Let $a,$ $b$ and $c$ are non-negative numbers. Prove that
$\sqrt[3]{\frac{(2a+b)(2b+c)(2c+a)}{27}}\geq\sqrt{\frac{ab+ac+bc}{3}}.$

Very nice Inequality, Arqady!
In my solution, we have to show that

$5\sum a^{3}b+\sum ab^{3}\ge 3\sum a^{2}b^{2}+3abc\sum a$,

which is true.

[quote="fjwxcsl"]Solly! Correctly,If $ 3/7<\equal{}k<\equal{}7/3, a,b,c>\equal{}0,$ then

              $ ((k*a\plus{}b)*(k*b\plus{}c)*(k*c\plus{}a)/(k\plus{}1)^3)^{1/3}>\equal{}((a*b\plus{}b*c\plus{}c*a)/3)^{1/2}.$[/quote]
I have obtained: $ \frac{1}{k_{0}}\leq k\leq k_{0},$ where $ k_{0}\equal{}\frac{3\sqrt[3]{4}\minus{}2\plus{}\sqrt{18\sqrt[3]2\minus{}12\sqrt[3]4}}{2}\equal{}2.333698853...$
Id est, indeed the following inequality is true too.

Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab\plus{}ac\plus{}bc\equal{}3.$ Prove that:
\[ (3a\plus{}7b)(3b\plus{}7c)(3c\plus{}7a)\geq1000\]

A proof of the original inequality see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=262970

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that
$ \sqrt [3]{\frac {(a \plus{} 2b)(b \plus{} 2c)(c \plus{} 2a)}{27}}\geq\sqrt {\frac {ab \plus{} ac \plus{} bc}{3}}.$[/quote]
Write our inequality as
$ (a \plus{} 2b)(b \plus{} 2c)(c \plus{} 2a) \ge (ab \plus{} bc \plus{} ca)\sqrt {3(ab \plus{} bc \plus{} ca)}.$
Applying AM-GM inequality, we have
$ \sqrt {3(ab \plus{} bc \plus{} ca)} \le \frac {1}{2}\left( a \plus{} b \plus{} c \plus{} \frac {3(ab \plus{} bc \plus{} ca)}{a\plus{}b\plus{}c}\right),$
Thus our inequality reduce to
$ 2(a \plus{} 2b)(b \plus{} 2c)(c \plus{} 2a)(a \plus{} b \plus{} c) \ge (ab \plus{} bc \plus{} ca)[(a \plus{} b \plus{} c)^2 \plus{} 3(ab \plus{} bc \plus{} ca)],$
which is, after expanding, equivalent to
$ \sum a^3b \plus{} 5\sum ab^3 \minus{} 3\sum a^2b^2 \minus{} 3abc\sum a \ge 0,$
or
$ \sum ab(a \minus{} 2b)^2 \plus{} \sum ab^3 \plus{} \sum a^2b^2 \minus{} 3abc\sum a \ge 0.$
By Cauchy Schwarz, we can easily show that
$ \sum ab(a \minus{} 2b)^2 \ge abc\sum a,$
Moreover, the AM-GM yields
$ \sum ab^3 \ge abc\sum a$, and $ \sum a^2b^2 \ge abc\sum a.$
Our proof is now completed. :)

I have found (and proved) the largest range of $ k$ such that the following inequality:
$ (a \plus{} kb)(b \plus{} kc)(c \plus{} ka) \ge (k \plus{} 1)^3,$
holds for all $ a,b,c \ge 0$ satisfying $ ab \plus{} bc \plus{} ca \equal{} 3$ is
$ \frac {3\sqrt [3]{4} \minus{} 2 \minus{} \sqrt {18\sqrt [3]{2} \minus{} 12\sqrt [3]{4}}}{2} \le k \le \frac {3\sqrt [3]{4} \minus{} 2 \plus{} \sqrt {18\sqrt [3]{2} \minus{} 12\sqrt [3]{4}}}{2} .$
:)

NIce proof, can_hang2007!
[quote="can_hang2007"]
I have found (and proved) the largest range of $ k$ such that the following inequality:
$ (a \plus{} kb)(b \plus{} kc)(c \plus{} ka) \ge (k \plus{} 1)^3,$
holds for all $ a,b,c \ge 0$ satisfying $ ab \plus{} bc \plus{} ca \equal{} 3$ is
$ \frac {3\sqrt [3]{4} \minus{} 2 \minus{} \sqrt {18\sqrt [3]{2} \minus{} 12\sqrt [3]{4}}}{2} \le k \le \frac {3\sqrt [3]{4} \minus{} 2 \plus{} \sqrt {18\sqrt [3]{2} \minus{} 12\sqrt [3]{4}}}{2} .$
:)[/quote]
Of cause! I also proved it by the same of the following way:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=262970


[quote="236factorial"]Prove that

$\lim_{x\to0}(1+rx)^{\frac{1}{x}}= e^{r}$

given that $\lim_{x\to0}(1+x)^{\frac{1}{x}}= e$.

[Do not use L'Hopital's rule.][/quote]

Let $u = rx$
$x = u/r$
$\lim_{u\to0}(1+u)^{\frac{1}{u/r}}=\lim_{u\to0}((1+u)^{\frac{1}{u})^{r}}= e^{r}$


I think we should have a post about how nm did in Purple Comet.

Our team, the Atomic City Deuterons, got an honorable mention in the MS Small school category, with 2 people!

New Mexico Red or Green, got an honorable mention in the MS large school category. Congrats!

These are the competitors on our team:
Alexandr Wang
Willie Zhao

These are the competitors i know of on NM R or G:
Robert Chen
Drew Brost
Geelon So
Joseph Griego

These are competitors i know of on New Mexico 6(MS):
Andy Chen

Just like on MC, my team made many "stupid" mistakes (most of them were made by me :( )

Congrats to all who participated, though.

my team got 11 too. We guessed on #14 and got it right, we got #12 because i didn't make a mistake with similarity (it has happened many times before), and we got #13 right because I didn't mess up in casework! :D  :lol:
All the other problems were pretty trivial (except 15), but my teammate misread #2 and #3, and made an arithmetic error on #7. :wallbash_red:

On the high school one we got 10/25... pretty terrible i know. we participated in both ( :lol: ), but i guess the word we is misused. my teammate didn't do any of the problems during the high school one, and kept saying "beef stroganoff," while drawing overly brawny caricatures of 18th century men on the whiteboard. Long story short, it didn't go well. I really liked the high school problems, though. They reminded me of AIME problems, but not as hard. Anyway, I think we did well! :D

Aw dang. We should have taken both MS and HS, too. I would have probably decided to try the HS myself, though. Again, Congrats to the NM Purple Comet Teams!

lol
We almost got 9, until i checked most of my partner's work. I'm so happy we got 14,13, and 12 right. especially 13, because my casework didn't really look right.

bleh... looked over the Purple Comet question, now I realized the I didn't need casework for the one that i'm so happy about (argh), and for the HS one, I could have gotten at least 5 more with my current knowledge (I was reeeeeeeeeeeaaaaaaaaaaaaaally bad at math before nats). urgh.


3.) Solve the system of equations:
5x-2y+7z = 46
x+5y+z = 13
5y-z = 1


AMC
AIME
probability
factorial
Pascal\u0027s Triangle

Three red counters, four green counters and five blue counters are placed in a row in random order. Find the probability that no two blue counters are adjacent.

[quote="mathkid's original post"]
the denominator is clearly 12! / (5!4!3!)

but to find the number of arrangements where no blue is together..
this is what i started doing.

letting * be a blue chip

*_*_*_*_*

since there must be at least one of the seven blue chips in between each..

*7*6*5*4*
where the digits are the number of chips that can be place in each

After that, the last three chips can be placed anywhere. there is a total of 10 places for the first, 11 for the next, and 12 for the last..

i don't know where i'm going with this.

if possible, try to explain how i can do it like i started with the blue chips already arranged.. if you can't, an alternate solution will suffice

thanks[/quote]

Instead of counting the combos where no two blue are adjacent, we shall count the ones that have 2 blue adjacent, 3, 4, and 5 although this might lead to PIE.

-interesting_move

My guess for the number of such arrangement : 8! * 7! / (3! * 4! * 5!)

[hide]

First I put the red and green counters : there are 7!/(4!*3!) way to do that. Then I insert the blue ones :

a1 b1 a2 b2 a3 b3 a4 b4 a5 b5 a6 where ai are the *number* of red or green counters put between the blue counters, bi the blue counters.

 :Sigma: ai = 7, a1, a6 >= 0, a2..a5 >= 1.

ai=bi+1 for 2<=i<=5, bi=ai otherwise.

 :Sigma: bi = 3, with bi >= 0.

By De Moivre formula the number of solutions is C(3+6-1,6-1) = 8!/3!*5!. Hence the result.

Correct ?[/hide]

Belenos, your answer appears to be correct. Another way: [hide]Continuing mathkid's line of reasoning from the *_*_*_*_*: we have 3 non-blue counters remaining, and 6 slots (count the sides) to put them in: C(8,3) ways. Then, distinguishing between red and green, we multiply by C(7,3). Same result as belenos, but a lot less work.[/hide]

[quote="Kamior"]a lot less work.[/quote]

At *lot* less work ???? You're kidding !

Quick counting method for the numerator, this is what I used to do as a kid before I knew about factorials and such (Think I was trying to work out lottery odds with like choose 6 numbers out of 48):

You have 12 "slots" and 5 blue chips. First arrangement is--
X _ X _ X _ X _ X _ _ _
locking the first 4 chips in place, there are 4 positions for the 5th chip that work (just visually). moving the 4th chip to the right one space, there are now 3 positions for the last chip, so for all positions of the last 2 (given the first 3), there are 4th triangle = 10 possible positions. Moving the 3rd chip one space to the right, there are now 3rd triangle = 6 positions for the other 2, so for all positions of the last 3, there are sum of first 4 triangles (call them tetrahedral numbers)=1+3+6+10=20 positions. Similarly, unlocking the 2nd chip gives the sum of the first 4 tetrahedral numbers=35, and first chip, the sum of the first 4 4-dimensional triangular numbers (sums of first n tetrahedrals)=56 for the numerator. All this can be read directly from Pascal's triangle if you know what you're looking for. Gives an answer of 1/495.


Each ring has 2 more squares on a side than the previous one. Since the first ring has 3^2-1^2, the second has 5^2-3^2, etc. we see that the nth ring has (2n+1)^2 - (2n-1)^2 = 8n. So for n = 100, we have 800 squares.

I too got 800, the amount of squares in a ring are multiples of 8.  so 100 * 8 = 800.

Hey Max, the e-mail address on your signature is wrong, I think.  You might wanna change it...

[color=cyan]I suspect that he is engaging in a common tactic to post ones e-mail without getting spammed.  Many programs are designed to search the web for things that fit the (blah)@(blah).(blah) form, adding them to a long list to recieve spam.  By slightly misspelling ones e-mail address in a way a human wishing to e-mail him might correct, or alternatively by writing something like repDELETETHISPART123max@(blah).(blah), he can let people know his e-mail without getting spammed.  Or, it might just be a typo, and all my elaborate explanation would be gone to naught.[/color]

hmmm...I've seen this before.  At LiveJournal, your e-mail address appears normal, but there's some sort of useless code in the middle of it so that spam-robots will get a garbled e-mail address that includes various nonsense letters or something....


With all these books and courses on "speed reading," I sometimes...

What is speed reading?
Is "speed reading" a hype, a myth, ... or just a market for people to sell courses and book about?

Would reading faster prevent you from understanding some of what you're reading?
Or is speed reading simply a collection of techniques to keep you focused on reading?

Are there comparable methods to actually better understand what you're reading rather than just "read" faster?

I took a speed reading course long time ago, but did not practice it since then. I suppose it may help if you are a law student and must read 10000 pages every week. But it doesn't help to learn mathematics faster. And if I read a book for pleasure, I don't try to finish it as soon as possible.

Did the speed reading course help you at all at anything useful?

The only time when I would be interested in reading faster is on the SAT verbal section... but even then I would need to read certain parts slowly to understand them better.

Is there any use in speed reading or is it more or less skimming? Can you retain information well if you speed read?

The ability to "speed read" is just the ability to skim effectively. A good course on speed reading will teach you how to quickly recognize the important information on a page without having to read all of the words. So yes, it's just skimming. But skimming is something you could be good or bad at. Reading 1/10 of the words in a paragraph at random will still give you an idea of what the paragraph is about, but if you read the right 1/10 of the words, you might get pretty much all of the useful info out of it.

That doesn't mean I have any good recommendations for you, though. I am sure lots of the stuff people sell about speed reading is not actually that helpful, and will either tell you obvious things or useless things. Just read free stuff about speed reading online. I'm sure if you google it and look around, you'll find a decent guide. Then you can practice (perhaps with SAT reading comp questions!) :)

I speed read before taking notes on a biology chapter (usually over 20 pages per chapter).This allows me to use effective note-taking skills and discern what would be important for my notes. (Should I annotate this passage about how they concluded DNA carries genetic info or should I save that space for later?).

