n}\} = \{k,k + 1,...,k + (n - 1)\}} \ \exists \ B\subset C = \{1,2,...,n\}$ such that $ \sum_{i\in B}a_{i} \in \mathbb{P}$

Any ideas?  :wink:[/quote]
Don't exist n .
We prove that there exist an integer k such that $ \forall B\subset C$ then
$ \sum_{i\in B}a_i$ is a composite.
$ \sum_{i\in B}a_i = |B|k + C$
Consider $ p_1 < .. < p_{2^{n}}$ such that $ p_1 > n$
Then for each subset B there exist an integer k such that
$ |B|_{i}k + c\equiv 0 (\mod p_i^2)$
From chines remainder theorem this system has a solution k .
It gives that $ \sum_{i\in B}a_i$ is an composite for any $ B\subset C$

[quote="polskimisiek"]You wanted
$ |B|_{i}k \plus{} c\equiv 0 (\mod p_i^2)$ right?
If yes then how you define c and why an existence of a system of congruences gives us that our sum is composite?[/quote]
I take $ p_i>|B|$ so $ \{|B|_i k\plus{}C\}_{k\equal{}1}^{p_i^2}$ is a complete reside mod $ p_i^2$
There exist $ k_i$ such that $ |B_i|k_i\plus{}C\equiv 0(\mod p_i^2)$
Consider the system $ x_i\equiv k_i(\mod p_i^2)$
It has a solution  :)

[quote="TTsphn"]
Then for each subset B there exist an integer k such that
[/quote]

Wait, we don't have the set B fixed. It is changing in terms of k, so we can't do sth like that. :huh:

Yes but I consider the $ |B|$
Consider
 $ \{c_1,..,c_n\} \equal{} \{0,1,...,n \minus{} 1\}$
Then $ \sum_{i\in B}a_i \equal{} |B|k \plus{} \sum_{i\in B}c_i$
Because $ |B|$ and $ \sum_{i\in B}c_i$ is constants .

Aaah, now I understand . Thank you  :) Beautiful solution :wink:


calculus
derivative
calculus computations

[quote="Thales418"]
${\sum_{n=1}^{\infty }\frac{n}{n!}=\sum_{n=0}^{\infty }\frac{1}{n!}=e}$
[/quote]
It's wrong...

read again

first we have

n=1

second we have

n=0

just changing the variables

[quote="Kent Merryfield"]Don't forget that $0!=1.$[/quote]
That's why I was wrong... Thanks... :blush:


Find all $ z$ such that $ 2/z \ge 1$.

I don't get why you can't just multiply by $ z$ on both sides.  Could someone help, please?

Thanks in advance.

You'll want to be a little rigorous when you solve it, so you'd want to consider two cases before multiplying by $ z$: $ z > 0$ and $ z < 0$. Note that $ z\neq0$, or else the L.H.S. is undefined.

If $ z > 0$, then $ \frac {2}{z}\geq1 \implies 0 < z\leq2$.

If $ z < 0$, then $ \frac {2}{z} < 0$ and $ \frac {2}{z}\geq1 > 0$, which is absurd.

Thus the latter case is not possible, and our solution is $ \boxed{z\in(0,2]}$.

EDIT: The post after me just disappeared!  :o


abstract algebra
superior algebra
superior algebra unsolved

Hello,

does someone know an example of a
Non-Noetherian ring having finite
Krull dimension?

The ring of integers in an algebraic closure of the field of $ p$-adic numbers is not noetherian and has dimension $ 1$.

Thank you! Do you know
an exposition of this example
somewhere in the literature or
on the web?

Is it essential to take an algebraic
closure of this field? Does it not work
with the field you mentioned itself?

.,.

$ \mathbb{Z}_p \equal{} \mathcal{O}_{\mathbb{Q}_p}$ is noetherian because it's a DVR, hence a PID. but $ \mathcal{O}_{\overline{\mathbb{Q}_p}}$ is not noetherian (since otherwise, since it's a valuation ring, it would be a DVR).

an easy example: $ (\mathbb{F}_2)^\infty$ is non-noetherian and of dimension $ 0$.


Let $ a,b,c,d$ be numbers such that:
\[ a\equal{}\sqrt{4\plus{}\sqrt{5\plus{}a}}\] \[ b\equal{}\sqrt{4\plus{}\sqrt{5\minus{}b}}\] \[ c\equal{}\sqrt{4\minus{}\sqrt{5\plus{}c}}\] \[ d\equal{}\sqrt{4\minus{}\sqrt{5\minus{}d}}\] Compute $ a\cdot b\cdot c \cdot d$

They, along with their negatives, are the roots of ((x^2-4)^2-5)^2=x^2.  The constant term is 121, so their product is 11 or -11.  a, b, c, and d are positive, so the product is 11.


inequalities
geometry
inequalities unsolved

[color=darkblue]Let $ ABC$ is triangle, let $ m_a, m_b, m_c$ are length of medians, $ h_a, h_b, h_c$ are length of altitudes, $ S$ is area of triangle.
Prove that: $ \boxed{(m_a^2 \plus{} m_b^2 \plus{} m_c^2)(h_a^2 \plus{} h_b^2 \plus{} h_c^2) \ge 27S^2}$.[/color]

It is well known that
\[ m_{a} \equal{} \sqrt{\frac{2b^{2} \plus{} 2c^{2} \minus{} a^{2}}{4}}
\textrm{.}
\]

Using similar formulas for $ m_{b}$ and $ m_{c}$ we get
\[ m_{a}^{2} \plus{} m_{b}^{2} \plus{} m_{c}^{2} \equal{} \frac{3}{4} \left( a^{2} \plus{} b^{2} \plus{} c^{2} \right)
\textrm{.}
\]

On the other side, from $ \displaystyle ah_{a} \equal{} bh_{b} \equal{} ch_{c} \equal{} 2S$ we get
\[ h_{a}^{2} \plus{} h_{b}^{2} \plus{} h_{c}^{2} \equal{} 4S^{2} \left( \frac{1}{a^{2}} \plus{} \frac{1}{b^{2}} \plus{} \frac{1}{c^{2}} \right)
\textrm{.}
\]

Now, AM-HM inequality
\[ \frac{1}{3} \left( a^{2} \plus{} b^{2} \plus{} c^{2} \right) \overset{AM\minus{}HM}{\geqslant}
3 \left( \frac{1}{a^{2}} \plus{} \frac{1}{b^{2}} \plus{} \frac{1}{c^{2}} \right)^{\minus{}1}
\]
gives
\[ \left( a^{2} \plus{} b^{2} \plus{} c^{2} \right) \left( \frac{1}{a^{2}} \plus{} \frac{1}{b^{2}} \plus{} \frac{1}{c^{2}} \right) \geqslant 9
\textrm{,}
\]
and we have
\[ \left( m_{a}^{2} \plus{} m_{b}^{2} \plus{} m_{c}^{2} \right)
\left( h_{a}^{2} \plus{} h_{b}^{2} \plus{} h_{c}^{2} \right) \geqslant
27S^{2}
\textrm{.}
\]

Equality holds if and only if $ a \equal{} b \equal{} c$.

Q. E. D.


Prove that there are infinite interger n such that
$ n|3^n\plus{}1$

[hide]
Let $ a_1 \equal{} 3^2 \plus{} 1$, and let $ a_n \equal{} 3^{a_{n \minus{} 1}} \plus{} 1$. Then $ a_1 \equiv 2 \pmod{4}$, and $ a_2 \equiv 3^{4k \plus{} 2} \plus{} 1 \equiv \minus{} 1^{2m} \plus{} 1 \equiv 2 \pmod{4}$ (where $ k,m \in \mathbb{Z}$), and in general, $ a_n \equiv 2 \pmod{4}$.

Now $ a_1 | a_2$. Suppose $ a_{n \minus{} 2} | a_{n \minus{} 1}$; we claim that $ a_{n \minus{} 1}|a_n$: As $ a_{n \minus{} 1} \equal{} 3^{a_{n \minus{} 2}} \plus{} 1$ and $ a_n \equal{} 3^{a_{n \minus{} 1}} \plus{} 1$, then $ a_{n \minus{} 1} | a_n$ if $ \frac {a_{n \minus{} 1}}{a_{n \minus{} 2}}$ is an odd integer (since $ x^m \plus{} 1 | x^{mn} \plus{} 1$ if $ n$ is odd). By our inductive assumption, $ \frac {a_{n \minus{} 1}}{a_{n \minus{} 2}}$ is an integer, and $ \frac {a_{n \minus{} 1}}{a_{n \minus{} 2}} \equal{} \frac {4k_1 \plus{} 2}{4k_2 \plus{} 2} \equal{} \frac {2k_1 \plus{} 1}{2k_2 \plus{} 1}$ must be odd.

Then the sequence $ a_n$ provides an infinite set of solutions.
[/hide]

Hello azjps I don't understand your induction step
[quote="azjps"]
...Suppose $ a_{n \minus{} 2} | a_{n \minus{} 1}$; we claim that $ a_{n \minus{} 1}|a_n$, and then the sequence $ a_n$ ...
[/quote]
Can you show me the full proof of this induction step please thx very much  :lol:


1. Two tangents are drawn to a circle from an exterior point A; they touch the circle at points B and C, respectively. A third tangent intersects segment AB at P and AC at R, and touches the circle at Q. If AB = 20, find the perimeter of triangle APR.

2. A circle is inscribed in a triangle with sides of lengths 8, 13, and 17. Let the segments of the side of length 8 made by a point of tangency be r and s, with r < s. Find the ratio of r:s.

3. Find the area of the inscribed circle of a triangle with sides of length 20, 21, and 29.

4. Segment AB is tangent at point A to the circle with center O; point D is interior to the circle, and DB intersects the circle at C. If BC=DC=3, OD=2, and AB=6, then find the radius of the circle.

Should that be AC in the first one?
And is the tangent on the major or minor arc of BC? If its on the outside, there is no one solution, and im not sure about the inside...

For problem 3, I think we can find it by using:
[hide]
$S=pr$[/hide]
And for problem 4, I get:
[hide]$\sqrt{22}$[/hide]


If q=2, any p is good (trivial),otherwise q is odd and (p+1)^q is a square iff
p+1 is a square iff p=3 (for p+1= n^2 <-> p= (n-1) (n+1) )

[quote]I think this should either be in Getting Started or Intermediate Topics. [/quote]
 :P  :lol:

[quote="BornForMath"][quote="Enrique Lesseps"]Find all primes $p,q$, for which $(p+1)^q$ is a perfect square.[/quote]
I think this should either be in Getting Started or Intermediate Topics.  :?[/quote]
it is not really suited for getting started... but should be intermediate topics.


Let $a_{1}\in (0,1)$ and $(a_{n})_{n\geq 1}$ the sequence of real numbers given by $a_{n+1}=a_{n}(1-a_{n}^{2})$ for any positive integer $n$. Find $\lim_{n\rightarrow\infty}\sqrt{n}a_{n}.$

It is easy, but long to prove.
 $\lim_{n\rightarrow\infty}\sqrt{n}a_{n}=\frac{\sqrt 2 }{2}.$

I can see that is the old topic, but I think my solution isn't very long.

