If n,a,b,c positive integers and a=>b=>c=>n proove that

  \[ \frac{a^n}{bc}+\frac{b^n}{ac}+\frac{c^n}{ab} \geq \frac{n b^{n-2}}{b}+\frac{n c^{n-2}}{c}+\frac{n c^{n-2}}{a} \]

Are you sure that is the correct form?Please look out at the last term.

Yes,I think there is a problem with this because it should be symetrical..and it`s not..but for the corrected version(if there is one) we can use re-arangement theorem..it`s just a guess.

don't worry...the problrm is ok..u can test any numbers u like...It doesn't metter if it's not symetrical...But remember...a=>b=>c=>n(it's a good hint)

_________________________________________________

Andrei u are right we are going to use the re-arangement inequality


MATHCOUNTS
AMC
AMC 8
USA(J)MO
USAMO
AoPS Books
BC. 8175205

I'm a mathcounts coach and kind of new to the scene. I have bright students but they don't know all of the tricks. To be honest neither do I. Would anyone be willing to share their tricks with me in solving some of the problems. I'm not looking to necessarily win this year, I would just like to not come in so far behind last place. Thanks.

The official mathcounts toolbox can be found here:
[url]http://www.mathcounts.org/webarticles/articlefiles/105-toolbox.pdf[/url]
,but tricks aren't really everthing:
[url]http://www.artofproblemsolving.com/Resources/AoPS_R_A_Prob_Solv.php[/url].

The AoPs books, classes, and mathjams should help your students a whole lot.

Nothing beats practice under test conditions with past tests.   The Warm-ups and all that other stuff that MATHCOUNTS produces each year are nice, but are sometimes (some coaches will say often) very unlike actual MATHCOUNTS tests.  Also, your students will have to learn to deal with issues like time budgeting and careless errors.  Try to get past contests either from MATHCOUNTS or other coaches in your area.  Occasionally have your students take a 'mock MATHCOUNTS test'.  You may also get mock MATHCOUNTS tests from our avid community members building them as they occasionally do (some of these may end up a little harder than MATHCOUNTS tests usually are, but generally, they're a fantastic source of practice and something I wish I'd had as a student).

[quote="Treething"]If you really want to know all of the tricks, do the Nationals tests and try to find detailed step-by-step solutions. That should help a lot, although it might make your head hurt at first...[/quote]

I disagree. Unless your students are extremely bright and all are capable of getting at least 28 or so on any give National test, you shouldn't give them Nationals to do. True, this may help them learn some new things, but they may forget them just as quickly. It's like giving 8th graders trying to improve for the AMC8 USAMO problems to do. State level problems would be good, mixed in with occasional National problems.

As Mr. R said, real tests are always good, to let your team members see what type of problems they may get in real competition.

True... I guess many of the problems in Nationals would be too hard for beginners...

But I advise giving out some of the first ten problems from past National competitions... those should be the equivalent of hard State problems...

Also, you may want to try browsing the Beginners section of the Problem Solving section of the AoPS forums.


IMO Shortlist
combinatorics
combinatorial geometry
convex polygon
polygon

Let  $ n > 2$.    Show that there is a set of $ 2^{n-1}$ points in the plane, no three collinear such that no $ 2n$ form a convex $ 2n$-gon.

Sorry for reviving this dead thread, but the WOOT group unsuccessfully tried this today... More importantly, the problem should actually be as follows:
[quote]Let  $ n > 2$.    Show that there is a set of $ 2^{n \minus{} 1}$ points in the plane, no three collinear such that no $ 2n$ form a convex $ 2n$-gon.[/quote]

[color=red]Original post now fixed by mod.[/color]

The famous Happy Ending theorem [url]http://en.wikipedia.org/wiki/Happy_Ending_problem[/url] conjectures that among $2^{n-2} + 1$ points in general position in the plane, there are $n$ being the vertices of a convex polygon. Erdös and Szekeres built examples with $2^{n-2}$ points for which no such convex $n$-gon exists.

So transposing this for $N=2n$, examples exist with $2^{N-2} = 2^{2n-2} = (2^{n-1})^2$, much better than what was requested.

Induct on $n$. Base $n=2$ is clear. For passing from $n$ to $n+1$ take a set $M$ of $2^{n-1}$ points without convex $2n$-gons and consider $M\cup M+x$, where $x$ is a small vector and direction of $x$ is of general position. Then three pairs $(a,a+x),(b,b+x),(c,c+x)$ are already not in a convex position, so the maximal number of points in convex position in the set $M\cup (M+x)$ is at most $(2n-1)+2$.

This is also an exercise in chapter 14 of "Combinatorial exercises" by Laszlo Lovasz,  Akademia Kiadó, 1979.


The sum of a set of distinct positive integers is 60. The average value of the integers is 20. What is the greatest integer that could be in this set of integers?

If the sum of the numbers is 60, and their average is 20, that means there are 3 numbers in total. The largest a number can be is $ 60 \minus{} 1 \minus{} 2 \equal{} \boxed{57}$.

the sum divided by the number of numbers is the mean, which means the sum divided by the mean is the number of numbers (i'll clarify l8r on)

so, there are $ \frac {60} {20} \equal{} 3$ numbers
since each number is different, we can have 1, 2, and 57, to get the greatest value
so our answer is 57

for my first statement....

$ \frac {\text{sum}} {\text{number of numbers}} \equal{} \text{mean}$

by multiplying the $ \text{number of numbers}$ to both sides, and dividing by $ \text{mean}$ we get....

$ \frac {\text{sum}} {\text{mean}} \equal{} \text{number of numbers}$

can anyone clarify on the last thing i said?

edit:beaten to it

Also, you should clarify that the answer could be $58$ if it didn't ask for distinct integers. Since it did, though, our answer is $60-1-2=\boxed{57}$.


quadratics
number theory unsolved
number theory

How to find the integer solutions of $ x^2\plus{}y^2\equal{}5z^2$?


On the blackboard there is numbers 1,2,3,4,5 .In one operation we can take any two numbers a and b and replace them with numbers a*b and a+b.and so on,,question:can we after finite number of these operations get numbers 21,27,64,180,540.?

The question doesn't seem clear. Am I correct in translating the question as:

If we take two numbers $ a$ and $ b$ out of the set {1, 2, 3, 4, 5}, and we can substitute $ ab$ for $ a$ and $ a\plus{}b$ for $ b$ repetitively, then are 21, 27, 64, 180, and 540 possible values of $ a$ and $ b$?

[quote="Mewto55555"]No, for example if I replace 1 and 2 with 2 and 3 we suddenly have 2 3 3 4 5. He is asking if there is every any way to have 21 27 64 180 540?[/quote]

yes you are right,

[quote="buzzer11"]I would look for invariants. I tried parity, and i found that no matter what, you cannot decrease the number of even numbers... But since the goal set has more evens than the original set, I don't think that will help...[/quote]

Then the answer would be no, right?

Probably, as it would be really tedious to actually get there. It's pretty difficult to find the invariant here though...

Yeah. (Can we move this out of classroom math?)

But these sort of problems can usually be tackled using invariants, and there still might be a way to solve it using invariants. That being said, there's probably other ways as well... What is the source of this problem?


Is there a forum for California? :huh:
If there is, I can't find it. :maybe:

Apparently $\mathcal{SAN DIEGO MATH CIRCLE}+\mathcal{BAY AREA}+\mathcal{SOUTHERN CALIFORNIA ARML}=\mathcal{CALIFORNIA}$...

Then where is Northern California?

I know California isn't Korea.

This old topic might be relevant: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=68756 .


Dear mathlink's members,
I have a problem book, it is a collection of over 200 problems with my own solutions. Also, it includes some of my new techniques in proving inequality.
Right now, I am a university student and my family is in a trouble, my parents are sacked and become "jobless". Maybe in the future, I cannot continue studying at the university...
Because of this, I would like to publish my book, but I don't know how to do it? In Vietnam, the publishing house said that the problems in my book are very hard and it is not suitable to publish.
Can someone help me? Thank you very much.

I understand your plight... I wish you will find out the solution early. As I know, you should contact to Lascu Mircea [b]<gil1993@zalau.astral.ro>[/b] who is president of GIL Publish House. Before you contact to him, I think you should find more imformation from hungkhtn because his book was published in here. Anyway, it is really hard to successful. I also knew your book is really very very excellent. If somebody don't know about this, you can see can_hang2007's post and download some attacked to forum (for example "toanhocmuonmaumain.PDF", etc...). Finnaly, Best wish for you.

I think someone will think bad about me when read this topic of mine. But I hope they will understand me if they put them in my situation.
Best regard.


$ ABCDEF\times 6\equal{}DEFABC$where A, B, C, D, E, F are integers. Find the original no. ABCDEF.

Do you mean digits?
[hide="If so..."]
Let $ ABC \equal{} x, DEF \equal{} y$, where $ x,y$ are integers between $ 1 and 999$, inclusive.
Then $ 6(1000x \plus{} y) \equal{} 1000y \plus{} x \iff 5999x \equal{} 994y \iff 857x \equal{} 142y$.
Clearly, the answer is $ x \equal{} 142, y \equal{} 857$, and our number is $ 142857$, so $ A \equal{} 1, B \equal{} 4, C \equal{} 2, D \equal{} 8, E \equal{} 5, F \equal{} 7$.[/hide]
edit: Noooo, got beaten!

i would suppose so, otherwise the problem becomes pointless
let ABC and DEF be 3 digit integers
we have 6000ABC+6DEF=ABC+1000DEF
or
5999ABC=994DEF
which reduces to
857ABC=142DEF and since ABD and DEF are 3 digit integers ABC=142 and DEF=857 and
ABCDEF=142857

stevenMEOW

See this:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=221017

The problems seem to be eerily similar.


Let $f$ be a function whose domain is all real numbers. If $f(x) + 2f(\frac{x+2001}{x-1}) = 4013 - x$ for all $x$ not equal to $1$, find the value of $f(2003)$


let$a,b,c$ be positive real
 prove that :
                 $\sum \frac{(2x+y+z)^2}{2x^2+(y+z)^2} \leq 8$

Posted again many times . See here
http://www.mathlinks.ro/Forum/viewtopic.php?t=53672


limit
real analysis
real analysis unsolved

Consider $ n$ is an even positive integer. Evaluate: $ \displaystyle\lim_{x\to \minus{}1}\left [ \frac{n(x^n\plus{}1)}{(x^2\minus{}1)(x^n\minus{}1)}\minus{}\frac{1}{(x\plus{}1)^2} \right ]$.

Since n is even, the required limit is equal to the limit:
$ \displaystyle\lim_{x\to1}\left[\frac {n(x^n \plus{} 1)}{(x^2 \minus{} 1)(x^n \minus{} 1)} \minus{} \frac {1}{(x \minus{} 1)^2}\right]$

Let $ x \equal{} (1 \plus{} s)$ where $ s\to0$.
the expression reduces to:
$ \displaystyle\lim_{s\to0}\frac {1}{s^2}\left[\frac {2 \plus{} ns\plus{} s^2(n)(n\minus{}1)/2}{(2 \plus{} s)(1 \plus{} s(n \minus{} 1)/2 \plus{} s^2(n \minus{} 1)(n \minus{} 2)/6)} \minus{} 1\right]$, considering only the terms which contribute to the limit.

this, on simplification gives $ \frac{n(n\minus{}1)}{2}\minus{} \left[\frac {(n \minus{} 1)}{2} \plus{} \frac {(n \minus{} 1)(n \minus{} 2)}{3}\right]$., or,
$ \frac{n^2\minus{}1}{6}$


Let $w$ be a polynomial of degree two with integer coeficients. Suppose that for each integer $x$ the value $w(x)$ is the square of an integer. Prove that $w$ is the square of a polynomial.