Thanks for responding, guys.
[b]
7h3.D3m0n.117, as you speed read the chapter, what particular things do you you look for when determining what is important?

Xevarion, do you consider speed reading something worth looking into?
[/b]

Yeah. It's really useful if you want to read about stuff but you just want an overview, or you want to get an impression of a book before reading it in detail. I'm so accustomed to the mathematical style of reading (read VERY slowly because everything is intricate and complicated and you need to understand the details) that I can't efficiently read anything else... :|

Xevarion, have you tried to learn how to speed read in the past? If you have, what would you recommend?
Right now, I am peeking around with google to find a guide or some general suggestions on it as you suggested.

I haven't tried. My roommate is really good at it, so I'm going to try to get him to teach me. After that happens, if you're still interested, I'll send you whatever tips I got from him. Anyone else who's also interested, PM me and I'll write to you too. But it might be a few weeks before this happens.

[quote="aufha"][b]
7h3.D3m0n.117, as you speed read the chapter, what particular things do you you look for when determining what is important?[/b][/quote]

Well first of all, general concepts are definately important but details that are already in another set of chapter notes should not be repeated (unless there is an extension). I normally make a habit of not recording notes that I already know. Notes should include study questions that not only tests your memorization of facts, but also application.

For example, the word "recombinant types"

First I would define what it is. Then I would write the question, "How is this different from parental types?" After that, I try to make a general connection. (Such as, "How does this apply to the rest of genetics?").

Thank you both for your responses, Xevarion and 7h3.D3m0n.117.

In my mind, it has no importance to read speedy, but it's important to do meaningly.

I think, it is important to be able to read speedy, e.g. at school exams (history, language...) you have to read and understand texts quickly to be done in time.

Speed reading has only helped me in the MathCounts State CD round '07. If I speed read a chapter in a book, I find that I have to read it again to remember/understand most of it.

Speed reading isn't just about skimming.  It's about how fast you can go while achieving close to 100% comprehension.  It comes in handy during mathcounts countdown rounds, but also has many other uses.  You should try to increase you reading speed as much as possible without sacrificing comprehension because that's what really matters.  JFK had a reading speed of over 1000 words per minute.


function
Functional Analysis
real analysis
real analysis unsolved

Given a Banach space $E$, does there always exist a differentiable function $f: E \rightarrow \mathbb{R}$ such that $f' \neq 0$ ? If $E$ is finite-dimensional, the answer is yes. What about the general case ?

there always exist linear functional isn't it ? (Hahn Banach)

Obviously it reduces to Hahn-Banach ...  :blush:  Thanks Alekk ...


floor function
logarithms
modular arithmetic
real analysis
real analysis unsolved

Prove that the following series converges:
\[
\sum_{n \geq 1} \frac{(-1)^{\lfloor n\sqrt{2} \rfloor}}{n}.
\]

Estimate its value.

this serie is convergent from Leibiniz criterion,but not absolutely convergent.

[quote="cezar lupu"]this serie is convergent from Leibiniz criterion[/quote] It is not that simple. Indeed, using the Abel transform, it is not hard to show that it suffices to prove that $\sum_{n=1}^m (-1)^{\lfloor n\sqrt 2\rfloor}=O(m^{1-\delta})$ for some $\delta>0$, but the only proof of that I know uses Fourier series and other "non-elementary" techniques. I wonder whether vess or someone else has a simpler approach. As to estimating the value, once we know the speed of convergence, we can give some rough bounds. I don't know if it is possible to find an exact value of the sum though.

I see a similatr problem as follow:

$\sum_{n \geq 1} \frac{(-1)^{\lfloor n \rfloor}}{n}$.

and the general will be true?

Be careful with generalizations here: you can easily find $\alpha>0$ such that $\sum\frac{(-1)^{\lfloor n\alpha\rfloor}}n$ diverges. Try $\alpha=2/3$, for example. You'll get the series $\frac 11-\frac 12+\frac 13 +\frac 14-\frac 15+\frac 16 +\frac 17 -\frac 18+\frac 19+\dots$. It is important that $\sqrt 2$ is irrational, and, morover, cannot be approximated by rationals too well.

sorry :blush:
and my problem should be :
$ \sum_{n \geq 1} \frac{(-1)^{\lfloor \sqrt n \rfloor}}{n}$ converges.
thanks fedja,can you prove your solution?

I finally found a more or less elementary solution. Let $S_m=\sum_{n=1}^m (-1)^{[n\sqrt 2]}$. If we show that $S_m=o(m)$, we can use the Abel transform to conclude that our series converges if and only if the series $\sum_{m\geqslant 1}\frac {S_m}{m(m+1)}$ converges and, in case of convergence, they converge to the same sum. We shall show that $S_m=O(\log m)$, which is more than enough for our purposes.

Let $(1+\sqrt 2)^\ell=A_\ell+B_\ell\sqrt 2$. The numbers $A_\ell$ and $B_\ell$ can be easily found from the recurrence relations $A_1=B_1=1$, $A_{\ell+1}=A_{\ell}+2B_\ell$, $B_{\ell+1}=A_\ell+B_\ell$. What is important for us is that $A_\ell$ are odd for all $\ell$, $B_{\ell+1}\leq 3B_\ell$ and $|B_\ell\sqrt 2-A_\ell|=(\sqrt 2-1)^\ell=\frac 1{A_\ell+B_\ell\sqrt 2}<\frac 1{2 B_\ell}$.

We shall also need a well-known fact that $|A-B\sqrt 2|\geq \frac1{100 B}$
for all integer $A$ and $B\geqslant 1$. It follows from here that, for any $B\geq 1$, any two integers $n$ for which the distance from $n\sqrt 2$ to the nearest integer is less than $\frac 1{2B}$ are at least $\frac B{100}$ units apart.

Let now $m$ be any positive integer. Choose the largest $B=B_\ell$ that is less than $\frac m2$. Let $m'=m-2B$. Note that $m'\leqslant \frac 23m$ (otherwise we could use $B_{\ell+1}$). We have $|S_m|\leqslant |S_{m'}|+\sum_{n=m'+1}^{m'+B}\left|(-1)^{[n\sqrt 2]}+(-1)^{[(n+B)\sqrt 2]}\right|$. Note now that, unless the distance from $n\sqrt 2$ to the nearest integer is less than $\frac{1} {2B}$, we have $[(n+B)\sqrt 2]=[n\sqrt 2]+A$ and, since $A$ is odd, the corresponding term in the sum is $0$. But there may be only 101 exceptional values of $n$ for which it isn't so. Thus, $|S_m|\leqslant |S_{m'}|+202$. Now do the same for $m'$ instead of $m$ and so on. We'll make at most $C\log m$ steps adding a constant during each step.

The estimate we got is quite rough: a direct computation shows that, for $m<200,000$ we have $|S_m|\leqslant 7$, so it is quite believable that the Abel transformation of the original series converges fairly fast. The sum of first 200,000 terms is $-0.5154\dots$ and it is quite believable that all digits presented are correct for the sum of the entire series.

@fedja : using the uniform distribution of $\lfloor n \sqrt{2}\rfloor \pmod{2}$ one can prove that
\begin{eqnarray*}\sum_{n = 1}^{m}(-1 )^{\lfloor n \sqrt{2}\rfloor}& = & o \left( m \right) \end{eqnarray*}
Is this enough to ensure convergence ?


[img]http://img241.imageshack.us/img241/913/adsz2nh3.jpg[/img]

what is the approximete length (in feet) of the longest ladder you can carry horizontally around the corner of the corridor shown here?round your answer down to the nearest foot.(thomas calculus  chapter 4 practice exercises question 72)

The same [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=24520]link[/url] I sent to someone else Tuesday.


Four distinct points, $A$, $B$, $C$, and $D$, are to be selected from $1996$ points evenly spaced around a circle. All quadruples are equally likely to be chosen. What is the probability that the chord $AB$ intersects the chord $CD$?

$\text{(A)}\ \frac 14 \qquad \text{(B)}\ \frac 13 \qquad \text{(C)}\ \frac 12 \qquad \text{(D)}\ \frac 23\qquad \text{(E)}\ \frac 34$

[hide="Solution"]Notice that all quadruples are equally likely likely to be chosen. Thus, we need only examine the six orderings of the points $A$, $B$, $C$, and $D$:

$ACBD$, $ADBC$ , $ABCD$, $ADCB$, $ABDC$, and $ACDB$.

Out of these six orderings, only two satisfy the intersection condition, so our answer is $\frac 26=\frac 13$.[/hide]

Clever problem!

[hide]Every set of four given points can be labeled A,B,C,and D in 24 different ways. In 8 of these possibilities, AB crosses CD. Therefore the probability is 1/3 and the answer is [b]B[/b]  :D [/hide]

Hexaditom - you don't need to consider all 24 pairings, because it's cyclic; you only need to consider six of them (fix one of the points).

[quote="me@home"]when I first looked at the problem, being lazy I wanted to say 1/2- either it crosses or it doesn't.
Then I solved the problem looking at the ABCD orderings and got 1/3
now, I know the fastest solution- we split the problem into three equally likely cases,
1) It intersects
2) Both A and B farther to the left (or right) than C,D. (This looks like two seperately removed self contained line segments)
3) B C (or AD) contained within the arc marked off by the other segment
These three are equally likely, so it is 1/3[/quote]

Cases 2 and 3 are very similar.  What if one chord is the diameter and the other chord does not intersect it?  Is that case 2 or 3?  Also, how do you know for sure that each case is equally likely?  Your answer is right, but your solution seems shaky.

I was simply looking for an intuitive approach, and judging from the answers possible this seems the best; also, as to your specific concerns, we can see that overall one case can't intuitively overpower another, with cancellations etc. The case 2 and case 3 options are just two subcases of non-intersect; they are meant to show that there are twice as many; it doesn't matter which case the diameter fits in really.

you set A as one point.
they intersect if and only iff b is opposite q, which is equally likelyas c or d, so it is 1/3


A uniform rope of weight W hangs between two trees. The ends of the rope are the same height, and they each make angle theta with the trees. Find

a) the tension at either end of the rope

b) the tension in the middle of the rope

Thanks for the help guys.

For part (a), imagine a rod of weight W suspended by two [i]massless [/i]ropes at each end, each of which makes an angle theta with the trees.  That will give you the same answer.

For part (b), the tension is zero.  Intuitively, since the rope makes less of an angle with the horizontal as you approach the middle, it becomes more "slack."

I think the part a) you have is correct, but the part b) answer doesn't seem to be. It is not zero because in the book I got the question from, it gave you a "hint" in which it gave you a definite angle (it was 45 degrees actually), and told you the tension for both parts (for the second part it was W/2). So I don't think it's zero.

Sorry, I didn't think that through very carefully.

What I should have said, for part (b), is: now imagine an infinitesimally short segment of massless rope connecting two rods, each of weight W/2, and which are each connected to the trees with two other segments of massless rope making angles of theta with the horizontal at their other ends.

For partb,horizontal component of tension at the is equal to tension in the middle.


On a chess board ($8*8$) there are written the numbers $1$ to $64$: on the first line, from left to right, there are the numbers $1, 2, 3, ... , 8$; on the second line, from left to right, there are the numbers $9, 10, 11, ... , 16$;etc. The $\"+\"$ and $\"-\"$ signs are put to each number such that, in each line and in each column, there are $4$ $\"+\"$ signs and $4$ $\"-\"$ signs. Then, the $64$ numbers are added. Find all the possible values of this sum.

[hide]Write each number in the form A + 8B, where A ranges from 1 to 8. A is constant along columns, B along rows. Sum As columnwise and Bs rowwise. The sum is 0. [][/hide]

On a chess board ($8\times 8$) there are written the numbers $1$ to $64$: on the first line, from left to right, there are the numbers $1, 2, 3, ... , 8$; on the second line, from left to right, there are the numbers $9, 10, 11, ... , 16$;etc. The ''$ + $'' and ''$- $'' signs are put to each number such that, in each line and in each column, there are $4$ ''$+$'' signs and $4$ ''$-$'' signs. Then, the $64$ numbers are added. Find all the possible values of this sum.


algebra
polynomial
function
calculus
derivative
real analysis

[quote="Moubinool"]Let $x_{0}<x_{1}<\ldots<x_{n}$ and $y_{0}<y_{1}<\ldots<y_{n}$. Prove that there exists a strictly increasing on $[x_{0},x_{n}]$ polynomial $p$ such that $p(x_{j})=y_{j}$, $j=0,\ldots,n$.
(A. Prymak)[/quote]

This was posted and solved in an earlier thread with postings from iura and fleeting_guest.

thank you fleeting guest

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46897

Is there more explicit formula for the polynomial, like

interpolation Lagrange or something else ?

Does anyone know some recents (between 1980-2006) articles, books
wich talked about this problem.

(The old  $\leq 1979$ I found on the google, all proof I read is nonconstructive )

thank you
m.

