limit
real analysis
real analysis unsolved

A number t>1 is called special if there exists a number k>0 such that the distance between $k\cdot t^n$ and the set of integers tends to 0 when n goes to infinity. Prove that the set of special numbers is infinite and at most numerable. A very very beautiful problem!

I think that the fact that this set is infinite could be proved like this:
Take  u_p=( int(sqrt(p))+1 )+sqrt(p)
and    v_p=( int(sqrt(p))+1 )-sqrt(p)  so 0<v_p<=1
where p is an integer.
Then (u_p)^n + (v_p)^n is an integer. We can conclude because v_p^n->0..

i've just realised that all integer numbers are special numbers ;-) .. but that shows there are others. If there are uncountable many such numbers then there is at least one accumulation point of this set of special numbers .. does this lead to a contradiction ?

I was searching through old problems and found this one. In my humble opinion, it deserves a second life. As usual, if nobody solves it in a couple of weeks, I'll post a solution :).

It seems you wish to show that the [url=http://mathworld.wolfram.com/PisotNumber.html]Pisot numbers[/url] are countable? Or can there be some Pisot numbers that aren't "special numbers" by your definition?

Can Pisot's result about equation (7) on that page give you that special numbers are algebraic and thus countable? It seems an interesting result to me that no transcendental number could be a special number. :) I wish I could be more helpful. :P

[quote="Xevarion"] It seems an interesting result to me that no transcendental number could be a special number. [/quote]
If you could prove this, you would become quite famous :). Fortunately, the problem asks for much less.

@fedja: Can you tell us, what this "famous-making" conjecture is ? ;)

It is exactly as Xevarion put it: every "special" number is an algebraic (and therefore Pisot) number. The best known result so far is that, if $\sum_n \text{dist}(kt^n,\mathbb Z)^2<+\infty$ and $t>1$, then $t$ is algebraic. The problem has remained open for almost 70 years (the result I cited is due to Pisot and, if I remember it right, goes back to late 1930's).

It's a pity nobody has solved it  :( . Anyway, here is the promised solution. The idea is very natural. We know that the set of positive integers is countable. Hence, it would be nice to associate some integers with every special number $t$. But we have some integers coming right from the definition of $t$! They are the closest integers $m_n$ to the numbers $k\cdot t^n$. It is easy to show that, if we know the sequence $m_n$, then $t=\lim_{n\to\infty}\frac{m_{n+1}}{m_n}$, so $t$ is uniquely determined by the sequence $m_n$. The only little problem is that the set of infinite sequences of integers is not countable. But the set of finite sequences is. So it is natural to ask a question whether the entire sequence $m_n$ is completely determined by its several first terms, or, more specifically, whether we can find $m_{n+2}$ if we know $m_n$ and $m_{n+1}$ at least for large enough $n$. Since $\lim_{n\to\infty}|k\cdot t^n-m_n|=0$, we can restrict ourselves to the values of $n$ such that $|k\cdot t^\ell-m_\ell|<\delta$ for all $\ell\ge n$ where $\delta>0$ will be chosen later (it will depend only on $t$). Now, how to find $m_{n+2}$ from $m_n$ and $m_{n+1}$? We know that $|m_{n+2}-k\cdot t^{n+2}|<\delta$, which would uniquely determine $m_{n+2}$ if we knew $k\cdot t^{n+2}$ and if $\delta$ were small enough. Unfortunately, we do not know $k\cdot t^{n+2}$, but we could find it if we knew $k\cdot t^{n+1}$ and $t$ by multiplying them. We do not know $k\cdot t^{n+1}$ exactly, but we have a fairly good approximation to it, namely, $m_{n+1}$. Also, we do not know $t$ exactly, but we can try to approximate it by $\frac{m_{n+1}}{m_{n}}$. Thus, the idea arises that $m_{n+2}$ must be just the closest integer to $\frac{m_{n+1}^2}{m_n}$. Now let's see if we can justify this claim. We have
\[ \left|m_{n+2}-\frac{m_{n+1}^2}{m_n}\right|\le |m_{n+2}-k\cdot t^{n+2}|+|k\cdot t^{n+1}-m_{n+1}|\cdot t+m_{n+1}\left|t-\frac{m_{n+1}}{m_n}\right| \]
The first two terms are easy to estimate: their sum does not exceed $(t+1)\delta$. The last term requires some more work:
\[ m_{n+1}\left|t-\frac{m_{n+1}}{m_n}\right|=\frac{m_{n+1}}{m_n} |tm_{n}-m_{n+1}| \]
The first factor tends to $t$ and, therefore, is less than $2t$ for large enough $n$. The second factor can be estimated as
\[ |tm_{n}-m_{n+1}|\le t|m_n-k\cdot t^n|+|k\cdot t^{n+1}-m_{n+1}|\le (t+1)\delta. \]
Thus, the whole difference does not exceed $(2t+1)(t+1)\delta<\frac 12$ if $\delta$ is chosen small enough. This means that, if we are far enough in the sequence $m_n$, we can uniquely determine the next term from two terms preceeding it and, therefore, the entire sequence $m_n$ is completely determined by its sufficiently long beginning subsequence. We are done.

This solution can be written in a shorter way, of course, but my aim was rather to try to show how one could find it than to make the most sleek and concise presentation  :)

That's really cool! Thanks fedja.

I just wonder: is it common belief that the conjecture about transcendentals is true? Or is it unknown? My intuition wants it to be true but I don't really have any reason to think that.

A vague idea: You proved that the special numbers are countable by looking at the corresponding sequences. How much information can we get if we start with the sequences and look at the corresponding $t$?

Somehow I think a solution would not be so elementary. Also, somehow I think I am not about to come up with the winning approach by randomly speculating on a topic with which I'm not familiar.

Thank you very much for this beautifull solution. It is always very interesting to see where the ideas comes from  :)


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inequalities unsolved
inequalities

I think you ineq will be true for only $1 \le k \le 5$

I'm sure that it's not true for n>=16.

I am very interested about how to disprove it for $n>9$

Put $a_i=\frac{1}{b_i}$  then $\sum \frac{1}{1+b_i}=n-1$

Now we search two positive real numbers $p$ and $q$ such that for every $a_i$ we have

$4b_i +a_i \geq p(b_i +1) +q(a_i+1)$

or

$(4-p)b_i +(1-q)a_i \geq p+q$

From $(n-1)^2b_i+a_i \geq 2(n-1)$ we deduce that

$p=\frac{-n^2+10n-5}{n^2} $ and $q=\frac{n^2-5}{n^2}$

Naturally $p$ and $q$ are positive only for $3 \leq n \leq 9$ so this is the condition on $n$ to make true the ineq.

Now we apply Cauchy in this way

$\frac{p^2(\frac{n}{n-1})^2}{p(b_i+1)} + \frac{(qn)^2}{q(a_i+1)} \geq \frac{(p\frac{n}{n-1}+qn)^2}{p(b_i+1)+q(a_i+1)} \geq \frac{(p\frac{n}{n-1}+qn)^2}{4b_i+a_i} $

and finally

$\sum \frac{1}{4b_i+a_i} \leq \frac{n(n-1)}{(n-1)^2+4}$

I found this solution on a Vietnam Mathematical Site which is not anymore available.

In my opinion it is OK but the first step is not natural to find.

I cannot to disprove it for $n \geq 9$

And for $n=2$ what does it happen?

In this case you can use trigonometric function for $a_1$and $a_2$


geometry
geometric transformation
reflection
trigonometry
real analysis
real analysis unsolved

How does one prove that $ \Gamma (z)\Gamma (1 \minus{} z) \equal{} \frac{\pi }{{\sin \pi z}}$ without complex analysis?

[quote="Hetidemek"]How does one prove that $ \Gamma (z)\Gamma (1 \minus{} z) \equal{} \frac {\pi }{{\sin \pi z}}$ without complex analysis?[/quote]
How does one define $ \Gamma (z)$ without complex analysis.

Ahhh well.

Talkin' with didilica, told me that such proof involves complex analysis anyway.

Thanks for your reply.

[quote="Hetidemek"]How does one prove that $ \Gamma (z)\Gamma (1 \minus{} z) \equal{} \frac {\pi }{{\sin \pi z}}$ without complex analysis?[/quote]

http://numbers.computation.free.fr/Constants/Miscellaneous/gammaFunction.html


Let $a,b,c$ be non-negative real numbers. Prove that
\[a^{3}+b^{3}+c^{3}+3abc+a^{2}+b^{2}+c^{2}+3 \ge a^{2}b+b^{2}c+c^{2}a+2(ab+bc+ca)+a+b+c.\]

\[****************\]

This is non-homogeneous, non-symmetric. I believe that this kind is so rare and unsuual every where.

This is to dedicated to VASC (my greatest adviser), Arqady (the most diligent inequality-solver) and N.T.Tuan (h..m the most diligent Vietnamese Mod) of Mathlinks. Every other Mathlinkers are well-come.  I would like to know if a Vietnamese member can crack it.

Thanks my friend...I cant solve it.Your inequality is very very beautiful. I hope someone (Vasc,....) posted nice proof... :D

Did you found new identity?

[quote="Sunjee"][quote="hungkhtn"]If you don't like the fraction $\frac{2}{3}$, what do you say about the following?
\[\frac{1}{4}(a^{2}b+b^{2}c+c^{2}a+9abc)+a^{2}+b^{2}+c^{2}+3 \ge a+b+c+2(ab+bc+ca) \]
[/quote]
nice job![/quote]

Thank you, but it is still unsolved.  :blush:

[quote="hungkhtn"]
Let a,b,c be non-negative real numbers. Prove that
\[a^{3}+b^{3}+c^{3}+\frac{9}{2}abc+a^{2}+b^{2}+c^{2}+3 \ge \frac{3}{2}\left(a^{2}b+b^{2}c+c^{2}a\right)+2(ab+bc+ca)+a+b+c. \]
[/quote]
[hide="hint"] Dear Hung, my solution uses the following
\[a^{2}b+b^{2}c+c^{2}a \le \frac{4}{27}(a+b+c)^{3}-abc \]
:)
[/hide]

[quote="hungkhtn"]This last one is really nice and hard
\[\frac{4}{3}\left(a^{2}b+b^{2}c+c^{2}a\right)+a^{2}+b^{2}+c^{2}+2 \ge 3(ab+bc+ca). \]
To VASC and Arqady. Could you present any solution for one of two above inequalities? Are they hard?

To N.T.Tuan. Any idea to solve it?  :lol:[/quote]

Divide it into two inequalities.
First case, $2abc+1 \ge ab+bc+ca$
$\frac{4}{3}\left(a^{2}b+b^{2}c+c^{2}a\right)+a^{2}+b^{2}+c^{2}+2$
$\ge 4abc+a^{2}+b^{2}+c^{2}+2$
$\ge 2abc+1+2(ab+bc+ca)$
$\ge 3(ab+bc+ca)$

Second case, $2abc+1 \leq ab+bc+ca$
$\frac{4}{3}\left(a^{2}b+b^{2}c+c^{2}a\right)+a^{2}+b^{2}+c^{2}+2$
$= \frac{1}{3}\left(a^{2}b+b^{2}c+c^{2}a\right)+\left(a^{2}b+b^{2}c+c^{2}a+a+b+c\right)+a^{2}+b^{2}+c^{2}+2-a-b-c$
$\ge abc+2(ab+bc+ca)+a^{2}+b^{2}+c^{2}+2-a-b-c$
$\ge 2(ab+bc+ca)+\frac{1}{2}(a^{2}+b^{2}+c^{2}+2abc+1)+\frac{1}{2}(a^{2}+b^{2}+c^{2}+3-2a-2b-2c)$
$\ge 3(ab+bc+ca)$

Thank you very much for your very nice solution, Soarer.

And THMM, I understand you solution, but how could you clarify it more for other Mathlinkers?

$(a^{2}+b^{2}+c^{2}+2abc+1) \geq 2(ab+bc+ca)$

I don't understand this step of Soarer proof, sorry.

[quote="pohoatza"][quote="manlio"]$(a^{2}+b^{2}+c^{2}+2abc+1) \geq 2(ab+bc+ca)$
[/quote]

Use Schur: $\frac{9abc}{a+b+c}\geq 2(ab+bc+ca)-(a^{2}+b^{2}+c^{2})$

And $(2abc+1)(a+b+c)\geq 9abc$ , by AM-GM.[/quote]

A general problem is posted here

http://www.mathlinks.ro/Forum/viewtopic.php?t=148944

Thank you for your problems. :)
[quote="hungkhtn"]To N.T.Tuan. Any idea to solve it?  :lol:[/quote]
It is not easy for me. I think that i must choose : these inequalities or B&S-D  :rotfl:

[quote="hungkhtn"]This last one is really nice and hard
\[\frac{4}{3}\left(a^{2}b+b^{2}c+c^{2}a\right)+a^{2}+b^{2}+c^{2}+2 \ge 3(ab+bc+ca). \]
To VASC and Arqady. Could you present any solution for one of two above inequalities? Are they hard?