[hide="Hint"] Complete the square. [/hide]


Find all integers $ n$ for which $ 4n\plus{}9$ and $ 9n\plus{}1$ are both perfect squares.


geometry
circumcircle
linear algebra
matrix
number theory
relatively prime
angle bisector

Day 1

Problem 1:
Let $O$ be the center of the circuncircle of an acute triangle $ABC$, let $P$ be any point inside the segment $BC$. Suppose the circuncircle of triangle $BPO$ intersects the segment $AB$ at point $R$ and the circuncircle of triangle $COP$ intersects $CA$ at point $Q$.
($i$) Consider the triangle $PQR$; show that it is similar to triangle $ABC$ and that $O$ is its orthocenter.
 ($ii$) Show that the circuncircles of triangles $BPO$, $COP$, $PQR$ have the same radius.

Problem 2:
Given several matrices of the same size. Given a positive integer $N$, let's say that a matrix is $N$-balanced if the entries of the matrix are integers and the difference between any two adjacent entries of the matrix is less than or equal to $N$.
($i$)Show that every $2N$-balanced matrix can be written as a sum of two $N$-blanced matrices.
($ii$)Show that every $3N$-balanced matrix can be written as a sum of three $N$-balanced matrices.

Problem 3:
Determine all pairs $(a,b)$ of integers different from $0$ for which it is possible to find a positive integer $x$ and an integer $y$ such that $x$ is relatively prime to $b$ and in the following list there is an infinity of integers:

$\frac{a + xy}{b}$, $\frac{a + xy^2}{b^2}$, $\frac{a + xy^3}{b^3}$, $\ldots$, $\frac{a + xy^n}{b^n}$, $\ldots$

Day 2

Problem 4:
A list of numbers $a_1,a_2,\ldots,a_m$ contains an arithmetic trio $a_i, a_j, a_k$ if $i < j < k$ and $2a_j = a_i + a_k$.
Let $n$ be a positive integer. Show that the numbers $1, 2, 3, \ldots, n$ can be reordered in a list that does not contain arithmetic trios.

Problem 5:
Let $N$ be an integer greater than 1. A deck has $N^3$ cards, each card has one of $N$ colors, has one of $N$ figures and has one of $N$ numbers (there are no two identical cards). A collection of cards of the deck is "complete" if it has cards of every color, or if it has cards of every figure or of all numbers.
How many non-complete collections are there such that, if you add any other card from the deck, the collection becomes complete?

Problem 6:

Let $ABC$ be a triangle and $AD$ be the angle bisector of $<BAC$, with $D$ on $BC$. Let $E$ be a point on segment $BC$ such that $BD = EC$. Through $E$ draw $l$ a parallel line to $AD$ and let $P$ be a point in $l$ inside the triangle. Let $G$ be the point where $BP$ intersects $AC$ and $F$ be the point where $CP$ intersects $AB$. Show $BF = CG$.

Thanks for sharing it with us. When and where did the olympiad took place? :)

it  was held  in  Campeche, Campeche  from 6 to  12  of November Dave  Vather  got a  41  Pablo  Soberon (bronce imo) got a  42    I   got  a  35!!  my  first  gold!!   :)

It's great to hear that! :) When is the IMO TST? (or how does it work)?

Please post the problems in the forum . There are quite interesting.

well  we will  be  going to practices  a week  of  every  month in wich 2 tsts  will  be applied. from  there a  couple  of guys will  be eliminated.  then  there  will  be a  period  of ten  days (give  or take) where we  will  be taking tst´s  every day.  then  the the imo team  will  be selected  according to our performance in the tests. the same thing  will  be applied  for selection of the  iberoamerican olympiad  only here  4  students are chosen, and  thats  it.   :)  hope to see you  in  Eslovenia!!!!!  :lol:

[quote="marko avila"]iPablo  Soberon (bronce imo) got a  42[/quote]
that's a-me  :roll:
and in the period of 10 days marco it talking about is one day with TST and one without.  it'll be held in may.  in that period the mexican team for the centroamerican olympiad will also be chosen.

[quote="bandrak"][quote="marko avila"]iPablo  Soberon (bronce imo) got a  42[/quote]
that's a-me  :roll:
[/quote]Congratulations! :)


group theory
abstract algebra
superior algebra
superior algebra theorems

Hello,
Here is a problem:
Assume that G is a finite group, and let H be a nonempty subset of G. Prove that H is closed iff H is a subgroup of G.

Proof:
1) Suppose H is a subgroup of G. Then H is also a group so that the closure property holds for H.

2) Suppose next that H is a nonempty closed subset of G. We wish to show that H is a subgroup of G.
Since H is nonempty and closed, if $a, a^{-1}$ in H, then $aa^{-1}= e$ is in H so that H has indentity.
Associativity is inherited from G. So H is a group.

I do not think the second part of the proof is correct because what I really needed to show is the existence of inverses in H, right? How do I show this, anyone?

Also was the assumption that "G is finite" necessary and if so where does one use it?

of course, you have to use that $G$ is finite!

use $g^{n}= 1$, where $n = |G|$, to see that the identity and inverse elements lie in $H$.

Thank you very much. It is now an easy proof.


Evaluate the sum

$ \sum_{i\equal{}0}^{n} \arctan(\frac{1}{i^2 \plus{} i\plus{} 1})$

hello, we have
$ \sum_{i\equal{}0}^{n}\arctan\left(\frac{1}{i^2\plus{}i\plus{}1}\right)\equal{}\arctan(n\plus{}1)$ by induction.
Sonnhard.

Ok here is my solution:
Set $ a_i= \arctan i$ and using the relation $ \arctan a- \arctan b = \arctan(\frac{a-b}{1+ab})$ we write $ \frac{1}{i^2 + i + 1} = \frac{i+1 - i}{1+i(i+1}$ which makes the summation telescopic yielding the final answer $ \arctan(n+1) -\arctan(0) = \arctan(n+1)$


linear algebra
matrix
calculus
function
Functional Analysis
real analysis
real analysis unsolved

Let $ X$ Hilbert space, $ A\in B_{h}(X)$, (hermitian linear bounded operator). Prove that there exists a unique $ B\in B_{h}(X)$ such that $ B^3\equal{}A$. Then $ B$ is often denoted as $ \sqrt[3]{A}$.

Any hints are welcome!!

In finite dimensions, a matrix is Hermitian if and only if it is unitarily diagonalizable, and that diagonal matrix is real. Is there an infinite-dimensional analogue?

If you know functional calculus on normal operators you're done.
Consider the $ C^*$ algebra generated by $ A$. $ A$ is self-adjoint, hence the algebra is commutative. By the Gelfand representation you can view your $ A$ as a continuous complex valued function $ f$ on a compact (or locally compact) space. Since $ A$ is self-adjoint your function $ f$ will actually be real valued (for functions the ajoint means their complex conjugate). You know how to extract that radical from a real valued function under the assumption that the result must be a real valued function (since you want $ B$ to be self adjoint also).

Is there an "elementary" solution to this problem, without using functional calculus?

It [b]is[/b] a problem about the construction of the functional calculus ($ \sqrt[3]x$ is not an analytic function in the neighbourhood of $ 0$, so it has no real advantages over an arbitrary continuous function, at least as far as the existence part is concerned). Note that functional calculus precedes the spectral theorem, so the solution based on the spectral theorem is even less elementary. Of course, you do not need normal operators, just self-adjoint, which simplifies matters a bit.

Thanks guys for all the help,[size=59]as always, Fedja nails the problem[/size]  :lol: ! You're right, using functional calculus, it's easy ti solve the problem, what confused me is that we were given this problem before we introduced functional calculus, and we we're told that it isn't that hard to solve the problem, I was concentrated on more elementary methods (what ever that means? ), cause we proved that for positive definite operator $ A$, there exist $ B$ such that $ B^2\equal{}A$, $ B\equal{}\sqrt{A}$, in a very simple way (truth be told, the proof was very long), so I though maybe it could be done the same for this problem.

Thanks for all the help!


Well ,guys what is your idea`s about this ?(I have already asked microsoft ,but it will somehow costs $45$ dollars so who wants to pay  :D?

Go to this Adress in your Windows (mine is XP professional):

 Control panel $\longrightarrow$ Administrative Tools $\longrightarrow$ computer managemnet $\longrightarrow$
system tools $\longrightarrow$local users and groups $\longrightarrow$ users:

You can see all of the acounts on your computer ,but there 2 more additional acounts which are not yours :!:  :!:

1.Remote Desktop Help Assistant Account   2.SUPPORT

Anyone knows what are these and how can i get logged on these ?(I tried by going in safe mode by i didnt managed :()

This is so those wonderful folks at Microsoft can commandeer your computer. :) :o

Those are 2 [b]disabled[/b] accounts. Nobody can enter your computer using those, unless you activate them :) (and usually folks do that so that others might remotely connect to the computer and solve their problem).

Interesting  :D ,I see that microsoft is somehow too smart (ha-ha they had made themselves hidden acounts ).
As Mr.Voricu pointed out those acounts are disabled .

So here
[quote]Control panel  Administrative Tools  computer managemnet
system tools local users and groups  users:
[/quote]
i clicked at these acounts ,and there was a place that you can check :
[quote]Acount is disable[/quote]
Well i deleted the check which means they are able right now , So,now i am searching all over my PC to find out where can i log on to these acounts.Any one knows where can i find them ?

ps.I restart my PC and pressed F8 before windows coming up & i entered with safemode(you can find the administator acount ,which usually you can not find in the normal windows :) )

Unless you can decode md5 I doubt you will be able to log in  :lol:

Well ,what do you say about this ?

The passwords of these acounts are somewhere in my Harddisk,
and i find out that each partition of the harddisk has an adress(Like Somethings like Multiple(x),row(x),...) ,So i was trying to find the adress of the partition which the password is coded in it.
Well ,If some how ,one manage to find that partition and read it ,wjat will be the results ?
He will recieve bunch of Ones and zeros ,?
and is there anyway that we can transalte back these ones and zeros to latters??? :!:

Lomos, I only have to say to please stop double posting. Edit your post if you have something extra to add and no one has posted before.

[quote="lomos_lupin"]hmm, md5,never heard of it before this,But when Mr.vornicu said it I started, searching
And right now i think it is so interesting (I wana learn how this MD5 works :idea:

[b]MD5=Message-Digest algorithm 5[/b] is a way of coding the informations designed by smart man Ronald Rivest
(Smart ,no he was the professor of MIT so he was Super smart ,

Which you can see him here :
[url]http://en.wikipedia.org/wiki/MD5[/url]

Well i think that smile of his means:
ha-ha, kid, no way you can decode this  :diablo:  :diablo:

But i wana learn every thing i can about how this $MD5$ ,So i really appreaciat any help?[/quote]

MD5 has already been cracked. As in, the time it takes to brute force one of those takes much less time than a "real" brute force.

There are several online dictionaries for reverse-lookup for md5. Those usually only work with dictionary words and IP addresses though.

[quote="kezyxes"]MD5 has already been cracked.[/quote]I'm sure they've already thought about something else by now :)

[quote]I'm sure they've already thought about something else by now[/quote]

Heard of SHA1 ? (Secure Hash Algorithm..)

What other forum? Can you tell me, please? :?

[quote="borz"]Can someone please explain to me why I can write in Latex on this Forum, but not on the others. I was on a forum that looks exactly like this one, but it doesn't except Latex???[/quote]

Because latex isn't supported on the other forum?

stock phpBB doesn't include latex support.


I'm moving this I don't think some people here can even understand the problem. Espscaily the way you stated it.

[quote="Rijul"]Hello all,

How is this problem solved?

If X raised to the power (2m+1) - 5X raised to the power of (m+1) + 4X  where m subset (\in) can be written in the form

(y+1) (y+2) (y+3) (y+4) (y+5) find the least value of m and the factors.

I have not yet gotten to use texnic hence the above text.[/quote]

Is this right?