Moubinool,  you are asking about efficient approximation of the Dirac delta function (derivative of our increasing step function) by positive ($\geq 0$) polynomials on an interval.  This must be a classical subject in books on approximation theory.

[quote="fleeting_guest"] This must be a classical subject in books on approximation theory.[/quote]

Sure but wich book ?
thank you.


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What are your opinions on creationism in the school curriculum?

[url=http://www.defconamerica.org/our-issues/Defending-Science/islands-of-ignorance/]www.defconamerica.org/our-issues/Defending-Science/islands-of-ignorance/[/url]

Just curious, which religious groups are generally pushing for Intelligent Design?

creationism and/or intelligent design can be taught in schools, just not incorporated with the science curriculum like some people want. :o  :o  that would be totally inappropriate :mad:  :mad: , no scientific evidence whatsoever that supports intelligent design. i also don't even understand the intelligent design theory: i mean, just look at the name; someone had to do the designing, who? aliens? E.T?

[quote="elston"] i mean, just look at the name; someone had to do the designing, who? aliens? E.T?[/quote]

A superior being.  What a lot of people would consider to be God.

I disagree with it being taught in school.  It just does not have a sufficient scientific basis.

what happened to separation of church and state?  the evolution theory has its weak spots but ID is just a way for religous groups to oppose evolution and influence the classroom. i've seen no proof for ID.
and notice how half the states pushing for ID are in the deep south, where God and NASCAR are unchallenged.  :D

Um, actually, furious, that statement is totally false, and I don't particularly appreciate it. Deep South=Georgia, Alabama, Mississippi.

But if you search for "intelligent design" on Google News, you come up with articles involving intelligent design in schools in Pennsylvania and Kentucky.

Yet again, I'd like to remind you that the creators of AoPS, two very intelligent (and quite liberal) men, are products of Alabama. I'd also like to tell you that it is very irritating when you make such obviously unintelligent generalizations.

Thanks!  :)

[quote="chess64"]The majority of Americans, by far, are monotheistic. In high school biology, "evolution" is taught, but not everyone believes in evolution by "natural selection." (How can you claim that species A is better than species B? There is no single yardstick by which it can be measured.)
[/quote]
Because species A lives longer and repoduces more.
[quote="chess64"]
Anyway, evolution should be taught as a THEORY, not as the truth. Intelligent design/belief in God should also be mentioned. I.e., "Evolution and intelligent design are two major theories about the creation of humans. ... [explain evolution], [explain intelligent design]" but don't declare either of them to be true. Just state that they are popular theories.[/quote]

Do I have to go into the definition of scientific theory? Theory of electricity anyone?

[quote="chess64"]The majority of Americans, by far, are monotheistic. In high school biology, "evolution" is taught, but not everyone believes in evolution by "natural selection." (How can you claim that species A is better than species B? There is no single yardstick by which it can be measured.)

Anyway, evolution should be taught as a THEORY, not as the truth. Intelligent design/belief in God should also be mentioned. I.e., "Evolution and intelligent design are two major theories about the creation of humans. ... [explain evolution], [explain intelligent design]" but don't declare either of them to be true. Just state that they are popular theories.[/quote]
The majority in America is not supposed to completely rule over the minority. Separation of church and state is separation of church and state no matter how few people are affected.
Scientists don't "believe" in evolution by natural selection. Scientists accept it based on the evidence.
Scientists don't claim that one species is better than another. Where do they claim that? A trait is favored if it results in more grandchildren. That's all there is to it.
Evolution IS taught as a theory, but a theory in science doesn't mean what you're using the word for. A theory in science is a well-supported explanation for our observations. It is as close to the "truth" as science can possibly get.
ID should not be taught in science class because it is pseudoscience. It only has support among the general public, not among scientists (of every religion, worldview, and political opinion there is, so don't claim conspiracy). The general public in general does not know that much about science and what is and isn't science. God is a supernatural entity, therefore its effects on the natural world cannot be predicted, therefore it cannot be part of a scientific theory or a science class.

[quote="bookaholic"]It only has support among the general public, not among scientists (of every religion, worldview, and political opinion there is, so don't claim conspiracy). [/quote]

Let's please not make these sweeping generalizations. There are definitely scientists who believe in Intelligent Design.

Even if they believe in Intelligent Design, those scientists cannot disregard the evidence for natural selection.

Scientists might believe in "Intelligent Design," but that doesn't mean that they don't completely agree with evolution. In fact that might support both - one in a scientific context, the other in terms of faith.

In terms of high school biology, evolution simply has no flaws. It explains absolutely everything, and it's also needed, it's a "necessary and sufficient" condition". Thus it needs to be taught in high school biology. A class in philosophy or religion class might be an acceptable place for ID.

[quote]Just state that they are popular theories.[/quote]

Evolution is popular among biologists, while ID is popular among the religious. One is sufficient for a theory to be taught as truth in a biology class. The other isn't.

[quote]In terms of high school biology, evolution simply has no flaws. It explains absolutely everything, and it's also needed, it's a "necessary and sufficient" condition". Thus it needs to be taught in high school biology. A class in philosophy or religion class might be an acceptable place for ID. [/quote]

Intelligent Design has no flaws either.  A Creator explains everything, especially the infinite details in all of biology.  Both theories should be taught for what they're worth and a student allowed to decide his or her opinion.
What I believe is biased is including parts of evolution in most subjects, not just science, and leaving ID without a glance.

One other thing, you can't really believe in one scientifically, and the other religiously.  They definitely contradict.

[u]Dr.[/u] Kent Hovind Believes the bible is completely true and believes in "In the beggining God created the hevens and the earts."
Penguinintegral-Do you really like the ACLU?

And think about it, Evolution is faith based too, there is no evidence that proves that macto-evolution actually happened. Don't say what about the Monkeys DNA, cause They're just similar that doesn't mean they both came from a StarFish.
And Pangea never existed millions of years ago, the earth is only about 6000 years old, and just because they seem to fit doesn't mean they did. For istance I bet your house and the nearest Mc'Donalds seem to fit together perfectly does that mean they were once connected but seperated and the rest of the town came up in between them?  And in Pangea where is Central America, and if India and Austrailia are on the same plate why didn't they move the same way away from this "Super Continent?"
The government cannot comand schools to teach one way or another because it states in the constitution the government shall have no part in schools.
Sorry if I don't answer your replies I don't come here often.
[quote]theory to be taught as truth [/quote] oh the oxymoron kills me. Theorys are not proven true or they would be facts.[/u]

[quote]oh the oxymoron kills me. Theorys are not proven true or they would be facts.[/quote]

Where's the oxymoron? Rarely does one call gravity the "theory of gravity" even if it is, in fact, a theory. In high school physics, it is taught as truth. The same is true of evolution - altough the use of the word theory has stuck, it is in fact use as a true or a fact.

[quote]Dr. Kent Hovind Believes the bible is completely true and believes in "In the beggining God created the hevens and the earts."
[/quote]

What does this prove? Dr. Kent Hovind is a creationist.

If you wish to debate the merits of a strict biblical interpretation, we could start another thread. For the moment, we are discussing ID vs. evolution.

[quote]One other thing, you can't really believe in one scientifically, and the other religiously. They definitely contradict.[/quote]

How?

[quote]Intelligent Design has no flaws either. A Creator explains everything, especially the infinite details in all of biology.[/quote]

It also gives absolutely no insight into biology.

Most people I know are already highly opinionated and I don't think teaching ID in school would do anything at all to sway most peoples opinions. Pretty pointless.  :|

Yes, but not all people entering science classes in public schools are knowledgeable about the subject matter at hand.   You can only argue about something if you know about it- people who had yet to take biology might not know the difference.

Thats even the problem: most Creatonists don't even now what evolution really is, they just deny it cause it is told to them...

Do you have justification for that statement? Just wondering.

mathnerd314 is back!!! :welcome: :(

[quote="mathnerd314"]Do you have justification for that statement? Just wondering. [/quote]When can we expect justification to be posted for your statements, such as the ones you were called on that appear to be complete fabrications?  Just curious.

Anyway, to answer your question to the other poster: if the public record of their statements about evolution is any indication, evolution deniers don't have much clue what "evolution" means.

They also ought to teach about the giant flying spaghetti monster
http://www.venganza.org/
There is just as much proof of that as ID

Apologies if others have delineated similar thoughts already, but this  happens to be a very long forum.

In any case, I wish to point out that the two theories are not necessarily mutually exclusive.  There is no reason for the impossibility of evolution after the creation of the universe by some divine being.  Similarly, evolution does not disprove the existence of a divine being.

Personally, I support evolution, but I get the impression that some people believe it in a fanatical/religious way (granted, I am guilty of believing in many of the physical laws in a similar way).  Science in general cannot be proven in the mathematical sense (given some axioms, results can be proven, but the validity of the axioms cannot be proven - ie Newton's Laws of motion).

In essence, I think there is more antagonism than is necessary.  Neither side will ever disprove the other and neither side is ever capable of proving their own point in the mathematical sense.

[quote="probability1.01"]Well I totally agree with intelligent design. First of all, it is your responsibility to prove that evolution is true. If you can't prove it, intelligent design must be true, assuming that I prove intelligent design to be true... wait I can't darn whatever just SHUT UP. Anyways, intelligent design is a perfectly reasonable theory, wait I just said theories must be discredited therefore I meant undeniable fact.

Plus, school is not supposed to go against people's religions, and it would be shaking my faith if you taught me some stupid elovution thingy or whatev. Wait, you say your religion doesn't believe in going to school? OK, fine.. HOLD ON NO IT ONLY COUNTS FOR MY RELIGION SO HAH!

So yeah schools will have a much better environment wait did you say that you wanted to teach something about green space monkeys controlling everything that goes on earth? Hmmm let's see NOPE YOU CAN'T TEACH THAT THAT'S HERESY I MEAN UNSCIENTIFIC WAIT HMMMM I'M CONFUSED DIE.

Anyway, we just want students to know that elovution is not the ONLY possibility.. in fact, 12 out of every 7 Americans believe in this alternate theory called intelligent design. So you see, we're very open-minded in alchemy class err wait I meant science class. Wait, you're saying scientific evidence tells me I'm wrong? Sorry, you didn't account for the 0.000000000000000000054324 chance that one of your pieces of evidence is inaccurate, therefore you're entire argument must die. Like, YOU'VE GOTTA HAVE FAITH, ok? JUST BELIEVE IT.

You know, I beleive in intelligent design more than ever now. In fact, we should teach intelligent design in physics and chemistry class too. After all, what proof is there for anything in those subjects? None. Therefore, everything must be controlled by God I mean an intelligent designer. Therefore, what I am writing here is controlled by a G-, no intelligent designer therefore I am divine therefore YOU MUST BOW DOWN TO ME OR ELSE.[/quote]

c'mon now, surely you realize that you can't have 12 out of 7 people believing something - that gives you a probability even higher than 1.01, which is evidently false; thus, the whole credibility of your post is undermined by this mathematical error  :(

I have to admit, though, that you make some other good points. I've never really noticed that neither physics nor chemistry have any solid basis. Now that you mention it, the assumptions do seem rather arbitrary.

[quote="MysticTerminator"]I have to admit, though, that you make some other good points. I've never really noticed that neither physics nor chemistry have any solid basis. Now that you mention it, the assumptions do seem rather arbitrary.[/quote]
Neither does mathematics, the incompleteness theorem states that no axiomatic system can prove its own consistency. Maybe MATH IS WRONG!!! So we shouldn't be teaching math in school without teaching an equally plausible theory, in which something is true iff God wants it to be true, and mathematical proof is heresy - after all, it's possible to PROVE that 1=2, which we all know is baloney. Math is the creation of the DEVIL!

[quote="MysticTerminator"]I have to admit, though, that you make some other good points.  [/quote]His parodic posting was good.  But the bar was set very high by the Flying Spaghetti letter to the creationist school board --- pure gold!

[quote]I've never really noticed that neither physics nor chemistry have any solid basis. Now that you mention it, the assumptions do seem rather arbitrary.[/quote]OF COURSE they are arbitrary, as they are constrained by real-world measurements.  The arbitrariness progressively decreases, as an ever-larger amount of data is derived from an ever-smaller set of assumptions (as is done in chemistry and physics, using ever more mathematics).  ID adds a meaningless assumption and no new conclusions, so is running in the wrong direction -- it increases the arbitrariness.

Just wondering... has anyone answered the question: What substance does ID have?

Because I was reading through this thread and I never found it :(.

In math/physix, most of the axioms are supported from evidence.  Everything else comes from mathematical derivation.  The way math is designed, it seems to be an inherent tautology.

I will assert again, that the two theories in question are not mutually exclusive.

In case anyone is interesting in knowing, the topic for all Public Forum Debaters in January has to do with this very topic. The Resolution is R: that the United States secondary school science curriculum should include study of the theory of Intelligent Design.

By the way, although I don't necessarily agree with the concept of Intelligent Design, I support it being taught, just as religion as being taught. Sort of as a theory, that hasn't been proven, and it should be given no more validation than what it deserves.