To N.T.Tuan. Any idea to solve it?  :lol:[/quote]
By Cauchy Schwarz, we have
\[\sum\frac{b}{a}=\sum\frac{a^{-1}}{b^{-1}}\ge \frac{(a^{-2}+b^{-2}+c^{-2})^{2}}{a^{-1}b^{-1}+b^{-1}c^{-1}+c^{-1}a^{-1}}=\frac{(ab+bc+ca)^{2}}{abc(a+b+c)}\]
Hence
\[a^{2}b+b^{2}c+c^{2}a \ge \frac{(ab+bc+ca)^{2}}{a+b+c}\]
It suffices to prove that
\[\frac{4(ab+bc+ca)^{2}}{3(a+b+c)}+a^{2}+b^{2}+c^{2}+2 \ge 3(ab+bc+ca) \]
Setting $x=a+b+c,y=ab+bc+ca,$ then $x^{2}\ge 3y$, the inequality is equivalent to
\[\frac{4y^{2}}{3x}+x^{2}+2 \ge 5y \]
By AM-GM inequality,
\[\frac{4y^{2}}{3x}+2 \ge 6\sqrt[3]{\frac{y^{4}}{9x^{2}}}\]
It suffices to prove that
\[x^{2}+6\sqrt[3]{\frac{y^{4}}{9x^{2}}}\ge 5y \]
which is clearly true since $x^{2}\ge 3y$. :)
We are done. Equality holds for $x=y=z=1$. :)

Dear Hung:
Note that
\[a^{2}b+b^{2}c+c^{2}a \ge \frac{(a^{2}+ab+b^{2})(b^{2}+bc+c^{2})(c^{2}+ca+a^{2})}{(a+b+c)(a^{2}+b^{2}+c^{2})}\]
and we have the stronger
\[\frac{1}{4}\left( \frac{(a^{2}+ab+b^{2})(b^{2}+bc+c^{2})(c^{2}+ca+a^{2})}{(a+b+c)(a^{2}+b^{2}+c^{2})}+9abc\right)+\sum a^{2}+3 \ge \sum a+2\sum ab \]
This is very hard and I want to delicated to you. :)

I think the following is really hard, and your initial way is not vaild, Can.

\[\frac{4}{15}(a^{2}b+b^{2}c+c^{2}a+12abc)+a^{2}+b^{2}+c^{2}+2\ge 3(ab+bc+ca).\]

This is a very very hard inequality, although the best constant (to replace $\frac{4}{15}$) is approximate $0.28337.$ I just hope you and others can find a nice solution as you have done to solve my previous inequalities.

[quote="hungkhtn"]I think the following is really hard, and your initial way is not vaild, Can.
\[\frac{4}{15}(a^{2}b+b^{2}c+c^{2}a+12abc)+a^{2}+b^{2}+c^{2}+2\ge 3(ab+bc+ca). \]
This is a very very hard inequality, although the best constant (to replace $\frac{4}{15}$) is approximate $0.28337.$ I just hope you and others can find a nice solution as you have done to solve my previous inequalities.[/quote]
But this inequality is not true, Hung. See $a=b=2.2,c=0$. :wink:

I am so sorry. I mean the following (h...m, I made some mistake in Computing)
\[\frac{1}{3}(a^{2}b+b^{2}c+c^{2}a+9abc)+a^{2}+b^{2}+c^{2}+2 \ge 3(ab+bc+ca). \]


Given that N is the middle point of CB, find the distance CM

Please explain it clear, easy and STEP by STEP writing all the theorems you use.

[img]http://i204.photobucket.com/albums/bb126/agerador/triangulo.jpg[/img]

Thanks! :lol:  :lol:

$ \frac{7x\plus{}5}{13\minus{}3x} \times \frac{1}{2} \times \frac{\frac{1\plus{}36x}{2}}{9x\plus{}\frac{1}{4}} \equal{} \frac{7x\plus{}5}{13\minus{}3x} \equal{} 1 \iff x\equal{}\frac{4}{5}$
$ \frac{y}{5x} \times \frac{9x\plus{}\frac{1}{4}}{\left(9x\plus{}\frac{1}{4}\right)\plus{}\frac{1\plus{}36x}{2}} \times \frac{1}{1} \equal{} \frac{y}{15x} \equal{} \frac{y}{12} \equal{} 1 \iff y\equal{}12$

Thus, $ CM \equal{} y\plus{}5x \equal{} 12 \plus{} 4 \equal{} \boxed{16}$.

[quote="Kouichi Nakagawa"]$ \frac {7x \plus{} 5}{13 \minus{} 3x} \times \frac {1}{2} \times \frac {\frac {1 \plus{} 36x}{2}}{9x \plus{} \frac {1}{4}} \equal{} \frac {7x \plus{} 5}{13 \minus{} 3x} \equal{} 1 \iff x \equal{} \frac {4}{5}$
$ \frac {y}{5x} \times \frac {9x \plus{} \frac {1}{4}}{\left(9x \plus{} \frac {1}{4}\right) \plus{} \frac {1 \plus{} 36x}{2}} \times \frac {1}{1} \equal{} \frac {y}{15x} \equal{} \frac {y}{12} \equal{} 1 \iff y \equal{} 12$

Thus, $ CM \equal{} y \plus{} 5x \equal{} 12 \plus{} 4 \equal{} \boxed{16}$.[/quote]A note to Chalao and inexperienced ones: The theorem used here is Menelaus' Theorem.


A hexagon$ ABCDEF$ has$ AB\equal{}CD\equal{}EF\equal{}m$,, $ BC\equal{}DE\equal{}FA\equal{}n$ .Prove that among $ AD,BE,CF$ ,there exists at least one which not bigger than ,$ (m\plus{}n),$

[quote="quykhtn-qa1"]A hexagon$ ABCDEF$ has$ AB \equal{} CD \equal{} EF \equal{} m$,, $ BC \equal{} DE \equal{} FA \equal{} n$ .Prove that among $ AD,BE,CF$ ,there exists at least one which not bigger than ,$ (m \plus{} n),$[/quote]
NO ONE INTERESTED IN THIS PROBLEM ? :(


On a  n × n  table we place natural numbers so that any two neighboring numbers
differ by at most 1. Prove that there is a number that appears on the table at least
n  times.

on number have 9 neighboring numbers or onl 4?!

We can make a beautifull solution of that using John Conway's Game of Life, don't we?

If so, what is the rules of the game?

@yagaron: I think 4 at most but I am not sure. You can solve it for 4 :)

if is $4$ the answer is very simple, i think

Let's a Example for $n=5$

${\left( \begin{array}{lllll}1 & \square & \square & \square & 1 \\ \square & \square & \square & \square & \square \\ \square & \square & 1 & \square & \square \\ \square & \square & \square & \square & \square \\ 1 & \square & \square & \square & 1 \end{array}\right)}$

The anothers spaces can be replaced by all numbers diferent from those picked, since there is no limitation for the numbers to place
And it's easy to see that $n>3$

I'm making the repeted number $1$ just to simplify, but could be another

We can generalize it to.

${\left( \begin{array}{lllllllll}1 & \square & \square & \square & \square & \square & \square & \square & 1 \\ \square & \ddots & \vdots & \vdots & \vdots & \vdots & \vdots & .\cdot{}^{\cdot}& \square \\ \square & \ldots & 1 & \square & \square & \square & 1 & \ldots & \square \\ \square & \ldots & \square & \square & \square & \square & \square & \ldots & \square \\ \square & \ldots & \square & \square & 1 & \square & \square & \ldots & \square \\ \square & \ldots & \square & \square & \square & \square & \square & \ldots & \square \\ \square & \ldots & 1 & \square & \square & \square & 1 & \ldots & \square \\ \square & .\cdot{}^{\cdot}& \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \square \\ 1 & \square & \square & \square & \square & \square & \square & \square & 1 \end{array}\right)}$

It's just valid for even $n$.
But with a similar construction we can make it for odd $n$'s too.
If we can't complete the numbers in those $2$ diagonals we can put is in another place distant from the both diagonals

PS: the number of neighboring could be 4 or 8. not 9

sorry


From Introduction to Counting and Probability

Exercise 2.3.4

In how many ways can 7 people be seated in a row of chairs if two  of the people, Wilma and Paul, refurse to sit next to each other?

Solution manual: The number of all seating arrangements is $7!$.  The number of seating arrangelemnts in which Wilma and Paul sit next to each other is $6!*2!$.  We can arrive at $6!*2!$ by considering WIlma and Paul as one person, arranging the "six" people first, then arranging Wilma and Paul.  Thus the number of acceptable arrangements is $7!-6!*2!=3600$.

I have never seen this strategy before.  Is it useful under many circumstances?  Is there a more standard way of doing this problem, still using complementary counting, but not grouping the two people into "one person"?

Furthermore:

Problem 2.5: The inhabitants of the island of Mumble (from Problem 2.2) decided to expand their alphabet to the standard Roman alphabet (26 letters, A through Z).  But they still limit their words to 3 letters or less, and for som e reason, they insist that all words contain the letter A at least once.  How many 3-letter words are possible?

In the Case with words with exactly 2 A's, there are 75 words.  However, my logic tells me that there would be 3 possible locations for the first A, two for the second, and 25 letters of the alphabet (which aren't A's), for 150 words...

I'd really appreciate any help.  :)

1st Problem:

Thats a great way to do the problem; count what you are NOT looking for then subtract it off. Often you will create "new" things to arrange such as a partition.

2nd one:

For two A's, just choose the position of the A's which is $3!/2!$ then the last one can be $25$ so thats $25 * 3 = 75$.

The way you went about it wouldn't work because you are saying that they are distinct. When you choose a place and you choose the 2nd, the way you do it won't differentiate between choosing the 2nd place first and choosing the 1st one 2nd. But the A's are not distinct so you need to divide by 2!.

What i would do for that problem though is find how many do NOT have A's.

For the second one, subtract the number of words which do not contain any '$a$'s. We have $26^3$ words without restrictions, and $25^3$ that contain no '$a$'s. Therefore the number of words is $26^3 - 25^3 = (26 - 25)(26^2 + (25)(26) + 25^2) = \boxed{1951}$.

Another "regular" way to do it would be to count the number of ways that are possible.

With two people with special needs the way to arrange the other 5 would be 5!.

Paul then could be placed in 6 different places with the other placed in 5 other places than Paul.

Thus 6*5*5!=3600


I was working on my (pretty low percent)% accuracy in Mastery of Counting Fundamentals. The same problem appears 7 times in a row, and now I passed the topic. This should probably be fixed.

Was the problem this-Julian tiled a 15 feet by 21 feet rectangular ballroom with one-foot square tiles. When he finished, he drew both diagonals on the floor connecting the opposite corners of the room. What is the total number of tiles that the diagonals pass through? Because I'm encountering the same thing, except it just gives me 44% every time I log back in. It seems to be the only problem in the subject Mastery of Counting Fundamentals Level 4.

[quote="Fidar"]I realized that a similar situation was appearing with Principle of Inclusion-Exclusion.
And in the report page, it says that I had answered about 10 questions for MAstery of Counting Fundamentals 2,3,4, and 5, whereas I remember answering dozens.[/quote]

I noticed that if I get a question wrong, and then get the question again later (and get it right), it only counts as doing the problem once.

Maybe that's the reason.

Glitches are now fixed. You should be getting a variety of problems, except for the two problems in Principle of Inclusion-Exclusion.


$ {(x\plus{}1)}^2\plus{}x^2\equal{}{(y\plus{}1)}^4\plus{}y^4$
solve in natural numbers

equation $ \Longleftrightarrow$ $ x(x\plus{}1)\equal{}y^2(y\plus{}1)^2\plus{}2y(y\plus{}1)$ $ \Longleftrightarrow$ $ x^2\plus{}x\plus{}1\equal{}(a\plus{}1)^2$, where $ a\equal{}y(y\plus{}1)$. But $ x^2<x^2\plus{}x\plus{}1<(x\plus{}1)^2$. Hence the equation has no solution in $ \mathbb{N}$


calculus
integration
function
inequalities
algebra
rational function
real analysis

Suppose [b]R[/b] is a rational function. Suppose [b]r[/b], [b]s[/b] are rational numbers such that [b]r + s[/b] is an integer. Find a suitable change of variable, such that  $\int R(x, (ax+b)^r, (cx+d)^s) \, dx$ is changed into the antiderivative of a rational function.

I can´t get it  :mad: . Some idea. Tks.

I'm afraid you are asking for something impossible. Indeed, let us consider the case when we have the expressions $x$, $x^{2/3}$ and $(x+1)^{1/3}$. Then your question is almost equivalent to asking to find a rational function $x=\rho(u)$ such that $x^{2/3}$ and $(x+1)^{1/3}$ are also rational functions. It immediately results in two rational functions $A(u)$ and $B(u)$ satisfying the relation $A(x)^3+1=B(x)^3$. But it is well-known that no such non-constant rational functions exist (I don't remember whose theorem it is :(, but it has been proved long ago that the Fermat Last Theorem is true for polynomials). Most likely, you misstated the problem and the comma between $(ax+b)^r$ and $(cx+d)^s$ is actually a multiplication sign, in which case the desired change of variable is $\frac{ax+b}{cx+d}=u^n$ where $n$ is the common denominator of $r$ and $s$.