$x^{2m+1}-5x^{m+1}+4x$ is the product of five consecutive integers, where m is an integer. Find the smallest value of m.

Hint: $x^5-5x^3+4x =  x(x^2-1)(x^2-4) =
 x (x - 1) (x - 2) (x + 2) (x + 1)$ .. four consecutive integers starting from  $x-2$

Hello RC-7th

Yes the equation you have written is correct. I am in class 8 and appearing for a test in maths and the question I put forth is from example paper. I have not understood the question. Kindly help.

Will help a great deal if I know how to type the equation using tex centre  :)

Hello Gyan,

Can you be a little more specific as I did not understand. Thank you.

Regards

Rijul

[quote="Rijul"]
Hello Gyan,

Can you be a little more specific as I did not understand. Thank you.
[/quote]

Gyan showed that if you let m = 2, then you can factorise the expression into $(x-2)(x-1)x(x+1)(x+2)$. Thats in the same form of what you need (5 consecutive numbers - what is y in this case?). So all you have to do now is show why the smaller values of m don't work.


Given: f(f(x)) = sqrt x

Find: f(x)

Can someone give me a hint as to where to start? Thanks.

Can we show that this is the only such $f$?  Or, can we find all such $f$?

For example, $f(x) = x^{-\frac{ \sqrt{2}}{2}}$ also works.


floor function
function
algebra
real analysis
real analysis unsolved

$\left[x\right]$ and $\left\{x\right\}$
Example on the first one: http://www.mathlinks.ro/Forum/viewtopic.php?t=89267

Most commonly, $[x],$ also written as $\lfloor x\rfloor,$ denotes the greatest integer function:

$\lfloor x\rfloor=\max\{n\in\mathbb{Z}: n\le x\}$

We should have $\lfloor x\rfloor\le x<\lfloor x\rfloor +1\,\forall\,x\in\mathbb{R}.$

$\{x\}$ then stands for the fractional part of $x:$

$\{x\}=x-\lfloor x\rfloor$


Let $ ABC$ be a triangle. Its incircle with center $ I$ touches the sides $ BC$, $ CA$, $ AB$ in $ D$, $ E$, $ F$. Denote the incenters of $ BDIF$ and $ CDIE$ by $ X$ and $ Y$. Prove that $ 2 \cdot \angle XAY \equal{} \angle BAC$.

The common internal tangent of the incircles of the tangential $ DIFB$ and $ DIEC$ (different from ID) goes through $ A.$ For a proof see the topic [url=http://www.artofproblemsolving.com/viewtopic.php?t=276013]Common tangent passes through A[/url]. If $ K$ denote the intersection of this tangent with $ BC,$ then the rays $ AX$ and $ AY$ bisect $ \angle BAK$  and $ \angle CAK$  internally $\Longrightarrow$ $ \angle BAC \equal{} 2\angle XAY,$ as desired.


geometry
trigonometry
inradius
geometric transformation
reflection
rectangle
conics

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14055[/img]

 :(

Assume (according to the attached figure) $ b = CA < c = AB.$ Let $ M_a \in BC$ be the midpoint of $ BC,$ $ U \in BC$ tangency point of the incircle $ (I),$ and $ D \in BC$ foot of the A-altitude $ AH.$ By the cosine theorem,

$ DM = c \cos B - \frac {a}{2} = \frac {c^2 - b^2 }{2a}$

$ DU = DM - DB = c \cos B - (s - b) = \frac {(c - b)(b + c - a)}{2a}$

$ UM = UB - MB = (s - b) - \frac {a}{2} = \frac {c - b}{2}$

where $ s$ is the triangle semiperimeter. The other tangent to $ (I)$ parallel to $ BC$ cuts the A-altitude at $ D'$ and $ \frac {AD'}{AD} = \frac {r}{r_a},$ where $ r$ is the inradius and $ r_a$ the A-exradius. Using the triangle area formulas $ S = \frac {ah_a}{2} = r_a (s - a),$

$ AD' = AD\ \frac {r}{r_a} = h_a\ \frac {r}{r_a} = h_a \frac {r(s - a)}{S} = \frac {2r(s - a)}{a}$

$ UM \cdot AD' = \frac {r(s - a)(c - b)}{a} = IU \cdot DU.$

Perpendicular to $ AH$ through $ I$ cuts $ AH$ at $ J$ and this is perpendicular bisector of $ DD'.$ Reflect $ A$ in $ IJ$ into $ A',$ so that $ AD' = A'D.$ Let $ OM, IU$ cut the perpendicular to $ AA' \equiv AH$ at $ A'$ at points $ N, V,$ so that $ VU = A'D = AD'.$ Then the rectangles $ VUMN, DUIJ$ have the same areas, $ UM \cdot UV = IU \cdot DU.$ Equivalently, the rectangles $ A'DMN, A'VIJ$ have the same areas, $ A'N \cdot A'D = A'J \cdot A'V.$ This means that that $ M$ lies on a rectangular hyperbola $ \mathcal H$ with known center $ A',$ known asymptotes $ A'A, A'N$ and passing through the given point $ I.$ Let $ AI$ cut the asymptote $ A'N$ at $ X.$ Since $ J$ is the midpoint of $ AA',$ $ I$ is the midpoint of $ AX$ and $ AX$ is the hyperbola tangent at $ I.$

$ AH, AO$ are isogonals with respect to the angle $ \angle CAB$; reflect $ AH$ is $ AI$ into $ AO$ ($ O$ is unknown). Since $ OM = \frac {AH}{2},$ parallel to $ AO$ through the midpoint $ K$ of $ AH$ passes through $ M.$ Let this parallel cut $ A'N$ at $ L.$ We have to construct intersection $ M$ of the line $ KL$ with the hyperbola.

One way to construct intersections of a hyperbola with a line is to project the line to another line and the hyperbola to a rectangular hyperbola; this step would not be necessary here. Take the perpendicular hyperbola asymptotes $ A'X, A'A$ as coordinate axes, take the line and hyperbola equations in this coordinate system, and solve the underlying quadratic equation for the coordinates of the intersections with compass and straightedge: [hide="Quadratic equation"]
$ \mathcal H: \ \ \ xy = A'V \cdot A'J = \frac {A'X \cdot A'A}{4}$

$ KL: \ \ \ \frac {x}{A'L} + \frac {y}{A'K} = 1$

Eliminating $ y$ yields a quadratic equation for the x-coordinate of $ M$:

$ x^2 - A'L \cdot x + \frac {A'X \cdot A'A \cdot A'L}{4\ A'K} = 0$

$ x^2 - 2 \varrho x + d^2 = 0$

Due to the positive sign of its absolute coefficient, it can be solved by Descartes method: Parallel to the x-axis $ A'X$ at the distance $ d = \sqrt {\frac {A'X \cdot A'A \cdot A'L}{4\ A'K}}$ cuts a circle with diameter $ \varrho = A'L$ at 2 points, their x-coordinates being the roots. Circle with center $ A'$ and radius $ A'X$ cuts the ray opposite to $ (A'A$ at $ Y.$ Circle wih diameter $ AY$ cuts the line $ A'X$ at $ P, P',$ such that $ A'P = A'P' = \sqrt {A'X \cdot A'A}.$ Circle centered at $ A'$ with radius $ AP$ cuts the ray $ (A'A$ at $ Q$ and $ A'Q = A'P.$ Parallel to $ KL$ through $ Q$ cuts the ray $ (A'X$ at $ R$ and $ \frac {A'R}{A'Q} = \frac {A'L}{A'K}.$ Circle with diameter $ P'R$ cuts the ray $ (A'A$ at $ S$ and $ A'S = \sqrt {\frac {A'X \cdot A'A \cdot A'L}{A'K}.}$ If $ T$ is the midpoint of $ AS$ then $ A'T = \sqrt {\frac {A'X \cdot A'A \cdot A'L}{4\ A'K}} = d$ is the desired distance. Parallel to $ A'X$ through $ T$ cuts the circle with diameter $ A'L$ at max. 2 points $ Z, Z'.$ Parallels to $ A'A$ through $ Z, Z'$ cut the line $ KL$ at $ M, M'$ and the line $ AO$ (with $ O$ unknown) at $ O, O'.$ For a solution to be acceptable, $ I, M$ resp. $ I, M'$ have to be inside the circumcircles $ (O), (O')$ with circumradii $ OA, O'A,$ respectively.[/hide]
Another way is to reduce the problem to a degenerated Apollonius problem to construct a circle passing through 2 given points and externally touching a given circle. The hyperbola foci are required for the construction. [hide="Apollonius problem"]Circle with center $ A'$ and radius $ A'V$ cuts the ray opposite to $ (A'A$ at $ W.$ Circle with diameter $ JW$ cuts the line $ A'X$ at $ P, P'$ and $ A'P^2 = A'P'^ = A'V \cdot A'J.$ Perpendiculars to $ A'X$ through $ P, P'$ cut internal bisector of the right $ \angle AA'X$ at $ Q, Q',$ the hyperbola vertices. Circle $ (A')$ with center $ A'$ and radius $ A'Q$ is the hyperbola pedal circle. The hyperbola tangent $ AIX$ cuts $ (A')$ at $ S, S'$ and perpendiculars to $ AIX$ at $ S, S'$ cut the hyperbola major axis at the foci $ F' F'.$ Reflect $ F$ in the line $ KL$ into $ E.$ Draw a circle $ (F')$ with center $ F'$ and radius equal to the major axis length $ QQ'.$ If we construct a circle $ (M)$ passing through $ E, F$ (centered on the line $ KL$) and externally tangent to the circle $ (F')$ at $ T,$ then $ MF' - MF = MF' - MT = TF' = QQ'$ and the center $ M \in KL$ of $ (M)$ is on the hyperbola. To construct $ (M),$ draw an arbitrary circle $ (N)$ through $ E, F$ intersecting the circle $ (F')$ at $ G, G'$ Radical axes $ GG'$ of $ (N), (F')$ and $ EF$ of $ (N), (M)$ meet at the radical center $ R$ of $ (F'), (M), (N).$ Tangent $ RT$ from $ R$ to $ (F')$ is the radical axis of $ (F'), (M)$ and $ M$ is intersection of $ KL, F'T.$ Of course, 2 different tangents $ RT, RT'$ exist from $ R$ to $ (F'),$ yielding 2 different circles $ (M), (M')$ and 2 intersections $ M, M'$ of the line $ KL$ with the hyperbola.[/hide]
 _____________________________________________

2nd solution:

Given are $ AH = 2R \cos A,$ where $ R$ is the triangle circumradius, $ AI = \frac {r}{\sin \frac {A}{2}},$ where $ r$ is the triangle inradius, and $ \angle HAI = \frac {C - B}{2}.$ Substituting these to the relation

$ \cos A + \cos B + \cos C = 1 + \frac {r}{R}$

and using some trigonometric formulas, we get

$ 1 - 2 \sin^2 \frac {A}{2} + 2 \cos \frac {C + B}{2} \cos \frac {C - B}{2} = 1 + 2 \frac {AI}{AH}\ \left (1 - 2 \sin^2 \frac {A}{2}\right) \sin \frac {A}{2}$

where $ \cos \frac {C + B}{2} = \sin \frac {A}{2}.$ Subtracting 1 from both sides and reducing by $ \sin \frac {A}{2}$ yields a quadratic equation for $ \sin \frac {A}{2}$:

$ \sin^2 \frac {A}{2} - \frac {AH}{2\ AI}\ \sin \frac {A}{2} + \frac {AH \cos \widehat{HAI} - AI}{2\ AI} = 0$

which can be solved with ruler and compass. With $ \frac {A}{2}$ known, the sidelines $ AB, AC$ can be drawn, then the incircle $ (I)$ and finally, the incircle tangent perpendicular to $ AH.$ The problem of this ugly solution is that sign of the absolute coefficient of the quadratic equation can be either positive or negative and the construction partially depends on the sign.


integration
calculus
calculus computations

Anyone know how to do this?