Lets sue them for not putting flying spaghetti monster on the curriculum :rotfl:


[quote="soruz"]Let  $ x_1,x_2,...,x_n$ be distinct positive integers, then prove
$ a)x{_1}^2 \plus{} x{_2}^2 \plus{} ... \plus{} x{_n}^2 \ge \frac {2n \plus{} 1}{3}(x_1 \plus{} x_2 \plus{} ... \plus{} x_n)$
[/quote]
Suppose that $ x_1<x_2 <...<x_n$,hence $ x_i \geq i$ for all $ i$
Put $ y_i\equal{}a_i\minus{}i$,the inequaluty becomes:
$ \sum^{n}_{i\equal{}1}b_i^2\plus{}2\sum^{n}_{i\equal{}1}ib_i\plus{}\frac{n(n\plus{}1)(2n\plus{}1)}{6} \geq \frac{2n\plus{}1}{3}\sum^{n}_{i\equal{}1} b_i\plus{}\frac{n(n\plus{}1)(2n\plus{}1)}{6}$
Note that $ b_1 \leq b_2 \leq ... \leq b_n$
Then by Chebyshev inequality we have:
$ 2\sum^{n}_{i\equal{}1}ib_i \geq (n\plus{}1)\sum ^{n}_{i\equal{}1} b_i \geq \frac{2n\plus{}1}{3}\sum^{n}_{i\equal{}1} b_i$
Done!


In year $N$, the $300^\text{th}$ day of the year is a Tuesday. In year $N+1$, the $200^\text{th}$ day is also a Tuesday. On what day of the week did the $100^\text{th}$ of year $N-1$ occur?

$\text{(A)}\ \text{Thursday}\qquad\text{(B)}\ \text{Friday}\qquad\text{(C)}\ \text{Saturday}\qquad\text{(D)}\ \text{Sunday}\qquad\text{(E)}\ \text{Monday}$

[hide]I got (D) Sunday

SInce there is a remainder of 6/7 when #of days between the 300th and 200th days is divided by 7, year N+1 is a leap year. So there are 565 days between the N-1 and N year and when 565/7, there is a remainder of 5 and 5 days after Tuesday is SUnday. [/hide]

ooooook.....
I also got the 565 days difference between the TUE of Nth year, and the day X of the year N-1
then I got that 567 days before the TUE was also TUE (567/7=81)
That means that 566 days ago (I mean, before) was WED,
and 565 days before was THU (A)
 :blush:

Oops. >.< I remember doing this on some practice test and getting Thursday...but yeah. Probably thursday then.

:rotfl: you atleat tried

I don't get what you are saying.

way to have a conclusive answer on this thread guys... anyway we can tell n is a leap year because day 200 of year N+1 is divisible by seven, meaning there must be 266 not 265 days between the two dates. Counting backwards we go back 565 days = 5 (mod 7) going back 5 days from tuesday we get to Thursday (A)

I got Thursday as well (mod 7 technique)
But mathemonster, please don't post on a 5-month-old topic.

its not 5 months, its 1 year and 5 months..

and i also got A by mod 7 teck

I got Thursday too.
I think they give you year N+1, which is a leap year, to prove N-1 is not a leap year. But you know that they would have to give you N+1 as a leap year or N-1 could either be a leap year or a normal one, so you wouldn't have to figure it out, right? :D

Year N is a leap year. Then year N-1's 100th day is 565 days before 300th day of year N. Then the 100th day of N-1 is Thursday.

[quote=Skupp3]Welcome to AoPS? Don't bump 10 year old posts...[/quote]

Okay

[hide=Solution]Note that this means the 6th day of year $N$ is Tuesday and the 4th day of year $N+1$ is Tuesday, so year $N+1$ is a leap year. There are hence $565$ days between the 100th of year $N-1$ and the 300th of year $N$, and $565 \equiv 5 \pmod 7$. Going backward five days gives $\text{Thursday} \implies \boxed{\text{(A)}}$.[/hide]

[b]End time:[/b] 4:15 PM

[b]Interruptions + Solution Writing + Non-Strict Environment:[/b] 15 minutes

[b]Projected End Time on Real Test:[/b] 4:00 PM (49 minutes)

[b]Satisfaction:[/b] 7/10

[b]Score:[/b] 145.5

[quote=OlympusHero][hide=Solution]Note that this means the 6th day of year $N$ is Tuesday and the 4th day of year $N+1$ is Tuesday, so year $N+1$ is a leap year. There are hence $565$ days between the 100th of year $N-1$ and the 300th of year $N$, and $565 \equiv 5 \pmod 7$. Going backward five days gives $\text{Thursday} \implies \boxed{\text{(A)}}$.[/hide]

[b]End time:[/b] 4:15 PM

[b]Interruptions + Solution Writing + Non-Strict Environment:[/b] 15 minutes

[b]Projected End Time on Real Test:[/b] 4:00 PM (49 minutes)

[b]Satisfaction:[/b] 7/10

[b]Score:[/b] 144[/quote]

Wait...did you speed run this AMC 10?? :o (You did a great job, BTW. :))

Yes, I did. I believe the question I missed was #18 (geometric probability except not really probability) for anyone curious.

Oh wait oops I got an answer of 47 which was not an answer choice so I actually would've left that blank so 145.5 not 144 :rotfl:

[hide=Solution]The $100^{\text{th}}$ day of the year $N-1$ is $565$ days before the $300^{\text{th}}$ day of year $N.$ Hence, the $100^{\text{th}}$ day of the year $N-1$ is the same day as the $-565\equiv-5\equiv2\text{th}\pmod7$ day after Tuesday, which is $\boxed{\textbf{(A) Thursday}}.$


function
real analysis
real analysis unsolved

We know that $ f: \mathbb{R} \to \mathbb{R}$, $ f$ is continuous on $ \mathbb{R}$ and for all $ x\notin \mathbb{Q} : f'(x) \equal{} 0$. Is it true that $ f \equal{} const$ ?

[quote="t0rajir0u"]Some modification of the [url=http://en.wikipedia.org/wiki/Cantor_function]Cantor function[/url] should be an appropriate counterexample.[/quote]

The Cantor set is uncountable.

One can show that if $ f\in C([a,b])$, $ f'$ exists for $ x\in [a,b]\setminus A$ where $ A$ is countable, and $ f'\in L^1([a,b])$, then $ f$ is absolutely continuous (and hence can be recovered from $ f'$ by integration).  In this case, one easily sees that we must have $ f$ constant.


function
calculus
derivative
algebra
functional equation
cyclic function

Please help me again with this one:

If F(x-3/x+1) + F(3+x/1-x)=x: x not=1; x not=-1, determine the function of F(x)

Posted in the wrong section - it's not linear algebra.

Do you mean $ F \left( \frac{x\minus{}3}{1\plus{}x} \right) \plus{} F \left( \frac{x\plus{}3}{1\minus{}x} \right) \equal{} x$

yes that's what i mean. i know it's in the wrong section and feel so sorry about it.

[hide="Hint"] Let $ G(x) \equal{} \frac{x \plus{} 3}{1 \minus{} x}$.  What is $ G^{\minus{}1}(x)$? [/hide]

yes i know that.
f(g) + f(inv g) = x
but i still can't get the answer. Would you please help me to solve it?

[hide="Second hint"] Compute $ g(g(x))$. [/hide]

:| sorry i have been evaluating so many time, but still can't get the answer.

Plugging $ x\rightarrow \frac {x \minus{} 3}{x \plus{} 1}$ we get
\[ f(x) \plus{} f(\frac {x \plus{} 3}{1 \minus{} x}) \equal{} \frac {x \minus{} 3}{x \plus{} 1}\ \ \ (i)
\]
and plugging $ x\rightarrow \frac {3 \plus{} x}{1 \minus{} x}$ we get
\[ f(x) \plus{} f(\frac {x \minus{} 3}{x \plus{} 1}) \equal{} \frac {3 \plus{} x}{1 \minus{} x}\ \ \ (ii)
\]
adding $ (i)$ and $ (ii)$ we get $ f(x) \equal{} \frac {x(3 \plus{} x)(3 \minus{} x)}{2(x \plus{} 1)(x \minus{} 1)}$.

[quote="ghjk"][quote="worthawholebean"][hide="another hint"]Try plugging in $ g(x)$ and $ g(g(x))$ as $ x$ to the original equation. Then solve for $ f(x)$.[/hide][/quote]
Nice approach, worthawholebean! Can you show me how can you think of it? Does it have any method to solve this kind of functional equation?[/quote]I used t0rajir0u's hints and got that $ g(g(g(x)))\equal{}x$, so I tried plugging in random things for a while and eventually tried $ g(x)$ and $ g(g(x))$, and I had a linear system.

In other words, $ g(x)$ is a [[cyclic function]] of order 3.  They are useful in that you can plug away end up with a usually solvable system of equations.
For a simple example, to find $ f(x)$ such that $ f(x) \plus{} 3f(1/x) \equal{} x\minus{}1$, we note that $ f(x) \equal{} 1/x$ is cyclic with order 2.  Then we plug in $ 1/x$ for $ x$ to get
$ f(1/x)\plus{}3f(x)\equal{}1/x \minus{} 1$.  Now we can solve this system.

[quote="archimedes1"we note that $ f(x) \equal{} 1/x$ is cyclic with order 2..[/quote]
I think it's incorrect since depending on the definition you provide, I guess the order of a cyclic function is n in oder to have: f^n(x)=x for some integer n>1.
Now we can't find any integer n>1 to satisfy:f^n(x)=1/x, so how can you state that f(x)=1/x is cyclic with oder 2? Am I wrong anywhere? :maybe:

Now I'm forgetting whether $ f^n(x)$ means $ f(f(f(...f(x))...)$ or $ (f(x))^n$.  According to AoPS Volume II, the first definition is correct, but according to some WOOT instructors the second is correct.  In either case, the definition in the wiki means $ f(f(...f(x))...)$, where $ f$ is iterated $ n$ times.

$ f(f(x)) \equal{} x$, so $ f(x)$ has order 2... right?

As far as I know, $ f^n(x)$ always means iteration (whenever $ f(x)$ is not a trigonometric function).  Incidentally, the "cyclic functions" we're talking about here are elements of finite order in the [url=http://en.wikipedia.org/wiki/Projective_linear_group]projective linear group[/url].

[quote="worthawholebean"]That ambiguous notation has always bothered me, but there's not much you can do about it - $ f^n(x)$ is iteration on any non-trigonometric function.$ f^{(n)}$ is differentiation as well.[/quote]
...at WOOT we were taught that $ f^{(n)}$ means iteration and $ f^n(x)$ is taking $ f(x)$ to the $ n$th power.  I guess that the definition in a problem is usually apparent from the context.

I remember something about that in one of the WOOT lectures, but the more common standard is that $ f^{(n)}$ denotes the $ n^{th}$ derivative.

its correct the function i found or not?  :(

Thank you friends, you just opened up my mind.
But the answer key is :
f(x) = 4x/(1-x^2) - 0.5
 :maybe:
i am sure it is getting so close now.


probability
linear algebra
matrix
probability and stats

As a simplified model for weather forecasting, suppose that the weather (either wet or dry) tomorrow will be the same as the weather today with probability p.

If the weather is dry on January 1st, what is the probability (in terrms of p) that the weather will be dry on February 1st?

That probability is $ \frac{1\plus{}(2p\minus{}1)^{31}}{2}.$

Unless $ p$ is very close to either $ 1$ or $ 0,$ that should be very close to $ \frac12.$

What you describe (besides bearing no resemblance to actual weather forecasting) is a two-state Markov chain whose transition matrix is

$ P\equal{}\begin{pmatrix}p&1\minus{}p\\1\minus{}p&p\end{pmatrix}.$

This matrix has eigenvalues $ 1$ and $ 2p\minus{}1$ and can be readily diagonalized. Using the diagonal form, we compute that:

$ P^n\equal{}\frac12\begin{pmatrix}1\plus{}(2p\minus{}1)^n&1\minus{}(2p\minus{}1)^n\\1\minus{}(2p\minus{}1)^n&1\plus{}(2p\minus{}1)^n
\end{pmatrix}.$

The answer above is the $ (1,1)$ entry of $ P^n$ for $ n\equal{}31.$


$a$ and $b$ are positive parameters.Find the minimum value $S = \sqrt{(x-a)^2+b^2}+\sqrt{(x-b)^2+a^2}$.

let there be $A(a,b)$ and $B(b,-a)$ as well as $P(x,0)$.

the we want to minimize $\overline{AP}+\overline{PB}$ and thats exactly when $A,P,B$ are colinear.

the line through $A$ and $B$ is $y=\frac{b+a}{a-b}(x-a)+b=\frac{a+b}{a-b}x -\frac{a+b}{a-b}a+b$

so $S$ is the distance between $A$ and $B$: $S=\sqrt{(a-b)^2+(b+a)^2}=\sqrt{2a^2+2b^2}$

peeta

great observation. i once did a similar problem which used the properties of ellipse!!!