Initial quiestion cames from http://www.math.ust.hk/%7Emamyan/ma204/lecture.pdf , exercise 3.3.5, page 160, and the comma is actually comma

[quote="fedja"]Well, it merely means that there is a misprint there, so the fault is not yours, but the professor's. Here is one more argument that the task is impossible. Consider the integral $\int x^{1/3}(x+1)^{1/3}\,dx$. According to the Chebyshev theorem cited in the very next exercise, it cannot be reduced to integration of a rational function. On the other hand, the integrand is $\frac{x(x+1)^{1/3}}{x^{2/3}}$, i.e., exactly of the kind you asked about.[/quote]

Certainly. Thank you, fedja


Is there a sequence of sets $ A_n\subset X$ such that there is no convergent subsequence $ A_{n_k}$? I know the answer when $ |X| \ge c$ (sequence $ A_n$ exists) or $ X$ is a finite set (there is no sequence $ A_n$), but if $ X$ is countable or $ X$ is countable and $ A_n$ is finite for all $ n$ then the problems are.

I solved the problem. There is no subsequence. Thanks for all.


If you search for "how many mathematicians" on wolframalpha.com, it gives about 3000 in the US (based no doubt on AMS figures). I would imagine there are ~5000 in the world, based on this. Is that correct?

What do you count as a "mathematician"? Do you mean people with Ph.D's in mathematics? Do you also count statistics or operations research? Or do you only count people who have published something "mathematical"? There are a few people with degrees in things like mechanical or electrical engineering who are doing what you would consider "mathematical" work.

well, i guess they are counting those people who graduate with Ph.D in math, whose job afterwards is concerned principally with research (i.e. they are at universities with small teaching loads, either teach:research= 1:1 or 2:1), or in industry doing mainly mathematics.

ya they count the people who did Ph.D in mathematics. By the way, do you consider all the people who graduate with maths or only doctorate in maths as an mathematician..

I would say that the count is based on more than having a math doctorate, since then the number would be much higher-just think of the shear number of people who write their doctoral thesis, graduate, and then maybe write one more paper that's similar to their thesis and never write another paper again.

Maybe the count is based on the number of people that have published a minimum amount of papers (say, 5) [i]and [/i] have a academic job.

Wolfram Alpha is grossly wrong.  As of 2007, there were over 20,000 full time math faculty in the United States, according to the [url=http://www.ams.org/employment/2007Survey-Third-Report.pdf]AMS[/url] (see page 2).  Of these, less than 4,000 did [i]not[/i] have doctorates.

[i]Edit[/i]:  I think the disparity may come from the Bureau of Labor Statistics, which may have tricked Wolfram Alpha by treating [url=http://www.bls.gov/oes/2008/may/oes152021.htm]"mathematicians"[/url] (2,770) and [url=http://www.bls.gov/oes/2008/may/oes251022.htm]"postsecondary mathematical science teachers"[/url] (45,700) as two separate (disjoint?) groups.

Hm, I have recently read the 5,000 mathematicians figure elsewhere (In a maths text). I'll provide the full reference and citation later. I do not think there are 20,000 math doctorates that are out there [b]regularly [/b]publishing papers, if in the Annals or Fibonacci Quarterly.

The number 5,000 for the US makes sense, too, if one thinks that the number of Tier I + Tier II schools is about 100. If one supposes each department has between 30 and 50 profs, then thats between 3,000 and 5,000. And then maybe a few hundred that are bouncing between academia and industry.

There are thousands of colleges out there, and if you count those guys, who may have doctorates in math but don't publish more than one or two papers after their Thesis, then the number would be about 20,000.

Anyway, I think it's important to remember that there are only 5,000 employed people out there publishing. It makes you realize how competitive things really are.

I still think that this 5,000 number is way too small.  I remain convinced that such estimates are due to book-keeping mistakes.

Perhaps you don't realize [url=http://www.ams.org/msnhtml/serials.pdf]how many[/url] math publications there are.  The list I linked gives a list of the serials refrenced by the AMS's [i]Mathematical Reviews[/i] (a listing of synopses of math articles published) as of November 2008.  There are a few repetitions, and some journals that overlap with fields such as physics, and a few look like advanced textbook series, but the listing still contains over 2700 entries.  Furthermore, the list is compiled by an American society, so it probably overlooks a few obscure journals published overseas---it even omits some publications here, such as the Harvard Math Review (admittedly an expository undergraduate math journal, but a math journal nonetheless).  Around 850 of the listed serials are read cover-to-cover for [i]Mathematical Reviews[/i].

I think it probably takes more than 5,000 authors to fill more than 2,000 journal-years of math every year. ;)

According to the AMS, United States universities have granted [url=http://www.ams.org/employment/2008Survey-Second-Report.pdf]around 1,000 math doctorates[/url] each academic year listed in the 2008 report (i.e., at least since 1998-9).

I also think you are underestimating the number of colleges out there.  There are [url=http://www.ed.gov/admins/finaid/accred/accreditation_pg4.html]more than 6,000[/url] post-secondary institutions accredited by bodies recognized by the U.S. department of education.  Sure, not all of them are high-brow universities, but most mathematicians are not world-renowned mathematicians, by definition. :)

In any case, I think it's misleading to look only at people who regularly publish in big-name journals.  To me, that seems like trying to guage the number of professional violinists in the world by looking only at people who sit on the front stand of internationally recognized orchestras.  My parents are not prolific authors, but as math professors, they've made enough of a living to raise children pretty comfortably and send at least one of them to college, so I'm not sure what they are if they're not mathematicians. :)

Well, I think the quote is in reference to the elite, the number of people at the forefront of industry and academia in the US. Im curious how they came up with that magic number

2,000 publications a year worldwide, right?

[quote="Boy Soprano II"]

[i]Edit[/i]:  I think the disparity may come from the Bureau of Labor Statistics, which may have tricked Wolfram Alpha by treating [url=http://www.bls.gov/oes/2008/may/oes152021.htm]"mathematicians"[/url] (2,770) and [url=http://www.bls.gov/oes/2008/may/oes251022.htm]"postsecondary mathematical science teachers"[/url] (45,700) as two separate (disjoint?) groups.[/quote]

There may be only 2,770 mathematicians who are not employed as "postsecondary mathematical science teachers", i.e. Math faculties in colleges.

The quote i was thinking of is from the (nice little) book Experimental Mathematics in Action, Bailey, et al.

[color=blue]Now is a wonderful time to be a (computational) mathematician. This is forcibly brought home by the Business Week cover storry of January 23, 2006. It comments

These slices of our lives now sit in databases, many of them in the public domain. From a business point of view, they're just beginning to be analyzed. ... The rise of mathematics is heating uup the job market for luminary quants, especially at the internet powerhouses where new math  grds with six/figure salaries and rich stock deals. Tom Leightom, an entrepreneur and applied math prof at mit, says "all of my students have standing offers at yahoo and google." Top mathematicians are becoming a new global elite. It's a force of barely 5,000 by some guesstimnates, but every bit as powerful as the armies of harvard university mba's who shook up corner suites a generation ago.[/color]

ok, it seems that i misread the quote, and i had to type it out (it seems) to realize that.


linear algebra
matrix
geometry
3D geometry
limit
superior algebra
superior algebra solved

After you have solved :
http://www.mathlinks.ro/phpBB/viewtopic.php?t=1421

try to solve the problem:

If A belongs to  M_2 (Z) with the property that det(A^2+X^2) is a perfect square for any matrix X which belongs to M_2 (Z). Prove that det(A^3+X^3) is a perfect cube for any matrix X which belongs to M_2 (Z). Is the converse true? Prove!!!

Cheers! :D :D

Let $a_1,a_2$ be the eigenvalues of $A$. Let $a,b\in\mathbb{Z}$ with $TrA=b$ and $\det A=a$.
Let $X=xI_2,\ x\in\mathbb{Z}$. Then $\det(A^2+X^2)=(a_1^2+x^2)(a_2^2+x^2)=x^4+(a_1^2+a_2^2)x^2+(a_1a_2)^2$ is a perfect square for all $x\in\mathbb{Z}$.
So $x^4+(b^2-2a)x^2+a^2=f^2(x)$ is a square for all integers x.
$4f^2(x)=(2x^2+(b^2-2a))^2+(4ab^2-b^4)$. Because $(x+1)^2-x^2$ tends to infinity, we get $b^2=4a$ or $b=0$.

If $b=0$ then $A^2=-aI_2$. Let $X$ be a diagonal matrix with integer diagonal entries $y,z$. Then $\det(A^2+X^2)=(y^2-a)(z^2-a)$ is a square for all integers $y,z$.
For arbitrary $y$ and $z=0$, we get that $(-a)(y^2-a)$ is a square.

Lemma: If $a\neq 0$ and $a\cdot n^2+b$ is a square for all $n\in\mathbb{Z}$ then  $b=0$.
Proof of lemma: For arbitrary $m,n\in\mathbb{Z}$ we get that $am^2n^2+b$ and $am^2n^2+bm^2$ are squares.
Assume $b\neq 0$ and fix $m>1$. Then $a(mn)^2+b$ and $a(mn)^2+bm^2$ are distinct squares with constant difference as $n$ tends to infinity. This is a contradiction with $\lim_{n\to\infty}((n+1)^2-n^2)=\infty$.

Back to the problem, we get $a=0$, so $A^2=0$, so $A^3=0$ so $\det (A^3+X^3)=\det (X^3)=(\det X)^3$ is a cube and we are done.

If $b\neq 0$ then $b^2=4a$ and $a_1=a_2=\alpha\in\mathbb{Z}$. We have $(A-\alpha I_2)^2=0$.
We don't need old $a,b$'s any more so I will use the symbols to denote something else. Don't let this confuse you.
It is easy to see that the nilpotent $A-\alpha I_2$ has the form $\left(\begin{array}{cc} a & b \\ -\frac{a^2}b & -a\end{array}\right)$.
Let $X$ be a diagonal matrix with integer diagonal entries $0,\alpha\cdot v$ for arbitrary $v\in\mathbb{Z}$.
Direct computations show that $\det (A^2+X^2)=v^2(\alpha^2+2\alpha a)+\alpha^2$. This must be a square for all $v\in\mathbb{Z}$. The lemma shows that $\alpha=0$.
So $A^2=0$ and same as before we are done.

The converse is true and actually easier to prove.
Assume $\det(A^3+X^3)$ is a cube for all $X\in M_2(\mathbb{Z})$. Let $X=A$. Then $\det(A^3+A^3)=4(\det A)^3$ is a cube, so $\det A=0$.
Let $0,a$ be the eigenvalues of $A$ and let $X=xI_2$. Then $\det (A^3+X^3)=(x^3+a^3)x^3$ is a cube for all $x\in\mathbb{Z}$. So $x^3+a^3$ is a cube for all $x\in\mathbb{Z}$. For $x=a$ we get $2a^3$ is a cube, so $a=0$, so $A^2=0$ and we are done.


my school doesn't offer the AMC B. we had the AMC 10/12 A on Feb. 6 like everyone, but i think i did terribly on it. so is there any way i could take the AMC 12 B and qualify for AIME? could i, say, go to another school to take the exam?

thanks

[quote="dkushnir"]Other schools might not like it for variety of reasons, some of them very stupid. It's all up to the school's administration.
But you can take it in one of the colleges listed on AMC site, even in a different state.[/quote]
where are the list of colleges listed?


Find the number of complex numbers $z$ such that $z^{19}=(\overline{z})^{99}$, where $\overline{z}$ is the conjugate of $z$.

[hide="Hint"] Consider the conjugate and the fact that $z\overline{z}=|z|^{2}$[/hide]

Couldn't this also be solved by writing it in expodential notation?

Indeed. However, you must be careful not to miss the extra solution $z=0$.

I don't know but I got the answer as $117$. Could anyone provide a full solution?

I made an arithmetic error before, but I have an alternate solution which is probably essentially the same just written differently.  Lets write $z=r(\cos \theta+\sin \theta)$ and $\overline{z}=r(\cos (360-\theta)+\sin (360-\theta)$.  Now we can see $r$ must be $0$ or $1$. Taking the case where $r=1$ we can write $99(360-\theta )=19\theta+360k$.  Solving we get $\theta = \frac{99*180}{59}-\frac{180k}{59}$.  Now for all the possible values of $\theta$ to be distinct, (since they will repeat modulo $2\pi$) $k \in \{-19,-10, \ldots , 97,98 \}$.  This gives a total of $118$ solutions but we have to add the case where $z=0$ for a total of $\boxed{119}$ solution.


I was thinking of applying to awesome math and was wondering how long are we expected (or how long do people typically write) to write for the prompts?

As for me, I can't seem to write more than 1 paragraph :(

Last year, I just wrote 1 paragraph for each prompt(and got in).
So it's ok to write short.

On the subject of the two prompts, I had a question about the first one.
When it says "Describe the nature and extent of your involvement," are we supposed to describe our involvement in every activity we participate in? Or just the one that is most important to us?

Possibly the latter, but certainly not the former. I would think writing a meaty paragraph showing your activities, passions, and achievements would be sufficient.

Here's what I did last year and this year for the prompts. On the first one, I sort of listed and shorly described the whys and hows and stuff like that, and on the second one, I just wrote a paragraph.