[img]http://i21.photobucket.com/albums/b256/lando-45/simpsons.jpg[/img]

We have the following formula by the Simpson's rule:
$ \int_{a}^{b} f(x)dx \approx \frac{h}{3} \sum_{k \equal{} 0}^{n \minus{}1} [f(a \plus{} 2kh) \plus{} 4f(a \plus{} (2k \plus{} 1)h) \plus{} f(a \plus{} 2h(k \plus{}1))]$
So for number one you set n = 2 and then use the taylor series formula.


Show that, for all integers $n$, $n^2+2n+12$ is not a multiple of 121.

[hide]
The given expression is same as
$(n+1)^2+11$.
Now we assume that this is divisible by multiples of $11^2=121$.
$(n+1)^2+11=11^2a$ for some positive integer $a$.
Rearranging the terms we get:
$11(11a-1)=(n+1)^2$.
However, $(11a-1)$ can never be a multiple of 11. Thus, $11(11a-1)$ can never be a perfect square. So $n^2+2n+12$ cannot be a multiple of 121.
[/hide]

[hide]Let's start by assuming the contrary: Let's say it is a multiple of 121.  So it would be a multiple of 11.
So we know that $11|n^2+2n+12$.  So we can also say that $11|n^2+2n+1$, or $11|(n+1)^2$.  Since we can also say that $11|n+1$, or $n\equiv10 \pmod{11}$.
Because of what we said above, we can rewrite $n$ as $11a+10$, where $a\in\mathbb{Z}$.  Replacing with what we originally had, we get $(11a+10)^2+2(11a+10)+12$.  After distributing everything out, we get $121a^2+242a+132$.  We can cancel out $121a^2$ and $242a$, because they are both multiples of 121 (by $a^2$ and $2a$, respectively).
So we're left with 132, which we have assumed to be a multiple of 121.  But it's not.  A contradiction!  So our assumption must have been false, so for any integer $n$, $n^2+2n+12$ can't be a multiple of 121.  Never.
$\text{Q.E.D.}$[/hide]
I tend to write my proofs as if I were talking.

[quote="boxedexe"]  ....We know that $n^2+2n+12\equiv n^2+22n+2n+11(12)+12\equiv (n+12)^2\pmod{11}$.  Therefore, $11|n^2+2n+12\iff n=11k-12$....[/quote]
Why make it  more complicated then necessary?
Eg this may be a little less simple: (Knowing 12 = 11+1)
[quote] $n^2+2n+12\equiv n^2+2n+1 \equiv (n+1)^2\pmod{11}$.  Therefore..[/quote]
[hide]You will see that the expression is $1-2+12 = 11 \bmod 121$ etc (With simpler calculation )...[/hide]

[quote="kimby_102"]Wait, is my way of approaching wrong??
did I make any false assumption or mistake?
Thanks,[/quote]
No.  Nothing wrong in your approach. You solution looks quite correct.  (And it is written very nicely).


Let $ABCD$ is convex quadrilateral.$\omega$ is inscribed cirle of $ABCD$.
$E\in{AB}\cap{\omega}$
$F\in{CD}\cap{\omega}$
$O\in{AC}\cap{BD}$
Prove that $\frac{AE}{CF}=\frac{AO}{CO}$
 ;)

This is Theorem 3 in the note "Circumscribed quadrilaterals revisited" on [url=http://www.cip.ifi.lmu.de/~grinberg/]my website[/url].

  darij

Thank you [b]Darij[/b],I proved this result when I tryed to solve the problem in this topic[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=410679#410679]www.mathlinks.ro/Forum/viewtopic.php?p=410679#410679[/url],and I never haven't seen it befor :(  :D,but I am glad that it is well know fact. :lol:

[quote="Virgil Nicula"][hide="Only for Tiks."] Sorry [i]Tiks[/i], with this easy property (1) proves the
 remarkable property of a circumscribed quadrilateral : $O\in EF\ !$
$(1).$ $\triangle ABC$, $\triangle A'B'C'$: $A=A'$, $B+B'=180^{\circ}$ $\Longrightarrow$ $\frac ab=\frac{a'}{b'}.$

[b]Remark.[/b] Tiks, you have posted here a many very fine problems. I think you tired. Sorry, this proposed problem must be at section $\boxed {\ \mathrm {High\ School\ -\ Getting\ Started\ !}\ }$[/hide][/quote]

[hide="Only for Virgil"]
[b]First[/b].The fact that $E$,$O$ and $F$ lie in same line is not very easy to prove,at lest for me.
[b]Second.[/b]I know that this problem is easy,but not very easy!!! :mad: ,and if you want to see hard problems(from my own problems),then in Geometry Proposed & Own Problems,ther are two unsloved problems([b]Geo ineq 5,Problem for Virgil Nicula and others.. :D[/b] ),try to solve them :D.[/hide]

[quote="Virgil Nicula"]Sorry, you didn't understand what I said. Again you are angry with me. I am reading  with much pleasure your problems...even and this. Henceforth I will speak with you only through private message. Maybe here I made a mistake.[/quote]
Dear Virgil,I never haven't angry with you,you my friend,so I only sayed that I don't agree with you that this one is problem for school :).

No Tiks, this problem is for the Getting Started ! What is your opinion about my words ? You and I are two stubborn persons ! And now come on to finish this matter. I advise you to read my motto under the topics:$\mathrm {Why\ ?}$

[u]Remark[/u]. The friendship redoubleth joys and cutteth griefs in halfs ([b]F. Bacon[/b]).

Sorry, Virgil and Tiks, but may I ask you to stop spamming? I have learnt through painful experience that "easy", "hard" etc. are hightly subjective attributes, and make sense only in connection with the person to whom the problem appears easy, hard or anything like that. A problem can seem easy to A and hard to B with A being much more experienced and successful in mathematical olympiads or mathematics in general than B, and conversely. If I say a problem is "easy", then the only possible meaning is that it was easy to me, that I had solved a similar problem before, that its solution used a method which I knew well, or that I was just lucky enough to find the solution quickly (so it tells more about me than about the problem). It should not be understood as if I claim that the problem was easy for everybody and had to be moved to Getting Started.

  Darij

O.K. Agreed !. It is 2-1 for Tiks. I stopped this matter.


geometry
power of a point
radical axis
geometry unsolved

Let $ d$ and $ d'$ be two parallel lines and $ \omega_1$, $ \omega_2$ two circles both tangent to $ d$ and $ d'$. Now let $ k$ be a line parallel to $ d$ such that $ d$ and $ d'$ lie on the same side of $ k$ and let $ P$ be a point on $ k$. Consider two circles:
(i) passing through $ P$ and such that $ \omega_1$ and $ \omega_2$ are internally tangent to it;
(ii) passing through $ P$ and externally tangent to $ \omega_1$ and $ \omega_2$.
Prove that these two circles are tangent or their intersection point lies on $ k$.

[hide="Small hint"]In my solution I reformulated the problem using inversion. Then it can by solved using composition of homotheties (but it's not so easy to find them  :P ).[/hide]

The problem is quite trivial by symmetry.Circles tangent internally or externally to two congruent circles (w1).(w2) have their center on the radical axis of both circles, two circles of this pencil  meet in two points that lie on a parallel line to the diametral axis of both circles. Then two circles are tangent only when P lies on the radical axis of (w1),(w2)


I wanted to take the AMC 12 this year, but my school has no interest, and thus the teachers are not keen on doing the contest. So, I contacted a close school, and the math department head said that I could do the contest there... However, he wasn't too clear on specifics, like what school I write for myself, and when/how I register... the teacher just said to bring $ \$5$ on the day of the contest... is this how its done?

Thank you for your assistance,

Nawahd Werdna

I'm doing the same thing and the math department head that I contacted didn't say anything about money and wasn't clear on the specifics as well.

If you look at a copy of the answer form,
http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2008-12a/media/2008_12B_af.pdf
you will see that there is no machine readable place on the student answer form for the name or ceeb number of the school.  All of the machine readable information is student identification information.  The blank box used for the written or stamped name of the school is for grouping together test papers in the extreme case that some get shuffled.  The school identification form filled out by the contest administrator separates groups of answer forms coming from different schools.

So the answer to your question is that you do not register as an individual, you take the contest in hte group with the other students at the non-usual school.  Your result will be reported with the other students at the non-usual school, It is the responsibility of the administering teacher to report your score back to you, the outside student, at his/her convenience and to disregard your score in the awarding of intra-school awards at his/her discretion.   The opportunity offered to take the contest at the non-usual school is a courtesy and a convenience offered to you by that teacher at the non-usual school.    Please treat it as the courtesy that it is.  In the event that you, the outside student, qualify for AIME, you should clear it with the non-usual school that you will be allowed to sit for the AIME at that school, since that is where our procedures will automatically send the AIME.  We can not stop our automatic procedures to count you, a single student out of 12,000, as a separate special case.

As for the 5 dollar charge on the day of the contest:  Some schools defray the cost of the activity by charging individual students enough to cover the costs of hte entire group activity.  Some schools do not, they pay for the group activity form some school or district fund available for such group activities.  It varies by school, and by district.  Both practices are common and both are acceptable.  This school apparently expects the involved students to defray the cost of the group activity.

This forum is not an appropriate place to ask and receive the answer to these two questions.  A responsible student would take the time to compose an email to the information email address that is printed on all of our brochures, tests, and informational material.  Another way to find an appropriate and informed answer to this question is to have the administering teacher call our office at the toll-free telephone number.

Steve Dunbar
MAA Director, American Mathematics Competitions


[quote="HiDN428"]perhaps educated guesses??

EDIT: the 2009 cd video.[/quote]

I hope, SOON. I need to see that. Like.right.now.

lol roughly 3/4 of the USA was not able to see the video live and the other 1/4 was either at nats or only managed to see the last 20 minutes of it.


Let $ X$ be the set of all positive integers greater than or equal to $ 8$ and let $ f: X \to X$ be a function such that $ f(x\plus{}y)\equal{}f(xy)$ for all $ x \ge 4, y \ge 4$. If $ f(8)\equal{}9$, determine $ f(9)$.

$ f(x\plus{}1)\equal{}f(x\times1) \equal{} f(x)$

$ f(9)\equal{}f(8\plus{}1)\equal{}f(8)\equal{}9$

I think I must be missing something

[quote="SimonM"]$ f(x \plus{} 1) \equal{} f(x\times1) \equal{} f(x)$

$ f(9) \equal{} f(8 \plus{} 1) \equal{} f(8) \equal{} 9$

I think I must be missing something[/quote]

You're missing that $ f(x \plus{} y) \equal{} f(xy)$ for all $ x \ge 4, y \ge 4$.

Right

New solution:

$ f(8)\equal{}f(4\plus{}4)\equal{}f(16)\equal{}f(8\plus{}8)\equal{}f(64)\equal{}f(4\plus{}16)\equal{}f(20)\equal{}f(4\plus{}5)\equal{}f(9)$

[quote="SimonM"]Right

New solution:

$ f(8) \equal{} f(4 \plus{} 4) \equal{} f(16) \equal{} f(8 \plus{} 8) \equal{} f(64) \equal{} f(4 \plus{} 16) \equal{} f(20) \equal{} f(4 \plus{} 5) \equal{} f(9)$[/quote]

Thanks!  :)


geometry
geometric transformation
LaTeX
calculus
calculus computations

[quote="FieryHydra"]It would be a lot better if you would write down the actual problem or at least scan and upload an image of it. Forum members may not necessarily always download the file and then open it to read the problem.

I did download the doc file (in this case) and I must say I am at a loss; what are "neighboring sequences"?[/quote]
Thank you for your advise. Well...as far as my problem writing is concerned I've just started learning LATEX.
It looks like my translation from French wasn't the best one:
We say "suites adjacentes", so it could be "adjacent sequences".