What's that problem? Can you post it here?


Each unit square of a 2 by n grid is to be colored with blue or green so that no 2by2 green square is obtained, let c_{n} be the number of different colorings. Determine max{k} so that 3^k divides c_{2001}

Let n>0 be a integer, s={1,...n}. Show that the nuber of permutations of s with no fixed points and the number of permuttions of s with exactly one fixed point differ by 1.

Let s={1,...n} T be the set of all subsets of s (including s and the empty set). One tries to choose 3 sets  T1, T2, T3from the set T so that T1 \in T3, T2\in T3. I howm any ways can this be done?

I think I have solved the first one.

Let $h_{n}$ be the number of good colorings, with two most-right squares green, and $g_{n}$ the numbers of good colorings, with two most-right squares colored either blue, either green-blue. (For example $h_{1}= 1, g_{1}= 3 , h_{2}= 3 , g_{2}= 12$).

We obviously have:
$c_{n}= h_{n}+g_{n}$
Now we think of a reccurency. If we want to create a good coloring of 2xN grid, we can put two squares colored in any way to the right end of a good colored 2x(N-1) grid, which has two most-right squares colored non-green.
We can also put two squares, which won't be both green, to the right end of the good 2x(N-1) grid, if the two most-right squares of that 2x(N-1) grid were both green. So we have:
$c_{n}= 3\cdot h_{n-1)+4 \cdot g_{n-1}= 4 \cdot c_{n-1}-h_{n-1}}$
Now we notice, that $h_{n}= g_{n-1}$
So we have:
$c_{n}= 4 \cdot c_{n-1}-h_{n-1}= 4 \cdot c_{n-1}-g_{n-2}= 4 \cdot c_{n-1}-c_{n-2}+h_{n-2}$
Now we continue, and we get:
$c_{n}= 4 \cdot c_{n-1}-c_{n-2}+c_{n-3}-...+(-1)^{n}c_{1}$
but there is also:
$c_{n+1}= 4 \cdot c_{n}-c_{n-1}+c_{n-2}-...+(-1)^{n+1}c_{1}$
adding up we get:
$c_{n+1}= 3 \cdot ( c_{n}+c_{n-1})$
Which is absolutely sattisfying us.
Now by checking that $c_{1}= 4 , c_{2}= 15$ (note that $c_{0}= 1$, so our reccurency works!!) we can easily find k =1000.

The second one is preety obvious, since from inclusion-exclusion we get, that the number of permutations with no fixed point is equal to:
$n! \cdot (\frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-...+(-1)^{n}\cdot\frac{1}{n!})$
now we choose the fixed point in n ways, and make a permutation with no fixed point with a set of ${1, 2, ... ,n-1}$. So we get:
$n \cdot (n-1)! \cdot (\frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-...+(-1)^{n-1}\cdot \frac{1}{(n-1)!})$
Thoose two obiously differ by $n! \cdot (-1)^{n}\cdot\frac{1}{n!}$

The third one is also not to hard. Every element can be either in $T_{3}$ and is not in any other subset, or it is in $T_{1}$ and $T_{3}$, or it is in $T_{2}$ and $T_{3}$, or in $T_{1}$ , $T_{2}$ and $T_{3}$. It can also not be an element of any of the subsets $T_{1}$ , $T_{2}$ and $T_{3}$. so there are $5^{n}$ such triples of subsets.

Sorry for double-posting, but I posted one solution, and after that I started thinking of the other.
I hope my solutions are clear and understandable (and true of course  :P )
And sorry for my awful english  :wink:

Theese were nice problems, I especially like the first one.

[quote="jgao"]Let n>0 be a integer, s={1,...n}. Show that the nuber of permutations of s with no fixed points and the number of permuttions of s with exactly one fixed point differ by 1.[/quote]

I think I have solved this one:

Define an $n$-permutation as a permutation of the set $\{ i | 1 \leq i \leq n \}$
Let the number of $n$-permutations with no fixed points be $a_{n}$
Let the number of $n$-permutations with one fixed point be $b_{n}$

Consider any $n$-permutation be $\{ p_{i}| 1 \leq i \leq n \}$
Let $G$ be a graph where there is a directed edge from vertex $i$ to vertex $j$ iff $p_{i}=j$
Let $q_{p_{i}}=i$
Then fixed points correspond to vertices each having an edge to itself
For any sub-permutation of the permutation with no fixed points,
Each vertex in the corresponding sub-graph has $1$ in-edge and $1$ out-edge
Thus the sub-graph is a set of disjoint cycles

Consider the graph corresponding to an $n$-permutation with no fixed points
Consider the vertex $n$
If $p_{p_{n}}\not=n$, removing vertex $n$ and replacing directed edges $(q_{n},n)$ and $(n,p_{n})$ with directed edge ${(q_{n},p_{n}}$ will result in an $(n-1)$-permutation with no fixed point,
And there are exactly $(n-1)$ $n$-permutations for each $(n-1)$-permutation because you can insert vertex $n$ after any of the $(n-1)$ vertices
If $p_{p_{n}}=n$, merging vertices $n,p_{n}$ will result in an $(n-1)$-permutation with one fixed point
And there is exactly $1$ $n$-permutation for each $(n-1)$-permutation

Thus $a_{n}=(n-1)a_{(n-1)}+b_{(n-1)}$

Consider the graph corresponding to an $n$-permutation with one fixed point
Removing the vertex corresponding to the fixed point will result in an $(n-1)$-permutation with no fixed points
And there are exactly $n$ $n$-permutation for each $(n-1)$-permutation

Thus $b_{n}=na_{(n-1)}$

So we have
$a_{n}=(n-1)a_{(n-1)}+(n-1)a_{(n-2)}$
$a_{n}-na_{(n-1)}=-a_{(n-1)}+(n-1)a_{(n-2)}$
$(a_{n}-b_{n})=-(a_{(n-1)}-b_{(n-1)})$

And it only remains to prove that $a_{1}$ and $b_{1}$ differ by $1$

Edit: Someone beat me to it :P but I still like my solution better, because in my opinion it uses intrinsic properties of the problem :lol:


How do you use the avatar you uploaded from your computer in the classroom? I looked through the faq and couldn't find anything.

If your avatar is a gif or a jpg, you should be able to change your avatar in the Classroom to your board avatar by just clicking on the little person by the input box.  After clicking that, scroll down and your avatar should appear.  The avatars for the classroom are refreshed every 24 hours, so if you upload something right now, for example, it won't be available tonight.


inequalities
absolute value
inequalities proposed

let $x_1,x_2 ... x_n$ be real numbers not equal to $1$ and with product $1$
show that :
$\left| \frac{x_1 x_2}{x_1-1}\right|+\left| \frac{x_2 x_3}{x_2-1}\right|+...+\left| \frac{x_n x_1}{x_n-1}\right| > \left[ \frac{n+1}{2} \right]$

here $\left[ x \right]$ denotes the greatest integer less than $x$ while $|x|$ the absolute value of $x$

Show that $\left[ \frac{n+1}{2} \right]$ can not be improved.


I saw lots of TST tests, but I have no clue what a TST is. I tried using the search(since everyone who doesn't seems to get attacked) but I don't think I'm to good with keywords. So could someone explain what they are?

Team Selection Test - theres another thread on this in the Other Topics forum.

A simple search on "what is tst" reveals the magic answer: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=what+is+tst&t=22017


polynomial
function
search
superior algebra
superior algebra unsolved

This is a problem from a previous IMO
Let $ a > b > c > d$ be positive integers and suppose that $ ac\plus{}bd \equal{} (b\plus{}d\plus{}a\minus{}c)(b\plus{}d\minus{}a\plus{}c)$.  Prove that ab + cd is not prime.

There exists a solution using the Eisenstein integers.

[quote="N.T.TUAN"]Use function search, i have http://www.mathlinks.ro/Forum/viewtopic.php?p=119217#119217[/quote]
Yes, but I am looking for the solution using Eisenstein integers ;)

[hide]
The given condition is equivalent to
\[ ac\plus{}bd\equal{}(b\plus{}d)^{2}\minus{}(a\minus{}c)^{2}\Rightarrow a^{2}\minus{}ac\plus{}c^{2}\equal{}b^{2}\plus{}bd\plus{}d^{2}\]
Let $ p \equal{} ab\plus{}cd$ and suppose $ p$ prime. Multiply above equation by $ b^{2}$ and replace $ ab$ with $ p\minus{}cd$ whenever it appears. We get
\[ (p\minus{}cd)^{2}\minus{}bc(p\minus{}cd)\plus{}b^{2}c^{2}\equal{}b^{2}(b^{2}\plus{}bd\plus{}d^{2})\]

\[ p(p\minus{}2cd\minus{}bc)\equal{}(b^{2}\minus{}c^{2})(b^{2}\plus{}bd\plus{}d^{2})\qquad (*)\]
Since $ p$ is a prime dividing a product of two integers, it divides one of them. But
\[ 0<b^{2}\minus{}c^{2}<b^{2}<ab<p\]

\[ 0<b^{2}\plus{}bd\plus{}d^{2}<ab\plus{}ab\plus{}cd<2p\]
so we must have $ b^{2}\plus{}bd\plus{}d^{2}\equal{} p$.
--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Now we get $ p\minus{}2cd\minus{}bc \equal{} b^{2}\minus{}c^{2}$ from (*), so $ p \equal{} b^{2}\plus{}bc\plus{}2cd\minus{}c^{2}$. Set $ p$ equal to $ b^{2}\plus{}bd\plus{}d^{2}$ again to get
\[ bd\plus{}c^{2}\plus{}d^{2}\equal{}bc\plus{}2cd\Rightarrow (c\minus{}d)^{2}\equal{}b(c\minus{}d)\]
which is ridiculous.
[/hide]
[hide="how the Eisenstein integers could be used"]
Cut off the part below the dotted line (hmm... how many computers will I ruin this way?), we then have
\[ p\equal{}(b\minus{}d\omega)(b\minus{}d\omega^{2})\equal{}(a\plus{}c\omega)(a\plus{}c\omega^{2})\]
where $ \omega \equal{}\frac{\minus{}1\plus{}i\sqrt{3}}{2}$ and each of the four factors, with norm $ p$, are irreducible in $ \mathbb{Z}[\omega]$, which has unique factorization. So each factor on one side is an associate of a factor on the other side. Since $ \mathbb{Z}[\omega]$ has 6 units, this will probably result in some casework.
[/hide]


modular arithmetic
algebra
binomial theorem

Happy New year to all !!!

Here are 3 questions to start this year :-)

$Question 1$
Imagine that all the numbers from $1$ to $1 million (both inclusive)$ are written in one line as under:

$123456789101112131415161171819202122...$

What is the the sum of the digits that appear in this large number? I am not looking for single digit sum since that can be easily found using $\mod(9)$.

$Question 2$
What would be the last three digits of ${2007}^{100}-{2003}^{100}$

$Question 3$
What would be the last three digits of ${2007}^{100}\times{2003}^{100}$

What?

[hide="That should be..."] $2007^{100}\equiv 7^{100}\bmod 1000$ and $2003^{100}\equiv 3^{100}\bmod 1000$

Now, $1000 = 8 \cdot 125$, where $\phi(8) = 4, \phi(125) = 100$, so when we solve $\bmod 8, \bmod 125$ and use CRT we obtain the result

$gcd(a, 10) = 1 \implies a^{100}\equiv 1 \bmod 1000$

This is a stronger result than the one obtainable with Euler's Totient Theorem, which tells you something about the need to understand the concepts behind theorems :)

Anyway, using this lemma we know that $7^{100}\equiv 3^{100}\equiv 1 \bmod 1000$ which solves both #2 and #3. [/hide]

I'm sort of confused.

[hide]
Why can't we do $\phi(1000)=400$ and then realize that 400 = 100+100+100+100.  Wouldn't that mean if $7^{400}$ gives a remainder of 1 then $7^{100}$ would have to as well?[/hide]

[hide="#1 solution"]
Let's see, looking at small patterns we see that:

$\sum_{k=0}^{10}d(k) = \sum_{k=0}^{9}k+1$

$\sum_{k=0}^{100}d(k) = 10\sum_{k=0}^{9}k+10\sum_{k=0}^{9}k+1$

$\sum_{k=0}^{1000}d(k) = 100\sum_{k=0}^{9}k+100\sum_{k=0}^{9}k+100\sum_{k=0}^{9}k+1$

With $d(k)$ representing the sum of the digits of k.

Now the pattern seems to be:

$\sum_{k=0}^{10^{n}}d(k) = n*10^{n-1}*\sum_{k=0}^{9}k+1$

$\sum_{k=0}^{10^{n}}d(k) = n*10^{n-1}*45+1$

$\sum_{k=0}^{10^{6}}d(k) = 6*10^{5}*45+1$

So our answer is $27,000,001$.