I think what they really want to know from these prompts is.. in the first one, what will you be doing at camp and what things are important to you besides math and how your life is.. and from the second one, what you will be interested in taking when you come to camp.


A racing car has a fuel capacity of 35 gallons of
fuel and an average mileage of 2.0 miles per
gallon. A family car has a fuel capacity of
17 gallons of fuel and an average mileage of
28 miles per gallon. What is the positive
difference, in miles, between the average
distance possible for each of these two vehicles
if each one starts with one full tank of fuel?


Let $ a,b,c>0$ satisfy $ a\plus{}b\plus{}c$≥$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}$.Prove that
$ a\plus{}b\plus{}c$≥$ \frac{3}{a\plus{}b\plus{}c}\plus{}\frac{2}{abc}$

[quote="SUPERMAN2"]Let $ a,b,c > 0$ satisfy $ a \plus{} b \plus{} c$≥$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}$.Prove that
$ a \plus{} b \plus{} c$≥$ \frac {3}{a \plus{} b \plus{} c} \plus{} \frac {2}{abc}$[/quote]
Because $ a \plus{} b \plus{} c\geq\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\geq\frac {3}{a \plus{} b \plus{} c} \plus{} \frac {2(a \plus{} b \plus{} c)}{ab \plus{} ac \plus{} bc}\geq\frac {3}{a \plus{} b \plus{} c} \plus{} \frac {2}{abc}.$
$ a\plus{}b\plus{}c\plus{}\frac{3}{a\plus{}b\plus{}c}\geq\frac{4}{abc}$ stronger ( with the same conditions ).

We have $ a \plus{} b \plus{} c\ge \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\ge \frac {9}{a \plus{} b \plus{} c}\Longrightarrow a \plus{} b \plus{} c\ge 3$

$ (a \plus{} b \plus{} c)^2 \equal{} \frac {1}{3}(a \plus{} b \plus{} c)^2 \plus{} \frac {2}{3}(a \plus{} b \plus{} c)^2\ge 3 \plus{} \frac {2}{3}(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})^2$

$ \Longrightarrow (a \plus{} b \plus{} c)^2\ge 3 \plus{} 2(\frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca})\Longrightarrow a \plus{} b \plus{} c\ge \frac {3}{a \plus{} b \plus{} c} \plus{} \frac {2}{abc}$ $ \square$

See [url]http://www.mathlinks.ro/viewtopic.php?p=838560#838560[/url]
I posted a nice proof there by breaking it up into $ abc \ge 1$ and $ abc \le 1$  :)


function
algebra
polynomial
modular arithmetic

$f(x)$ is a polynomial function of degree n.
$f(2)$ is odd and $f(3)$ is odd. Prove that there is no integer $x$ such that $f(x) = 0$.

A condition is missing or implied here. The coefficients are also integers.

So this is a parity problem, eh?

[hide="Apparently"]
Let $x$ be an arbitrary integer. We have:

$f(x)=\sum_{k=0}^{n}a_{k}x^{k}$

$x^{k}\equiv x \implies f(x) \equiv \left( \sum_{k=1}^{n}a_{k}\right) x+a_{0}\pmod{2}$

$f(2) \equiv 1 \implies a_{0}\equiv 1 \implies f(x) \equiv \left( \sum_{k=1}^{n}a_{k}\right) x+1 \pmod{2}$

$f(3) \equiv 1 \implies \left( \sum_{k=1}^{n}a_{k}\right)+1 \equiv 1 \implies \sum_{k=1}^{n}a_{k}\equiv 0 \implies f(x) \equiv 1 \pmod{2}$
[/hide]

You seem to be doing a lot of extra work.  How about using [hide="magical fruit"] the fact that $a-b | f(a)-f(b)$ [/hide] instead?

Mmmmmmm...
that's true that's true, my bad

[hide="Good solution"]Assume that there is an integer $a$ such $f(a)=0$, then, by $a-b|f(a)-f(b)$, $2-a|f(2)$ and $3-a|f(3)$. If $a$ is even, $2-a$ is even, and then an even number divides $f(2)$ and thus $f(2)$ is even, but this contradicts the givens, so $a$ is odd. If $a$ is odd, then $3-a$ is even, resulting in the same contradiction as before with $f(3)$, implying that $a$ is even. However, $a$ cannot be both even and odd, so it must not exist. QED.[/hide]


calculus
integration
trigonometry
derivative
function
calculus computations

\[ \begin{array}{l} Find{\rm{ }}the{\rm{ }}integral \\ \int {\sqrt{Sin^2 \left( {\frac{\pi }{4} - x} \right)} } {\rm{ }}dx \\ \end{array}  \]

The answer in that book was
$2\sqrt{2}\cdot[ \frac{t}{\pi} %Error. "Rght" is a bad command.
]+ \sqrt{2} \text{sgn(t)}\cdot \{ \cos\frac{t}{\pi}-\cos(t) \} \text{ where }t: =x-\frac{\pi}{4}$ and [] is the integer part and i cant come to that answer

you know of course that $f(x) = -\cos(\frac{\pi}{4}-x)$ has derivative $\sin(\frac{\pi}{4}-x)$. but you need the function $F$whose derivative is $|\sin(\frac{\pi}{4}-x)|$, a function which takes on non-negative values. so what $F$ should be is $F(x) = f(x)$ on intervals where $f(x)$ is increasing, and $F(x) = -f(x)$ on intervals where $f$ is decreasing; $F'$ will then take on the same values as $f'(x)$, except always with a positive sign, which is exactly what you want. now it's just a matter of finding those intervals, and writing the appropriate definition of $F$...go from there...

[quote]$2\sqrt{2}\cdot[ \frac{t}{\pi} %Error. "Rght" is a bad command.
]+ \sqrt{2} \text{sgn(t)}\cdot \{ \cos\frac{t}{\pi}-\cos(t) \} \text{ where }t: =x-\frac{\pi}{4}$[/quote]

what is $sgn(t)$? Is it $sin(t)$?

no. it's the sign of the number t... plus or minus...


What is the remainder when $ 2^{20}$ is divided by  $ 7$?

[hide="Solution"]By Fermat's Little Theorem, $ 2^6 \equiv 1$ modulo $ 7$, so $ 2^{20} \equiv 2^2 \equiv \boxed{4}$ modulo $ 7$.[/hide]

step by step solution

We know $ 2^3 \equiv 1 (mod7)$

then    $ (2^3)^6 \equiv 1^6  (mod 7)$

           $ 2^{18} \equiv 1  (mod 7)$

            multiplying both sides by $ 2^2$  we get

              $ 2^{20} \equiv 4  (mod 7)$


trigonometry
function
real analysis
real analysis unsolved

Let $ f$ be absolutely continuous in the interval $ [\varepsilon,1]$ for each $ \varepsilon>0$. Does continuity of $ f$ at 0 imply that $ f$ is absolutely continuous on $ [0,1]$? What if $ f$ is also of bounded variation on $ [0,1]$ (in addition to being continuous at 0)?

For the first question, no. Consider $ x \sin \frac{1}{x}$

Yes. In fact, I had shown earlier that the function $ f$ defined by $ f(0)\equal{}0$ and $ f(x)\equal{}x^{2}\sin(1/x^{2})$ for $ x \neq 0$ is continuous at $ 0$, absolutely continuous on $ [\varepsilon,1]$ for $ \varepsilon>0$ but not of bounded variation on $ [0,1]$, thus not absolutely continuous on $ [0,1]$.

Any comments for the second part?

I would like to suggest the following approach:
First show that the total variation $ T_{0}^{\eta}(f)$ of $ f$ on $ [0,\eta]$ tends to 0 as $ \eta$ tends to $ 0^{\plus{}}$. Then $ T_{0}^{\eta}(f)$ will be small for a suitable $ \eta$. This together with the absolute continuity of $ f$ on $ [\eta,1]$ will lead to the absolute continuity on $ [0,1]$.

Comments anyone?


AMC
USA(J)MO
USAMO
geometry
blogs
articles
real analysis

[quote="Valentin Vornicu"][quote="d.lam.86"]Have you considered the AoPS books? What type of materials are you planning to explore on?[/quote]If she want's to reach USAMO and do well on that (and also on CMO) I would suggest she takes some of the AoPS Olympiad classes and maybe [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=40675]WOOT[/url]   in addition to spending more time in the [url=http://www.mathlinks.ro/Forum/index.php?f=217]Olympiad Forums[/url] ;)[/quote]

WOOT sounds awesome...only a little pricey though.  Best of luck to all aspiring olympians.

[quote="Valentin Vornicu"][quote="d.lam.86"]Have you considered the AoPS books? What type of materials are you planning to explore on?[/quote]If she want's to reach USAMO and do well on that (and also on CMO) I would suggest she takes some of the AoPS Olympiad classes and maybe [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=40675]WOOT[/url]   in addition to spending more time in the [url=http://www.mathlinks.ro/Forum/index.php?f=217]Olympiad Forums[/url] ;)[/quote]

The W00T classes are costy but I think the classes are worth the payment.

Oh no kidding Woot will be an awesome program.  Considering who is putting it together, and the fact that Naoki from Canada is helping too it will be amazing.

I still think the WOOT classes cost too much. If the price lowers by a great number, then I'll likely attend. Not every household can afford to spend so much on online math-olympiad classes. I probably could but I would be doing a bad deed to my parents—relying on books is my strategy at this moment.  :oops:

The WOOT classes are very reasonably priced (dirt cheap might be a better comparison).  You know, the people at AoPS have to earn some money from this, and the price of WOOT is cheaper than a dozen hours of tutoring from a decent tutor (assuming you could find someone with olympiad-level knowledge to tutor you, which is virtually impossible).

Besides, nothing you have ever posted gives the impression that you have the competency necessary for WOOT.  It's all fine and dandy to give random opinions, but you should be careful when passing judgment on something that you really know very little about.

You don't have to butcher me because I know I'm not at the Olympiad level—I'm not claiming to be. All I'm saying is that not everyone (including me) are financially ready to pay 495 dollars. This Olympiad training will only welcome the ones that can pay and not the ones that wish they can. "dirt cheap" is in your opinon; not everyone can say that 495 dollars are "dirt cheap". This is indeed state-of-the-art teaching since barely any tutors do teach about olympiad-writing competitions but honestly, don't you think that the majority of the participants in the training camp will be of the first-world countries and/or the richer families of the world?

I think that the majority of olympiad level participants are from families above the median income in their respective countries, yes.  There is a positive correlation between socioeconomic status and academic performance, and in certain areas (say, doing research, or olympiad training, or whatever) having money facilitates training that would not otherwise be available.  Of course, being sufficiently good can also open doors to an individual (I do not, off hand, know if the three US IMOers who attended Exeter last year were on financial aid or not).

Your point about the first world and third world is, however, a straw man argument inasmuch as the instructors for the program buy their books in North America, pay for their housing in North America, buy their groceries in North America, and so on.  I will note that I just bought four math texts from amazon.ca - three texts on real analysis, one on dynamical systems, and I now am roughly 425 dollars poorer than I was before.  Spending hundreds of dollars on academic resources is incredibly easy to do.  On a similar note, I tutor.  At 30 dollars an hour (my going rate) and accouning for the exchange rate, WOOT costs roughly the same as 17 hours of tutoring from me.  I know which one is the better deal (and I'm not tooting my own horn here).

My point, in any event, is that you should be especially careful about passing judgment about the pricing of AoPS classes, since you are using a vehicle that the AoPS instructors have provided free of charge (that is, the forums) to criticize company policy and directly or indirectly discourage individuals from financially supporting the company.

Your second paragraph has convinced me to think likewise. I, personally, think that the WOOT camp should offer financial aid to the ones that do qualify. Possibly the WOOT instructors could conduct an examination to find out—out of the ones that require financial aid—how experienced one is. Finding out who has the knowledge for the olympiad training camp, and that need financial aid, will allow many lower-incomed members of the world to participate in the camp due to their hard work and perseverence for preparation into their IMO team. In my opinion, one of the reasons why I think the majority of the same countries place high within the IMO is because of their financial status, which is why I think the idea of an aid to the ones that qualify—under the circumstances of the examination—is a good one.
If I have criticized the AoPS company (policy?  :huh: ), I am sorry—but I hope my opinion has suggested some good points that will hopefully turn out to be a positive outcome.

Hmm, just had a thought.  Maybe we could get people to sponser a person if they qualify.....

Heh... my family is definitely NOT above the median income... no way.

Personally, although I know there are lots of people who would benefit from something like WOOT, I think you can do just fine if you look at past problems and solutions and just practice different techniques.  There are people on here with enough problems to last you forever... I never had any training and I did just fine.  You don't need special training to succeed on olympiads.

This is probably the wrong forum to be trashing the WOOT program on though.  You should watch what you say.  There are lots of nice people working very hard to make WOOT happen.

sorry to but-t in a canadian community forum without proper acknowledgement but arent these replies going off topic? the topic is "Trying to find a math buddy..." also, WOOT is a "take it or leave it" type of thing. think about how gold medalists did in IMO before WOOT was created... they never had WOOT!!! so my point is if you can afford WOOT, it gives you an extra edge on problem solving, if not just shrug your shoulders, pass it and look for other alternatives since you cant afford it. lets not jump into conclusions. stay on topic. and sorry to interrupt the canadian community, please forgive me and move on!