1st of all, that is easy to show that $ \{a_{n}\}$ is bounded and decreasing, so it's convergent and it has a limit $ \lim_{n\to\infty}a_{n}=0$. And now:

$ \lim_{n\to\infty}\sqrt{n}a_{n}=\lim_{n\to\infty}\sqrt{\frac{n}{\frac{1}{a_{n}^{2}}}}$
Lets compute now:
$ \lim_{n\to\infty}\frac{n}{\frac{1}{a_{n}^{2}}}$, by Cesaro-Stolza lemma we have:
$ \lim_{n\to\infty}\frac{1}{\frac{1}{a_{n+1}^{2}}-\frac{1}{a_{n}^{2}}}=...=\lim_{n\to\infty}\frac{(1-a_{n}^{2})^{2}}{2+a_{n}^{2}}=\frac{1}{2}$, therefore:
$ \lim_{n\to\infty}\sqrt{n}a_{n}=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}$


geometry
trapezoid
rectangle
geometry unsolved

The quadrilateral $ABCD$ is inscribed in a circle with center $O$. The diagonals $AC,BD$ cut at $M\neq O$. The line through $M$ perpendicular to $OM$ intersects sides $AB,CD$ at $X,Y$. Show that $AB=CD\leftrightarrow BX=CY$.

If $AB = CD$ then the given quadrilateral is a trapezium (it can not be a rectangle since $M\neq O$). In this case trivially $XY\parallel BC\parallel AD$ and the desired conclusion is obvious. Another way round, let us at first bear in mind the theorem of butterfly which suggests that $XM = MY$. One then easily verifies that the quadrilaterals $OMYC$ and $OMXB$ are equal and hence $MC = MB$. But $MA\cdot MC = MB\cdot MD$ and hence $\triangle MDC = \triangle MAB$ so that $CD = AB$ as desired.


quadratics
number theory proposed
number theory

Solve the following diophantine in distinct natural numbers:
(x_1)^2+(x_2)^2+...+(x_2002)^2=1335(x_1+x_2+...+x_2002)

Maybe express the equation as a quadratic in x_k (like with Markov equations).

Observing that 1335=(2*2002+1)/3, we obtain the equality case in a problem given in a romanian team selection test by Laurentiu Panaitopol:
    If x_1<...<x_n are natural numbers then  \sum x_k 2  \geq (2n+1)/3* \sum x_k.


calculus
integration
logarithms
real analysis
real analysis unsolved

i asked this question for homework earlier this morning...

$ \int_{0}^{1}\int_{0}^{1}\;\;\frac{1}{\left(\ln x\;\plus{}\;\ln y\right)\cdot\left(xy\;\minus{}\;2\right)}\;\;\textbf dx\;\;\textbf dy\;\;\equal{}\;\;\boxed{\ln 2}$

[ PS: it can be transformed into a single integral by a clever change of variables :| ]

$ I\equal{}\int_{0}^{1}\int_{0}^{1}\frac{1}{\ln(xy)\, (xy\minus{}2)}dx\, dy$

For the inner integral substitute $ x\equal{}\frac{u}{y}$ then $ I$ becomes

$ \int_{0}^{1}\int_{0}^{y}\frac{1}{\ln(u)\, (u\minus{}2)}\frac{du}{y}\, dy$ . After it integrate in reverse order.

$ I\equal{}\int_{0}^{1}\int_{u}^{1}\frac{1}{\ln(u)\, (u\minus{}2)}\frac{dy}{y}\, du\equal{}\int_{0}^{1}\frac{\minus{}1}{u\minus{}2}du\equal{}\ln 2$


$ a,b,c,d,e,f,g,h,i$ are $ 9$ different single digit integers, such that the sum of $ \frac {a}{bc}$, $ \frac {d}{ef}$ and $ \frac {g}{hi}$ is $ 1$.

What are the digits represented by the letters $ a$ to $ i$?

There are at least two solutions.

[quote="tdl"]I have a solution:
$ \frac {1}{3.6} \plus{} \frac {5}{8.9} \plus{} \frac {7}{2.4} \equal{} 1$[/quote]

how do you arrive at this?


function invariant quadratics algebra polynomial Vieta induction

Show that there are infinitely many pairs of positive integers $ (m,n)$ such that
\[ \frac{m\plus{}1}{n}\plus{}\frac{n\plus{}1}{m}\]
is a positive integer.

[quote="nayel"]Show that there are infinitely many pairs of positive integers $ (m,n)$ such that
\[ \frac {m \plus{} 1}{n} \plus{} \frac {n \plus{} 1}{m}
\]
is a positive integer.[/quote]
it's fale
we must have $ mn| m\plus{}n$ so if $ n\le m$ we have $ n\le 2$......

Why should $ mn\mid m\plus{}n$? What if $ m\equal{}35, n\equal{}14$? Maybe you made a mistake somewhere, so check again. :wink:

The given condition implies that $ \frac{m(m\plus{}1)}{mn}\plus{}\frac{n(n\plus{}1)}{mn}$ is an integer, hence $ E \equal{} \frac{m^2\plus{}n^2\plus{}m\plus{}n}{mn}$ must be an integer.  If this is true, then obviously $ E\plus{}2$ must be an integer.  But $ E\plus{}2 \equal{} \frac{(m\plus{}n)^2\plus{}(m\plus{}n)}{mn} \equal{} \frac{(m\plus{}n)(m\plus{}n\plus{}1)}{mn}$.

So we really have to show that there are infinitely many pairs $ (m,n)$ such that $ \frac{(m\plus{}n)(m\plus{}n\plus{}1)}{mn}$ is an integer.  There are three general attack strategies that I see.

1. Show that as $ m$ varies over some infinite (not necessarily proper) subset of the positive integers, there always exists an $ n$ for which the given equation is an integer.

2. Create a function $ f(m,n) \equal{} (m',n')$ closed over the pairs of positive integers with an invariant $ g(m,n)$ such that $ (m', n')$ satisfies the condition if $ (m,n)$ does, and $ g(m',n') > g(m,n)$.  Also give a pair that satisfies this relation.

3. Find one $ m$ such that there are infinitely many $ n$ for which the relation holds.

Of these, I think that (2) is hardest but most likely to work, (1) is manageable but creating the set is hard, and (3) is easy to do, but unlikely to be true.

First, note that the pair $ (m,n) \equal{} (1,1)$ works.  In this case,
\[ \frac{m \plus{} 1}{n} \plus{} \frac{n \plus{} 1}{m} \equal{} 4.\]
In other words, $ (m,n) \equal{} (1,1)$ is a solution of the quadratic equation
\[ m^2 \minus{} 4mn \plus{} n^2 \plus{} m \plus{} n \equal{} 0.\]
Then...
[hide]use Vieta flipping.[/hide]

Define recurrences $ m_0\equal{}1$, $ n_0\equal{}1$, and $ m_k\equal{}4n_{k\minus{}1}\minus{}1$ and $ n_k\equal{}m_{k\minus{}1}$ for all $ k\in\mathbb{Z}^\plus{}$.

Then we have to go about proving that

i) both sequences are strictly increasing sequences of integers (use induction with $ m_k,n_k\ge 1$)
ii) the pair $ (m_k,n_k)$ for all $ k\in\mathbb{Z}^\plus{}$ makes that fraction equal to $ 4$ (another easy induction)

I should've thought of that. This is what I did:

[hide]We have $ \frac{m\plus{}1}{n}\plus{}\frac{n\plus{}1}{m}\in\mathbb N\iff \frac{(m\plus{}n)(m\plus{}n\plus{}1)}{mn}\in\mathbb N$. Let $ m\equal{}a(a\plus{}b),n\equal{}b(a\plus{}b)$. Then $ \frac{(m\plus{}n)(m\plus{}n\plus{}1)}{mn}\equal{}\frac{(a\plus{}b)^2\plus{}1}{ab}$. Hence we need $ k\equal{}\frac{a^2\plus{}b^2\plus{}1}{ab}$ to be an integer. Let us fix $ k$. It is well known that if there exists $ (a,b)$ such that $ k\equal{}\frac{a^2\plus{}b^2\plus{}1}{ab}$ is an integer then $ k\equal{}3$, and also that the equation $ \frac{a^2\plus{}b^2\plus{}1}{ab}\equal{}3$ has infinitely many solutions. Thus we conclude that there are an infinite number of intended pairs $ (m,n)$.

[hide="Proofs of the above facts"]Let $ (a,b)\equal{}(a_0,b_0)$ the pair satisfying $ \frac{a^2\plus{}b^2\plus{}1}{ab}\equal{}k$ and $ a_0\plus{}b_0$ is minimal, and $ a_0\ge b_0$. We will show that $ a_0\equal{}b_0$. Note that since $ (a_0,b_0)$ is a solution to $ a^2\minus{}kab\plus{}(b^2\plus{}1)\equal{}0$, then so is $ \left(\frac{b_0^2\plus{}1}{a_0},b_0\right)$. Hence by our assumption, $ b_0^2\plus{}1>a_0^2\ge b_0^2$, which implies $ a_0\equal{}b_0$. Thus $ 2a_0^2\plus{}1\equal{}ka_0^2$ implies $ a_0\equal{}1\equal{}b_0$, and thus $ k\equal{}3$.

Now to show that $ a^2\plus{}b^2\plus{}1\equal{}3ab$ has infinitely many solutions, verify that if $ (a,b)$ is a solution then so are $ (b,a)$ and $ (3b\minus{}a,b)$. With these, it's not difficult to show that we can generate infinitely many solutions.[/hide][/hide]

[hide="Thanks nsato."]
Assume to the contrary there is a finite set $ S$ of solutions to $ \frac {m \plus{} 1}{n} \plus{} \frac {n \plus{} 1}{m} \equal{} k$ for positive integer $ k$. By AM-GM, we have that $ k \equal{} \frac {m \plus{} 1}{n} \plus{} \frac {n \plus{} 1}{m} > \frac {m}{n} \plus{} \frac {n}{m}\ge 2$, so $ k\ge3$. Then we have there must be some solution $ (M,N)\in S$ that maximizes $ m \plus{} n$. WLOG assume that $ M\ge N$. Now consider the equation $ \frac {M \plus{} 1}{x} \plus{} \frac {x \plus{} 1}{M} \equal{} k$. This simplifies to:
\[ x^2 \plus{} (1 \minus{} kM)x \plus{} (M^2 \plus{} M) \equal{} 0
\]
Obviously, one root is $ x_1 \equal{} N$. By Vieta's, the sum of the roots is $ kM \minus{} 1$, so the other root is thus $ x_2 \equal{} kM \minus{} 1 \minus{} N$ and $ x_2$ is an integer. Similarly, $ x_2 \equal{} \frac {M^2 \plus{} M}{N}$, so $ x_2$ must be positive as well. Thus, $ x_2 \equal{} kM \minus{} 1 \minus{} N$ and $ M$ are positive integers which satisfy the given equation, so $ (x_2,M)\in S$. Since $ (2,2)\in S$, we have that $ M \plus{} N\ge4$ to maximize $ m \plus{} n$, and since $ M\ge N$,
\[ 2M\ge M \plus{} N\ge4 \\
M > 1
\]
Since $ (M,N)$ maximizes $ (m \plus{} n)\forall (m,n)\in S$, $ kM \minus{} 1 \minus{} N \plus{} M\le M \plus{} N$ or $ kM \minus{} 1\le 2N$. However, we have that $ M \minus{} 1 > 0$, $ 2M > 2N$, and $ kM\ge 3M$, so $ 3M \minus{} 1 > 2N$ and thus $ kM \minus{} 1 > 2N$, a contradiction, so our assumption must be false and there are infinite $ (m,n)$ which satisfy the given equation.

The problem is relatively straightforward using Vieta flipping. Nayel's solution uses it as well[/hide]

[quote="Altheman"]Define recurrences $ m_0 \equal{} 1$, $ n_0 \equal{} 1$, and $ m_k \equal{} 4n_{k \minus{} 1} \minus{} 1$ and $ n_k \equal{} m_{k \minus{} 1}$ for all $ k\in\mathbb{Z}^ \plus{}$.

Then we have to go about proving that

i) both sequences are strictly increasing sequences of integers (use induction with $ m_k,n_k\ge 1$)
ii) the pair $ (m_k,n_k)$ for all $ k\in\mathbb{Z}^ \plus{}$ makes that fraction equal to $ 4$ (another easy induction)[/quote]Where did you come up with those sequences?