$S_{n}=\sum\frac{(-1)^{k}}{k}$

where $n\geq 1$

a) Find limits
$(S_{n2})_{n\geq 1}$  and  $(S_{2n-1})_{n\geq 1}$
b) Show that one of these $S_{n}$ is increasing and the other - is decreasing.

Here I've tried to paraphrase the same problem thanks to Latex( first try), that I've already posted.

[color=green][Enclose your LaTeX code between dollar signs - mod][/color]


algebra
function
domain
topology
calculus
advanced fields
advanced fields unsolved

Is there a nontrivial model category structure on the category of affine varieties? (Or, equivalently, finitely generated k-algebras that are also domains)?

What is a model category structure? I have no clue.  :maybe:

For lack of a better reference:

http://en.wikipedia.org/wiki/Model_category

Basically, a category with classes of maps that have solutions to lifting and extension problems. I was wondering in the spirit of my earlier post http://www.artofproblemsolving.com/Forum/viewtopic.php?t=304126, a corollary of the generalization at the end would be that, for certain classes of maps of affine varieties, epimorphisms are necessarily surjections. I was wondering if these "nice" classes are actually a class of cofibrations in some model category structure -- it wouldn't be too surprising ... and model category structures are nice ways of viewing general lifting and extension problems, as well as defining a "homotopy category" where weak equivalences are inverted.

Thanks. I had a short glance at the link. I think to have lifting properties the Zariski topology is too coarse. Your example makes it necessary to switch to another topos on the category of varities in order to have a a lot finer sense of open set. This is extensively done in algebraic topology by moving to the realm of Grothendieck topologies and their homology theories. E.g., the étale site yields a formal lifting property which could serve for your purposes. If you have a look at that you will find out that the analogies to real calculus and the properties of local diffeomorphism are striking.


Ok. I've had the program working for over a week and a half, but I'm having issues with getting the timing down for test case 8 (where N is 13).  My program runs for 12 in about 0.54 seconds, and 13 is still over 2 seconds...PLEASE if anyone has any tips on how to get the time down, I'd [b]really[/b] appreciate it!  I'll post what I have so far if needed. Thanks!

i'm sure this questions been asked before...
post your code and i'll see what i can do (even though i'm not that good of a programmer)

First, make sure that the time limit doesn't need to be longer for that problem by somehow checking with someone, like Rob Kolstad.

My code is attatched in a doc file.  Thanks so much for trying to help!

I don't have a lot of time now, but a couple of things. I'm not a computer expert, so I don't know how much this will help...

everytime you use the placepiece function, you're checking if n%2==0, which can definitely be avoided...this should help a little.

also, you keep track of the rows visited in memory, but i think you can speed it up if you also keep track of the diagonals...

that's it for now...

Thanks.  :)  On the problem description page, one of the hints says to try to keep track of the diagnols in a 2*N+1 array...how one would store the diagnols like that confused me, if anyone could explain, I'd really appreciate it.

note that to spots are in the same diagonal if the sum of their row and column are the same (N - row + column for the other direction of diagonals)

Thanks! The diagnol thing got test case 7 (12) down to .42 seconds, but when N=13 it's still over 2 seconds... :? back to staring at solution grids for patterns! :P  :huh:

If you post your code again, I could probably come up with something else...

I don't know how much this will help, but if I did this I would move the first few lines of code in your placepiece function outside of the function. That way you only check if row==0 once, not every single time you call the function.

My ugly solution to that task was to write a special routine for N=13. (with heavy optimisations, like loop unrolling and such things ;) ). Actually, my solution is faster for N=13 than for N=12. :P

Yeah, I was contemplating having to do 13 differently.  I'm still relatively new to programming in general, and this is my first time really actually using recursion to solve a problem (factorial, etc. doesn't count  :P ), so it's definately twisting my mind in circles (which is good, it means I'm learning...I think.. :? )  Anyway, thanks again!  And here's what I have so far, with quite a few new tweaks...
And sorry for bothering everyone, I REALLY appreciate everyone's help...as  I said, I'm new to this stuff, I'm obsessed, and I suddenly hit a stride and did the problems from 1.2, 1.3, and the other two in 1.4 in about 4 days (which involved a weekend of me not sleeping much) and then I hit this one, which has taken over two weeks, and I'm anxious and excited to get onto chapter 2.  So thanks again!

While we're on the topic of USACO...

How exactly does the US Open work? Is it the same as the other contests throughout the year (e.g. you submit the programs online) except with a proctor?

[code]
void placePiece(int row){
for(int col=0; col<N; col++){
        	bool isValid = true;
        	d1=row+col;
        	d2=N-row+col;
        	for(int x=0; x<row; x++){
            	if(col==pieceAtRow[x]){
                		isValid = false;
              	 	break;
            	}
            	else{
[/code]

That bit of code is unneccessary complicated...

The "for(int x=0; x<row; x++){" part can be modified to a "isColumnOccupied" (which results in about N times or so faster solution :D).

Just for fun, I did the task again... You can check my new solution if you want. :)

Thank you SO much everyone!!!!!!! I got it working!!!  :lol:  Thank you thank you thank you!
Now onto chapter 2!


calculus
integration
trigonometry
algebra
partial fractions
calculus computations

$\int_{-1}^0 \sqrt{\frac{1+x}{1-x}} = \int_{0}^1 \frac{4u^2}{u^4+2u^2+1}$ (Substituion $x=\frac{u^2-1}{u^2+1}$)

Partial fractions

$\frac{-4}{(1+u^2)^2}+\frac{4}{1+u^2}$

Simple integration now gives $-1 + \frac{\pi}{2}$

Any nicer ways?

[quote="kunny"]Your answer is correct, dheardie. :)

You can set $1+x=2\sin ^ 2 t$.[/quote]

Lovely. I can never 'see' trig substituions

We can avoid "substitutions". Remembering that \[\int \frac{dx}{\sqrt{1-x^{2}}} = \sin^{-1} x + c\] and noting that \[ \frac{d}{dx} \sqrt{f}=\frac{f'}{2 \sqrt{f}} \] we may write \[ \int_{-1}^{0} \sqrt{\frac{1+x}{1-x}}\, dx = \int_{-1}^{0} \frac{1+x}{\sqrt{1-x^{2}}}\,dx = \left[ \sin^{-1} x - \sqrt{1-x^{2}}\, \right]_{-1}^{0}=\frac{\pi}{2} -1.  \]

Yes, your solution is exactly same as one of my solutions, jgor2861. :)

kunny


Prove that the product of 4 consecutive natural numbers cannot be a perfect cube.

If the numbers are $ n\minus{}1,n,n\plus{}1,n\plus{}2$
we get
$ (n^2\plus{}n\minus{}1)^2\equal{}y^3\plus{}1$
which by http://www.mathlinks.ro/viewtopic.php?p=643038#643038
tells us $ y\equal{}2$ and $ n^2\plus{}n\minus{}1\equal{}3$ but this last one has no solution in integers  :)

Well the equation $ p^2\plus{}1\equal{}q^3$ is a lot simpler, as being a special case of the link you posted.

$ q^3\equal{}(p\minus{}1)(p\plus{}1)$

1) The RHS factor are relative prime
$ p\minus{}1\equal{}a^3,p\plus{}1\equal{}b^3$
$ a^3\plus{}2\equal{}b^3$, no solutions easily

2) $ (p\minus{}1,p\plus{}1)\equal{}2$, then those numbers divided by 2 are relative prime and consecutive but this time we have only solution $ p\equal{}3,q\equal{}2$ with the same process like above

So, $ n^2\plus{}n\minus{}1\equal{}3$, no solutions

:lol: it's not like I used Fermat's last theorem or anything

[quote=CantonMathGuy][quote=songssari]I think ${k+1, k+2, k+3, k+4}$ is a complete residue system $mod 4$.
So it is the product of those numbers are $2 mod 4$ so it cannot be a perfect square. q.e.d[/quote]

No, it's not $2\pmod{4}$.[/quote]

Oh gods sorry

$n(n-1)(n-2)(n-3) = k^3$, let $n^2 - 3n = p$, then we have $p(p+2) = k^3$ which is equivalent to $(p+1)^2 = (k-1)(k^2 + k + 1)$, but note that $n(n-1)(n-2)(n-3) \equiv 0 \pmod{3} \implies p \equiv 0,-2 \pmod{3}$, so $p+1$ is not a multiple of $3$. We have $gcd(k-1, k^2 + k + 1) = gcd(3, k-1) = 1$, which means both $k-1$ and $k^2 + k + 1$ are perfect squares. let $k = m^2 + 1$ then $k^2 + k + 1 = m^4 + 3m^2 + 3 \equiv -1 \pmod{4}$ and $-1$ is not a QR modulo $4$, so there's no solutions.


I have this song where an entire page has R.H. 3 equal notes to a beat and L.H. 2 equal notes to a beat.  :( I'm half confident that I'll be able to do it eventually. But then there's a part in the middle where R.H. is 3 equal notes to a beat and L.H. 4 equal notes notes to a beat.  :(  :(

Almost no other instrument will give the student these kinds of troubles...

Any tips? (And yes, I've tried setting the metronome super fast to (3,4)=12 ticks a beat - it's still too hard)

i remember playing a two against three section (Tchaikovsky's Novemeber).  I started real slow, counting in 6, until i got a feel for the rhythm, then sped it up.  As for 3 against 4, i've never played it, but I think you could approach it the same way

You're having problems? With wierd rhythms, they usually have the notes written where they're supposed to be relative to each other, so if you have like, 8 in treble and 6 in bass or something, count in the 8s and stick the 6s in between them.

[quote="leoxnlin"]Yeah, the problem is, the 6's won't be exactly in the middle of the 8's. So you'll have perfect 8's and funky 6's.[/quote]
Obviously they won't be in the middle. I dunno for me it was really easy to do that from the start, but if you have trouble then go slowly and try to get the hang of the time.

I like to put it to some kind of syllabic chant. :)

* - - * - - * - - * - -
* - - - * - - - * - - -
Not  Ve-ry Diff-fi-cult

Hahaha I had to deal with these in Chopin's Fantasy-Impromptu and Beethoven's Piano Concerto No. 5

After a while you'll get the hang of it. Just my two cents.

[quote="HydraPyros"]Well, there's usually a lot of those in Chopin. Practice the right hand by itself, then the left hand by itself. Then try to play them together. (Slowly of course) When you feel confident, you can gradually increase the speed. There's one in the Chopin ballade no. 1 that i'm working on. Since most music scores are "drawn to scale" draw lines from the right hand notes down so you know when the left hand notes play in respect to the right.[/quote]
Yup. My little brother played a nocturne a while ago that had that rhythm...

[quote="thepianistalex"]Not Ve-ry Diff-fi-cult [/quote]
...he used "not difficult".


If you have [i]X[/i] coins, what is the probability that at least [i]Y[/i] are heads?
How do you do these types of problems (Is there a fast way that is suitable for sprint rounds)?
EDIT: LET x=8 and y=2
t0rajir0u, I understand your post; You dont need to post again with 8 and 2 substituted for X and Y. :)

I don't think there is a real general formula.  It won't help at all, because you probably should just understand the idea of probability.

EDIT: There's one that's too annoying to remember.