I've moved on. I don't really remember how this turned out to be a WOOT conversation but oh well. How come some are so angry all of sudden? Seriously, going off-topic is typical—it's not like I haven't seen it anywhere else.

Having talked to the US team at IMO, I feel that one of the things that they have in the States that we lack is the abundance of sponsors. They are financially equipped to give out huge scholarships, run the MOP program, and provide significant incentive for the competitors (although I'm sure that no one is competiting solely for the sake of money).

Maybe in the future, more corporations and organizations will be interested in supporting math contests in Canada. But that will never happen unless we do something about it.

真想学好自己狠狠努力就行了。。。

[quote="xlntyork"]真想学好自己狠狠努力就行了。。。[/quote]

Haha yeaup.  But I still think that it is helpful to have a direction.  However, I would prefer the Independent Study over WOOT because you get 1 on 1 time.

[quote="Elyot"]Personally, although I know there are lots of people who would benefit from something like WOOT, I think you can do just fine if you look at past problems and solutions and just practice different techniques.  There are people on here with enough problems to last you forever... I never had any training and I did just fine.  You don't need special training to succeed on olympiads.

This is probably the wrong forum to be trashing the WOOT program on though.  You should watch what you say.  There are lots of nice people working very hard to make WOOT happen.[/quote]I dare to disagree with you. That is why the Canadian team did worse than the Romanian, Russia, US, Iran, Korean and many other teams: lack of training.

If you think that you don't need special training to succeed on olympiads I dare you to give me one single example of a gold medalist which hasn't trained at all. I think you will rarely find a person that qualified for the IMO without any special training.  You can ask Yufei how much he trained to earn his gold medal.

PS There are two Canadians already enrolled in WOOT (one of which registred on the site especially for that). We are very sure that many of the returning IMO students from abroad will also join woot. We are also looking (and I think we can do it) for a sponsor to get all Romanian MOPpers enrolled.

Hello, I used to live in Canada for about 6 years.

I assume that WOOT is intended for american citizens (is it?). For such training, expecially over the web, 495 dollars isn't notoriously high. Plus, where (and how) else can you train like this. A focused kind of training well benefit your skills better than just the problems posted on this site.

I have to say, even though I'm not quite at the IMO level, it works the same for me in studying for other levels of competitions. Just after 2 months of studying hard and doing endless amounts of problems, I improved greatly in problem solving skills. Then I went on some three week break, and it greatly damaged my math skills. Some problems that were simple for me before no longer came to me quickly.

The reason I assume the WOOT program for US citizens is because when you convert 495 dollars to another currency, like let's say chinese, it suddenly changes to about 4000 dollars, which with my knowledge from visiting China recently, is equivalent to 20-25 math books, and 30-50 hours of tutoring. While it isn't as bad in Canada, it is a great price for many international students.

Still, you have to balance it out. For you, which has the better effects: the costs (and not only money) or the training?

Let me make a few points here.

First, feel free to express your opinions about WOOT, or anything else that we do for that matter.  We can take the criticism, and often constructive criticism can help us improve AoPS.  Of course you should always disagree (whether with us at AoPS or with other AoPS members) in a civil and respectful manner, and avoid personal attacks.

Second, it is not correct that WOOT is intended primarily for American citizens.  The "W" in WOOT stands for "Worldwide".  It is true that there are a large number of Americans in the program, which is due in large part to our sponsorship from D. E. Shaw group.  (And don't kid yourself -- WOOT would never have even gotten off the ground if we had not received this sponsorship.)  But we hope to have a worldwide enrollment.

Third, about the cost: WOOT is a very labor-intensive program to run.  Although we have fantastic instructors (and we will be adding a new one today -- watch the WOOT page!), it is unrealistic to expect them to teach WOOT for free.  We also need people to provide the feedback on all of the WOOT practice tests.  We will be getting ex-Olympiad participants for this, but again it is unrealistic to expect them to provide their time without compensation.  Therefore, we have to charge tuition.  We actually think that our tuition is very reasonable given what we are providing, but this is a free market: if you feel that our tuition is too high, then you are free not to enroll in the program.

Fourth, regarding financial aid: to manage financial aid would be a logistical nightmare for us.  How can we possibly determine who would qualify for financial aid and who wouldn't?  Should we just take an applicant's word for it, in which case what is to stop unscrupulous people from demanding financial aid that they don't really need?  Also, where is the funding for our financial aid going to come from?  So this is really not an option for us.  Your best option, if you're interested in enrolling in WOOT but are put off by the tuition, is to try to find sponsorship, either individually or perhaps at a national level.  We know of at least one student who was able to pay his tuition to WOOT merely by writing his neighbors, describing WOOT, and asking for donations: see [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=1817]rrusczyk's blog[/url] for more details and some helpful suggestions.

Finally, to the original poster in this thread: I think that finding a "math buddy" to study with is a great idea!  You might also see if another high school in the Toronto area has a math club that you might be able to join, and if there's a math circle or something similar at a local university like UT.  (Your fellow Toronto-area residents in this forum would know 100 times more about this than I.)  Also, if you're feeling ambitious, you might try to start a program at your own school.  I recommend reading [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_MATHCOUNTS.php]this article[/url] if you'd like some inspiration.

[quote="DPatrick"]Third, about the cost: WOOT is a very labor-intensive program to run.  Although we have fantastic instructors (and we will be adding a new one today -- watch the WOOT page!), it is unrealistic to expect them to teach WOOT for free.  We also need people to provide the feedback on all of the WOOT practice tests.  We will be getting ex-Olympiad participants for this, but again it is unrealistic to expect them to provide their time without compensation.  Therefore, we have to charge tuition.  We actually think that our tuition is very reasonable given what we are providing, but this is a free market: if you feel that our tuition is too high, then you are free not to enroll in the program.
[/quote]

I understand the high costs, I know it is very hard for you to spend a lot of time online training, and thinking more while offline. I think it's very generous of all of you for first of all providing the membership as a free thing, as it's very hard work to manage such a big site. I'm very grateful I've found this site, it has changed my view of math.

After some thought about the exclusiveness of the WOOT program (sample tests, lectures, feedback, work problems), I think that students devoted to math problem solving should take this training. It's definetly worth it and it will help you for possibly the rest of your life. No where else can you recieve this kind of excellent training (unless you spend a ton of money on Olympiad classes, which isn't offered in my area).

I'm still at the MOEMS level, a while from the International Olympiads.

no d.lam.86 im not angry at you or anyone, maybe i didnt make myself clear last post. the last post i wanted to say to stay on topic but it seems discussion about the world general population. i have a request, and that is moving the discussion about WOOT to appropriate forum if it applies. why?

1) its going off topic (ok i know many of you guys dont agree with me)

2) many people who wants to input their 2 cents are not able to see this topic because this is Canadian forum, maybe The Round Table or WOOT forum shall appeal  more constructive (ok more people still dont agree with me)

3) if this is WOOT discussion regarding Canadian population, then i apologize and i leave right now, never to come back to Canada forum, and sorry to interrupt. if not, then the whole world should have access to this discussion, not just Canada alone (all of you dont agree with me)

i say you all dont agree with me because ... well you guys like to disagree me anyways :D  :stretcher:

edit: d.lam.86 if everywhere else people stay off topic, we should all go off topic too? chaotic world we live in  :P

Well, Palytoxin, I agree this isn't on the math buddy topic, but I can't move topics, so it will stay here until an administrator comes along and splits this thread.  ;)

I think, for those people who are at the Olympiad caliber, getting into WOOT financially should not be a roadblock. The kind of training that already goes in amounts to much more than WOOT's cost, which is in my opinion, way cheaper than it actually would be without all the Sponsorship. And if WOOT is something that you think would really help you, I think even a family with a little less than median income could manage it. I've seen sacrifice when sacrifice is called on for the better...

Please think before criticizing.

[hide]I feel really bad...[hide]You know why?[hide]It only 'costs' me two essays... I guess Americans do have an 'advantage' if you went to MOP.[/hide][/hide][/hide]

[quote="236factorial"]Essays? YOu have to write essays to get into WOOT? At least it isn't 495 dollars  :P[/quote]Due to our sponsorships from D. E. Shaw group and Google, all students (other then graduating seniors) who attended the U.S. Math Olympiad Summer Program are able to join WOOT for free.  The only catch for them is that they are required to contribute to the WOOT blog.  So that's what was meant by "essays".

We are strongly encouraging other countries to try to set up their own similar sponsorship arrangements.

Personally I believe that WOOT is well worth the money and is VERY cheap for what you get in return. However I know that some students may be in a financial squeeze (as I am right now) so they might find it expensive. As has been metioned earlier I suggest writing to your school district/neighbors/businesses seeking sponsorship (I have already but no luck yet). If anyone succeededs in being sponsored please give the rest of us some more ideas!  :P

[quote="DPatrick"]Let me make a few points here.

Third, about the cost: WOOT is a very labor-intensive program to run.  Although we have fantastic instructors (and we will be adding a new one today -- watch the WOOT page!), it is unrealistic to expect them to teach WOOT for free.  We also need people to provide the feedback on all of the WOOT practice tests.  We will be getting ex-Olympiad participants for this, but again it is unrealistic to expect them to provide their time without compensation.  Therefore, we have to charge tuition.  We actually think that our tuition is very reasonable given what we are providing, but this is a free market: if you feel that our tuition is too high, then you are free not to enroll in the program.

[/quote] Could we simply take WOOT without the feedback or just buy the Materials. We would still be willing to pay reasonably.


A wooden model of a square pyramid has a base edge of 12 cm and an altitude of 8 cm. A cut is made parallel to the base of the pyramid that separates it into two pieces: a smaller pyramid and a frustum. Each base edge of the smaller pyramid is 6 cm and its altitude is 4 cm. How many cubic centimeters are in the volume of the frustum?

Volume of the big pyramid is 12*12*8/3=384
Volume of the small pyramid is 6*6*4/3=48
Volume of the frustum is 384-48=336

well, the volume of the frustum is just the volume of the original pyramid minus the volume of the smaller pyramid

the volume of the a pyramid is $ \frac {1}{3}Bh$ where B=the area of the base and h=the height

therefore, the volume of the original pyramid is $ \frac {1}{3}(12)(12)(8)$ (make sure you see why i multiplied 12 two times)

$ \frac {1}{3}(12)(12)(8)$

$ \equal{} \frac {1}{3}(1152)$

$ \equal{} 384$
the volume of the original pyramid is 384

the volume of the new pyramid is $ \frac {1}{3}(6)(6)(4)$

$ \frac {1}{3}(6)(6)(4)$

$ \equal{} \frac {1}{3}(144)$

$ \equal{} 48$

now we said the volume of the frustum is just the volume of the original pyramid minus the new pyrimid

so, $ 384 \minus{} 48 \equal{} 336$

the volume of the frustum is $ 336$

edit: beaten

so the frustum volume is (1-1/8) = 7/8 of the total volume. Thus the answer is 1/3*12*12*8*7/8 = 48*7 = 336


As $2+h(1) = 2h(2)$, we would like to proceed by induction. It remains to prove that $1+h(n) = (n+1)h(n+1)-nh(n)$, $\forall n\in\mathbb{N}\setminus\{0,1\}$. But this is clearly true, because \[\begin{aligned}(n+1)h(n+1)-nh(n) & = (n+1)(h(n+1)-h(n))+h(n) = (n+1)\frac{1}{n+1}+h(n) \\ & = 1+h(n)\end{aligned}\]


when is y'll last day for school?

my is this Thursday 19th
yaaaaaaaaaaaa
i'm in GA :lol:  :lol:  :lol:

Mine is on June 3rd(I think).  This week is CRTs(end of level testing), but I don't know what's going on next week.  I'm not sure if this applies to any of the other Utah AoPS memebers because they all live in SLC and I don't.

Next Wednesday the 25th...but I only have exams for two classes so friday is basically my last day except for one exam.

[quote="fanaticsm"]Wow, June 27th here....[/quote]
Poor you...

[quote="byrdgyrl"]May 27, but I'm leaving early (the 24th) to travel.  Yippee![/quote]

Aren't the last 3 days finals?

May 25th.  But the last 4 days are easy, since many people are exempt from certain finals.  :D

May 31....that's right. the tuesday after a three day weekend. finals after a three day weekend. [sarcasm]really lookin forward to that one[/sarcasm]

June 23... from the west coast here... how are you guys out already? though i have to say it is a better idea now that ap tests are over, what's the point of school? :D

i get outta school on the 20th (fri) we have some schools in our county who got out last fri. though... =/ lol

I'm not out till June 21, then followed by a week of exams.

[quote="Sly Si"]June 7 here.  That's a little early, I think because of the way the days of the week fall.  We generally end around the 10th or so, but always on a Tuesday (for seniors) or Wednesday (for everyone else).[/quote]

Seniors only a day earlier? That sucks... our seniors get out usually 1-2 weeks early.