I think he used Vieta to find them. :wink:

Since if $ (m,n)$ is a solution then so are $ (4m \minus{} 1\minus{}n,n)$ and $ (n,m)$.

Woops it should be $ m_1\equal{}n_1\equal{}1$.
$ m_{k\plus{}1}\equal{}n_k$ and $ n_{k\plus{}1}\equal{}4n_k\minus{}m_k\minus{}1$.

$ n_{k\plus{}1}\equal{}4n_k\minus{}n_{k\minus{}1}\minus{}1$

We can still prove everything...

[quote="nayel"]I think he used Vieta to find them. :wink:

Since if $ (m,n)$ is a solution then so are $ (4m \minus{} 1 \minus{} n,n)$ and $ (n,m)$.[/quote]Heh, I missed that...

Another solution (without Vieta jumping):

The Pell's equation $a^2-5b^2=1$ have infinitely many solutions in positive integers, hence there exist infinitely many $b$ such that $5b^2+1$ is a perfect square. Let $c=2b$. There exist infinitely many positive integers $c$ such that $5c^2+4$ is a perfect square. Let $n=\frac{-c+2+\sqrt{5c^2+4}}2$ and $m=n+c$. So $(2n+c-2)^2=5c^2+4$, $4n^2+4n(c-2)+(c^2-4c+4)=5c^2+4$, $n^2+n(c-2)-(c^2+c)=0$, $c^2+c+2n=n^2+cn$. Then we have \begin{align*}\frac{m+1}n+\frac{n+1}m&=\frac{m^2+m+n^2+n}{mn}\\&=\frac{n^2+2cn+c^2+n+c+n^2+n}{n^2+cn}\\&=2+\frac{c^2+c+2n}{n^2+cn}\\&=2+1\\&=3\end{align*}Therefore there exist infinitely many pairs $(m,n)$ such that $\frac{m+1}n+\frac{n+1}m=3$.

It will suffice to show that $(m+1)/n+(n+1)/m=3$ has infinitely many positive integer solutions. Indeed, it is not hard to verify that $m=2s_n(5s_n-p_n)$ and $n=2s_n(5s_n+p_n)$ works, where $p_n,s_n$ are given by $p_n+s_n\sqrt{5}=(2+\sqrt{5})^{2n-1}$ for all positive integers $n$, that is, all positive integer solutions of the negative Pell, $p^2-5s^2=-1$.

Generally, there are infinitely many $m,n\in \mathbb{Z}_+$ such that

$\frac{m+1}{n}+\frac{n+1}{m}=x$

holds for all positive integers $x$ such that the negative Pell's equation with coefficient $x^2-4$ is soluble. We will use a lemma very useful for solving Diophantine equations. (See attached.)

Multiplying through by $mn$ we have

$m^2+m+n^2+n=xmn$,

which, after writing $m^2+n^2=(m+n)^2-2mn$ and collecting like terms becomes

$(m+n)^2+m+n=(x+2)mn$,

which after factorization is

$(m+n)(m+n+1)=(x+2)mn$.

Then by the lemma, there exist $a,b,c,d,e,f\in \mathbb{Z}_+$ such that

$\begin{Bmatrix}
  m+n=abc, \\
	m+n+1=def, \\
	x+2=ad, \\
m=be, \\
n=cf.
 \end{Bmatrix}$

Further, since $x+2$ is a constant we might as well set $a=1$ and $d=x+2$ to get

$\begin{Bmatrix}
m+n=bc,\\ m+n+1=(x+2)ef,\\ m=be,\\ n=cf.
 \end{Bmatrix}$

Now, in order for the first equation to hold we must have $be+cf=bc$, or $cf=b(c-e)$; hence the lemma entails $u,v,w,z\in \mathbb{Z}_+$ such that

$\begin{Bmatrix}
c=uv,\\ f=wz,\\ b=uw,\\ c-e=vz.
 \end{Bmatrix}$

Therefore, substituting $c=uv$ into the last equation and rearranging we get $e=v(u-z)$, and so the previous system is equivalent to

$\begin{Bmatrix}
m+n=u^2wv,\\ m+n+1=(x+2)(u-z)vwz,\\ m=uwv(u-z),\\ n=uvwz.
 \end{Bmatrix}$

Further, combining the first two equations we get

$u^2wv+1=(x+2)(u-z)vwz$,

whence $vw$ divides $1$, giving $v=w=1$. Consequently, the system is transformed into

$\begin{Bmatrix}
m+n=u^2,\\ m+n+1=(x+2)(u-z)z,\\ m=u(u-z),\\ n=uz,
 \end{Bmatrix}$

and the equation is transformed into

$u^2-(x+2)uz+(x+2)z^2+1=0$.

Clearly the discriminant of this quadratic in $u$ must be a perfect square, hence

$(x+2)^2z^2-4((x+2)z^2+1)=t^2$,

which rearranges to give

$t^2-(x^2-4)z^2=-4$.

Thus, taking $t=2t_1$ and $z=2z_1$ we have

$t_1^2-(x^2-4)z_1^2=-1$,

which is a negative Pell's equation with coefficient $x^2-4$. Further, if this equation is soluble for the coefficient (e.g. for $x=3$), then it is well-known that all positive integral solutions are given by $(t_1,z_1)=(p_n,s_n)$, where $p_n,s_n$ are generated from the minimal solution $(p_0,s_0)$ by the formula

$p_n+s_n\sqrt{x^2-4}=\left(p_0+s_0\sqrt{x^2-4}\right)^{2n-1}\ \forall\ n\in \mathbb{Z}_+.$

Thereafter we will have $(t,z)=(2p_n,2s_n)$, and solving the quadratic in $u$ (substituting for $t$ and $z$ in the process) yields

$u=p_n\pm (x+2)s_n$.

Therefore, substituting this value back into the last system above we obtain

$\begin{Bmatrix}
m=(p_n\pm xs_n)(p_n\pm (x+2)s_n),\\ n=2s_n(p_n\pm (x+2)s_n).
 \end{Bmatrix}$

Lastly, since

$p_n^2=(x^2-4)s_n^2-1<(x^2+4x+4)s_n^2=(x+2)^2s_n^2$,

or what is the same thing,

$p_n<(x+2)s_n$,

we must take $\pm=+$ to ensure $m,n>0$. It follows that the pair

$\begin{Bmatrix}
m=(p_n+ xs_n)(p_n+ (x+2)s_n),\\ n=2s_n(p_n+ (x+2)s_n)
 \end{Bmatrix}$

generates infinitely many positive integer solutions to the problem, and we are done. [b]Q.E.D.[/b]

My solution:(can anyone please check it)
Let it be equal to some positive integer $k$.  We can also see that $(m, n) =(2,1);(1,2);(1,1)$ are a few solutions.  Let's assume that the number of solutions are finite. So we can take some $(M, N)$ from that set such that $M+N$ is maximum. Without the loss of generality assume that $N \ge M$ So the quadratic in $m$ is $x^2+x(1-kN)+N^2+N$.  It's roots are $M$ and $m_1$. Now $M+m_1=kN-1$ and $Mm_1=N^2+N$.  Second equation proves that $m_1>0$ and first one proves that $m_1$ is an integer. Now $m_1= (N+1) \frac{N}{M}$ since $N \ge M$ So $\frac{N}{M} \ge 1$ or $m_1 \ge N+1 > M$ So we get $m_1$ such that $m_1+N>M+N$ which are contradicts our assumption so the equation must have infinite number of solution in positive integers. :)


linear algebra
matrix
vector
linear algebra unsolved

Assume $ \alpha_{1},\alpha_{2},...,\alpha_{n}$ is a orthonormal basis on Euclid space $ \mathbb{R}^n$

and its Gram matrix is:

$ G\equal{}\left(\begin{array}{cccc}(\alpha_{1},\alpha_{1}) & (\alpha_{1},\alpha_{2}) & ... & (\alpha_{1},\alpha_{n})\\ (\alpha_{2},\alpha_{1}) & (\alpha_{2},\alpha_{2}) & ... & (\alpha_{2},\alpha_{n})\\ ... & ...& ... & ...\\ (\alpha_{n},\alpha_{1}) & (\alpha_{n},\alpha_{2}) & ...&(\alpha_{n},\alpha_{n})\end{array}\right)$

show that:

$ det(G)\equal{}det(\alpha_{1},\alpha_{2},...,\alpha_{n})$
 :roll:

Well, if in two-dimensional case one takes $ \alpha_1 \equal{} (0,1)$ and
$ \alpha_2 \equal{} (1,0)$, then $ det(G)\equal{} \minus{} det(\alpha_{1},\alpha_{2})$. So the correct statement of the problem is:

Demonstrate that $ |det(G)|\equal{} |det(\alpha_{1},\alpha_{2},...,\alpha_{n})|$.

The easiest solution is given in the message by Carcul. There is a general  statement which explains the geometric meaning of this exercise. Namely, for arbitrary vectors $ \alpha_{1},\alpha_{2},...,\alpha_{n} \in \mathbb{R}^n$  the volume of parallelepiped $ P$ with edges $ \alpha_{1},\alpha_{2},...,\alpha_{n}$ is equal to  $ |det(\alpha_{1},\alpha_{2},...,\alpha_{n})|$.

[quote="Carcul"]Orthonormality implies $ \langle \alpha_i, \alpha_j \rangle \equal{} \delta_{ij}$, which in turn means $ G \equal{} I_n$. On the other hand, if $ Q \equal{} (\alpha_1, \alpha_2, ..., \alpha_n)$ then $ QQ^T \equal{} G \equal{} I_n$, from which it follows that $ [\det(Q)]^2 \equal{} \det(G)$. But I am not seeing how we get $ \det(Q) \equal{} \det(G)$.[/quote]

If $ Q \equal{} (\alpha_{1},\alpha_{2}, ...,\alpha_{n})$

then $ Q^{T}Q \equal{}$$ \left(\begin{array}{cccc}\alpha_{1}^{T}\alpha_{1} & \alpha_{1}^{T}\alpha_{2} & ... & \alpha_{1}^{T}\alpha_{n} \\
\alpha_{2}^{T}\alpha_{1} & \alpha_{2}^{T}\alpha_{2} & ... & \alpha_{2}^{T}\alpha_{n} \\
... & ... & ... & ... \\
\alpha_{n}^{T}\alpha_{1} & \alpha_{n}^{T}\alpha_{2} & ... & \alpha_{n}^{T}\alpha_{n}\end{array}\right)$

But why $ Q^{T}Q \equal{} G \equal{} I_{n}$?, you probably say: $ (\alpha_{i},\alpha_{j}) \equal{} \alpha_{i}^{T}\alpha_{j}$

but i think for different inner product, the orthonormal basis are not the same ,in other words,

under the inner product $ \alpha_{i}^{T}\alpha_{j}$, $ \alpha_{1},\alpha_{2},...,\alpha_{n}$ probably not

a orthonormal basis.....

so we can not get the conclusion:

$ Q^{T}Q \equal{} I_{n}$ :roll:

In general, when talking about an orthonormal basis in R^n we assume the usual Euclidean inner product: $ \langle \alpha_i, \alpha_j\rangle \equal{} \alpha_i^T \alpha_j$, in which case my post above holds. But I see now that you are referring to an arbitrary inner product in R^n relative to which the $ \alpha_i's$ are an orthonormal basis. Is that it? Well, at least in that case we still have $ G \equal{} I_n$, by definition of orthonormality.