The probability that [i]exactly[/i] $ Y$ are heads is

$ \frac { {X \choose Y} }{2^X}$

Why?  The denominator is the total number of ways to flip $ X$ coins, and the numerator is the number of ways to pick exactly $ Y$ of them to be heads.  Therefore, the probability that at least $ Y$ coins are heads is

$ \frac { {X \choose Y} \plus{} {X \choose Y \plus{} 1} \plus{} {X \choose Y \plus{} 2} \plus{} ... \plus{} {X \choose X} }{2^X}$

Edit:  There is an important special case.  If $ X \equal{} 2Y\minus{}1$, the probability is $ \frac{1}{2}$.  This is because

$ {2Y\minus{}1 \choose Y} \plus{} {2Y\minus{}1 \choose Y\plus{}1} \plus{} ... \plus{} {2Y\minus{}1 \choose 2Y\minus{}1} \equal{} {2Y\minus{}1 \choose 0} \plus{} {2Y\minus{}1 \choose 1} \plus{} ... \plus{} {2Y\minus{}1 \choose Y\minus{}1}$,

that is, the number of ways to flip at least $ Y$ heads is equal to the number of ways to flip at least $ Y$ tails (because either we flip $ Y$ or more heads or we flip $ Y$ or more tails and the two are symmetric).

[quote="alanchou"]I don't think there is a real general formula.  It won't help at all, because you probably should just understand the idea of probability.[/quote]
Ummm..... I'm not looking for a formula...... I'm looking for an approach to this problem; In other words, how would you do this problem? would you use combinations, complementary counting, etc.

[quote="andrewliu"]would you use combinations, complementary counting, etc.[/quote]

Both.  If $ Y < \frac{X\plus{}1}{2}$ you can use the straight combinatorial approach, and if $ Y > \frac{X\plus{}1}{2}$ it'll be faster to use complementary counting.  The reason this works is because we have the identity

$ \left( {X \choose 0} \plus{} {X \choose 1} \plus{} ... \plus{} {X \choose Y} \right) \plus{} \left( {X \choose Y\plus{}1} \plus{} {X \choose Y\plus{}2} \plus{} ... \plus{} {X \choose X} \right) \equal{} 2^X$,

in other words, the sum of the number of ways to flip $ 0, 1, 2, 3, ... Y, Y\plus{}1, Y\plus{}2, ... X$ heads is the total number of ways to flip $ X$ coins (of course!).

THere is an really easy way.  Use Pascals Triangle.  Say you want to know the prob. of 2 heads and a tail.
               1
              11
             121
            [b]1331[/b]

Left 1 stands for 3 heads, left 3 for 2 heads and 1 tail, right 3 for 2 tails and one head, and right 1 for 3 tails.  add 1 and 3 and 3 and 1.  You get 8, since you want 2 heads and 1 tail, the Number would be 3, so 3/8 is the answer.

PM if not understood

[quote="xpmath"]Let's say the number was big, like 11 tosses. I wouldn't want to write up to the 11th row of Pascal's triangle. In ajain's example, there are $ \binom{3}{1}$ ways to choose the tail, so it's 3/8.[/quote]
So if it was 11 tosses and you want the probabillity of having at least 3 head is $ \frac{\binom{11}{3}}{2^{11}}$? Sorry.. I don't think I got it.

[quote="t0rajir0u"]That's the probability of [i]exactly[/i] three heads.  Refer back to my posts.[/quote]
Hmm.. So would it be $ \frac {1 \minus{} \binom{11}{2} \plus{} \binom{11}{1}\plus{}\binom{11}{0}}{2^{11}}$? Blah..I am sorry.

Close.  You want

$ 1 \minus{} \frac{ {11 \choose 2} \plus{} {11 \choose 1} \plus{} {11 \choose 0} }{2^{11}}$

[quote="DonkeyKong"][quote="xpmath"]Let's say the number was big, like 11 tosses. I wouldn't want to write up to the 11th row of Pascal's triangle. In ajain's example, there are $ \binom{3}{1}$ ways to choose the tail, so it's 3/8.[/quote]
So if it was 11 tosses and you want the probabillity of having at least 3 head is $ \frac {\binom{11}{3}}{2^{11}}$? Sorry.. I don't think I got it.[/quote]
I was just talking about ajain's post, where you had to choose 2 heads, and a tail.

[quote="t0rajir0u"]Close.  You want

$ 1 \minus{} \frac { {11 \choose 2} \plus{} {11 \choose 1} \plus{} {11 \choose 0} }{2^{11}}$[/quote]
Oh, I see. Thank you.

dude seriously complement count. find the probability that you get  0 or 1 head and subtract that from 1.
there are a total of $ 8$ coins and the number of possbilities for each one is $ 2$.
so the denom. of our fraction is $ 2^8$. now we find the number of ways we can have 0 heads which is just $ 1$. Now we find the number of ways to have exactly $ 1$ head. which is $ \binom{8}{1}$. so if you add the 2 numbers you get $ \frac{9}{256}$, but now you need to do $ 1\minus{}\frac{9}{256}$, which is $ \fbox{247/256}$. probly wrong answer but this an awesome way to do "at least" problems.


Daca $|a|=|b|=|c|$ si $\sum ab(|c-a|+|c-b|)=0$ atunci a,b,c sunt afixele unui triunghi echilateral

[hide="Solutie"]Notăm cu $ABC$ triunghiul ale cărui vârfuri au afixele $a$, $b$, $c$.
Mai întâi, din $|a|=|b|=|c|$, deducem că relaţia din ipoteză se scrie
\[\sum{a(|a-b|+|a-c|)}=0. \]
Ţinând cont că $|a-b|=AB$, $|b-c|=BC$, $|c-a|=CA$, obţinem
\[\sum{(AB+CA)a}=0, \]
scrisă echivalent
\[(AB+BC+CA)(a+b+c)=BC\cdot a+CA\cdot b+AB\cdot c, \]
încă echivalentă cu
\[\frac{BC\cdot a+CA\cdot b+AB\cdot c}{AB+BC+CA}=a+b+c. \]
Ultima egalitate se scrie
\[z_{I}=3z_{G}, \]
altfel spus, dacă ţinem cont că $H, G, O$ sunt coliniare şi $HG=2GO$,
\[z_{I}=z_{H}. \]
Deducem astfel că $I=H,$ ceea ce asigură că triunghiul $ABC$ este echilateral.
(Am notat cu $z_{I}, z_{G}, z_{H}$ afixele punctelor $I, G, H$).[/hide]


[quote="shadysaysurspammed"]P is any point on the circumcircle of a triangleABC.PL,PN,PM are perpendiculars drawn to sides BC,AB,AC respectively.Prove that L,M,N are collinear.
P.s-eminem is prohibited from replying to htis post,no hints as well(cuz he knows the solution and he did not get it on his own)[/quote]
this is exactly Simson theorem,and it is easy to prove it. :D


[url]http://news.bbc.co.uk/2/hi/middle_east/5215366.stm[/url]
  This time it's not due to the outdated map or something. Israel doesn't even bother to find an excuse for it.
  I was also shocked when I heard this. After so many days' hard thinking, I still can't get a clue why they did this. Why? What good will this act do to Israel? How will things go?
[url]http://www.cnn.com/2006/WORLD/meast/07/26/mideast.main/index.html[/url]
Seems like US is backing Israel up as usual.

A zest of intelligence would compell you to realize it is not in Israel's interest to purposedly bomb down a UN center; the observers are everywhere, and will remain omnipresent. The only effect this had was to highlight how inaccurate their hits can be.

Buffalo you might be interesting in doing some inquiries about the 1996 Qana shelling during Operation Grapes of Wrath (yes this isn't the first time it happens in Qana  :o ).

Dutch general van Kappen was just a few hours ago interviewed in the Dutch television programme "Nova".  He ran the investigation back then.  He said there was footage by a Norwegian soldier (such images where shown to the public) showing an Israeli aircraft above the site, which would indicate that the Israelis should have known the place was full of civilians.
General van Kappen explains that he is almost sure it couldn't have been an accident, but that he doesn't think it was ordered by the Israeli top.  He claims that the Israeli army is very motivated but lots of people follow their instinct, and this could have been the result of some commander going out of line intentionally.  He said it was possible this was the case again.

Targeting civilians is not nice but in war (which is never meant to be "nice") it can be an object, in order to affect public opinion (but it can go all ways after you do that).  You will find sites claiming the USA intentionally targeted Serbian civilians in 1999 (those sites are usually quite obviously biased).

Israel apologizes but does not deny the bombing (how could they) but justifies it by claiming they saw a truck belonging to Hezbollah driving in.

If your choice is between (i) killing 10 million civilians in the process of achieving security, but actually achieving that goal, versus (ii) being driven into the sea, do you really have a choice?  (In the past, there probably have been groups who actually chose option (ii).  We have a word for them -- "extinct".  For better or worse, even if you consider option (ii) "correct", evolution only keeps those who choose option (i).)  If you are forced into that "choice", who really deserves the blame for the dead civilians, you, or whoever forced you?

The proper moral calculation considers terrorists to be [i]killing their own civilians[/i].  As long as you kill less civilians than the enemy had the ability to kill if his goal was to maximize that death toll, you have performed some humanitarian action.  (And the only reason even that constraint exists is because of the risk of spoofs -- e.g. someone who secretly wants a race exterminated organizes a significant terrorist act knowing that the target will retaliate with genocide of the desired race, when he couldn't have come close to achieving said genocide himself.)

It is proper to be self-critical about your humanitarian abilities and relentlessly try to improve them.  It is better to kill 10,000 civilians than 10 million.  (This also vastly reduces your vulnerability to spoofs.)  But it is ridiculous for others to condemn you for not achieving a perfection that often isn't possible.

Do you agree with the following: the anti-semitism and holocaust in Europe(tell me, which country is innocent?) caused what happened there.
Why should Middle East-ian pay for European's fault? Why didn't Jews build their country somewhere in Germany? Just Because Jerusalem is Holy Land of Jews, the capital of King David three thousand years ago? Jerusalem is no longer the Holy Land of Moslems?
BTW, the land that UN promised to Israel is 15,000 km^2, tell me how much land they control now.

Well, I was taught the naive view in high school that after WWII the entire world gave a gift to the Jews and they all wanted to go live there, and then some badass Arab nations started annoying them.

It's hard to answer that question.  Jewish influx was happening long before the end of WWII, in the late thirties an earlier attempt was done to establish a Jewish nation there.

There are several theories :
1.Christian conservatives in the USA (either out of guilt or under the influence of powerful Jewish groups) backed the establishment of Israel as it was a way to appease Jewish refugees without having to face the burden in their own country

2.Most of the Palestinian refugees are not descendants of locals , but of neigbouring Arab nations who were attracted by the economic improvements, actually due to the growing Jewish presence in the twenties or thirties.

3.Palestinians weren't thrown out of the current state of Israel , they were told by other Arab leaders to get out of the way of their armies, and were promised the right of return later. (Many pro Palestinian sites claim that many Arab leaders at several points urged Palestinians to hold on to their ground).

4. many of the current Palestinians (the Arabs) are actually Jewish too, but converted to Islam.  In general, the ethnicity is interesting :

5. Israel refuses to declare its official borders in favour of possible future expansion.  Ben Gurion would have said that the establishment of Israel in 1948 was only a beginning, not satisfactory on a long term basis.  This is a map of the proposed Greater Israel (warning! map is from [u]pro Palestinian [/u]site!)

[img]http://www.nakbaonline.org/Maps/New/Map_GreaterIsreal.gif[/img]

(Do note the Lebanese part)

These are just theories!  Don't attack me, I won't defend them as I shouldn't/wouldn't want to/can't.

I don't think many Jews wanted to stick around in the country that attacked them.

Buffalo, the UN partition plan was refused by Arab leaders, many Palestinians will acknowledge this themselves.
So comparing with the UN partition plan is useless.

I do wanna warn for the danger of going offtopic, we started by discussing a single bombardment in a single current conflict, now we are back discussing the legitimity of the state of Israel  :maybe: .

Seeing that gff is locked for the night, I am forced to come here.

Firstly, the Israeli bombers are doing nothing for the good of their country. You should have noticed that. But having said that, can you explain to me what the US ... er ... UN observers were doing there?