Not sure about it though...
[/hide]

[quote="t0rajir0u"][quote="Potato Theory"]Wouldn't that mean if $7^{400}$ gives a remainder of 1 then $7^{100}$ would have to as well?[/quote]

Absolutely not.  It just means that $7^{100}$ gives a remainder $a$ such that $a^{4}\equiv 1 \bmod 1000$.  We could have $a = 1,-1$, or two other numbers.[/quote]

Okay, I see what you mean.  Your solution works well then!

[hide="1"] Let the "digit inverse" of a digit $d$ be the digit $d^{-1}$ such that

$d+d^{-1}= 9$

Now, take a given number from $0$ to $999999$ inclusive and group it with the number where each digit (including starting $0$s; for example, $13 = 000013$) of our given is replaced by its digit inverse.  The sum will be $9+9+9+9+9+9 = 54$ for each such pair, and there are $500000$ such pairs, plus an extra $1$ for $1000000$, making our final answer

$500000 \cdot 54+1 = 27000001$ [/hide]

for #3 i have a different opinion...
2007^100 * 2003^100=(2007*2003)^100=21^100(mod 1000)
21=20+1 , so using binomial theorem we know that the last 3 terms is 400,20,1 so add them together we have the last 3 digit 421

[quote="Potato Theory"][quote="t0rajir0u"]Hmm.  Are you sure about your pattern?

[hide="1"] Let the "digit inverse" of a digit $d$ be the digit $d^{-1}$ such that

$d+d^{-1}= 9$

Now, take a given number from $0$ to $999999$ inclusive and group it with the number where each digit (including starting $0$s; for example, $13 = 000013$) of our given is replaced by its digit inverse.  The sum will be $999999$ for each such pair, and there are $500000$ such pairs, plus an extra $1$ for $1000000$, making our final answer

$500000 \cdot 999999+1 = 4999995000001$ [/hide][/quote]

That's seems quite high.  Consider just the numbers 1-99.

We have lists:

00    10
01    11
02    12
03    13
04    14
05    15
06    16
07    17
08    18
09    19

(Rather simplistic; I know)

Anyway, for the ten's place we see ten 0s in the first list.  Ten 1s in the second place, and so on, so 10 * 45 gives a total of 450 for just the ten's place digits.  For the one's, there are ten sums of 0-9, so once again 10 * 45 gives 450.  450+450+1 = 901.  Your method gives 99*50+1 = 4951, which is almost as high as the sum itself (5050).

For your example $000013$ + $999986$ gives a digit total of 54 (like JavaMan said.)

Perhaps I'm missing something; it's late.  :([/quote]

The sum of the digits should be 18 * 50 + 1, I think t0r forgot to add the digits of 999999 (which in this case is 99, 9+9=18). You're finding the sum of the numbers from 1 to 100, where as the original question asked for the sum of the digits of the numbers when written as a string.

tor's method (with the correction):  $\frac{10^{n}}{2}*n*9+1$ is equivalent to $n*10^{n-1}*45+1$

[quote="JavaMan"][quote="t0rajir0u"]Hmm.  Are you sure about your pattern?

[hide="1"] Let the "digit inverse" of a digit $d$ be the digit $d^{-1}$ such that

$d+d^{-1}= 9$

Now, take a given number from $0$ to $999999$ inclusive and group it with the number where each digit (including starting $0$s; for example, $13 = 000013$) of our given is replaced by its digit inverse.  The sum will be $999999$ for each such pair, and there are $500000$ such pairs, plus an extra $1$ for $1000000$, making our final answer

$500000 \cdot 999999+1 = 4999995000001$ [/hide][/quote]

Isn't the sum of each pair's digits 9 * 6 = 54?[/quote]

Yes.  Yes, it is.   :oops:

Does anyone know the source of these 3 problems?


In the figure, the line segments AN, AM, and BC are tangent to a circle at points N, M, and P, respectively. The points B and C are on segments AN and AM, respectively. Angle BAC is 30 degrees and angle ABC is a right angle. If the length of AN is 12, then what is the area of triangle ABC?

[hide]
Let $ BN \equal{} x$. Then $ OP \equal{} BN \equal{} x$, because they are parallel and intersected $ NO$ and $ BP$, which are perpinducular to them. $ NO, BP, CM \equal{} x$ as well.

$ AB \equal{} 12 \minus{} x$ and because triangle ABC is a 30-60-90 triangle, $ BC \equal{} \frac{12 \minus{}x}{\sqrt{3}}$ and $ AC \equal{} \frac{2(12\minus{}x)}{\sqrt{3}}$.

$ PC$ and $ CM$ are both tangent to the circle, and because they intersect, they are equal. $ BC \minus{} BP \equal{} PC \equal{} CM \equal{} AM \minus{} CM \Rightarrow PC \equal{} \frac{12 \minus{}x}{\sqrt{3}} \minus{} x \equal{} 12 \minus{} \frac{2(12\minus{}x)}{\sqrt{3}}$

Solving this says that $ x \equal{} 24 \minus{} 12\sqrt{3}$. Now, legs AB and BC can be found and the area can be found. $ AB \equal{} 12 \minus{} (24 \minus{} 12\sqrt{3}) \equal{} \minus{}12 \plus{} 12\sqrt{3}$ and $ BC \equal{} \frac{\minus{}12 \plus{} 12\sqrt{3}}{\sqrt{3}}$.

 [/hide]

I just could not get started before, but thanks to gullibelu I have a start. I agree as far as
[hide][quote="gullibelu"]
Let $ BN \equal{} x$. Then $ OP \equal{} BN \equal{} x$, because they are parallel and intersected $ NO$ and $ BP$, which are perpinducular to them. $ NO, BP \equal{} x$ as well.
.... triangle ABC is a 30-60-90 triangle[/quote][/hide] but then
[hide]I agree that $ ABC$ is a 30-60-90 triangle and $ OPBN$ is a square, and the idea of calling the side of that square (also radius of the circle) $ x$ is promising.
However, I do not think that $ CM\equal{}x$ too; that would make $ OPCM$ a square too.[/hide] and talking about 30-60-90 triangles,
[hide]$ OPC$ and $ OMC$ are congruent and are 30-60-90 triangles too.[/hide] After that, I keep making mistakes in the calculations.

A solution that seems to check
[hide]From the leg ratios of 30-60-90 triangles $ ABC$ and $ OPC$ we know:
$ BP = OP = x =\sqrt {3}PC$ and
$ AB = \sqrt{3}BC$
Knowing that $ BC=BP+PC=\frac {1+\sqrt {3}}{\sqrt {3}}x=\frac {3+\sqrt {3}}{3}x$,and
$ AB =12 - x$,
$ AB = \sqrt{3}BC$ turns into
${ 12 - x=(1+\sqrt {3}})x$,
which can be solved to get
$ x=12(2-\sqrt{3})$,  so
$ BC=\frac {3+\sqrt {3}}{3}x=\frac {3+\sqrt {3}}{3}12(2-\sqrt{3})=4(3-\sqrt {3})$, and
$ AB = \sqrt{3}BC=\sqrt{3}*4(3-\sqrt {3})= 12(\sqrt{3}-1)$
Then the area of $ ABC$ is
$ AB*BC/2=6(\sqrt{3}-1)*4(3-\sqrt {3})=48(2\sqrt{3}-3)$[/hide]
or did I mess up somewhere?

[quote="sept.23"]How can you tell that triangle OPC is also a 30, 60, 90, triangle.[/quote]
I believe that the line that goes through the intersection of two tangents and the center bisects that angle, therefore OCP would be 60.

[quote="sept.23"]How can you tell that BP = OP.[/quote]
$ OPBN$ is a square. The problem says the angle at B is a right angle. It has right angles at $ N$ and $ P$ (tangents to a circle are perpendicular to the radius at the point of tangency) and the contiguous sides $ OP$ and $ ON$ are congruent (radius of the circle).

I can first tell that triangles $ OPC$ and $ OMC$ are congruent right triangles. They have right angles at $ P$ and $ M$ (tangents to a circle are perpendicular to the radius at the point of tangency) and 3 pairs of congruent sides (more obvious for some of the pairs than for others). Their congruent angles at $ O$ added to the right angle $ PON$ add up to the central angle $ NOM$ that is part of the quadrilateral (kite) $ ANOM$, along with two right angles and a 30 degree angle. The measure of $ NOM$ is 360 degrees-(90+90=30) degrees = 150 degrees. $ PON$ measures 150 degrees - 90 degrees = 60 degrees. Both $ POC$ and $ MOC$ measure half of that, or 30 degrees.

Or did I mess up in this alphabet soup?


The problem version is the following:

max(cos(x)cos(y)+cos(x)cos(z)+cos(y)cos(z)), where

a) x+y+z=180 degrees;
b) x,y,z are the angles of a triangle.

Please HELP!

hello, we get with the help of Lagrange multipliers
$ \cos(x)\cos(y)\plus{}\cos(x)\cos(z)\plus{}\cos(y)\cos(z)\le\frac{3}{4}$, the equal sign holds if $ x\equal{}y\equal{}z\equal{}\frac{\pi}{3}$.
Sonnhard.

Thank you for your reply!

Is it possible to post the full solution for this problem?

Thank you!

I outlined one in your [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=285419]other thread[/url]. By the way, please, do not post the same question twice in the future. It does not increase your chance of getting an answer, just irritates people and creates confusion ;).


blogs
topology
calculus
calculus computations

Show that [i]every[/i] subset of a metric space $ M$ can be written as the intersection of open sets.

If you mean 'arbitrary intersection', then it is quite easy. In metric space, every one point set is closed. Thus for each $ x \in M$, $ U_x \equal{} M \minus{} \{x \}$ is open.

Now let $ A \subset B$ a proper subset of $ M$. (Note that $ M$ itself is open, so we may consider proper subset.) Then

$ A \equal{} \bigcap_{x \in M\minus{}A} U_x$

This proves the assertion.

In fact, this holds for any topological space where every one point set is closed.

Related challenge/exercise: below is an argument intended to solve the given problem.  Explain why it doesn't work in general.  Also, give a necessary and sufficient condition on $ X$ in order for this argument to work.

Let $ X$ be the subset of $ M$ that we're interested in, and define for every $ r > 0$ $ X_r \equal{} \bigcup_{x \in X} B_r(x)$, where $ B_r(x)$ is the ball centered at $ x$ of radius $ r$.  Thus, $ X_r$ is what we get by taking a small ball around every element of $ X$, and so $ X_r$ is clearly open.  I claim that $ I \equal{} \bigcap_{r > 0} X_r \equal{} X$.  Certainly $ X \subset X_r$ for every $ r$ and so $ X \subseteq I$.  Suppose we have some point of $ M$ that's not in $ X$, say $ y$.  For any $ x \in X$, we know there is some $ r$ such that $ r < d(x, y)$, so $ y$ will not be in a ball around $ x$ with radius $ r$.  Thus, for small enough $ r$ we have that $ y \not\in X_r$ and so $ y \not \in I$, so $ X \equal{} I$, as desired.

[quote="JBL"]$ X_r$ is what we get by taking a small ball around every element of $ X$, and so $ X_r$ is clearly open. [/quote]
This isn't my specialty, but $ B_r(x)$ need not be open if $ r$ is small and $ x$ is isolated, so I believe a necessary and sufficient condition is that $ X$ contain no isolated points.

A general point like this was recently raised on Tim Gowers' blog - namely, that the weakest point in an argument is usually when a word like "clearly" is being used :P

$ B_r(x)$ is the open ball in $ M$ around $ X$, and $ X_r$ is a union of these open balls and so open.  So that step is actually fine.  (But you're right that as a stylistic matter I shouldn't have used "clearly.")

I believe the problem is here:[quote="JBL"]For any $ x \in X$, we know there is some $ r$ such that $ r < d(x, y)$, so $ y$ will not be in a ball around $ x$ with radius $ r$.  Thus, for small enough $ r$ we have that $ y \not\in X_r$ and so $ y \not \in I$.[/quote]Though the first statement is true, the conclusion does not to follow.  But the argument should work if $ X$ is nowhere dense in $ M$.

The problem with the reasoning, as JoeBlow showed, is that r depends on x so the infimum of such r's could well be 0. sos440 is right that such a statement holds in topological spaces with strong topologies. What is an example of a topological space in which it does not hold?

[quote="§outh§tar"]What is an example of a topological space in which it does not hold?[/quote]Take the simplest example of a topological space $ X$ in which not every point is closed: $ X\equal{}\{1,2\}$ with the topology $ \{\emptyset,\{1\},\{1,2\}\}$. Then $ 1$ is not a closed point and $ \{2\}$ is not the intersection of open subsets.

[quote="JoeBlow"]But the argument should work if $ X$ is nowhere dense in $ M$.[/quote]  Yes, this is sufficient, but actually there is a simpler necessary and sufficient condition: [hide="Answer to my question"]This argument works if and only if $ X$ is closed in $ M$.  (So, oddly enough, this argument "usually" fails to write open sets as intersections of open sets.  Also, a nice example of a spectacular failure of this argument (suggested by JoeBlow's comment) is when $ M \equal{} \bf R$ and $ X \equal{} \bf Q$.)[/hide]

The notion of $ G_\delta$-sets applies here.