[quote="Carcul"]In general, when talking about an orthonormal basis in $ R^n$ we assume the usual Euclidean inner product: $ \langle \alpha_i, \alpha_j\rangle \equal{} \alpha_i^T \alpha_j$, in which case my post above holds. But I see now that you are referring to an arbitrary inner product in $ R^n$ relative to which the $ \alpha_i's$ are an orthonormal basis. Is that it? Well, at least in that case we still have $ G \equal{} I_n$, by definition of orthonormality.[/quote]

:D
I know that if $ \alpha_{i}'s$ are an orthonormal basis, we will have $ G\equal{}I_{n}$ by definition

So, there is actually only one inner product on $ R^{n}$

Thank you very much

[quote]So, there is actually only one inner product on $ R^n$[/quote]

No, that's false (unless you have something else on your mind). The weighted Euclidean inner product is another example: if $ x,y \in \mathbb{R}^n$, with $ x \equal{} (x_1,x_2,...,x_n)$ and $ y \equal{} (y_1,y_2,...,y_n)$, and $ 0 < \alpha_1,\alpha_2, ..., \alpha_n \in \mathbb{R}$, then

$ \langle x,y \rangle \equal{} \sum_{i\equal{}1}^{n} \alpha_ix_iy_i$

is an inner product in $ \mathbb{R}^n$.


$ k$ black pieces are placed on $ k$ consecutive squares of top row starting from upper left of a $ 2\times 5$ board. We are placing white pieces on empty squares one by one in arbitrary order. Two squares is said to adjacent if they have common vertex. When a white piece is placed on a square, the pieces on adjacent squares change their color. For which $ k$, when all the squares are filled, it is possible that color of every piece is white?

$\textbf{(A)}\ 0  \qquad\textbf{(B)}\ 1  \qquad\textbf{(C)}\ 2  \qquad\textbf{(D)}\ 3  \qquad\textbf{(E)}\ \text{None}$

[hide="Solution"]For $ k\equal{}1$,
[code]S _ _ _ _
_ _ _ _ _
[/code]

Now put W's to the  first row consequtively.

[code]W W _ _ _
_ _ _ _ _

W B W _ _
_ _ _ _ _

W B B W _
_ _ _ _ _

W B B B W
_ _ _ _ _

[/code]

Now write down number of neighbours of cells.

[code]
2 3 3 3 2
_ _ _ _ _
[/code]

In any permutation of bottom row placement, upper row will all convert to white.
Now let's find a permutation for bottom row. (2,4,1,3,5) is a desired permutation.

[code]
W B B B W
_ _ _ _ _

B W W B W
_ W _ _ _

B W B W B
_ W _ W _

W B B W B
W B _ W _

W W W B B
W W W B _

W W W W W
W W W W W
[/code]

So $ k \equal{} 1$, answer is $ B$.
[/hide]


Synopsis: The year is 2031. The state-owned [i]National Plagiarism Board[/i] has been disbanded by the Republican government then in place, leaving the door wide open for private, start-up ventures to begin. As the CEO of one of them, will you be able to build a successful, private company who sells exam cheating services?

Players:

Forum name:
Company name:
Headquarters:
Funds: (You all start out with 9999 credits)
_________________________________________

Forum name:egglyv999
Company name: AoPS
Headquarters:California
Funds:9999
_________________________________________

Units:

Driving license: Allows staff to drive a vehicle, 5 credits

Copier: Copies work from other students in class, but at a risk of detection inversely proportionnate to their experience, 5 credits

Thief: Physically steals work from other companies/their sources, but you always run the risk of detection, 15 credits

Telepath: Less efficient than a thief but undetectable, 25 credits

Security agent: Allows you to detect thieves if the police cannot arrive in time, 30 credits

Parking slots: Required to park your vehicle fleet, at the expense of a plot of land, 40 credits for 1 slot

Photocopier: Allows you to mass-manufacture papers, at the expense of an office space, 50 credits

Computer: Allows you to mass-manufacture cheating material on electronic media, at the expense of an office space, 65 credits

Warehouse space: Allows you to store papers, at the expense of a plot of land, 75 credits for a space

Hybrid car: Can carry 5 cheaters or their weight in cargo; a driving license is required to drive them and takes up a parking slot, 100 credits

Office space: Required to house photocopiers, computers, warehouse spaces and rotating presses, 125 credits for a space

Alarm system: Allows you to catch the thieves in combination with security agents; alerts the police if enemy thieves steal your equipment, 250 credits complex-wide

Rotating press: Allows you to mass-manufacture papers more efficiently than a photocopier, 400 credits

Hybrid bus: Can carry 50 cheaters or their weight in cargo over land; a driving license is required to drive them, also requires 2 parking slots, 750 credits

Yacht: Can carry an hybrid car and 30 cheaters or their weight in cargo overseas (or 60 without the hybrid car), 800 credits

Ferry: Can carry an hybrid bus and 50 cheaters or their weight in cargo overseas (or 120 without the bus, 90 if a car is onboard), 1500 credits

Land: Required to build your complex; variable costs depending on locations within states
_____________________________________

A state is considered fully captured by a company if said company is the only one to operate in the state. However, if two or more companies share the same state, they get income proportionnate to their market share.

Example: Gwori Cheating Industries and Texas Chainsaws have Texan market shares of 27% and 73% each. The quarterly income of the state is 4800 credits; Gwori Cheating Industries gets 1296 and Texas Chainsaws gets 3504 credits.

Two players cannot start in the same state. Only starting in a player's 2nd quarter of the game that one can invade another player's turf.
______________________________________

States:

The 54 "states" (for the purposes of this, the Virgin Islands, Guam, Washington DC and Porto Rico are considered states) are grouped into 6 Regions:
________________________________

Midwest:

Illinois (base quarterly income: 2600):

Indiana (base quarterly income: 1300):

Iowa (base quarterly income: 600):

Kansas (base quarterly income: 600):

Michigan (base quarterly income: 2000):

Minnesota (base quarterly income: 1100):

Missouri (base quarterly income: 1200):

Nebraska (base quarterly income: 400):

North Dakota (base quarterly income: 200):

Ohio (base quarterly income: 2300):

South Dakota (base quarterly income: 200):

Wisconsin (base quarterly income: 1200):
________________________________

Northeast:

Connecticut (base quarterly income: 700):

Delaware (base quarterly income: 200):

Maine (base quarterly income: 400):

Maryland (base quarterly income: 1100):

Massachusetts (base quarterly income: 1300):

New Hampshire (base quarterly income: 400):

New Jersey (base quarterly income: 1700):

New York (base quarterly income: 3900):

Pennsylvania (base quarterly income: 2400):

Rhode Island (base quarterly income: 300):

Vermont (base quarterly income: 200):

Washington D.C. (base quarterly income: 200):
_________________________________

Outside the contiguous 49:

Alaska (base quarterly income: 200):

Guam (base quarterly income: 100):

Hawaii (base quarterly income: 300):

Puerto Rico (base quarterly income: 800):

Virgin Islands (base quarterly income: 100):

__________________________________

Southeast:

Alabama (base quarterly income: 1000):

Arkansas (base quarterly income: 600):

Florida (base quarterly income: 3700):

Georgia (base quarterly income: 2000):

Kentucky (base quarterly income: 900):

Louisiana (base quarterly income: 900):

Mississippi (base quarterly income: 600):

North Carolina (base quarterly income: 1900):

South Carolina (base quarterly income: 900):

Tennessee (base quarterly income: 1300):

Virginia (base quarterly income: 1600):

West Virginia (base quarterly income: 400):
__________________________________

Southwest:

Arizona (base quarterly income: 1300):

New Mexico (base quarterly income: 400):

Oklahoma (base quarterly income: 800):

Texas (base quarterly income: 4800):
__________________________________

West:

California (base quarterly income: 7300):
AoPS 100%

Colorado (base quarterly income: 1000):

Idaho (base quarterly income: 400):

Montana (base quarterly income: 200):

Nevada (base quarterly income: 600):

Oregon (base quarterly income: 800):

Utah (base quarterly income: 600):

Washington (base quarterly income: 1400):

Wyoming (base quarterly income: 200):
________________________

Note: As chaos in this game, I will only have to report events as they occur. I will not have a company for myself. Also, base incomes can rise according to randomly decided inflation indexes.

This is not the starting funds you have! But you got the rest right.


Can someone describe how to rotate stuff, and also how to use the picture commands? I'm having some trouble...

What is 'stuff'? You need to be much more specific. One way of rotating text is to use:
\usepackage{rotating}
\rotatebox{90}{text}
will rotate text by 90 degrees anti-clockwise.

For the picture environment, see [url=http://tug.ctan.org/tex-archive/info/lshort/english/lshort.pdf]The Not So Short Introduction to LaTeX2e[/url] section 5.2


geometry
trigonometry
quadratics
trig identities
Law of Sines

$ A$, $ B$, and $ C$ are points on a circle of radius $ 3$ with $ \angle ACB \equal{} 30^\circ$. If $ AC \equal{} 2$, find length $ BC$.

[hide="my excruciating solution"]Call the center of the circle $ O$, with $ AO$ intersecting $ BC$ at $ X$. $ \angle AOB \equal{} 2\cdot30 \equal{} 60$.

Let $ \angle AXC \equal{} \angle BXO \equal{} \theta$. Let $ XO \equal{} h$.

Law of sines: $ \frac3{\sin\theta} \equal{} \frac {BX}{\sin60} \equal{} \frac {h}{\sin(60 \plus{} \theta)}$

$ \frac2{\sin\theta} \equal{} \frac {3 \minus{} h}{\sin30} \equal{} \frac {CX}{\sin(30 \plus{} \theta)}$

$ \implies \frac {3\sin(60 \plus{} \theta)}{\sin\theta} \equal{} h$

$ \frac {2\sin30}{\sin\theta} \equal{} 3 \minus{} h$

$ 3\sin(60 \plus{} \theta) \plus{} 2\sin30 \equal{} 3\sin\theta$

Writing everything in terms of $ \sin\theta$ gives the quadratic

$ 9\sin^2\theta \minus{} 3\sin\theta \minus{} 23/4 \equal{} 0\implies\sin\theta \equal{} \frac {1 \plus{} 2\sqrt6}6\implies\cos\theta \equal{} (1/6)\sqrt {11 \minus{} 4\sqrt6}$

From $ \frac3{\sin\theta} \equal{} \frac {BX}{\sin60}\implies \frac2{\sin\theta} \equal{} \frac {BX}{(3/2)\sin60}$

$ \frac2{\sin\theta} \equal{} \frac {CX}{\sin(30 \plus{} \theta)} \equal{} \frac {BX}{(3/2)\sin60} \equal{} \frac {BC}{\sin(30 \plus{} \theta) \plus{} (3/2)\sin60}$

where $ BX \plus{} XC \equal{} BC$

From here we just substitute the found values for $ \sin\theta,\cos\theta$ - it's gonna be very very messy... but is my solution correct so far?[/hide]

I've forgetten most of my sine and cosine but I would start from AB = 3 (equilateral triangle) and then use sine law on triangle ABC.

Right. Noticing that fact would yield a MUCH faster solution. I got $ \sqrt{11 \plus{} 4\sqrt6}$ by the way.
... not sure if my first solution is correct, but can this be solved with pure geometry without trig?

#$ \triangle ABO$ is equilateral and we get $ AB\equal{}3$.
Applying Cosine relation in $ \triangle ABC$ we get $ BC$#

Let O be the center of the circle. ACO is isoceles, so $ \angle ACO \equal{} cos^{ \minus{} 1}(1/3) > 30$. So BC separates A and O. So $ \angle BCO \equal{} cos^{ \minus{} 1}(1/3) \minus{} 30$. We have $ cos(BCO) \equal{} \frac {BC}{6}$ since BCO is isoceles, so $ BC \equal{} 6cos(cos^{ \minus{} 1}{1/3} \minus{} 30) \equal{} \sqrt {3} \plus{} 2\sqrt {2}$.