June 15.. I think. That's after finals. When did you all start school??

school REALLY ends next wednesday, but for our class, with APs and everything, school ended... when APs were over! (granted i'm still having AP related nightmares, but there you are)

Aug 16 - June 7 (I can't believe I still have to take finals for AP classes! Grrr..)
The Ministry of Education here requires sth like 180+ school days a year..  :(

Seniors at my school have the same final exam schedule unless they exempt (93% average or higher, and depending on teacher), plus graduation is on June 11th or sth.... I don't envy them :P

[quote="xxreddevilzxx"]June 24... crap.[/quote]
Ouch!!! :(  When'd you start school?

Our last day is the 2nd of June, which is a Thursday. What's up with that? We started on a Tuesday and end on a Thursday. Whatever happened to Mondays and Fridays?

The last week will be kind of pointless, though. We'll probably have to empty our lockers on Tuesday(no school Monday), and then all we can bring Wednesday and Thurday are a pencil or pen and our yearbooks.  :(

But that's ok. Do any of your schools plan activities for the last day/week or whatever? I know that on this Friday all of the 8th graders have their own dance and fun night, and then on the last Wednesday, we go to this place that has batting cages, mini golf, bowling, this maze to crawl in, and go karts.

[quote="bubala"]June 24 or 25 or 26.  Can't remember.

You lucky duckies!!!

Us poor NYers...[/quote]

ur not the only one... we have 2 end skool in late June :( *sigh*

wow i had no idea school ended so late for you guys... that must suck a lot  :(

Our high school starts at the last week of August or first week in September

School ends June 16

June 17 is graduation day.

[quote="mkkool64"]Wow everyones schools end early... mine doesn't end until June 13... :([/quote] which is my birthday. :D

My school ends when May ends.

Last day is June 9th. This school year started August 23rd or 24th for me (can't remember which) and next school year will start August 22nd. We've been starting one day earlier each year for the last maybe 4 or 5 years, because the district's planning a second spring break, a ski week, starting next year. I personally would rather have more time in summer vacation and not have a ski week, but whatever. I'm in CA.

June 2nd for most people, but May 25th for me  :D

yaaaaaaaaaaaaaaaaa
i just got out
i'm so excited
i thought i failed my spanish2 finals(superhard), but actually got a 74 :D  :roll:  :D  i was so happy

offically may 25
but all i have is a precal and english exam to take and i've only had those two classes since may 11.
what fun
senior next year. yay.
 :ninja:  (random ninja)

June 7 - but after APs May is pretty easy-going except for the occassional final project and final exam studying.

[quote="byrdgyrl"][quote="BornForMath"][quote="mkkool64"]Wow everyones schools end early... mine doesn't end until June 13... :([/quote] which is my birthday. :D [/quote]
Hey, we've got the same b-day :gleam:[/quote]
cool, mine is August 6

finals end on june 16th
graduation on the 24th
report card on the 28th
bleh, why couldn't everything happen together so that i could travel, this completely messes up the month of june for anyone who wants to do anything :mad:

June 3rd, just the day before ARML, so i'm gonna miss that day and of course miss all the finals, which i will have to take before hand. sigh...


[color=darkblue]Slove this equation:

[hide="equation 1"]1) $ \sqrt {x^2 \minus{} 2x \plus{} 1} \plus{} 3y^2 \equal{} \minus{} 4(3y \plus{} 1)$[/hide]

[hide="equation 2"]2) $ \sqrt[4]{x^2\minus{}2x\plus{}2}\equal{}\frac{7}{y^2\minus{}2y\plus{}6}$[/hide][/color]

1) $ 3y^2 \ge \minus{}12y \minus{}4$ so $ 3y^2\plus{}12y\plus{}4 \ge 0$ Solving for y, it equals 0 when $ y\equal{}\frac{\minus{}12\pm\sqrt{96}}{2}$ so when y is not in between those 2 values, we can find the 2 values of x symmetric about x=1.

1) $ |x\minus{}1|\plus{}3(y^2\plus{}4y\plus{}4/3)\equal{}0\to \minus{}2\minus{}\sqrt{8/3}\le y\le \minus{}2\plus{}\sqrt{8/3}, x\equal{}1\pm \sqrt{\minus{}3y^2\minus{}12y\minus{}4}.$
2) $ x^2\minus{}2x\plus{}2\ge 1$, therefore $ 0<y^2\minus{}2y\plus{}6\le 7 \to 1\minus{}\sqrt 2 \le y\le 1\plus{}\sqrt 2$.


Vieta
geometry
3D geometry
algebra
polynomial
sum of roots
product of roots

If a,b, and c are the roots of the equation $ f(x) \equal{} x^3 \plus{} 2x^2 \plus{} 5x \minus{} 9$ find $ a \plus{} bc \plus{} c \plus{} ac \plus{} abc \plus{} b \plus{} ab$

Okay, first I found the  product of the roots (abc) by taking the constant which is -9, and dividing it by the coefficient of the 1st term, which is 1, to get $ \minus{} 9$.

Then, I found a+b+c by the sum of the roots, which is 2/1, or b/a, which is $ 2$.

Then, I found ac+bc+ab, which is the sum of the mixed products of the roots BUT ONe, which is -(the next to last coefficient)/ the first term's coefficient to get -5/1 = $ \minus{} 5$.

Therefore, I got an answer of $ \minus{} 9 \plus{} 2 \minus{} 5 \equal{} \minus{} 12$. What did I do wrong?

The answer key says the answer is 12.

If $ a,b,c$ are the roots of the equation,

$ (x\minus{}a)(x\minus{}b)(x\minus{}c)\equal{}x^3\minus{}(a\plus{}b\plus{}c)x^2\plus{}(ab\plus{}bc\plus{}ca)x\minus{}(abc)\equal{}0$

So $ abc\equal{}9$, $ ab\plus{}bc\plus{}ca\equal{}5$, $ a\plus{}b\plus{}c\equal{}\minus{}2$

Looks like you weren't careful with the signs...

what is the positive negative rule, as in with the terms? I think I have it mixed up, could someone elaborate?

like i thought the product of the roots is c/a, whic his -9/1??

maybe it begins with the 1st term, which is positive, and then second, which is negative... ahh.

also, for $ x^3\minus{}4x^2\minus{}2x\plus{}10$, can you infer that there are 3 roots a,b, and c for that equation, and find the sum of the cubes of the roots?

Well if you tried to multiply out for polynomials of degree 2, 3, 4, 5, etc, you can construct your own patterns and find out this 'positive-negative' rule for yourself. I can tell you my opinion of this 'rule' but in order to understand things, you'll have to do it your way.

Here is my way:

Sum of roots is always negative

The rest are just alternating negative, positive, negative, positive

Product of roots is positive if degree is even, negative if degree is odd

$ (x \minus{} a)(x \minus{} b) \equal{} x^2 \minus{} (a \plus{} b)x \plus{} ab$

$ (x \minus{} a)(x \minus{} b)(x \minus{} c) \equal{} x^3 \minus{} (a \plus{} b \plus{} c)x^2 \plus{} (ab \plus{} bc \plus{} ca)x \minus{} (abc)$

$ (x \minus{} a)(x \minus{} b)(x \minus{} c)(x \minus{} d) \equal{} x^4 \minus{} (a \plus{} b \plus{} c \plus{} d)x^3 \plus{} (ab \plus{} bc \plus{} cd \plus{} da \plus{} bd \plus{} ca)x^2 \minus{} (abc \plus{} bcd \plus{} cda \plus{} dab)x \plus{} (abcd)$

See a pattern?

-+
-+-
-+-+
and so on...

EDIT: Yes you can find the sum of the cubes of the roots. Use the fact that $ f(a)\equal{}f(b)\equal{}f(c)\equal{}0$, in other words, plug in $ a,b,c$ and find $ a^2\plus{}b^2\plus{}c^2$ from $ (a\plus{}b\plus{}c)^2$.

[quote="10000th User"]

The rest are just alternating negative, positive, negative, positive

Product of roots is positive if degree is even, negative if degree is odd

-+
-+-
-+-+
and so on...

EDIT: Yes you can find the sum of the cubes of the roots. Use the fact that $ f(a) \equal{} f(b) \equal{} f(c) \equal{} 0$, in other words, plug in $ a,b,c$ and find $ a^2 \plus{} b^2 \plus{} c^2$ from $ (a \plus{} b \plus{} c)^2$.[/quote]

do you mean the product of roots is negative if the degree is even, and positive if the degree is odd? Because $ x^2\plus{}5x\minus{}6$ has roots 6 and -1, and their product is -6, with a degree 2?

[quote="Bicameral"]do you mean the product of roots is negative if the degree is even, and positive if the degree is odd? Because $ x^2 \plus{} 5x \minus{} 6$ has roots 6 and -1, and their product is -6, with a degree 2?[/quote]That's the mistake you have that I was referring to.

$ \minus{}(a\plus{}b\plus{}c)$ LOOKS negative but that doesn't mean NEGATIVE value. For example...

$ x^3\plus{}5x^2\minus{}3x\minus{}4$ has sum of roots equal to $ \minus{}5$ because $ \minus{}(a\plus{}b\plus{}c)\equal{}5$ and $ a\plus{}b\plus{}c\equal{}\minus{}5$. Do you see how to use the negative sign now?

When I said even degree the product is 'positive' means that you don't have to change the sign...

Following your example, the product is $ \minus{}6$ because there is no need to change the sign...

Look at this equation ------>$ x^2\minus{}(a\plus{}b)x\plus{}ab\equal{}0$. The product of the roots is $ ab$.

Now for a cubic equation, the last term is $ \minus{}abc$ so the product $ abc$ is the negative of the last term.


Let $A$ be a subset of $\mathbb {R}$ which staisfies the three following properties :
1) $1 \in A$
2) If $x \in A$ then $x^2 \in A$
3) If $(x-2)^2 \in A$ then $x \in A$.

Prove that $2004 + \sqrt {2005} \in A$.

Pierre.

Heres what i get:

[hide]
We have that $x \in A \Rightarrow x^2=(2+x-2)^2 \in A \Rightarrow 2+x \in A$
Thus given that $1 \in A$ we know that all odd numbers are also in A.  So $2005 \in A$.

Also we have that $x>0;\ x \in A \Rightarrow (2+\sqrt{x}-2)^2 \in A \Rightarrow 2+\sqrt{x} \in A$
But then $2+\sqrt{2005} \in A$ and using the first result succesively proves $2004+\sqrt{2005} \in A$[/hide]

That's exactly what I've done... :D

Pierre.


i am sad.
i know this is not the better place to post this
because i can make the difference between cs and coding.
every single day one more techonolgy appears
ajax,activex,javascript,php....i see now ajaxstl.
when are people going to realise these are only...bullsh**.
wtf do they have to concentrate on these things ?
we have better things to do!
computers are not meant for these purposes.
we seem to have forgot why computers were made!
i look around and see the computers and their world
has been corrupted by some weird and useless stuff.
what is with all these formalizations,all these different
ways of printing stuff on the screen...all these classes
and that kind of stuff really freaks me out.
i was a programmer...i gave up somewhere...
in the process of learning and not because it was too
much for me...just because it was really useless
math is much more important ,cs is also important

If I'm understanding you correctly, I think you are absolutely right on one level. All these new languages, it's not worth so much learning all of them. Theory is much much much more important. But it is also good to know at least one language suited to whatever sort of stuff you plan to be doing. Then if it ever becomes obsolete it isn't that hard to learn another newer one like it.


conics
geometry
circumcircle
ellipse
hyperbola
geometric transformation
reflection

Given triangle $ ABC$ with an arbitrary point $ M$, $ A'B'C'$ is cevian triangle of $ M$ let $ A_{b},A_{c}$ be circumcenter of triangle $ MAC',MAB'$ resp, similarly we have $ B_{a},B_{c},C_{a},C_{b}$ is circumcenter, prove that six points $ A_{b},A_{c},B_{a},B_{c},C_{a},C_{b}$ lie on a conic and that conic is circle when $ M\equiv G$.
Similar problem of six points [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=133767]here[/url]

The hexagon $ A_bB_aB_cC_bC_aA_c$ has the opposite sides parallel, for example, $ A_bA_c \parallel B_cC_b$ (both $ \perp AA'$), etc. Let $ a,\ a'$ be normals to $ AA'$ at $ A,\ A'$, let $ b,\ b'$ be normals to $ BB'$ at $ B,\ B',$ and let $ c,\ c'$ be normals $ CC'$ at $ C,\ C'.$ Define $ A_b' \equiv a \cap c',\ A_c' \equiv a \cap b',$ and similarly $ B_c' \equiv b \cap a',\ B_a' \equiv b \cap c',$ $ C_a' \equiv c \cap b',\ C_b' \equiv c \cap a'.$ Hexagons $ A_b'B_a'B_c'C_b'C_a'A_c' \sim A_bB_aB_cC_bC_aA_c$ are centrally similar with center $ M$ and coefficient 2. The triangle vertices $ A,\ C',\ B,\ A',\ C,\ B' \in (O)$ are feet of normals from $ M$ to $ a \equiv A_b'A_c',\ c' \equiv A_b'B_a',\ b \equiv B_c'B_a',$ $ a' \equiv B_c'C_b',\ c \equiv C_a'C_b',\ b' \equiv C_a'A_c'$ $ \Longrightarrow$ $ a,\ c',\ b,\ a',\ c,\ b'$ are tangent to a conic $ \mathcal T'$ (ellipse or hyperbola) with pedal circle $ (O),$ center $ O$ and focus $ M.$ By Brianchon's theorem, the main diagonals $ A_b'C_b',\ \ B_c'A_c',\ C_a'B_a'$ concur. Since the opposite sides are parallel, the concurrency point is the common midpoint of the hexagon diagonals, identical with the tangent conic center $ O,$ and the main diagonals of the original hexagon $ A_bB_aB_cC_bC_aA_c$ concur at the midpoint $ P$ of $ OM.$ (The centrally similar hexagons $ A_b'B_a'B_c'C_b'C_a'A_c' \sim A_bB_aB_cC_bC_aA_c$ have the opposite sides not only parallel, but also equal.) A unique conic $ \mathcal K$ with center $ P$ passes through 3 points, say $ A_b,\ B_c,\ C_a$ and this conic also passes through their reflections $ C_b,\ A_c,\ B_a$ in the conic center $ P.$