[quote]Firstly, the Israeli bombers are doing nothing for the good of their country. You should have noticed that. But having said that, can you explain to me what the US ... er ... UN observers were doing there[/quote]?

They are UNIFIL, the UN mission for Lebanon.  As you might know Lebanon suffered from a civil war from 1975 until 1990.  They were asked to stay after Israel ended its  18 year long occupation in 2000 by the Lebanse government with a mandate expiring ironically on 31th july 2006.

Well, even people of the UN themselves have claimed that Hezbollah like to shoot missiles in the proximity of camps or UN posts, using them us a human shield.
Is that a justification for Israel?  I think not.

Let us not forget that also in Khiam (where the four observers were killed), there is a prison where for 18 years, people were tortured.  As it is likely to have been executed by christian proxy men, it is hard to prove what exactly Israel did there.  Anyway, many sites claim that Israel is now seizing the opportunity and trying to bomb the prison, to erase the past.

[img]
http://photos1.blogger.com/blogger/2871/232/400/aaa.93.jpg
[/img]

About the down with USA and Israel comment, I would advise not using language like that.  There are Mathlinks members from Israel and many many from the USA, and personal attacks are usually frowned upon.
I do think it should be possible to criticize every nation's actions here.

USA always liked to support Israel.  I wonder when they will cut their country in half and give one part to the native Indians.

[quote="Buffalo"] Why should Middle East-ian pay for European's fault?
Why didn't Jews build their country somewhere in Germany?[/quote]

As a minority of about 1 percent in Germany, the Jews were apparently quite unsafe. What do you imagine their future in Germany would have been like under your proposal, after the proportion fell to 0.1 percent, living among an angry and Nazified postwar population?   Their civilization was destroyed, so it was not so simple to just return as though from a vacation --- what would they go back to except isolation and danger?  Survivors of the death camps were murdered by the dozens in Poland shortly after the war.   It is not  for [i]you[/i] to determine what is safe or "safe enough" for [i]them[/i].

[quote="JBL"]Dog of Justice, your first paragraph seems to presuppose that blowing up civilians is an important part of Israeli national defense, i.e. that if Israel were to not blow up civilians, Hezbollah would suddenly overwhelm them (or push them into the sea, or something).  This seems badly unsupported to me.  Israel is most likely to secure its long-term security by having stable, relatively non-aggressive and aggrieved neighbors.  Bombing Lebanon and especially Lebanese civilians creates destabilized, aggrieved neighbors and shores up support for the most aggressive groups in those nations (e.g. Hezbollah).[/quote]

I should have mentioned that the "terrorists are killing their own civilians" principle matters a lot more in the long term than it matters in the current war.

The idea is that you announce this principle, do the minimum it takes to convince all relevant parties that you really believe it, then pretty much just sit back.  The incentives now work in your favor; [i]civilians are now compelled to stop terrorist action on their turf[/i].  They will do much of the work for you, as soon as you can convince them that it is them that the terrorists are really killing.

As for the current war, I don't claim there is any risk of Hezbollah suddenly overwhelming Israel.  But that's irrelevant, if they're never going to stop, if it will never be any easier to separate them from civilians in the future, and if they're not going to disappear on their own if we just wait long enough.  Of course, that last "if" is not obviously true, and the last several decades of diplomacy in the Middle East have been focused on trying to create conditions to falsify it.  But how much progress has really been made?  We know, from the fall of Eastern European communism, that it is in fact possible for things to suddenly and completely turn in our favor after decades of gloom.  However, there is little evidence that we can count on something similar to happen in the Islamic world.  Unlike Eastern Europe, the popular revolutions have not been against our enemy, they have been against us.

On a final note, I don't think generals should be beyond criticism.  But the criticism should be constructive.  Complain about excessive civilian casualties -- if you have an idea for a plan that avoids them and still can be expected to achieve the current objectives.  Most, possibly all, of the current critics cannot offer any alternative [i]that can actually be expected to work[/i].

I thought there are four major world powers: USA, Islamic fundmaentalists, Neo Communists, and Neo Nazis (in different forms). So that adds a fifth: Jewish fundamentalists.

Cor! Things are getting really complicated.

Despite of the humanitarianism or some other moral stuff, could Israel find a "final solution" to their state safty? Sword definitely won't do the work, the more they killed, the more land they controlled, the more hate they are gaining, more and more "terrorists" will come up and kill those "innocent" civilians, and more Israel want to kill, this is a dead loop. The native americans failed because they're too outdated(not only in Sci. & Tech., they don't know strategy either, although they knew some about tactics), divisive and helpless, the colonist could easily destroy them one by one, or buy some tribes to fight the other tribes.

Side question, why isn't Israel's bombing the civilians called terrorism?

The same reason why US bombings of civilins in Iraq and Afghanisthan were not called terrorism.

Maybe some Israeli's would have to teach their kids some manners (and their soldiers too) :

[url]http://www.mpacuk.org/content/view/2274/34/[/url]
This site is obviously biased, but the video you can download (about 8.3 MB, viewable with Quicktime) was made by Human Right Workers
:
[url]http://images.indymedia.org/imc/washingtondc/media/video/11/stonedagain15june2006.mov[/url]

[quote="probability1.01"][quote="Dog of Justice"]The idea is that you announce this principle, do the minimum it takes to convince all relevant parties that you really believe it, then pretty much just sit back.  The incentives now work in your favor; [i]civilians are now compelled to stop terrorist action on their turf[/i].  They will do much of the work for you, as soon as you can convince them that it is them that the terrorists are really killing.
[/quote]

I seriously doubt that this would work. The following appears to be your strategy:

Person A punches you. You punch some random person B, tell B that the reason you punched him was because A punched you, and therefore B should get rid of A for you.

Now tell me, whom would person B most likely help?[/quote]

A more accurate analogy would have person A continuously hiding behind person B without the latter's consent, so we can't retaliate against person A without doing something about person B.  The logic isn't really any different from negotiating with hostage-takers; in this case terrorists hold their home country's civilians hostage.

My strategy would not work if we weren't more powerful than them.  But we are, overwhelmingly so, and we're perfectly capable of killing even millions of person B's that "call our bluff" and react by joining the terrorists.  After such a massacre, nobody will call our bluff again, and as a result we won't kill any more of them.  In the long run lives are probably saved.

The alternative is that we refuse to ever use our power.  Then we will eventually lose the conflict.  Sometimes this is acceptable, e.g. Vietnam.  But I think Israel is worth defending, and if we decide to defend it, we should decide to succeed.

[quote="bubka"]@ Dogs of Justice:

What do you mean by 'we'?[/quote]

I agree with Bubka on this one : who is "we"?  USA?  The divine axis of purity USA-Israel?  Occassionally there is a non USA'er who dares to pop up here  :o .

[quote="bubka"]@ Dogs of Justice:

What do you mean by 'we'?[/quote]

Those who think Israel is worth defending against terrorists.  Of course, the USA and Israel itself are both in that crowd, hence the overwhelming power advantage.

I am actually open to being convinced that the Vietnam analogy applies and it isn't really worth defending, that if the Israelis really desire peace they should arrange to relocate somewhere else with less historical baggage and let the Arabs have Palestine.  But as long as we are actually defending Israel, we should do it right.  Even Jacques Chirac, who was strongly opposed to the Iraq invasion, is on the record saying that [url=http://www.washingtonpost.com/wp-dyn/content/article/2006/01/19/AR2006011903311.html]he's perfectly willing to engage in nuclear (!!) retaliation against terrorism[/url].

[quote="fredbel6"]Maybe some Israeli's would have to teach their kids some manners (and their soldiers too) :  [links to Islamist web site] [/quote]

The European mania concerning Jews and Israel is quite disturbing, considering the absence of actual Jews in Europe.

Regarding the Jewish settler children described at the web site, note that the settlers' Arab neighbors frequently murder the settlers, and the children of those Arabs are (universally and openly) taught slogans about exterminating the Jews.   It is a typically European priority in this situation to worry about plastic bags of water.

[quote="fredbel6"]Source?

They're also throwing the rocks.  The thing that disturbs me is that the soldier doesn't do anything.[/quote]

I can't load the video, but:

The so-called "human rights workers" generally are agonists who perpetuate the violence in many ways, whatever the value of the specific work they may do.   This causes more deaths in Israel and it is hardly surprising that an Israeli soldier doesn't stop the settlers from delivering a message to the workers: GO HOME.

Some of those workers, you like to put between "'s try to provide water.
You know what you are saying : that is perfectly wise to let people get stoned if they are not supposed to be there in your opinion.

I am not saying it is just as bad as blowing yourself up in in Tel Aviv or something but it's injust in my eyes.

And if the majority of Israelis think this is the way any kid of theirs should be raised, they should be ashamed.

[quote="fredbel6"]Some of those workers, you like to put between "'s try to provide water.  [/quote] Providing water is one thing.  Stoking the violence is another.

Whether or not these particular "human rights workers" were in the first category or the second is unclear, and I would not take the word of an Islamist web site on this question without further investigation.  Many people in the Occupied Territories are claimed to be Human Rights Workers, ranging from innocent helpers to heavily armed Hamas "charity" or "medical" organizations.   The term Human Rights has a high propaganda value and is abused for that reason; it is naive to just accept it at face value in this conflict.

[quote] You know what you are saying : that is perfectly wise to let people get stoned if they are not supposed to be there in your opinion.
[/quote]
I can speak for myself without your assistance.   The point was more that "people who contribute to trouble should not be surprised when the trouble finds them".  If these were just innocent water-delivering human rights workers, how is it that they were carrying a videocamera in the course of their water works?

[quote] it's injust in my eyes. And if the majority of Israelis think this is the way ... they should be ashamed.[/quote]

The real issue is, given that this video had absolutely nothing to do with the thread on Lebanon-Israel conflict, why did you find it important to post it here?   Some people never miss an opportunity to criticize Israel.  When those people live in ethnically cleansed Jewless parts of the world, that is in very poor taste.

About posting the video, it was a reply to the comment just above.  I won't deny that it can be see as off topic however.  THAT is a weakness I must admit. :|

About the comment from ethnic cleansing : I can assure you there are still influential Jews in my country, but they aren't a big part of our population.  So I can't criticize Israel's actions now?  Am I mistaking, or are you in some way using the same old argument that Zionism is justified by the holocaust?

About the cameras, maybe that is experience?
One could also ask if living with 500 Jews in a place like Hebron isn't asking for trouble.  Human Right Workers can't bring cameras, but those 500 Jews can bring 500 armed soldiers with them?

Note : I was talking about other workers, I'm not saying these were water workers.


At the Town School, 99% of the 100 students have glasses, but only 98% of the students in 8th grade have glasses. Which of the following could be the number of students in grade 8 who have glasses?

A) 98
B) 97
C) 49
D) 48

The first sentence tells us that only one person doesn't have glasses.
[hide="Treething says hide this"]
Because 100 isn't an answer choice, we know that the person w/o glasses is in 8th grade.
Therefore 98/100 = x/(x+1).
98/100 = 49/50
49/50 = x/(x+1)
49x+49 = 50x
x = 49
If there are 50 students in 8th grade and 49 of them have glasses, then the answer is 49[/hide]

Correct!

KBabe-I guess following RC-7th's work can help you

[quote="dinoman"]sorry that's not correct
maybe it would help to show your work so I could see what you did wrong[/quote]

my fault, i really didn't know how to solve it, just took a guess.....sry

I remember this problem! The first time I heard of it (although in a different version, the goldfish version :D ) was when I was in middle school and a friend of mine couldn't solve it and asked me if I could help him (let me see, that was in ... 1994, I think). I still remember the solution. Click below to see my solution, if anyone is interested or needs additional explanation.

[hide=""My solution""]99% of 100 students have glasses. This means 99 have glasses and 1 doesn't.

Next the key to this problem is to decide whether the student who doesn't have glasses is in 8th grade or not.