A nicer solution without trig:
Drop perpendicular $ AK$ to $ BC$. AKC is a 30-60-90 triangle, so $ KC \equal{} \sqrt {3}$. $ \angle AOB \equal{} 2\angle ACB \equal{} 60$, so AOB is equilateral, so $ AB \equal{} 3$, so by Pythag $ BK \equal{} \sqrt {8} \equal{} 2\sqrt {2}$. So the answer is $ \sqrt {3} \plus{} 2\sqrt {2}$


Let $ x_0 \equal{} 1$, and $ x_{n\plus{}1} \equal{}\frac{3 \plus{} 2x_n}{3 \plus{} x_n}$

Prove that $ lim  x_n$ exists, and find its value.

Do the same for $ x_{n\plus{}1} \equal{}\frac{1}{2 \plus{} x_n}$ provided $ x_0 \equal{} 1$

I'm not entirely sure how to prove that the limit exists...

But if the limit does exist, it is the point where recursing the sequence one more time cannot change the value.  For the 1st sequence, we have $ x \equal{} \frac {3 \plus{} 2x}{3 \plus{} x} \implies 3x \plus{} x^2 \equal{} 3 \plus{} 2x$ so $ x^2 \plus{} x \minus{} 3 \equal{} 0 \implies x \equal{} \frac { \minus{} 1 \pm \sqrt {13}}{2}$.  But only the plus holds since, from starting with a positive number, the recursive formula cannot generate a negative.

So $ \lim_{n \to \infty} x_n \equal{} \frac { \minus{} 1 \plus{} \sqrt {13}}{2}$.

For the second sequence, we do the same: $ x \equal{} \frac {1}{2 \plus{} x} \implies x^2 \plus{} 2x \minus{} 1 \equal{} 0$ so $ x \equal{} \minus{} 1 \pm \sqrt {2}$.  Again, we start with a positive number so only the positive value can hold: $ \lim_{n\to\infty} x_n \equal{} \sqrt {2} \minus{} 1$.  To prove that this exist, we can see that the recurrence gives the continued fractions $ [1]$, $ [0;2,1]$, $ [0;2,2,1]$, $ [0;2,2,2,1]$, etc. and it is well known that all periodic continued fractions (which we see the limit is) converge.

I'm not sure if this works and it might be complete bs but...

\[ \frac{3\plus{}2x}{3\plus{}x} < \frac{9\plus{}3x}{3\plus{}x} \equal{} 3
\]

Thus all terms are less than 3 and the sequence is strictly increasing so it must have a limit.

It's not complete BS. That's a good approach to show that there exists a limit. But why not use 3 + 2x < 6 + 2x?

x0  is positive .If  xn  be positive then xn+1 will be positive ,so every xn is positive by induction .

1<=   xn+1 = 1 + xn /(3+xn)  <= 2

xn+1- xn = ( - xn ^2 – xn +3 )/(3+ xn )

xn+2- xn+1 = xn+1 /(3+ xn+1) - xn /(3+xn)

xn+2- xn+1 = (3+2 xn)/(3(3+xn)+ (3+2 xn ) ) -  xn /(3+xn)

xn+2- xn+1 = (3+2 xn)/(12+5 xn )  -  xn /(3+xn)

xn+2- xn+1 = ( -3 xn ^2 – 3  xn +9 )/((12+5 xn ) (3+xn))

-3 xn ^2 – 3  xn +9 = 3  ( - xn ^2 – xn +3 )

The last relations show that if xn+1- xn  be positive then xn+2- xn+1 will be positive .Since  x1- x0 is positive we result by inductions that the sequence is increasing .An increasing and upper bounded sequence has a limit .

Perhaps the most direct way of showing that $ \lim_{n \rightarrow \infty} x_n$ exists is to find a closed formula for $ x_n$.

As shown in previous posts, this limit is likely to be $ r = (-1 + \sqrt{13})/2$.  So, let $ a_n = x_n - r$.  Then
\begin{align*}
a_{n + 1} + r &= \frac{3 + 2(a_n + r)}{3 + (a_n + r)} \\
\Rightarrow \quad a_{n + 1} &= \frac{3 + 2r + 2a_n}{3 + r + a_n} - r \\
\Rightarrow \quad a_{n + 1} &= \frac{3 + 2r + 2a_n - 3r - r^2 - ra_n}{3 + r + a_n} \\
\Rightarrow \quad a_{n + 1} &= \frac{3 - r - r^2 + (2 - r)a_n}{3 + r + a_n} \\
\Rightarrow \quad a_{n + 1} &= \frac{(2 - r)a_n}{3 + r + a_n}.
\end{align*}

Thinking that $ \lim_{n \rightarrow \infty} a_n = 0$, we can re-write the last equation as
\[ \frac{1}{a_{n + 1}} = \frac{3 + r + a_n}{(2 - r) a_n} = \frac{3 + r}{2 - r} \cdot \frac{1}{a_n} + \frac{1}{2 - r}.\]
You can then try to finish the rest.

By the way, I think this problem is a bit too difficult for "Intermediate Topics".

Let $ b_n \equal{} \frac {1}{a_n}$, $ c \equal{} \frac {3 \plus{} r}{2 \minus{} r}$ and $ d \equal{} \frac {1}{2 \minus{} r}$.  Then we have $ b_{n \plus{} 1} \equal{} cb_n \plus{} d$.  Starting with $ b_0$, we have $ b_1 \equal{} cb_0 \plus{} d$, $ b_2 \equal{} cb_1 \plus{} d \equal{} c^2b_0 \plus{} cd \plus{} d$, $ b_3 \equal{} cb_2 \plus{} d \equal{} c^3b_0 \plus{} c^2d \plus{} cd \plus{} d$, so in general $ b_n \equal{} c^{n}b_0 \plus{} d\cdot\frac {c^{n} \minus{} 1}{c \minus{} 1}$.

Then we have $ a_n \equal{} \frac {1}{c^n b_0 \plus{} d\cdot\frac {c^n \minus{} 1}{c \minus{} 1}}$ and thus $ x_n \equal{} \frac {c \minus{} 1}{(c \minus{} 1)c^n b_0 \plus{} d\cdot(c^n \minus{} 1)} \plus{} r \equal{} \frac {c \minus{} 1}{(cb_0 \minus{} b_0 \plus{} d)c^n \minus{} d} \plus{} r$

Now since $ c \approx 6.17129273 > 1$, $ \lim_{n \to \infty} \frac {c \minus{} 1}{(cb_0 \minus{} b_0 \plus{} d)c^n \minus{} d} \equal{} 0$ so $ \lim_{n\to\infty} x_n \equal{} r \equal{} \frac { \minus{} 1 \plus{} \sqrt {13}}{2}$

Note that in the other case, where $ r \equal{} \frac { \minus{} 1 \minus{} \sqrt {13}}{2}$, $ c \approx .162040604$ so the sequence's limit would be $ \frac {c \minus{} 1}{ \minus{} d} \plus{} r \equal{} \minus{} (3 \plus{} r \minus{} (2 \minus{} r)) \plus{} r \equal{} \minus{} (1 \plus{} 2r) \plus{} r \equal{} \minus{} 1 \minus{} r \equal{} \frac { \minus{} 1 \plus{} \sqrt {13}}{2}$ - the exact same value!


If you had 100% bank interest (doubled money every year) and you compounded it every second you would get a little more than 2.7 times the money.

1.0001^10000=2.71814592682522486403766467491315
1.00000001^100000000=2.71828181486763621765297724300918

(e)

How about 6% interest compounded every second?

[hide]
this is the original defination of $e$.[/hide]

[quote="probability1.01"]I thought the original definition was $\frac{d}{dx} e^x = e^x$, but anyhow, I don't think anyone here could possibly prove the formula.  It is an interesting fact, however.[/quote]
i don't know this definition before but i think it involves Calculus ($dx$ thing), isn't it??
One of my book says that $e$ was first discovered the way jimli does, and then Euler named it, and define it as the upper bound of:
\[a_n=(1+\frac1n)^n\]

but my book did not say about the exact year, so, does anyone know it??


trigonometry
geometry
3D geometry
pyramid
number theory
prime factorization

I know a couple handy formulas, but don't know the proofs for them. They are

1. Trig addition subtraction  - sin(x+y) = sinxcosy + sinycosx
2. Sum of proper factors of an integer x = x-1
3. Volume of a pyramid is 1/3 base * height

Can anyone prove any of them?

The second one isn't true..
eg 9: 1+3 = 4
10: 1+2+5 = 8..

What is a "proper factor"? Any factor that is not the number itself?

Correct. I still don't get why 0=-1 though...

I think he meant (sum of proper factors of x) = x-1..

Hum. Scratch that.

What is the expression for the sum of a numbers factors? I should know this, but I dont.

if a number's prime factorization is 2^a*3^b*5^c, for example, the number of factors is (a+1)(b+1)(c+1).  So for example 12 = 2^2*3, has (2+1)(1+1)=6 factors.  verify: 1, 12, 2, 6, 3, 4

He is asking for the sum of the factors though, not the number of factors.

I think its something like s = (1+p_1+p_1^2+...+p_1^k_1)(1+p_2+p_2^2+...+p_2^k_2)...(1+p_n+p_n^2+...+p_n^k_n)

woops, my bad.

thats right, if the prime factorization is p_1^k_1*p_2^k_2...*p_n^k_n

use the sum of a geometric series to simplify.

sorry :(
You can find a complete proof that sin(x+y) = sinxcosy+sinycosx at[url]http://www.cs.uleth.ca/~holzmann/notes/trigadd.pdf[/url]

bookworm,

I think the OP meant '-' as a dash, not as a minus or negative sign. For example, a dash as when giving a definition of a word:

Run - to walk rapidly


[quote="shfdfzhjj"][quote="shfdfzhjj"]a problem :)[/quote]
Any idea? :)[/quote]
If you post the original problem, I think many people will like it. With this looking, it seems to be ugly and not interesting.


inequalities
integration
geometry
geometric transformation
calculus
analytic geometry
function

let $1 \leq p < \infty$ be a real, $f: \mathbb{R}^{n}\to \mathbb{R}$ a $C^{\infty}$ function. For any ball $B(x,R)$, define
$f_{B}=\frac{1}{|B|}\int_{B}f(y) dy$
Show that there exists a constant $C$, depending only on $n$ and $p$ such that for any ball of radius $R$ the following inequality holds:
$\|u-u_{B}\|_{p}\leq C R \| \nabla u \|_{p}$

\[|u(x)-u_{B}|\leq \frac{(2R)^{n}}{n}\frac{1}{m(B)}\int_{B}\frac{\nabla u(y)}{|x-y|^{n-1}}\text{ d}y \]
implies
\[|u(x)-u_{B}|\leq c{n}\frac{1}{m(B)}\int_{B}\frac{|\nabla u(y)|}{|x-y|^{n-1}}\text{ d}y. \]
Now from the potential estimate we get
\[\int_{B}|u(x)-u(y)|^{p}\text{ d}y\leq c(n)^{p}(2R\omega_{n-1})^{p}\int_{B}|\nabla u(y)|^{p}\text{d}y. \]
where $\omega_{n-1}$ is the area of $\partial B_{n}(0,1)$.