When $ M$ is on one triangle side, say $ M \in BC,$ then $ B' \equiv C,\ C' \equiv B.$ The $ \triangle MBC' \equiv \triangle MBB,\ \triangle MCB' \equiv \triangle MCC$ are degenerated, their circumcenters $ B_a,\ C_a$ are the midpoints of $ MB,\ MC,$ respectively. The circumcenters $ A_b,\ B_c$ of the $ \triangle MAC' \equiv MAB,\ \triangle MBA'$ are on the perpendicular bisector $ s_b \perp BC$ of $ MB$ and the circumcenters $ C_b,\ A_c$ of the $ \triangle MCA',\ \triangle MAB' \equiv MAC$ are on the perpendicular bisector $ s_c \perp BC.$ Then $ B_a,\ A_b,\ B_c \in s_b$ are colliner and $ C_a,\ C_b,\ A_c \in s_c$ are collinear and the conic $ \mathcal K$ degenerates into 2 parallel lines $ s_b \parallel s_c.$ It is also obvious that $ \mathcal K$ is an ellipse for $ M$ inside the $ \triangle ABC,$ but no claims about $ \mathcal K$ can be made for $ M$ outside the triangle.

If $ G$ is the triangle centroid, then the claim about $ \mathcal K$ being a circle is false, easily disproved by considering an isosceles right triangle with $ BC \equal{} 2$ and $ \angle A \equal{} 90^\circ.$ In this case, the circumcenters are $ A_b \in AB,\ A_c \in AC,$ such that $ AA_b \equal{} AA_c \equal{} \frac {AB}{3} \equal{} \frac {AC}{3}.$ The circumcenters $ B_a,\ C_a$ are midpoints of the median segments $ BG,\ CG,$ and the circumcenters $ B_c,\ C_b$ are symmetrical to $ A_b,\ A_c$ with respect to the line $ B_aC_a.$ We can easily calculate $ B_aC_a \equal{} 1,$ $ A_bB_a \equal{} \frac {\sqrt {10}}{6},$ $ A_bC_a \equal{} \frac {\sqrt {34}}{6},$ $ B_aC_a^2 < A_bB_a^2 \plus{} A_bC_a^2,$ $ \angle B_aA_bC_a < 90^\circ,$ the circumcenter of the isosceles trapezoid $ A_bB_aC_aA_c$ is not the midpoint of $ B_aC_a \Longrightarrow$ $ B_c,\ C_b$ do not lie on its circumcircle.

[quote="yetti"] ...If $ G$ is the triangle centroid, then the claim about $ \mathcal K$ being a circle is false, easily disproved by considering an isosceles right triangle with $ BC \equal{} 2$ and $ \angle A \equal{} 90^\circ.$ In this case, the circumcenters are $ A_b \in AB,\ A_c \in AC,$ such that $ AA_b \equal{} AA_c \equal{} \frac {AB}{3} \equal{} \frac {AC}{3}.$ The circumcenters $ B_a,\ C_a$ are midpoints of the median segments $ BG,\ CG,$ and the circumcenters $ B_c,\ C_b$ are symmetrical to $ A_b,\ A_c$ with respect to the line $ B_aC_a.$ We can easily calculate $ B_aC_a \equal{} 1,$ $ A_bB_a \equal{} \frac {\sqrt {10}}{6},$ $ A_bC_a \equal{} \frac {\sqrt {34}}{6},$ $ B_aC_a^2 < A_bB_a^2 \plus{} A_bC_a^2,$ $ \angle B_aA_bC_a < 90^\circ,$ the circumcenter of the isosceles trapezoid $ A_bB_aC_aA_c$ is not the midpoint of $ B_aC_a \Longrightarrow$ $ B_c,\ C_b$ do not lie on its circumcircle.[/quote]

  Now this casts some shadows on many proofs (?) posted by Yetti here.
  If Darij did not respond the way he did, we would have been talked into a false claim,
  may be for ever. I suggest Darij goes over all Yetti's proofs with a fine comb, to
  rid the ML pages of those bugs and other insects very possibly still hiding there.

  Thank you.
  M.T.

[quote="darij grinberg"][quote="yetti"]If $ G$ is the triangle centroid, then the claim about $ \mathcal K$ being a circle is false,[/quote]

No, the claim is true. See http://www.mathlinks.ro/viewtopic.php?t=48059 , http://www.mathlinks.ro/Forum/viewtopic.php?t=1536 .

The original problem is not completely correct, either - the 6 points also lie on one circle if $ M$ is the orthocenter of triangle $ ABC$. See [url=http://forumgeom.fau.edu/FG2003volume3/FG200305index.html]Alexei Myakishev and Peter Woo, [i]On the circumcenters of cevasix configurations[/i], Forum Geometricorum 3 (2003) pp. 57-63[/url].

  darij[/quote]

Thanks for pointing out my error, I misunderstood the problem. I considered the $ \triangle A'B'C'$ to be circumcevian triangle of the $ \triangle ABC$ WRT the point $ M,$ not the [b]cevian[/b] triangle.


Given non-negative reals $p,q,r$ such that $p+q+r = 1$, prove that

$7(pq+qr+rp) \le 2+9pqr$.

[hide="hint"] [url=http://wangsblog.com/jeffrey/?p=18]Homogenize[/url][/hide]

Homogenize,
\[ 7(\sum pq)(p+q+r) \leq 2(p+q+r)^3+9pqr \]
Expand to get:
\[ (p^2q+q^2p)+(p^2r+r^2p)+(q^2r+r^2q)\leq (p^3+q^3)+(q^3+r^3)+(r^3+p^3) \]
Since: $a^3+b^3 \geq a^2b+b^2a$, so the above inequality it true.

Or by schur you can do this :
$7(pq+qr+rp)=4(pq+qr+rp)+3(pq+qr+rp)\leq 1+9(p+q+r)+(p+q+r)^2=2+9(p+q+r).$

I still don't know what you mean by homogenizing. :?

[quote="varun"]what do you mean by "Schur"
If it is an inequality how do you use it. Please explain.[/quote]

Let \[ T_{x,y,z}[a,b,c]=x^ay^bz^c+x^ay^cz^b+x^by^az^c+x^cy^az^b+x^by^cz^a+x^cy^bz^a  \]

Then [b]Schur  inequality[/b] says that
\[ T_{x,y,z}[2a+b,0,0]+T_{x,y,z}[a,b,b]\geq 2T_{x,y,z}[a+b,b,0]  \]

m@re

Another version of Schur is

$a^n(a-b)(a-c)+b^n(b-c)(b-a)+c^n(c-a)(c-b) \ge 0$.

Homogenizing means basically making the total power (degree) on each term equivalent:

For example, $pq+qr+rp \le p^3+q^3+r^3$ is not homogenous (the LHS has degree two, the RHS three), but if we multiply by $p+q+r = 1$, we get

$(pq+qr+rp)(p+q+r) \le p^3+q^3+r^3$ which has a degree of three on each term, so it is homogenous.

When an inequality is homogeninze, can we set the sum of the variables to a constant number like $1$?

[quote="shobber"][quote="M@re"]And if we can, why is that?[/quote]
Let $\sum a =k$, then $\sum \frac{a}{k}=1$, put it back into the inequality and $k$ cancels out if it is homogenize.[/quote]

Niiiiiiiiiiiiiiiiiiiiiiiiiiiiiice   :D

i think i can solve it very simple:
i prove abc>=(a+b-c)(b+c-a)(c+a-b)
after that abc>=(1-2c)(1-2b)(1-2c)
abc>=1-2(a+b+c)+4(ab+bc+ac)-8abc
so:9abc>=-1+4(ab+bc+ac)
9abc+2>=1+4(ab+bc+ac)>=7(ab+bc+ac)(because 1=(a+b+c)^2)
nice?


group theory
abstract algebra
search
superior algebra
superior algebra unsolved

Suppose that $ G$ is a group of which all elements different from the neutral element have infinite order, and $ G$ has a cyclic subgroup of finite index. Show that $ G$ is cyclic.

I found it: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=601886877&t=43409 . A harder problem of this type, see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=252425 .


function
combinatorics unsolved
combinatorics

A={1,2,3,4}

B{1,2,3}

how many functions $f$ are there of $B$ into $A\times A$ so that for every $a\in A$ there is an $x\in B$ so that $a$ is the right or the left element in the set $f(x)$

Can you please explain in words

i dont understand :(

each function goes from B to AxA

meaning for every element in B there is a subset of AxA

however, they are looking for every subset in AxA there is element in B

i.e an overall function

example

f(1) = (1,2) f(2) = (3,4) f(3) = (1,1) is fine

g(1) = (1,2) g(2) =(2,1) g(3) = (1,1) isnt ok because there is no refrence for any 3's or any 4's

You are correct

your analysis is good, the answer is really bad

the way you wrote it made me think

basically you wrote 123411

its a 6 digit number that must have 1,2,3,4 in it

so basically find all combinations that have 1,2,3,4 in them (the other 2 numbers dont matter)

and that would be the correct answer!

(other examples, 431232 or 133442)

You could generalize to the problem of finding the number $f(n,x)$ of $n$-element sequences where each element is from the set $S = \{ i | 1 \leq i \leq x \}$ such that there is at least one of each element of $S$ in the sequence.

$f(n,x) = \sum_{i=0}^{x}{ (-1)^{i}\binom{x}{i}(x-i)^{n}}$
because $\binom{x}{x-i}(x-i)^{n}$ is the number of sequences which have at most $(x-i)$ different elements

So we have

$f(6,4) = 4^{n}-4(3^{n})+6(2^{n})-4$

[quote="Lena"]just when i thought i had it figured out, you've managed to  confuse me again

the book asks for a "numeric answer"

the answer is 1560[/quote]

Sorry, I gave the answer to the general problem, but when I put in the values for your problem I forgot to replace $n$.

It should be $f(6,4) = 4^{6}-4(3^{6})+6(2^{6})-4 = 1560$

Though I suggest you look at general problems rather than specific examples though, because otherwise the method you use to solve a specific problem may not be adaptable for the general case.

For the general case, I gave a simple method to find the answer, so for example if we wanted the number of 10-element ordered sets $\in \{1,2,3,4\}^{10}$ with at least one occurrence of each element in $\{1,2,3,4\}$ then the answer would be $f(10,4) = 4^{10}-4(3^{10})+6(2^{10})-4$

thank you!

now i understand it better

much appreciated


This problem is taken straight out of the Intro to Geometry book:

My dad was going to cut down a dead tree in our yard one day, but he was afraid it might hit some nearby power lines. He knew that if the tree were over 45 feet tall, the tree would hit the power lines. He stood 30 feet from the base of the tree and held a ruler 6 inches in front of his eyes. He lined the bottom of the ruler up with the base of the tree, and saw that the top of the tree lined up with a point 8 inches high on the ruler. He then knew he could safely cut the tree down. How did he know?

One of my favorite problems! :)

This is just similar triangles. Quite easy.


Solve this and please write the solutions there.        G :roll:  :blush: d luck


So, does the countdown round at nationals decide who gets first (10th in written could get first?)

What would be good exercises to practice for the countdown round?

No, I don't think so.

I'd say to get a Mathlete and challenge him/her to a past practice countdown round, practice makes perfect. Also it might help you see the trends in the questions.

There are three distinct levels of winning at mathcounts -- Masters Round Winner, Written Round Winner, and Overall Winner.  (Those are not the exact names, but it's the right idea.)  The overall champion is the winner of the countdown round, who almost always was number 1 or 2 in the written round.  The masters winner is chosen by a seperate competition from the top four individuals after countdown (I think -- might be after written.  This was a long time  ago for me.)

? are we going to forget the team competition? that gets a $2000 scholarship and a lotta stuff too... (ya go NC GOT 5TH PLACE STICK THAT IN YOUR CROCODILE AND SMOKE IT)

blah. you definitely need to practice countdown rounds if you want to win mathcounts.


solve this :
$ sin(x)\plus{}sin(2x)\plus{}sin(3x)\equal{}1$
I try so many time but I still can't solve it.any idea please??

We can write the whole thing in terms of $ sin x\equal{}n$, we get
$ n\plus{}2n\sqrt{1\minus{}n^2}\minus{}4n^3\plus{}3n\equal{}1$, equivalent to $ 4n^2(1\minus{}n^2)\equal{}(4n^3\minus{}4n\plus{}1)^2$, I don't get a nice answer.