Not to mention the argument misses the case for empty sets  :rotfl:

 :maybe:


The sequence $ S\equal{}a_{1}a_{2}...a_{N}$, where $ a_{i}\in \{A, B\}$  is not palindromic. Let $ a_{i}^{j}, i\leq j$ denote the subsequence $ a_{i}a_{i\plus{}1}...a_{j}$. For any subsequence $ a_{i}^{j}$, let $ F(a_{i}^{j})\equal{}A^{k}B^{j\minus{}i\plus{}1\minus{}k}$, where $ k$ is the number of A's and $ j\minus{}i\plus{}1\minus{}k$ is the number of B's in the subsequence. Thus, for any length N sequence, we get $ \frac{N*(N\plus{}1)}{2}$ F's, one for each subsequence. Let this collection of F's be called $ D(S)$. If two sequences $ S_{1}$ and $ S_{2}$, have same D, i.e,  $ D(S_{1}) \equal{} D(S_{2})$, then show that $ S_{1}$ must be of the form $ Sx_{1}Sx_{2}Sx_{3}...Sx_{k}S$ and $ S_{2}$ in that case will equal  $ Sx_{k}Sx_{k\minus{}1}Sx_{k\minus{}2}...Sx_{1}S$, for some fixed sequence $ S$ (same for both) and $ x_{i}\in\{A, B\}$.

Sorry, a small part of the qn is missing. The last time will read - If two sequences $ S_{1}$ and $ S_{2}$, have same D, i.e,  $ D(S_{1}) \equal{} D(S_{2})$, then show that $ S_{1}$ must be of the form $ Sx_{1}Sx_{2}Sx_{3}...Sx_{k}S$ and $ S_{2}$ in that case will equal  $ Sx_{k}Sx_{k\minus{}1}Sx_{k\minus{}2}...Sx_{1}S$, for some fixed sequence $ S$ (same for both) and $ x_{i}\in\{A, B\}$, OR $ S_{1}$ is the reversed sequence $ S_{2}$. Does anyone have any ideas about this problem and how to go about it? Thanks.


Find the greatest positive constant A which satisfies:

x/\sqrt(y2 + z2) + y/\sqrt(x2 + z2) + z/\sqrt(x2 + y2) > A

for any positive reals x, y and z.

[Well, I was wondering: shouldn't it be \geq A? Or am I totally wrong?]

Agree the strict inequality is strange.

A = 3/\sqrt 2 no ?

. We can suppose \sum x = 1 since it's homogeneous.

. C.S : \sum x / \sqrt(y+z) * \sum \sqrt(y+z) >= (\sum \sqrt x)  >= 3 (last by concavity of \sqrt)

. C.S again : (\sum \sqrt(y+z)) <= 6 \sum x <= 2 (last by convexity of x)

In fact, A=2 and the problem is quite simple. All you have to do is to observe that taking a=b and c very small the expression tends to 2 and to apply the Arithmetic-Harmonic mean:
  x/sqrt(y^2+z^2)=sqrt(x^2/(y^2+z^2))>=2x^2/(x^2+y^2+z^2). Sum up these inequalities and observe that we cannot have equality.


Is there a reason why the beginner level problems are in "Getting Started", medium level questions are in "Intermediate [b]Topics[/b]", and the hard level questions are in "Advanced [b]Problems[/b]"?
Just out of curiosity...

It's all subconscious.  Heh, you noticed something that Mr. Rusczyk probably didn't when he renamed the forums...

Well, it seems somewhat natural to me..."Getting Questions" wouldn't make much sense, right?

Yes, but shouldn't Advanced Problems be Advanced Topics since the others are called "... Topics"

Well, no...he can name then whatever he wants as long as they get the message across.

But why does it matter?

I actually find the deviation in the naming pleasant. But seriously, go worry about other things.


Hi. I was looking for the 2008 Nat Cd Round. They took it down from the mathcounts website. Do any of you know how to access it now?  :huh:

here , u can have it 4 free. :D I'm willing to!

I think he was talking about the 2008 web cast? if so, it's just gone from the Mathcounts website forever. Every year, they take off the old webcast forever to replace it with the new one.

well... i was looking for the webcast..   :maybe:
but thank you cubex!! :lol:

Maybe we can share our answers if anyone else tried to solve these? Please post your answers here.

If you want a solution for a particular problem, it's probably best to create a new thread.  :wink:

EDIT: Not to be a hall monitor for MATHCOUNTS, but is it okay to post a single test on a thread (like CubeX did)?

Ok here are the solutions for the 2008 Nationals:

could u give me 2008 national mathcounts sprint.
Please

CuBeX, I'm pretty sure it's illegal to post full MATHCOUNTS tests on the Internet.

Catalog is [url=http://mathcounts.org/Document.Doc?id=294]here[/url] and order form is [url=http://mathcounts.org/Document.Doc?id=295]here[/url].


Hi,
I guess I should start some olympiad preparation now. If someone can recommend me some nice preparation books(national olympiad level), that'll be great!
Thanks!

[quote="boxedexe"]Chen Sun (if that is correct) recommends [i]Problem Solving Strategies[/i] by Arthur Engel.[/quote]

Yes, that is correct  :clap2: . And you, boxedexe, are?

His name is in his signature -- Masoud Zargar. ;)

Oops, just notice a typo in the title :oops:

[quote="sunchips"]engel  :lol:[/quote]
Thanks! What kind of questions does the book cover? I mean the difficulty. >=national olympiad?

[quote="Phelpedo"]I would call it pre-olympiad, but it has a lot of (the easier) olympiad problems.[/quote]
 :P  hmm By what percentage? roughly
Easy:     Medium:    Hard:

anyone?
 :) i just want to make sure because I did mail back a book which i ordered several weeks ago and turned out to be fairly easy and "meaningless" when I got it. I had to pay the postage anyways:wink:

Thanks! i know it's hard to say because everyone is different. i just want to have a general idea.

It ranges from trivial problems to IMO level problems.

P.Solving strategies is a good book but u have to learn most of the topics from the problem its not an ideal book for learning theory.
If u want to b a good problem solver u should read the book the art nd craft of problem solving by paul zeitz
>moon
i love math i love bangladesh

[quote="andy_tok"]Thanks! Can I buy it somewhere in Canada?
Or I may have to order online on amazon.com?[/quote]
Online: http://www.amazon.com/Problem-Solving-Strategies-Problem-Books-Mathematics/dp/0387982191/sr=8-1/qid=1169909691/ref=sr_1_1/102-9944810-3528962?ie=UTF8&s=books
It might exist in a big bookstore like Chapters or Indigo.

Here is a web url:
http://www.kalva.demon.co.uk
here u may get a lot of probs of different math olympiads.
Nd for learning better u may read some books
1.103 trigonometry problems.
2.The art and craft of problem solving
nd u should practice probs thats d only way 2 b grt math wizard
>moon

[1] Mathematical Olympiad Challenges, Second Edition by Titu Andreescu

[2] Challenging Problems in Geometry by Alfred S. Posamentier

[3] Mathematical Olympiad Treasures by Titu Andreescu

[4] 102 Combinatorial Problems by Titu Andreescu

[5] Problem-Solving Strategies (Problem Books in Mathematics) by Arthur Engel

[6] 103 Trigonometry Problems: From the Training of the USA IMO Team by Titu Andreescu

[7] A Path to Combinatorics for Undergraduates: Counting Strategies (Paperback)
by Titu Andreescu, Zuming Feng

[8] Principles and Techniques in Combinatorics (Paperback)
by Chen Chuan-Chong, Koh Khee-Meng

[9] Mathematical Miniatures (New Mathematical Library) (Paperback)
by Svetoslav Savchev, Titu Andreescu

[10] Elementary Number Theory (Hardcover)
by David M. Burton

sigh... i wish Titu Andreescu had included similar commentary ahead of 102 problems in Combinatorics, just like he did for 103 problems in Trigonometry.

I could really use some training for combinatorics


A seven digit positive octal integer N is formed by each of the nonzero octal digits from 1 to 7 exactly once, such that N satisfies all the following conditions:

(A) The sum of the digits 1 and 2 and all the digits between them is equal to the octal number 12.

(B) The sum of the digits 2 and 3 and all the digits between them is equal to the octal number 23.

(C) The sum of the digits 3 and 4 and all the digits between them is equal to the octal number 34.

Given that the first digit of N is greater than the last digit, what are the possible value(s) of N?

[hide="solution"]1+2+3+4+5+6+7=34 (octal). Therefore 3,4 have to be the end digits. Since 4>3, 4 is the first digit and 3 is the last.
4xxxxx3
The 2~3 segment excludes 4 so the maximum possible is 30. Since we are aiming for 23, we must subtract 5 additional. There are several ways to do so: (5), (4,1), (3,2), but only (5) is valid.

452xxx3

Now we work on the 2~1 segment. It excludes 4,5,3 so the maximum possible is 20. We are left with 1,6,7 and must subtract 6, thus 6 is excluded.

4527163 is our desired number.[/hide]

[quote="james4l"]The maximum possible is 30. Since we are aiming for 23, we must subtract 5 additional.[/quote]

I believe you mean 7.


geometry
perimeter
calculus
integration
national olympiad

find the no. of all integer sided isosceles obtuse angled triangles with perimeter 2008

hi I've solved it in this way
suppose in triangle ABC ,C be the obtuse angle.
now 90<c/2<45
1<sin(c/2)<1/sqrt( 1/2)
but sin (c/2)=sqrt[(s-b)(s-a)/ab]
=s-b/b          .....since a=b;
1<s-b/b<sqrt(1/2)
this will give u the valid value of b
that will be the no. of ans but remember only take integral value as per question.
I remember I got 86 triangles.
if u don't like the answer tell I have some more methods.

please give the answer to my question
i am not satisfied with the previous answer

The answer is 86.  If you are looking for ordered triples, you should take 86*3 = 258.

$ 2008  \equal{} 2x \plus{} z$ and $ \sqrt{2}x < z < 2x$ which gives us $ 832 \leq z \leq 1002$. Note that $ z$ should be even. Hence 86.


advanced fields
advanced fields unsolved

Show that a line bundle is trivializable if and only if it admits a nowhere-zero section.

Thanks.

If $ s$ is such a section of the line bundle $ \pi: L\to M$, then try the $ M$-bundle morphism $ \mathbb R\times M\to L$ defined by $ (\lambda,p)\mapsto \lambda s(p)$.


When the Earth, Moon, and Sun form a right triangle with the Moon located at the right angle, as shown in the figure below, the Moon is approaching its third quarter. (The Earth is viewed here from above its north pole.) Find the magnitude and direction of the net force exerted on the Moon. Give the direction relative to the line connecting the Moon and the Sun.

the picture is shown in [/img][u]Physics[/u] by James S. Walker p. 389, #8 under problems

The force exerted by the Earth on the Moon has the direction of line that links the centers of Earth and Moon, sense from the Moon to the Earth, and magnitude

$ F_{E,M} \equal{} \frac{GM_EM_M}{d^2}$

where d is the distance between Earth and Moon.

The force exerted by the Sun on the Moon has the direction of line that links the centers of the Sun and Moon, sense from Moon to the Sun, and magnitude

$ F_{SM} \equal{} \frac{GM_SM_M}{r^2}$

where r is the distance between Moon and Sun. These are the only two forces acting on the Moon, and as they form a right triangle then the net force $ F$ exerted on the Moon will have magnitude

$ F \equal{} \sqrt{F_{E,M}^2 \plus{} F_{S,M}^2}$.

The angle $ \theta$ that F makes with the line connecting the Moon and the Sun is given by

$ \theta \equal{} \arctan \frac{F_{E,M}}{F_{S,M}}$.


Well the problem goes like this

"When a school play charges $ \$$2 for admission, an average of 100 people attend. For each 10cents increase in admission price, the average number decreases by 1. What charge would make the most money?"

I need to make an equation in the form : a(x-h)^2 + k.

If you can do this it would be of great help =P

[hide]meh I have way too much spare time..

remember for every $ x$ time the price increases by 10 cents, there are $ x$ less people.

[hide="still clueless?"]Cost $ \equal{} (2 \plus{} .10x)(100 \minus{} x)$ because for every $ x$ time 10 cents increase in charge, an avg. of $ x$ number of people decrease.

so solving,

$ \equal{} \minus{}0.1x^2 \plus{} 8x \plus{} 200$

completing the square

$ \equal{} \minus{}\frac{1}{10}(x^2 \plus{} 80x \plus{} 1600) \plus{} 200 \minus{} 160$

$ \equal{} \boxed{\minus{}\frac{1}{10}(x\plus{}40)^2 \plus{} 40}$
[/hide][/hide]


[quote="zaya_yc"]a,b,c>0
$ \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}$ find min[/quote]
The minimum does not exist.  :wink:

What do you mean?('The minimum does not exist?)Why?
I know that $ \sum\sqrt{\frac{a}{b+c}}>2$ But i don't know is this one true:$ \sum\sqrt{\frac{a}{b+c}}\geq\frac{3}{\sqrt{2}}$ :maybe: ?