I don't know what is this potential estimate, but if you admit the first inequality you have stated, this is just Hölder  :)

I just copied from my lecture notes. There was said the following: let
\[v\geq 0,v\in L^{p}(B(a,R)),p\geq 1 \]
and
\[u(x)=\int_{B(a,R)}\frac{v(y)}{|x-y|^{n-1}}\text{ d}y,\quad x\in B(a,R). \]
The potential estimate is the following inequality:
\[\int_{B(a,R)}u(x)^{p}\text{ d}u\leq (2R\omega_{n-1})^{p}\int_{B(a,R)}v(x)^{p}\text{ d}x. \]

[quote="puuhikki"]
\[|u(x)-u_{B}|\leq \frac{(2R)^{n}}{n}\frac{1}{m(B)}\int_{B}\frac{\nabla u(y)}{|x-y|^{n-1}}\text{ d}y \]
[/quote]

Proving this requires some calculation. By the translation invariance, we may assume that $x=0$. (Then the center of $B=B(a,R)$ is not the origin). Let $y\in B,y\ne 0$. Let $\gamma(t)=(1-t)y$ be a line segment from $y$ to $0$, $y\subset B$, $0\leq t\leq 1$. Now
\[u(0)-u(y)=\int_\gamma \nabla u\cdot d\bar{s}=\int_{0}^{1}\nabla u(\gamma(t))\cdot \gamma'(t)=-\int_{0}^{1}\nabla u(\gamma(t))\cdot y\text{ d}t. \]
Let $1-t=\frac{s}{|y|}$ so that $-\text{d}t=\frac{\text{d}s}{|y|}$. Now the integral equals
\[\int_{|y|}^{0}\nabla u\text{ d}s. \]
Integrate this over $B$:
\[\left |u(0)m(B)-\int_{B}u(y)\text{ d}y\right |\leq \int_{B}\int_{0}^{|y|}\left |\nabla u \left (\frac{y}{|y|}s\right )\right |\text{ d}s\text{ d}y. \]
Let $V(z)=|\nabla u(z)|$ as $z\in B$, $V(z)=0$ as $z\in\mathbb{R}^{n}\setminus B$.
Now the integral equals
\begin{eqnarray*}&&\int_{B}\int_{0}^{|y|}V\left (\frac{y}{|y|}s\text{ d}s\text{ d}y\right )\\ &\leq&\int_{B(0,2R)}\int_{0}^{2R}V\left (\frac{y}{|y|}s\text{ d}s\text{ d}y\right ) \end{eqnarray*}
By Fubini over the set $B(0,2R)\times [0,2R]$, this integral equals
\[\int_{0}^{2R}\int_{B(0,2R)}V\left (\frac{y}{|y|}s\right )\text{ d}y\text{ d}s. \]
The inner integral can be transformed to spherical coordinates, so we get the integral
\begin{eqnarray*}&&\int_{0}^{2R}\int_{0}^{2R}\int_{\partial B(0,1)}V(se)t^{n-1}\text{ d}S(e)\text{ d}t\text{ds}\\ &=&\frac{(2R)^{n}}{n}\int_{0}^{2R}\int_{\partial B(0,1)}V(se)\text{ d}S(e)\text{ ds}\\ &=&\frac{(2R)^{n}}{n}\int_{0}^{2R}\int_{\partial B(0,1)}V(se)\frac{S^{n-1}}{S^{n-1}}\text{d}S(e)\text{ d}s\\ &=&\frac{(2R)^{n}}{n}\int_{B(0,2R)}\frac{V(y)}{|y|^{n-1}}\text{ d}y\\ &=&\frac{(2R)^{n}}{n}\int_{B}\frac{|\nabla u(y)|}{|y|^{n-1}}\text{ d}y \end{eqnarray*}
Now divide both sides by $m(B)$.

Note. It was pretty hard to read the handwriting, so I hope I didn't do many typos.

something related:
Suppose that $f: \mathbb{R}^{n}\to \mathbb{R}$ is a $C^{\infty}$ function whose support lies in a ball $B$ with radius $R$. Show that:
$\| u \|_{p}\leq C(n,p) R \| \nabla u \|_{p}$


ARML probability LaTeX symmetry trigonometry geometry incenter

My Mock ARML Individual is finished.  Many, many, many, many thanks to CatalystOfNostalgia for his VERY helpful feedback on how to improve the problems.  :)

You all know how this works.  10 minutes per pair, no calculators.  Would a deadline for answer submission of Wednesday night work for people?  It would be nice to have this finished up before ARML.

While not required, a brief comment on how you solved the problem and what you thought of it would be appreciated.  Also, feel free to work on the problems after the time limit and send me your thoughts, as long as you clearly denote what your official answers are.  This is especially true of #8, which I expect very few people to solve within the time limit.

I'm not that good at making nice PDF's, so if someone could make one, that would be great.  :)

[hide="Set 1"]
1. Compute all real values of $ x$ such that $ \sqrt{\sqrt{x\plus{}4}\plus{}4}\equal{}x$.

2. A positive integer $ n$ is a yo-yo if the absolute value of the difference between any two consecutive digits of $ n$  is at least $ 7$ . Compute the number of $ 8$-digit yo-yos.
[/hide]

[hide="Set 2"]
3. In regular hexagon $ ABCDEF$ with side length $ 1$, $ AD$ intersects $ BF$ at $ G$, and $ BD$ intersects $ EC$ at $ H$.  Compute the length of $ GH$.

4. There are $ 4$ black balls and $ 1$ white ball in a hat. A turn consists of picking a ball from the hat and replacing it with one of the opposite color. Compute the probability that, after a sequence of turns, there are $ 5$ black balls in the hat before there are $ 5$ white balls.
 [/hide]

[hide="Set 3"]
5. The positive real numbers $ x_1, x_2, \ldots, x_{10}$ are in arithmetic progression in that order.  They also satisfy

$ x_1^2\minus{}x_2^2\plus{}x_3^2\minus{}\cdots\minus{}x_{10}^2\equal{}x_1\plus{}x_2\plus{}\cdots\plus{}x_{10}$.

Compute the common difference of this arithmetic progression.

6. Square $ ABCD$ has side length $ 2$.  $ M$ is the midpoint of $ CD$, and $ N$ is the midpoint of $ BC$.  $ P$ is on $ MN$ such that $ N$ is between $ M$ and $ P$, and $ m\angle MAN\equal{}m\angle NAP$.  Compute the length of $ AP$.
[/hide]

[hide="Set 4"]
7. Compute the number of $ 3$-digit base-$ 5$ positive integer multiples of $ 7$ that are also divisible by $ 7$ when read in base $ 10$ instead of base $ 5$.

8. For positive real numbers $ a,b,c,d$,

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

$ 2a^2\plus{}\sqrt{(a^2\plus{}c^2)(a^2\plus{}d^2)}\equal{}2cd$

$ 2a^2\plus{}\sqrt{(a^2\plus{}d^2)(a^2\plus{}b^2)}\equal{}2db$

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

Compute $ ab\plus{}ac\plus{}ad$.
[/hide]

Looking back, I believe that #8 has a fatal flaw. I PM'ed e^{pi*i}=-1, for now I guess don't open set 4  :P Eh I wish I had realized this earlier.

Oh well I'll try to actually make myself useful but I probably made a few typos in the LaTeX document.

Has this "fatal flaw" been fixed?

I've only done the first three sets. I thought it took a bit too long: I only finished one problem per set. If I hadn't messed up/had had the right idea immediately, I could probably have finished #4 and #6, but even so, it seemed to take a lot longer than, say, last year's individual round.

I don't think there is a fatal flaw.  Go ahead and do the last set.

Yeah, that's my fault. I messed up, as usual...

The period for submitting answers has ended.

[hide="Answers"]
1: $ \frac {1 \plus{} \sqrt {17}}{2}$

2: $ 1233$

3: $ \frac {\sqrt {39}}{6}$

4: $ \frac {5}{8}$

5: $ \minus{} 1$

6: $ \frac {5\sqrt {5}}{3}$

7: $ 2$

8: $ \frac {\sqrt {3}}{2}$
[/hide]

Results:

1st Place: (Tie)  Xantos C. Guin and perfect628, with scores of 6.  :winner_first:

2nd Place: (Tie)  azjps and ghjk, with scores of 5.   :winner_second:

Number of Participants: 14.

Average Score: 3.86.

The number of participants (out of 14) who solved each problem:

1: 6
2: 2
3: 11
4: 1
5: 14
6: 6
7: 10
8: 4

. . . which is pretty strange.   I sort of suspected that #5 was easy (I should have placed it in the first set, as I originally intended), but I didn't think everyone would solve it.  I didn't think #2 and #4 would be so hard - I guess that the techniques used aren't as well-known as I thought.  And #8 was unfortunate.  I should have realized that there would be an easier solution that the one I intended.

Please feel free to post your solutions here.

[hide="fuzzy #8"]By symmetry we say $ b\equal{}c\equal{}d$. Now solve for $ a$ and get that the answer is $ \frac{\sqrt{3}}{2}$.[/hide]

Yeah usually when you have symmetry like that i guess it would be safe to assume that b=c=d, but you can never be sure that it's the only solution...or at least I don't think you can

I'm sorry i don't really know how to use LaTex well but is this the intended solution for #8?

Since a,b,c, and d are postive reals let's assume that there exist a triangle with sides
$ sqrt(b^2 \plus{} a^2) , sqrt(a^2 \plus{} c^2)$, and (b + c), and also a 120 degree angle opposite the side with length (b +c).
Applying the law of cosines to the triangle yields the first equation. Creating similar triangles to suit the other equations, we notice that the final equation represents an the sum of the areas of the triangles with (1/2)sin120 factored out. We also notice that the expression we wish to find is also an area expression in terms of the triangles. So summing up cyclicly we get that the desired expression is equal to the sin120 or $ (sqrt3) /2$ as desired.

That's the basic idea.  What I did was take an equilateral triangle, let the distances from the vertices to the incenter be $ x$, $ y$, and $ z$, and the tangents to the incircle be $ b$, $ c$, and $ d$.  Then use Law of Cosines to express the sides in terms of $ x$, $ y$, and $ z$, and Pythagorean Theorem to express $ x$, $ y$, and $ z$ in terms of $ b$, $ c$, $ d$, and the inradius $ a$.  This gives you the first three equations.  The fourth is the result of the sine area formula for the three small triangles, and gives the area as $ \frac{\sqrt{3}}{2}$. The desired expression is $ rs$, which is also the area, so the answer is $ \frac{\sqrt{3}}{2}$.

My solution for #7:
From the hypothesis, we will have: 25a+5b+c is divisible by 7(1)
                                               100a+10b+c is divisible by 7(2)
Now substracting (2) to(1) we will have: 75a+5b is divisible by 7, which is equivalent to: 2a+2b is divisible by 7. Since 1=<a+b=<18 and(2,7)=1, we will have:
a+b=7(there are 6 pairs of positive integers (a,b))
a+b=14(there are 5 pairs of posivtive integers (a,b))
Considering each cases to count the number of solutions, I got the result is: 16,which i already checked satisfying both (1) and(2). But I still got it wrong :(. Can anyone interest in showing own right solutions or help me by pointing out my mistake in this solution? I'm very aprreciated about it.
@e^PI^(-1): Well, now I see the reason why I got #2 wrong! The problem said:"absolute difference", not just the difference, so it makes this a little bit harder than i did in the first time. Anyway, a nice problem! :)

Interesting-even though I used correct methods on every thing, I somehow got some wrong...

anyways, use recursion on #2, and states on #4.

[quote="azjps"]Some solutions can be found on the wiki: [[2008 Mock ARML 1 Problems/Problem 7]] (I hope you don't mind, E^(pi*i)=-1?).[/quote]
Thank you very much, azips! I didn't know the rule that when a number is written in the base-n, we just have n digits to express it: 0,1,2,...n-1. :(

[quote="perfect628"]Ugh, I made a calculation mistake on #4 and got some obscene answer...Also I missed #6 as geo is evil.[/quote]

I didn't even use geo except for pythagorean thereom. Mostly trignometery.

Eh math comps have become so that trig is more a tool of geo.

[quote="azjps"]Some solutions can be found on the wiki: [[2008 Mock ARML 1 Problems/Problem 7]] (I hope you don't mind, E^(pi*i)=-1?).[/quote]

Great!

[hide="Solution to #4"]
Note that the notation here is an improvement over mine, suggested by a couple of participants.