This can be decided quickly: if that student were not in 8th grade, we would have 100% of 8th grade students wearing glasses. Why? We got 100 students and the number of 8th grade students cannot be more than 99.

So that remaining student has to be an 8th grade student.

Let's assign $x=\texttt{number of 8th-graders with glasses}$

So $\frac{x}{\texttt{number of 8th-graders}}=\frac{x}{x+1}=\frac{98 \% }{100 \% }$

Solve the equation like in RC-7th post and we should get $x=49$.[/hide]

[Edit]:o Oops, looks like I have violated a Forum policy: I'm way over my age group!:blush: Sorry. Please delete my post if any moderators or anyone feels intimidated by post. Thx

Funny problem...

Once again, I would use ratios..

there were 100 people in that school
and 1 person doesn't have glasses

and sice there is -2% of people who doesn't have any pair of glasses,

I would state it like this

1:2 = x:98

x would equal 98/2 which is 49...

[quote]my fault, i really didn't know how to solve it, just took a guess.....sry[/quote]

this problem could be solved easily by using Ratios....

Hey, KBabe, are you in 6th grade? (If you are, you probably haven't learned about ratios yet)

yea... i'm in sixth grade.... :(  :?  :?  :(


[hide]it can be

XXXX
++++
/XX\
\XX/
/++\
\XX/
X/\X
+/\+
X\/X
+\/+

[b]10 designs[/b][/hide]

[quote="mhisno1"]Since you can only use three squares, it means if you use one /, you must use the other \.  If you use one without the other, the design is automatically not semetrical.  They must be placed on opposite sides to maintain symetry.

XXX
X+X
+X+
+++

/X\
/+\
\X/
\+/

There are 8 possibilities.[/quote]

there are more according to the answer

[hide]
XXX
X+X
+++
+X+
/+\
\+/
/X\
\X/
X++   These have symmetry verically.
XX+
++X
+XX
Is it 12?[/hide][/hide]

[hide]Let's call the squares, from left to right, 1, 2, 3, 4.
Start with designs with a vertical line of symmetry.
Only # 1 and 2 can be the middle tile; for each there are 4 choices for the outer tiles (x_x, +_+, /_\, \_/, where _ is replaced by x or +). So so far we have 4*2=8 possibilities.
Next, find designs with a horizontal line of symmetry.
Neither / nor \ work, so for each of the 3 spaces we have two choices: + or x. Therefore 2^3=8 choices.
However, we double-counted designs with both vertical [i]and[/i] horizontal symmetry: XXX, X+X, +X+, +++. So the answer should be 8+8-4=[b]12[/b].
However, since this problem has relatively few possibilities, you're probably better off just counting.
[/hide]

[hide="answer"]
XXX
X+X
\X/
/X\
/+\
\+/
+++
+X+
+XX
XX+
++X
X++
horizontal and vertical symmetry.
the answer is 12.
[/hide]

[hide]1+X+
2X+X
3/X\
4\X/
5/+\
6\+/
7+++
8XXX
9XX
  X
10++
    +
11++
    X
12XX
    +
13/\
    X
14\/
   X
15\/
   +
16/\
   +
edit:for the last 6, imagin that the things on the bottom are in the middle of the two above it.
sixteen solutions.[/hide]


1. Does time matter? That is, is there any incentive to do a problem faster rather than slower?

2. Does number of tries matter? Is there any evidence of the difference between getting it on the first try and getting it on the second try?

1. No.

2. Yes.
EDIT: My bad on number 2. :)

1. No.
2. No. But you are limited to 2 tries.


This is where you discuss the solution to MathFiend's question:
What five letter word completes this equation?

99+9=9 _____

I apologize for the inconvinience, for I have mis-split a thread once again.

no, it's not 'whurm.' As a matter of fact, I can't remember what it is right now, but when you get the answer, you'll know.

[quote="MysticTerminator"][quote="Tare"]please, [b]please[/b] don't tell me it's "WHURM" :)[/quote]
you are yet untrained in the ways of the whurmitude. WHURM!=14[/quote]
shush up :) , please.

no clue as to the problem. Do you mean stuff like 9*679 or do you actually mean letters.

my point exactly. Maybe they are like "seven" or maybe each letter represents something.

you don't add the letters! It's just that when you insert a 5 letter word into the blank and then read the equation, it sounds true.

wait, so isn't this more of an English question than a math question?

yeah, kinda, but it was given to me in Algebra so...w/e

that's not very nice...I'm still going at it.

It's a real word chinaboi, and I can't remember the answer, I'll ask my teacher tomorrow

hmmm...how 'bout lolho? it's a splicing of lolt and the famous lolbahhohumbugbag

...MT, must I remind you that you're a comoderator on a daily basis? :)
Seriously speaking, can you give us a hint? I have no clue either.

excellent, cano, liu kang, raden, johnny cage, scorpion, sub-zero, sonia. MORTAL KOMBAT!

I asked my teacher, and he told me that the answer is [hide]99+9=9 dozen[/hide]

ugh, i thought it asked for 5 words. so i was adding up random numbers. give me the address of this teacher of yours

no, it says 5 letters and I don't think I'm at liberty to give out that information if I want to remain in MathCounts


So when I construct a list like {5,4,2,6,4}, how do I get a function that grabs a value given the list name and the index of the list? (for example, 4*element[1] + 5*element[2] + ...)

I'm not entirely sure what you mean ... [code]L[[i]][/code] (where L is a list and i is a positive integer) gives you the i-th element of L.  If that's not what you're asking, could you perhaps give a particular example of the sort of function you want?

Actually thanks! That helps. yeah, I used L[i] but didn't double bracket it, explaining the confusion

[quote="Poincare"]Is there ANY(legal) way i can get Mathematica for free becoz i don't want to spend almost 2000 dollars!!

thanks,
dhaivat pandya[/quote]

Well you can get an education version for around $ \$$150 or something like that.  Or you can win ARML.  Or go to MOP and get a copy for free.  Or score high enough in the USAMTS.


calculus
ratio
function
algebra
binomial theorem
algebra proposed

Find the value of this series without using calculus method.(I know the method of using calculus...)

$ 1 \plus{} \frac {1}{2}p^2 \plus{} \frac {1\cdot 3}{2\cdot 4}p^4 \plus{} \frac {1\cdot 3\cdot 5}{2\cdot 4 \cdot 6}p^6 \plus{} \frac {1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}p^8 \plus{} \cdots(0 < p < 1)$ (Convergence is guaranteed by the ratio test).

Thanks.

[hide]The $ k^{\rm th}$ term is given by

$ a_k p^{2k} \equal{} \frac {1 \cdot 3 \cdot \ldots \cdot (2k \minus{} 1)}{2 \cdot 4 \cdot \ldots \cdot 2k} p^{2k} \equal{} \frac {(2k)!}{(2 \cdot 4 \cdot \ldots \cdot 2k)^2} p^{2k} \equal{} \frac {(2k)!}{(k!)^2} (p/2)^{2k}$

$ \equal{} \binom{2k}{k} (p/2)^{2k}$.

This suggests that we should look at the binomial theorem to sum the series.

[hide]Recall that

$ \binom{r}{k} \equal{} \frac {r!}{k!(r \minus{} k)!} \equal{} \frac {r(r \minus{} 1)(r \minus{} 2)\ldots(r \minus{} k \plus{} 1)}{k!}$

which also holds when $ r$ is a non-integer.  In particular, when $ r \equal{} \minus{} 1/2$, we find

$ \binom{1/2}{k} \equal{} \frac {( \minus{} \frac {1}{2})( \minus{} \frac {3}{2})\ldots( \minus{} \frac {2k \minus{} 1}{2})}{k!} \equal{} ( \minus{} 1)^k \frac {1 \cdot 3 \cdot \ldots \cdot (2k \minus{} 1)}{2^k k!}$

$ \equal{} \frac {( \minus{} 1)^k}{2^{2k}} \binom{2k}{k} \equal{} ( \minus{} 1)^k a_k$.

But since

$ (x \plus{} y)^r \equal{} \sum_{k \equal{} 0}^\infty \binom{r}{k} x^k y^{r \minus{} k}$,

with the choice $ r \equal{} \minus{} 1/2$, $ y \equal{} 1$, $ x \equal{} \minus{} p^2$, we obtain

$ (1 \minus{} p^2)^{ \minus{} 1/2} \equal{} \sum_{k \equal{} 0}^\infty \binom{ \minus{} 1/2}{k} ( \minus{} p^2)^k \equal{} \sum_{k \equal{} 0}^\infty a_k p^{2k}$,

from which it follows the given series is equal to $ (1 \minus{} p^2)^{ \minus{} 1/2}$.[/hide][/hide]

There is an identity we have from the calculation of Catalan numbers by generating function: $ \frac {1}{\sqrt {1 - 4x}} = \sum_{n = 0}^{\infty} {{2n} \choose {n}}x^n$.

For the proof of the above, we will calculate the coefficient of $ x^n$ in the left-hand side. Indeed:

$ [x^n]\frac {1}{\sqrt {1 - 4x}} = [x^n](1 - 4x)^{\frac { - 1}{2}} = {{\frac { - 1}{2} \choose {n}}( - 4)^n = \frac {( - \frac {1}{2})( - \frac {3}{2})\cdots ( - \frac {2n - 1}{2})}{n!}\cdot ( - 4)^n}$
$ = \frac {2^n\cdot 1 \cdot 3 \cdots (2n - 1)}{n!} = \frac {(2\cdot 1)(2\cdot 2)...(2n)(1\cdot 3\cdots \cdot (2n - 1))}{(n!)^2}$
$ = \frac {(2n)!}{(n!)^2} = {{2n} \choose {n}}$

So the identity is proved.

Now put $ x = \frac {p^2}{4}$, we have $ S = \frac {1}{\sqrt {1 - p^2}}$ where $ S$ is the sum we have to calculate.


solve for a. (a^2)^2 = 10000.

Now find out what is (a^23)-4a+82*(sqrt/64)

If it's easy, please do it yourself and not spam the forums... Please. And second of all, that is not easy for some people. Third, how are you contributing to the forums by putting un-LaTeX form problems that we can't even understand? Fourth: I think the answer is $ 10^23\plus{}82^8$ Which I highly doubt that you can calculate that yourself.

hello, why don't you use $ \text{\LaTeX}$? Do you mean $ a^{23}-4a+82\cdot\sqrt{64}$?
Sonnhard.

No, but I think u meant this $ 10^{23}$ not $ 10^23$[/code]

Yeah, make sure you write 10^{23}, not 10^23


An auto dealer purchases a car at dealer cost, adds a $ 30\%$4 markup to arrive at the list price, later has to discount that price by $ 20\%$, and ends up selling the car for $ \$12,480.00$. What was the net profit of this transaction for the dealer?

Let the original price be $ x$.  Then we have $ x \times 1.3 \times 0.8\equal{}12480 \implies x\equal{}12000$ so the profit is $ \boxed{\$ 480}$.


modular arithmetic
number theory unsolved
number theory

Prove that $(\frac{D}{m})=-(\frac D{m+D/2})$, for $D \equiv 0 \pmod 4$, where $(\frac{}{})$ is the Jacobi symbol.


Twenty-five hobbits split plot 5x5 to 1x1 plots. No hobbit does not want to be the enemy for a neighbor (or even that his land touches the tops of the neighbors). It is known that none of them clashed with more than three other hobbits. Prove that they can space out so that everyone was just surrounded by friends. (hint -> prove that if the neighbors are there any enemies can reduce number of such neighbors).

[quote="aktyw19"]or even that his land touches the tops of the neighbors[/quote]

clarify, please. and what's that strange word in your hint?

[quote="pleurestique"][quote="aktyw19"]or even that his land touches the tops of the neighbors[/quote]

[/quote]
have no common points