[quote="dgreenb801"]We can write the whole thing in terms of $ sin x \equal{} n$, we get
$ n \plus{} 2n\sqrt {1 \minus{} n^2} \minus{} 4n^3 \plus{} 3n \equal{} 1$, equivalent to $ 4n^2(1 \minus{} n^2) \equal{} (4n^3 \minus{} 4n \plus{} 1)^2$, I don't get a nice answer.[/quote]
So we can't solve this by hand???thanks :D


For $ x > 0$, the area of the triangle with vertices $ (0, 0), (x, 0)$ and
$ (x, 5)$ is 30 square units. What is the value of $ x$?

We see that $ 6 \times 5 \equal{} 30$. However, the area of a triangle is $ \frac{1}{2} \times b \times h$. So our answer is $ 6 \times 2 \equal{} \boxed{12}$.


We can obviously put 100 unit balls in a $10\times10\times1$ box.
How can one put $105$ unit balls in? How can we put $106$ unit balls in?

Normally, a unit ball is considered to have radius r = 1, but this problem calls for unit balls with diameter d = 2r = 1 and radius $r = \frac 1 2$. If the balls are stacked so that the centers of 4 adjacent tangent balls form a square, then the distance between the center lines of unit ball rows is 2r = 1. For 10 rows 10 balls each has 10 between the center lines at the distance 2r = 1 plus 2 extra radii for the 1st and the 10th rows, the total is

$r + 9 \cdot 2r + r = \frac 1 2 + 9 + \frac 1 2 = 10$

which fits into the box. If the balls are stacked so that the centers of 3 adjacent tangent balls form an equilateral triangle, then the distance between the center lines of the balls rows is the altitude of an equilateral triangle with the side 2r = 1, i.e., $h = \dfrac{\sqrt 3}{2}$. For 11 rows 10 balls each odd row and 9 balls each even row at the distance h plus 2 extra radii for the 1st and the 11th rows, the total height of the stacked balls is

$r + 10h + r = 10h + 2r = 10 \dfrac{\sqrt 3}{2} + 1 \doteq 8.66 + 1 = 9.66$

which fits into the box. The total number of balls is $6 \cdot 10 + 5 \cdot 9 = 60 + 45 = 105$. If we replace one 9-ball row anywhere in this sequence by a 10 ball row (with the center lines 2r apart from the adjacent 10-ball rows), for example, the 2nd row, the total height of the stacked balls is

$r + 4r + 8h + r = 8h + 6r = 8 \dfrac{\sqrt 3}{2} + 3 \doteq 6.93 + 3 = 9.93$

which still fits into the box. The total number of balls is now $7 \cdot 10 + 4 \cdot 9 = 70 + 36 = 106$.

Which is the maximum number of such balls that someone can put in a $10\times10\times2$ box or in a $10\times10\times10$ box?
Quite intriguing...


algebra
polynomial
superior algebra
superior algebra unsolved

If $a_1, \ldots, a_{n-1} \in \mathbf{Z}$ then the polynomial

$\frac{X^n}{n!} + a_{n-1}\frac{X^{n-1}}{(n-1)!} + \ldots + a_1X \pm 1$
is irreducible over $\mathbf{Q}[X]$.

I tried some ANT but couldn't succeed in solving this monster (?). I'm sure there are elementary solutions too.

you can take $a_n=\infty$ (is it possible to do that?) and use the fact that

\[      e^x=\sum_{k=0}^{\infty} \frac{x^k}{k!}=1+\frac x1+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots     \]

and do some computations...

it's $\displaystyle e^x=1+\frac x1+\frac{x}{2!}+\frac{x^3}{3!}+\cdots$

$\displaystyle e^x=1+\frac{x}{1}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$

l am very sorry to have encumbered this forum.

Arthur you can edit to correct your post
 ;)

l've made two times the same error :\fracx1 doesn't yields $\displaystyle\frac{x}{1}$

about the problem, what does "irreductible" for a polynomial mean ? that we can't factorize it ?

This result is attributed to I. Schur. The original proof rests on an irreducible criterion of Dumas, which is formulated in terms of the so-called Newton polygons.

Here is a paper in which a similar result is proven; you may easily modify the proof techniques to obtain the desired irreducibility theorem of Schur.

http://www.math.sc.edu/~filaseta/papers/schurpapthird.pdf

Here is another link:

http://arxiv.org/PS_cache/math/pdf/0406/0406307.pdf

Although this latter paper addresses a slightly different problem, it contains a lavish introduction to the theory of Newton polygons.

--Vesselin


Melissa is driving a sports utility vehicle along the interstate at a constant speed of 55 mph. A sports car one-half mile behind her that is moving at a constant speed passes her in 60 seconds. How many miles per hour is the speed of the sports car?

The sports car "catches up" at a rate of a half-mile per minute, or 30 mph. Thus it moves 30 mph faster, so its speed is 30+55=85 mph.


How many two-digit positive integers have at least one 7 as a digit?

My method used complementary counting:
There are 90 numbers in the list 10, 11, 12...88, 89, 90. There are 8x9=72 ways to create a number [b]without[/b] a 7 in it (the first digit can have anything other than 7 and 0, and the last digit can have anything but 7.) 90-17=18, so there are 18 ways to create a 2-digit number with at least one 7 in it.

[hide=Solution]there are 9 that end in 7, and 10 that start in 7, but we overcounted 77, so there are 9+10-1=18[/hide]


Given an isosceles $\triangle ABC$ such that $CB=CA$ and $\angle C>60^{\circ}$. Select two points $A_{1}$ and $B_{1}$ on side $AB$ such that $\angle A_{1}CB_{1}= \angle CAB$. A circle is externally tangent to the circumcircle of $\triangle A_{1}B_{1}C$ and also tangent to rays $CA$ and $CB$ at $A_{2}$ and $B_{2}$ respectively. Let's call the point that the two circles are tangent at, $P$. Show that the extension of $A_{2}P$ and $B_{2}P$ intersects the circumcircle at points $B_{1}$ and $A_{1}$.

Assume $A, A_{1}, B_{1}, B$ follow on the line AB in this order. Let $(O_{1})$ be the circumcircle of $\triangle A_{1}B_{1}C$ and $(O_{2})$ the other circle tangent to it at P. Quadrilateral $A_{1}PB_{1}C$ is cyclic with the circumcircle $(O_{1}),$ $\angle A_{1}PB_{1}= 180^\circ-\angle A_{1}CB_{1}= 180^\circ-\angle CAB.$ The isosceles triangles $\triangle CA_{2}B_{2}\sim \triangle CAB$ are centrally similar with similarity center C, $A_{2}B_{2}\parallel AB \equiv A_{1}B_{1},$ $\angle CA_{2}B_{2}= \angle CAB.$ $CA_{2}$ is a tangent to $(O_{2})$ at $A_{2},$ $\angle B_{2}PA_{2}= 180^\circ-\angle CA_{2}B_{2}.$ It follows that $\angle B_{2}PA_{2}= \angle A_{1}PB_{1}.$ The tangency point P of the circles $(O_{1}),\ (O_{2})$ is their internal similarity center. The parallel chords $A_{1}B_{1}\parallel B_{2}A_{2}$ are centrally similar with the same similarity center P and with the same similarity coefficient, because $\angle A_{1}PB_{1}= \angle B_{2}PA_{2},$ hence $P \in A_{1}B_{2},\ P \in B_{1}A_{2}.$

EDIT: Posting a figure.

For some reason, I seemed to have found an extremely simple proof which I did not notice before. Is there some flaw to it?

Using [b]Yetti's[/b] assumptions of the position of the points and the use of $O_{1}$ and $O_{2}$. The quadrilateral $CA_{2}O_{2}B_{2}$ is cyclic as $\angle CA_{2}O_{2}= \angle CB_{2}O_{2}= 90^{\circ}$ therefore $\angle A_{2}O_{2}B_{2}= 2 \angle CBA$. Now, since angles subtended from an arc to the the origin of the circle is always two times greater than the angle formed when the same arc is subtended onto a point on the circumference. We call a point $P'$ on the circumference of the circle with $O_{2}$ as the origin ($P'$ does cannot be a point on the arc $APB$). $\angle A_{2}P'B_{2}= \frac{1}{2}\angle A_{2}O_{2}B_{2}$, we have $\angle A_{2}PB_{2}= 180^{\circ}-\angle CBA= \angle A_{1}PB_{1}$ therefore $A_{1}$, $P$, $B_{2}$ are collinear and $A_{2}$, $P$, $B_{1}$ are collinear.


trigonometry
limit
inequalities
function
real analysis
real analysis unsolved

[quote="kunny"]Rewrite the recurrence as $u_{n+1}=-\frac{1}{2}\cos [2(u_{n}+11)]-2007$, [/quote]
Very fast! But i think you are wrong here :D

[quote="kunny"]Rewrite the recurrence as $u_{n+1}=\frac{1}{2}-\frac{1}{2}\cos [2(u_{n}+11)]-2007$, let $c=\frac{1}{2}-2007$, we have $u_{n+1}=-\frac{1}{2}\cos [2(u_{n}+11)]+c$.

Thus since $b=-\frac{1}{2}\cos [2(b+11)]+c$, we have $u_{n+1}-b=-\frac{1}{2}(\cos [2(u_{n}+11)]-\cos [2(b+11)])$, use $|\cos \alpha-\cos \beta|\leq |\alpha-\beta|$,  which is given by M.V.T.

$\therefore |u_{n+1}-b|=\frac{1}{2}|\cos [2(u_{n}+11)]-\cos [2(b+11)]|$

$\leq \frac{1}{2}|2(u_{n}+11)-2(b+11)|=|u_{n}-b|$,  $\therefore 0\leq |u_{n}-b|\leq |u_{1}-b|.$[/quote]
OK for recurrence relation, but you NOT complete problem. From $|u_{n+1}-b|\leq |u_{1}-b|$ we can't have $u_{n}\to b$. Try again :P

I do know your indication, N.T.TUAN . Maybe estimation of inequality is weak. :|

[quote="N.T.TUAN"]The sequence of numbers $u_{1},u_{2},...$ is defined by $u_{1}=a,u_{n+1}=\sin^{2}{(u_{n}+11)}-2007(n=1,2,...).$ Where $a$ is a given real number. Prove that
a)Equation $\sin^{2}(x+11)-x=2007$ has unique a real root. Denote it by $b$.
b)$\lim_{n\to\infty}u_{n}=b$.[/quote]

a) Let $f(x)=\sin^{2}(x+1)-x-2007$, we have $f'(x)=\sin 2(x+1)-1\leq 0$ and $0\leq \sin^{2}(x+11)\leq 1 \Longrightarrow-x-2007\leq f(x)\leq-x-2006$.

Moreover $\lim_{x\rightarrow \infty}f(x)=-\infty, \lim_{x\rightarrow-\infty}f(x)=\infty$, thus $f(x)$ is monotone decreasing function, which completes the proof.

b) From the result of a), we have $b=f(b)$. Since we can rewrite the given reccurence as $u_{n+1}=f(u_{n})$,

the mean value of theorem gives $u_{n+1}-b=f(u_{n})-f(b)=\frac{f(u_{n})-f(b)}{u_{n}-b}(u_{n}-b)=f'(c)(u_{n}-b)$, $c$ is on the interval between $u_{n}$ and $b$.

That is to say, $|u_{n+1}-b|=|\sin 2(c+11)-1\parallel u_{n}-b|<r|u_{n}-b|\ (|r|<1)$, I wish I could find such this constant, but I couldn't. :(


Pyridine is a heterocyclic six-membered ring with 5 C and one Nitrogen. It also has three double bonds, so a total of 6 electrons in the p-orbitals of the ring atoms. However, what does the lone electron pair of the Nitrogen contribute to total number of electrons in the p-orbitals, or is it already included? If so, doesn't that make 8 electrons? (2 lone pair from N with 6 from double bonds?)

Then why does my book say it is aromatic?

The lone pair of nitrogen is not needed to complete the six pi electrons of the aromatic system: all atoms are sp2 hybridized, so each have one electron on a p orbital to contribute to the aromatic ring. As the lone pair of N is not used, it remains on an sp2 orbital orthogonal to the pi system, therefore without participating in it. Note how this is in direct contrast with pyrrole, in which the lone pair is needed to complete the aromatic sextet (compare pyridine with benzene and pyrrole with cyclopentadiene anion). That's also the reason why pyridine is a weak base but pyrrole is not, for pratical purposes, even a base.


function
real analysis
real analysis unsolved

Help me solve this differential equation of the string.
Starting speed is 0.
I need to seperate w(movement along y axes) in function of time and position (W(x)*T(t).
x-position along x axes
t-time
c,b,h,L are constants.

w(x,t)=?
W(x)=?
T(x)=?

What have you tried? This is a fairly standard seperation of variables

$w(x,t) = W(x) T(T)$

Then the DE becomes

$c^2 T(t) \frac{d^2 W(x)}{dx^2}-b T(t) \frac{dW(x)}{dx} = \frac{d^2T(t)}{dt^2}W(x)$

Bring all the T(t)'s to one side, and the W(x)'s to the other and apply standard technqiues