Let $ f(n)$ be the probability that we obtain $ 5$ blacks before $ 5$ whites, starting with $ n$ blacks.  When we have $ 4$ blacks, we move to $ 5$ blacks with probability $ \frac{1}{5}$, and to $ 3$ blacks with probability $ \frac{4}{5}$.  So

$ f(4)\equal{}\frac{1}{5}(1)\plus{}\frac{4}{5}f(3)$.

When we have $ 3$ blacks, we move to $ 4$ blacks with probability $ \frac{2}{5}$, and to $ 2$ blacks with probability $ \frac{3}{5}$.  So

$ f(3)\equal{}\frac{2}{5}f(4)\plus{}\frac{3}{5}f(2)$.

By symmetry, $ f(2)\equal{}1\minus{}f(3)$.  So, substituting and removing fractions, this equation becomes

$ 5f(3)\equal{}2f(4)\plus{}3\minus{}3f(3)\Rightarrow 8f(3)\equal{}2f(4)\plus{}3$.

The first equation becomes

$ 5f(4)\equal{}1\plus{}4f(3)\Rightarrow 8f(3)\equal{}10f(4)\minus{}2$.

Subtracting our equations, we see that

$ 8f(4)\minus{}5\equal{}0\Rightarrow f(4)\equal{}\frac{5}{8}$.

So the answer is $ \frac{5}{8}$.
[/hide]

[quote="Anaxerzia"]Really? You guys used trig? It was just LoC and Angle Bisector Theorem... I thought it was a lot easier than some of the others[/quote]

I drew a perpendicular from A to MN and used AP = length of perpendicular/cos

Question about problem 4:

If I tried to find $f(1)$, what would be the expression for that? I thought that $f(1)=\frac{4}{5}f(2)$, but if I use that we eventually get $f(4)=\frac{22}{25}$, which is wrong...

For #1, I got the answer, AND $\dfrac{-1 + \sqrt{13}}{2}$. The AoPSWiki agrees with me, but the problem writer doesn't. I just want to confirm with someone

all problems have been posted in separate threads in HSM  , I just created the related post collectionw

[url=https://artofproblemsolving.com/community/c3731763_2008_mock_arml_1_individual]here[/url]

enjoy  / start solving

more USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more ARML mocks [url=https://artofproblemsolving.com/community/c2442143_arml_mocks]here[/url],
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)


geometry
geometric transformation
search
national olympiad

ARO (= All-Russian Olympiad): http://www.mccme.ru/olympiads/vmo/30/zadachi.htm

The proposers of the problems are in the post below.

Here are the link of the problems,

9.1. http://www.mathlinks.ro/viewtopic.php?p=17839
9.2. http://www.mathlinks.ro/viewtopic.php?p=17842
9.3. http://www.mathlinks.ro/viewtopic.php?p=17843
9.4. http://www.mathlinks.ro/viewtopic.php?p=17845
9.5. http://www.mathlinks.ro/viewtopic.php?p=17846
9.6. http://www.mathlinks.ro/viewtopic.php?p=17847
9.7. http://www.mathlinks.ro/viewtopic.php?p=17848
9.8. http://www.mathlinks.ro/viewtopic.php?p=17849

10.1. = 9.1.
10.2. = 9.2.
10.3. http://www.mathlinks.ro/viewtopic.php?p=17850
10.4. = 9.4.
10.5. http://www.mathlinks.ro/viewtopic.php?p=17851
10.6. http://www.mathlinks.ro/viewtopic.php?p=17852
10.7. http://www.mathlinks.ro/viewtopic.php?p=17854
10.8. http://www.mathlinks.ro/viewtopic.php?p=17451

11.1. = 10.1. = 9.1.
11.2. http://www.mathlinks.ro/viewtopic.php?p=17918
11.3. http://www.mathlinks.ro/viewtopic.php?p=17915
11.4. http://www.mathlinks.ro/viewtopic.php?p=17916
11.5. http://www.mathlinks.ro/viewtopic.php?p=17919
11.6. http://www.mathlinks.ro/viewtopic.php?p=17924
11.7. http://www.mathlinks.ro/viewtopic.php?p=17926
11.8  http://www.mathlinks.ro/viewtopic.php?p=17450

Many thanks to Darij for helping me with the translation !  :)

9.1. proposed by Sergei Berlov
9.2. proposed by L. Emeljanov
9.3. proposed by Jury
9.4. proposed by Sergei Berlov
9.5. proposed by I. Bogdanov
9.6. proposed by O. Poblipskij
9.7. proposed by Sergei Berlov
9.8. proposed by A. Smirnov
10.1. = 9.1.
10.2. = 9.2.
10.3.  proposed by Sergei Berlov
10.4. = 9.4.
10.5. proposed by A. Protopopov
10.6. proposed by D. Karpov, A. Smirnov
10.7. proposed by W.Dolnikov
10.8. proposed by E. Tschernjschov, I. Bogdanov
11.1. = 10.1. = 9.1.
11.2. proposed by A. Akonjan, T. Emeljanov
11.3. proposed by A. Bjstrikov
11.4. proposed by I. Bogdanov, G. Tschelnokov
11.5. proposed by W. Senderov
11.6. proposed by O. Poblipskij
11.7. proposed by W.Dolnikov
11.8. proposed by S. Boltschjonkov

[quote="iura"]Did any mathlinker participate at ARO? (maybe Myth)
Also does anyone know the results?[/quote] :D Indeed, Myth did participate at ARO, just not this year as a student at least. He is a graduate student. He won IMO gold in 1997 in Argentina :P

Also Fedor Petrov and Fedor Bakharev are old ARO winners, which are part (or so I think) of the ARO committee. Maybe they can shed some light into this ... :?

Hi!
Orl, I wanted to ask, where from do you know so many things about olympiads, I mean new, just happened olympiads?
I also search the net, but I don't find them  :?

if you know Russian you can find them easily :) orl knows a little bit of russian, so he manages. :)

Here are the results:

http://www.mccme.ru/olympiads/mmo/2004/rez/ros-results.htm

Петухова Надежда Александровна and Дубашинский Михаил Борисович from Sankt-Petersburg go to IMO.  :)

You can download the problems as .pdf-file now.

Hi all.

As far as I know, no Russian high school student is a MathLinks member. And we use a lot of problems from here on our tournaments, like Kolmogorov memorial tournament.

As for this year olympiad, I consider problems worse than usually. Only 9.4=10.4 is more or less nice, other are either obvious or ugly.

As for our IMO team, I may say in private messages, how is "creative potential", which is the main criterion for form national IMO team, calculated. You would never guess yourself.

If our result will be better than in 2003, it will be a great astonishment for me.

Mathlinker andrew is a member of russian team.

[quote="Fedor Petrov"]As for this year olympiad, I consider problems worse than usually. Only 9.4=10.4 is more or less nice, other are either obvious or ugly.[/quote]

I don't think so, I personally find 9.2 nice, too. As you have said, 9.4 is beautiful, but it is a bit depressing, as one sees the solution which is so simple...

And if you wanna see really UGLY problems, then look at the ones of the [url=http://www.mathlinks.ro/viewtopic.php?t=5398]German NMO[/url] (especially the first three ones of classes 11-13). They claim that they already have prepared the problems for the next 2-3 years, but actually it is so obvious that they have not enough good problems...

  dg

[quote="Fedor Petrov"] As far as I know, no Russian high school student is a MathLinks member. And we use a lot of problems from here on our tournaments, like Kolmogorov memorial tournament. [/quote]

I feel Saint-Petersburg (SP) are appreciated around the world due to its beautiful and challenging problems. I think that we should further improve our friendship with our Russian friends. Maybe even more Russian students join MathLinks (ML) in the case you provide a link for ML on the SP link site: http://www.pdmi.ras.ru/~olymp/links.html  :)

[quote="orl"][quote="Fedor Petrov"]If our result will be better than in 2003, it will be a great astonishment for me.[/quote]

Hmm. Your remark seems to be surprising. As far as I can see the second Russian team performed well on the Bulgarian NMO. I assume the first Russian team must be even better. Maybe you can tell us your reasons why you think this way ?[/quote]

I just know these guys well. They are not too strong, indeed. And those six chosen are not the strongest six, in my opinion.

Why Russia does not choose the strongest six for IMO? It's really strange. How does Russia choose their IMO contestants?

"Six most loved by team leader (Nazar Agakhanov)"

[quote="logalexxxx"]I don't understand the point of view of the person who don't like Nazar Agakhanov. I think that he hasn't ever seen him.[/quote] Fedor represented Russia in IMO, so I would think that he has seen him a couple of times :)

[quote="Valentin Vornicu"]I have a question: does 12th grade exist in russia and you do not select people from it, or does it not exist at all in school ? :)

PS This year you made very well at the IMO :)[/quote]

Well, Russia has 11 grades. Before about 1990 they just even had 10 grades. Maybe their students being clearly younger than 20 is an disadvantage for their team. But on the other hand it depends what you get taught in school in the respective grades. Darij is rather a youngster too but did very well at IMO 2004...  :)

What is the real name of the Russian team member above ?

[quote="logalexxxx"]I don't understand the point of view of the person who don't like Nazar Agakhanov. I think that he hasn't ever seen him. But I have. He is a good and clever man.

So, and what about the criterias. In my opinion persons who have much talant should have some expirience if they want to be a winners. Therefore they should go to IMO not in 11 form at first, but in 9 form! Even there are persons from 11 form who are more clever, but have not an experience. So usually the team is consist of several persons from 11 form, 10 form and 9 form. Certainly we may make a team that will be consist only of persons from 11 form. And we will have a great results at IMO. But next year these persons will not be able to take part in IMO. And we will have not persons with experience. Therefore next year we will have a bad result. Anyway we should give students an opportunity to take part in IMO and have an experience.[/quote]

I don't agree. In my opinion always the best six contestants shall be chosen. Surviving all the mathematical rounds in the Russian Team Selection is quite challenging. People getting in the team will have enough experience.  :)

[quote="zortech"]I think a person who does not even know who Fedor Petrov is knows just NOTHING
about the Russian Team selection process.[/quote]

Yes, Fedor Petrov is one of the most important people in the Russian IMO Preparation. He proposed problems for ARO and some of the hardest problems for SPb MO. I think Nazar does not take Russian TST results too much into consideration but rather subjective feelings, China NMO results etc. What is the use of Russian TST then ? Maybe it is the best to take everything into account: ARO, Russian TST, China NMO etc. What do you think, especially the Russian guys ?  :)

[quote]
 I have a big doubt that F. Petrov is F. Petrov.
[/quote]
   F.Petrov is F.Petrov. I know him well. On the other hand, I am sure that logalexxxx is not logalexxxx. It is an impersonation, and pretty unconvincing at that  :)

[quote="Anonymous"]Yes, but I don't want to tell you my true name. I'm not a mad such as F. Petrov.[/quote]

You do not seem to be sufficiently self-confident. Everybody can state one's opinion whether other people agree or don't agree.

First of all, it's pity, that such avatars as zortech's are acceptable, the only thing that it has is the encoded dirty Russian word. :(  I think that English members would not like to see the avatar with the most popular word from their slang.

Looking from the side, my opinion is that the team leader must be the past IMO winner (such as e.g. S.Konyagin). Results are good now, but not the best. I think Russia is able to beat China. :D

[quote="nonIMO"]First of all, it's pity, that such avatars as zortech's are acceptable, the only thing that it has is the encoded dirty Russian word. :( [/quote]

ой, чёрт, я бы сам и не заметил... а это вправду то знаменитое слово, стоящее на заборах... ну да, конечно можно было бы выбрать картинку покультурней, но всё-таки так уж прямо в глаза не бросается...

  dg

[quote="logalexxxx"]Ok. I'm from Russia. And I know who F. Petrov is. Also I know who N. Agakhanov is. But I didn't believe that F. Petrov may have such an opinion. I have a big doubt that F. Petrov is F. Petrov.