[quote="Erken"]What do you mean?('The minimum does not exist?)Why?
I know that $ \sum\sqrt{\frac{a}{b+c}}>2$ [/quote]
You are right and the value $ 2$ does not hold.
[quote="Erken"]But i don't know is this one true:$ \sum\sqrt{\frac{a}{b+c}}\geq\frac{3}{\sqrt{2}}$ :maybe: ?[/quote]
It's wrong: try $ a=b=1$ and $ c=0.07.$

[quote="Erken"]But it means that the answer is $ 2+\frac{1}{\infty}$,isn't it?[/quote]
I am sorry, what is this? This is not number. :wink:

[quote="Erken"]What do you mean?('The minimum does not exist?)Why?
I know that $ \sum\sqrt{\frac{a}{b+c}}>2$ [/quote]
$ 2$ is called infimum of that express  :lol:

[quote="Erken"]I know that $ \sum\sqrt {\frac {x}{y\plus{}z}}>2$. [/quote]

If $ x,y,z > 0,\frac {49 \minus{} 9\sqrt {17}}{32}\equal{} 0.371\cdots< k<\frac {49 \plus{} 9\sqrt {17}}{32}\equal{} 2.69\cdots$, then

\[ \sqrt {\frac {x}{y\plus{}kz}}\plus{}\sqrt {\frac {y}{z\plus{}kx}}\plus{}\sqrt {\frac {z}{x\plus{}ky}}>\frac {2}{\sqrt [4]{ k}}\].
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=105343]I have not an elementary solution to these inequalities.[/url]

[quote="zaya_yc"]a,b,c>0
$ \sqrt {\frac {a}{b \plus{} c}} \plus{} \sqrt {\frac {b}{a \plus{} c}} \plus{} \sqrt {\frac {c}{a \plus{} b}}$ find min[/quote]
If $ a,b,c\ge0$,we have $ Min \equal{} 2 \Leftrightarrow a \equal{} b,c \equal{} 0$and itspermutations :)
If $ a,b,c > 0$,we have a stronger result is
$ \sqrt {\frac {a}{b \plus{} c}} \plus{} \sqrt {\frac {b}{a \plus{} c}} \plus{} \sqrt {\frac {c}{a \plus{} b}}>\sqrt {4 \plus{} \frac {(a \minus{} b)^2} {(c \plus{} a)(c \plus{} b)}}$
Comprehend:
$ \sqrt [n]{\frac {a}{b \plus{} c}} \plus{} \sqrt [n]{\frac {b}{a \plus{} c}} \plus{} \sqrt [n]{\frac {c}{a \plus{} b}}> 2 \plus{} \sqrt [n]{\frac {b \plus{} c}{a \plus{} c}}(\sqrt [n]{\frac {a \plus{} c}{b \plus{} c}} \minus{} 1)^2$

I think that $ 2$ is the largest number such that the inequality holds for all positive numbers $ a,b,c$
Indeed, when $ a\equal{}b$, $ c$----> 0 , the left hand side of the inequality come to $ 2$ . So, we can't replace $ 2$ by any another number greater than $ 2$.
On other hand, we can't find the minimum value of given express.

[quote="ngtl"]I think that $ 2$ is the largest number such that the inequality holds for all positive numbers $ a,b,c$
Indeed, when $ a \equal{} b$, $ c$----> 0 , the left hand side of the inequality come to $ 2$ . So, we can't replace $ 2$ by any another number greater than $ 2$.
On other hand, we can't find the minimum value of given express.[/quote]
Please read my discussion of this problem with arqady above.


This should go in another forum, I think...

i think it still applies... it's still kinda under the category of american mathematics competitions.

I dont know how to solve it, but here are just some details that may help....

the primes <=23 are 2,3,5,7,11,13,17,19,23... so all the 1985 numbers can be written as a product of powers of these numbers

i think there was a similar problem in art and craft.... and it talked something about parity, where any of these two numbers' product had to be a perfect square, so their powers had to be both even or both odd.

[quote="darktreb"]Pigeonhole the parity, done.[/quote]

 :?: could you go deeper into it please?

I see the approach, too, but am having trouble putting pidgeon-holing it. Then again, it is an IMO problem.

[quote="paladin8"]Let's see if this works...

So every number is of the form $2^{a_1}3^{a_2} \cdots 23^{a_9}$. Considering all the $a_i \mod{2}$ we have $2^9 = 512$ "different" numbers. But since $1985 > 3(512)$ there must be $4$ of the "same" number; aka four that have the same parity on all the exponents. Then multiplying these four together, we see that the new number has all exponents that are $0 \mod{4}$.[/quote]

Be careful. 5, 5, 5, and 125 are all the "same" number under your pigeon hole, but do not have a perfect fourth-power as their product.

[quote="DPopov"]Consider a set of 1,985 positive integers, not necessarily distinct, and none with prime factors bigger than 23. Prove that there must exist 4 integers in this set whose product is equal to the fourth power of an integer.[/quote]

I saw this problem and decided to register :).  Anyway, this solution took me several hours to produce :oops:

I worked backwards (but I'll present the solution forwards for ease of reading).  First, map each integer n to a 9-tuple a(n) = (a0, a1, ..., a9), where ai is the exponent of the ith prime in n's prime factorization.  Considered mod 2, there are 512 possible tuples (which will be our buckets).  If m and n map to the same bucket, then a(mn) = a(m)+a(n) = 2*a(m) has entries that are all 0 mod 2, and thus in {0, 2} mod 4.  There are at least (1985-512)/2 = 737 such pairs.  Two of them map to the same tuple (mod 4), and since their entries are all in {0, 2}, they add to the 0 tuple, which is our desired 4th power.

I'm pretty sure that works.

excuse me, what is pigeon-hole(as you can see, im bad at number theory, actually i never learn about it...)

I don't think pigeon hole's quite number theory, it has a huge number of uses in all subjects.
http://mathworld.wolfram.com/DirichletsBoxPrinciple.html


for a given positive integer a_0 define the squence a_n by:
a_{i+1}= a_{i}/2 if a_{i} even
a_{i+1}= 3a_{i} -1 if a_{i} odd
prove that: if n is a natural number such that a_n=a_0 then 2^n>a_0

It was already posted and if I remember correctly no solution was proposed. The solution I had for that on the exam (or actually sent as it appeared during first round so the problem was to be dealt at home) took 6 pages and it was quite unappealing - I don't know any better solution.

How about thinking the binary expension system?


integration
trigonometry
calculus
parameterization
function
complex analysis
calculus computations

This can be done by contour integration ([url=http://en.wikipedia.org/wiki/Fresnel_integral]here[/url]), but there is another approach, by introducing a parameter into the integral... I wish I could remember it.  :(

[quote="mathnerd314"]$\int_{0}^\infty \frac{\sin x}{x}=\frac{\pi}{2}$

Prove or disprove.[/quote]

That's correct . A classical way is the   the Fourier transformer of the gate function .
 :cool:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=73681]Link 1[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=41288]Link 2[/url]

I've got my own favorite method in link 2.

This has probably appeared in other places on Mathlinks/AoPS as well; I was just looking for my own posts.

Yesterday I was having a look in the Complex Analysis - Computations section. I found something very interesting [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=44440]there[/url].

It seems just pure luck that we stumble upon this integral.


let $f: [0,\infty)\longrightarrow R$ be increasing function ,such that $f(x)-3x\ ,\ f(x)-x^3$ are increasing too.
prove that :$f(x)-x^2-x$ is an increasing function.
 :oops:  :blush:

Another proof.  Use the identity
\[ 3 [f(x) - x^2 - x] = 2[f(x) - 3x] + [f(x) - x^3] + (x-1)^3 + 1 \, .  \]
Each of the four terms on the right is an increasing function.  Hence the left side is increasing.

We didn't need the hypothesis that $f$ itself be increasing, because it follows from $f(x) - 3x$ being increasing.


Hi, we were assigned a problem to do a poster on in Chemistry but I can't seem to figure this out.

The problem is
[code]
A mole of marbles would cover the continental united states by a depth of how much?

A regular marble is .5 inch across
The continental united states is 2,959,064.44 square miles
That is 1.562386024 x 10^10 feet
That is 1.874863229184 x 10^11 inches.[/code]

My initial thought was to get area covered by a marble (2 dimensional area) and multiple it by 6.02 x 10^23 and divide it by the area of the USA. Is this correct or should I approach another method? :ninja:

I would find the volume of 1 mol of marbles then use volume equals area of the base multiplied by height where the area of the base is the area of the continental US.

I would look up the ratio of empty to total space in a compact arrangement of spheres, just in case we'll be covered to a high depth.
Here we go. Check my math. I've been known to make mistakes.
Let's use a packing density of 0.74.
The volume of a marble is:
(pi/6)diameter^3=(pi/6)0.125 cu. in.=0.6545 cu. in.
Considering the density (gaps between marbles), each marble will account for
0.06545/0.74 cu. in. = 0.08845 cu. in.
6.02 * 10^23 (1 mole) marbles (and the gaps in between) would occupy
(6.02 * 10^23)(0.08845 cu. in.) = 5.32 * 10^22 cu. in.

From the other end, if the continental united states is 2,959,064.44 square miles
That is (2.959 * 10^6 sq mi)(5280 ft/mi)^2 = 8.25*10^13 sq ft
(1 mile = 5280 ft, but 1 sq mile = 5280^2 sq ft)
It is also (8.25*10^13)*144 sq in = 1.79*10^16 sq. in.

(5.32 * 10^22 cu. in.)/(1.79*10^16 sq. in.) = 4.48*10^6 in is the height of the marble covering. Dividing by 12 in/ft we find it's 3.74*10^5 ft. Further dividing by 5280 ft/mi, we find it's 70.7 miles.


inequalities
number theory theorems
number theory

I'm looking for a proof of Liouville Inequality (in algebaric number theory).
Is there some proof online ? At worst, can anyone post it ?

Do you mean this one: For all algebraic $\alpha$ of degree $\geq2$ there exists a positive constant $c$ such that for all rational $p/q$ we have \[ \left|\alpha-\frac{p}{q}\right|\geq \frac{c}{q^n}? \] because I have a proof of that one.

At the time I meant a different Liouville Inequality ... (I've seen the proof of yours)
Unfortunately I forgot which :( (But I guess there is not too much of them :))
As fast as I retrieve it, I will make my question more precise.


Hello :)

How are you?

I'm new to this board.

Jason

:spam:

Don't spam, please. But anyway, hi.  :P

.......As it is more than three months after the original post and you feel the need to respond, you yourself are spamming. Did you not see that no one answered him?


geometry
power of a point
radical axis
geometry proposed

[quote="Virgil Nicula"][color=darkred]For $k\in\overline {1,2}$ the circle $C_k=C(O_k,r_k)$ is interior tangent to the circle $C=C(O,r)$ in the points $T_k$ and $|r_1-r_2|\le O_1O_2$. Denote: $A_1\in C_1$, $A_2\in C_2$ for which the line $A_1A_2$ is common exterior tangent of the circles $C_1$ and $C_2$; the intersection $R\in T_1A_1\cap T_2A_2$. Prove that:
$1^{\circ}.\blacktriangleright\ R\in C$ and $OR\perp A_1A_2\ ;$
$2^{\circ}.\blacktriangleright$ The point $R$ belongs to the radical axis of the circles $C_1$ and $C_2\ ;$
$3^{\circ}.\blacktriangleright\ O_1\in C_2\Longrightarrow$ the line $A_2B_2$ is tangent to the circle $C_1$, where for any $k\in \overline {1,2}$ the point $B_k\in C_k\ ,$ $B_k\ne A_k$ and the line $B_1B_2$ is the second common exterior tangent of the circles $C_1$ and $C_2$.[/color][/quote]

Consider $r_1>r_2$:
1º)The common tangent at $T_1$ cut the line $A_1A_2$ at $P$, now the triangle $PA_1T_1$ is isosceles then $<PT_1A_1=<PA_1T_1$. Also $T_1A_1$ meet the circle $C$ at $R_1$ and $A_1A_2$ meet at $B_1$  (between $A_1, P$) and $B_2$. Therefore $R_1$ is midpoint of $arc B_1R_1B_2$ (In anti clock wise sense).Similarly for the circle $C_2$, now: $R_2=R_1=R$.Therefore $OR$ is orthogonal to the chord $B_1B_2(A_1A_2)$.
2º) We have: $arc T_1B_1+ arc RB_2 = arc T_1B_1+ arc B_1R$, i.e. $<RA_1B_2= <PA_1T_1=<RT_1P= <RT_2T_1$, so: $A_1A_2T_2T_1$ is cyclic, therefore $RA_1.RT_1=RA_2.RT_2$, now $R$ is on radical axis of $C_1$ and $C_2$.