So, F. Petrov didn't have much influence in Russian olympiad komitee, and there was very few problems that he had proposed to ARO.

I have proposed some problems to ARO too. And what is it implies?

What about russian system of education. There are ten forms in Russia now: 1, 2, 3, 5, 6, 7, 8, 9, 10, 11. But there is no 4 form. So, many years ago there was 4 form and so we have forms: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Only the names have changed but not the sense.

I don't understand the question: What will do N. Agakhanov if he read the words of F. Petrov?[/quote]

I hope, he would not send an assassin. So, whoever are you, please do not worry. I am not still the member of ARO jury and comission, hence I have no common business with Nazar. In a sense, it may be bad for my further probable students, who would become candidates - you understand, yeah, but on the other hand it's a challenge for me to teach them better.

[quote="Anonymous"]If USSR did not break up into Russia, Belarus, Ukraine, then USSR can beat China in 2000. ;)

In IMO 2000, China has 218 points, Russia has 215 points. But USSR will have an amazing score of 237 (!!) if Russia, Belarus and Ukraine are united under USSR.
Alexandr Usnich (BLR) 42
Alexandr Gaifoulline (RUS) 42
Alexei Poiarkov (RUS) 42
Iouri Lifchits (RUS) 38
Artsiom Kochan (BLR) 38
Maksym Fedorchuk (UKR) 35[/quote]

Yes, it could. And the last time Russia/USSR "really and officially" beat China (if we don't take into account a tie 182-182 in 1999 and IMO 1996, where both countries performed badly, Russia was 4th, China was 6th :))  was USSR in 1991 (the last year of it's existence): 241 points (China had 231).

Eugenia Malinnikova 777777 42
Alexandr Perlin 777777 42
Andris Ambaynis 777777 42
Mikhail Tyomkin 777777 42
Vsevolod Zhukhovitsky 673777 37
Vladimir Nekrashevich 673767 36

Russia didn't have such nice numbers after that, did it?  :D

[quote="nonIMO"]And the last time Russia/USSR "really and officially" beat China (if we don't take into account a tie 182-182 in 1999 and IMO 1996, where both countries performed badly, Russia was 4th, China was 6th :))  was USSR in 1991 (the last year of it's existence): 241 points (China had 231).

Eugenia Malinnikova 777777 42
Alexandr Perlin 777777 42
Andris Ambaynis 777777 42
[b]Mikhail Tyomkin 777777 42[/b]
Vsevolod Zhukhovitsky 673777 37
Vladimir Nekrashevich 673767 36[/quote]

Hmm, which Mikhail Tyomkin do you mean now? I know one Mikhail Tyomkin, he comes from Russia or Ukraine (I don't remember exactly), he was in the German IMO team last year and got a silver medal...

  Darij

Both are Михаил Темкин 's :) I don't know if they are relatives, but they are different people for sure :D One graduated from the Moscow 57th secondary school in 1991. The other, as I've known just now, was in the 8th form of Kharkov (Ukraine) PhML 27 in 1999.

[quote="nonIMO"]The other, as I've known just now, was in the 8th form of Kharkov (Ukraine) PhML 27 in 1999.[/quote]

Yes, this is very likely. At least, he is an undergraduate student in Munich now.

  Darij


Using estimation and/or rounding, can you determine if the product
in A) below is closer to 6000 or 7000?  Also, by the same way, can you tell if the product in B) below is closer to 6000 or 7000?  For either product
(or both), you might wind up electing to guess the answer.

Only check the "exact" values (for decisive closeness between the two numbers, that is) by using a calculator/computer (or longhand on paper) afterward to see if you are correct for either A) or B) or both.

A)  4.084*5.089*5.089*6.089*10.094

B)  4.171*5.171*5.171*6.171*10.171

[hide]A) i would say closer to 6000
B) i think this is a trick, so i'm gonna say 7000[/hide]

[hide="the real answers"]
A) is abt 6500.7, which is closer(just barely) to 7000.
B) is 7000.1, which is obviously closer to 7000[/hide]

If you had been presented with only product B) to be asked about, your answer
may have gone the other way because you wouldn't have had A) to compare it to, in my opinion.  The chance that you may have gone with 6000, though, may not have been "overwhelming."

I wouldn't have taken a second look at product A). So I would have blown it off, gone on my way, and thought I was correct about its value.  If I had not constructed it (or discovered it, if you will) myself, I would not believe it.  I really like A), and at the same time, it makes me mad (due to a lack of understanding on my part).

I looked at these using calculus-style estimation in my head; the first is $ 4\cdot 5^{2}\cdot 6\cdot 10\equal{}6000$ plus about (2+5+1)% (the 5% groups the three middle terms). This is obviously close to 6500, and thus needs more precision to answer; you've actually tuned it so close that it needs calculator-level precision.
That makes it... not a very good question, unless all you're trying to do is stump people. I'd use something that was at least 1% away from the boundary; that's within reason for a person's ability to estimate

The second is far easier- it's about twice as far away from 6000, which makes it close to 7000.

There is no "obviously" about it; that's one of many non-mathematical subjective words and phrases.

"...unless all you're trying to do is stump people."

I am inclined to stump because this section contains puzzles and brainteasers,
and I am, in part, a puzzler of certain kinds of mathematics.

No, my questions were of the type "CAN you determine" using one or both of those methods?  It doesn't state that with an undestanding you will be able to
successfully be able to do it with either method with either product.

This "1% away" desired level is another subjective quantity.

Amended (preferred) question:

The following product is closer to 6000 or to 8000?

4.171*5.171*5.171*6.171*10.171


Could the moderators of the "Other US Contests & Programs" forum and the "High School-Intermediate" forum, or the admististators, please move the obviously ARML-related topics over here? There's been quite a bit posted recently, and this would seem to be the obvious home for those topics.

I just moved the ones in Intermediate that I could find.  If there are any more PM me or one of the other moderators of Intermediate the name of it.

Btw, thank you admins for creating an ARML forum  :)

I was wondering why there wasn't any ARML forum...

[quote="DPatrick"]Since the moderator of the larger "Other US Contests" forum is also the moderator of this forum, I'll let him handle it.[/quote]

Done!


[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \geq\ \frac {a}{b \plus{} 2c} \plus{} \frac {b}{c \plus{} 2a} \plus{} \frac {c}{a \plus{} 2b}$[/quote]
An expanding gives $ \sum_{cyc}(2a^4c\plus{}a^3c^2\plus{}2a^3bc\minus{}5a^2b^2c)\geq0,$ which is obvious.

[quote=Hong Quy]We  have to prove:
$ \frac{a^2c}{b\plus{}2c}\plus{} \frac{b^2a}{c\plus{}2a}\plus{}\frac{c^2b}{a\plus{}2b}  \geq  abc \left(\frac{1}{b\plus{}2c}\plus{}\frac{1}{c\plus{}2a}\plus{}\frac{1}{a\plus{}2b} \right)$

The other hand, $ LHS \geq \frac{a^2c\plus{}b^2a\plus{}c^2b}{3}\cdot \left(\frac{1}{b\plus{}2c}\plus{}\frac{1}{c\plus{}2a}\plus{}\frac{1}{a\plus{}2b} \right)\geq RHS$[/quote]

 $ LHS \geq \frac{a^2c\plus{}b^2a\plus{}c^2b}{3}\cdot \left(\frac{1}{b\plus{}2c}\plus{}\frac{1}{c\plus{}2a}\plus{}\frac{1}{a\plus{}2b} \right)\geq RHS$ ?

[quote=nguoivn]Given $ a, b, c > 0$. Prove that: $ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \geq\ \frac{a}{b\plus{}2c}\plus{}\frac{b}{c\plus{}2a}\plus{}\frac{c}{a\plus{}2b}$[/quote]

[img]http://s2.sinaimg.cn/middle/006ptkjAzy77HD9pDPz41&690[/img]

[quote=xzlbq]$x,y,z>0$,prove that

\[{\frac {x}{y+2\,z}}+{\frac {y}{2\,x+z}}+{\frac {z}{x+2\,y}}\geq \frac{1}{3}\,{
\frac { \left( x+y+z \right) ^{2}}{xy+zx+yz}}\]

\[{\frac {x}{y+2\,z}}+{\frac {y}{2\,x+z}}+{\frac {z}{x+2\,y}}\geq \frac{1}{2}\,{
\frac { \left( \sqrt {{x}^{3}+xyz}+\sqrt {{y}^{3}+xyz}+\sqrt {{z}^{3}+
xyz} \right) ^{2}}{ \left( xy+zx+yz \right)  \left( x+y+z \right) }}\][/quote]
both of them is T2!
but the third is hard!;)

[quote=nguoivn]Given $ a, b, c > 0$. Prove that: $ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \geq\ \frac{a}{b\plus{}2c}\plus{}\frac{b}{c\plus{}2a}\plus{}\frac{c}{a\plus{}2b}$[/quote]
Because
\[\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} - \sum \frac{a}{b\plus{}2c} = \frac{\displaystyle \sum ca(2a+3b+c)(a-b)^2}{(ab+bc+ca)(a+2b)(b+2c)(c+2a)} \geqslant 0.\]

[quote=nguoivn]Thank you very much, arqady  :lol:
Indeed, I haven't try to expand. But I'm surprise that after expand, we obtain an easy ineq.[/quote]

How do you think these are made? Take a weird binomial expansion, shuffle it up to make it symmetric. Shuffle it up a it more. And you got an inequality. By the way FE's are similarly easy to make. Get a random trivial algebraic identity and add random $f()$ to terms. For example, x+y+xy=x+y(x+1) gives $f(x)+f(y)+f(xf(y))=f(x)+yf(x+1)$


Set $S$ consists of the first 20 natural numbers. How many non-empty subsets of $S$ are there such that the members of each subset have no common factors?

[hide="hint"]
There can only be one even number.
The rest have to be odd multiples of the odd primes.
[/hide]

[quote="mathcountsfreak"][hide]there aren't any because 1 is a common factor of all natural numbers[/hide][/quote]
Any common factors > 1.

[hide]lol ok: well we can start with some rules:
I. we can have 1 or 0 even numbers
II. we can have as many primes as we want as long they don't conflict with the even number (ie the prime number is not a factor of the even number)
III. we can choose whether or not to include 1
IV. odd multiples of primes are included if the prime(s) that they are multiples of are not.
V. or it could just be a set of 1 number

so we have 20 sets according to V

you can have the primes and 1 (there are 8 primes, so we have 9 total numbers that we can interchange, so there are $2^{9}= 512$ subsets, subtract 10 for the empty set and for the 9 sets of only 1 number, so we have 502.

if we use other even numbers:
using 4 we can still use the other primes there are 512 like before, we subtract 10 and then we have to subtract 256 because we have already counted all those involving just the 8 primes leaving us with 246.

with 6 we can't use 3 either, so we have $2^{8}-2^{7}-9 = 119$

with 8 it's the same as 2 and 4: 246

10 is like 6 except instead of not counting 3 we can't count 5: 119

12 is like 6 also: 119

14: 119

16 is like 2,4,8: 246

18: 119

20: 119

Now if we don't use any even numbers we have to use odd non-primes like we were using evens before

start with 9: if you use 9 you can't count 3, so there are still 9 numbers in our set (7 primes, 9, and 1), so there are $2^{9}-10-2^{8}= 246$

15: you can't count 3 or 5, so we have 6 primes, 1, and 15, so there are $2^{8}-9-2^{7}$...wait a second I just realized I'm subtracting the empty set and the 1 number sets twice...huh well I'm gonna work on this problem a little more and I'm gonna edit this to be right...[/hide]

im here, but i have no clue where to begin my answer

This is just annoying casework/brute force; the problem in essence isn't really that hard.


trigonometry
advanced fields
advanced fields unsolved