How to differentiate $\ln x$ and $\log x$? It doesn't say in my high-school textbook.

[quote="shobber"]lnx part:
I remember proving it by using the fact that $e=\lim_{n\to \infty}{(1+\frac{1}{n})^n}$.[/quote]

How?

And what about $\log x$?

[quote="shobber"]lnx part:
I remember proving it by using the fact that $e=\lim_{n\to \infty}{(1+\frac{1}{n})^n}$.[/quote]

Weird.

The standard proof goes as follows... We know that $f(g(x))' = f'(g(x)) \cdot g'(x)$ (chain rule). Take $f(x) = e^x$ and $g(x) = \ln{x}$. Obviously $f(g(x)) = x$ so that $f(g(x))' = 1$. And $f'(g(x)) = x$ since $f'(x) = f(x)$. So $g'(x) \cdot x = 1$ and $\left(\ln{x}\right)' = \frac{1}{x}$.

Arne: I think shobber meant proving like that that $(e^x)' = e^x$, and from there on like you did.

But it can again go much shorter, this is my own proof:

$(ln(x))' = \lim_{h=0}\frac{ln(x+h)-ln(x)}{h} = \lim_{y=y_0}\frac{y-y_0}{e^y-e^{y_0}} = \frac1{(e^y)'} = \frac1{e^y} = \frac1x$

Which can without adaption be used for complex ln as well :)

No, it can be done with that identity.

$\frac{(\ln(x+h)-\ln(x))}{h}=\frac{1}{h}*\ln(\frac{x+h}{x})=\ln((1+\frac{h}{x})^\frac{1}{h}).  Substituting $h'=1/h$ and taking the limit as h' goes to infinity (that is, h goes to 0) we get the desired result.


[i]Prove that $ a\equal{}b\cos{C}\plus{}c\cos{B}$[/i]
I have that

$ \cos{C}\equal{}\frac{x}{b}\implies x\equal{}b\cos{C}$

$ \cos{B}\equal{}\frac{a\plus{}x}{c}\implies x\equal{}c\cos{B}\minus{}a$

Hence,

$ a\equal{}c\cos{B}\minus{}b\cos{C}$

Why a difference instead of a sum??

If C is not acute, as your picture shows,
$ \cos(\pi\minus{}C) \equal{}\minus{}cos(C) \equal{}\frac{x}{b}$,
if C is acute, the result is obvious.

yeah

$ \cos(C) \equal{}\minus{}cos(\pi\minus{}C) \equal{}\minus{}\frac{x}{b}$

hey guys

     be careful of <C .

     you should know which <C .

     the first is wrong.

cagula   (math.haiyanh.com)

It is not the law of cosines, but from

$ a^{2}\equal{} a(b\cos{C}\plus{}c\cos{B})$

You should be able to derive it.


Let $ x,y,z\ge 0,x \plus{} y \plus{} z \equal{} 3$
Find the min of  $ \sqrt {2x \plus{} y^2 \plus{} z^3} \plus{} \sqrt {2y \plus{} z^2 \plus{} x^3} \plus{} \sqrt {2z \plus{} x^2 \plus{} y^3}$

Minimum is 6 because for example the 1st root can be expressed as:
2x+y^2+z^3 =6-2(y+z)+ y^2+z^3 = (y-1)^2+z(z-1)^2+2z^2+4 >= 4 so:
Min = 3*4^(1/2)=6


I run a math club for 3rd, 4th, and 5th graders. This is a wonderful site. I got some good advice when I was planning the math club this Summer and I have recommented this site to everyone in my club. Now I am back with another question.

We usually end the math club with 15 minutes of math games. We have tried logic puzzles but I am hoping to get more suggestions from folks here. Are there books or websites for this type of activity for elementary school kids?

Regards,
rakesh sinha

around the world

they sit at desks and the first two people stand up, you hold up a problem first one to blurt it out moves in and the loser sits down, once a player completes a lap around the room, he wins

You have a certain numbers of objects (like coins, pencils etc.), and two players.  Each player can pick anywhere from 1 to a certain number of objects for each turn.  And then the person who picks the last object either wins or loses.

Depending on how you decide on the rules of the game, there's a strategy so that either the first person or the second person who goes can win.

I dunno if this helps, but maybe you could watch Cyberchase to get some ideas.  It's a tv show about math, aimed for 4th graders, and is shown on PBS.

Thanks for the suggestions. I will make sure to check out 'Cyberchase' but unfortunately some of the more creative games, e.g., optimal arrnagement of chairs in cafeteria,  are not feasible in my setting. We have approx 25 kids per room so we are more or less restricted to doing paper and pencil exercises.

Any pointers to puzzle books or websites?


geometry
geometric transformation
homothety
ratio
geometry proposed

Cant anyone help me about this question?From Where can i learn more on homothety?
Please help

[quote="Moonmathpi496"]Can anyone please explain,why-
"If we concider a homothety of a triangle with ratio -1/2 and center G (centroid) then it sends the triangle to the medial triangle."
I have proved it in a different way but cannot understand the matter "center G"
and plz explain about the cases when the center of homothety is inside a figure.[/quote]

 I think that the ratio must be $-\frac{1}{4}$

 Babis


Let $d \geq 3$ a prime such that $d \mid 2^{2^n}+1$ and $d \mid n$.
We have $2^{2^{n+1}}\equiv1[d]$ so the order $\alpha$ of $2$ in $\left(\mathbb{Z}/d\mathbb{Z}\right)^*$ divides $2^{n+1}$. But $2^{2^n}\equiv-1[d]$, so $\alpha=2^{n+1}$. By Fermat little theorem we obtain $\alpha \mid d-1 \Rightarrow 2^{n+1}<d$. But $d \leq n \Rightarrow$ impossible.

In fact for all $3 \leq d \leq 2^{n+1}$ , $d$ and $2^{2^n}+1$ are coprime.


Let L be a six digit (base ten) integer whose last digit is not 0, and let M be the number formed by moving this last digit to the front. For example, if L=354268, then M=835426. Find all possible L such that L=3M. What is the sum of all the digits of all possible L?

Answer is 54, but I don't know how to get that result.

Cheating on the same exam?  :wink:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=308869

(Searched for "last digit front.")

Thanks for the link. The other user and I are in the same math club at school, and this was a practice problem.


This is very easy, but u guys can do it as a warm up problem

If a,b,c are positive reals prove that:

$ (a\plus{}b\minus{}2c/b\plus{}c)$ + $ (b\plus{}c\minus{}2a/c\plus{}a)$ + $ (c\plus{}a\minus{}2b/a\plus{}b)$ greater than or equal to zero.

can anyone post the solution.... m unable to crack this.... :(

[hide]
Hint:

WLOG a>=b>=c

Try to replace all the denominators with the max of $ a\plus{}b$,$ b\plus{}c$,$ c\plus{}a$

and then everything falls in place[/hide]

Excellent way da befuddlers....
But even without WLOG it can be done....but quite a long soln....Use Darij Grinberg's nice lemma related to an Iranian inequality!!!

@ aravind
can u please provide the link to this lemme.....

[quote="befuddlers"][hide]
Hint:

WLOG a>=b>=c

Try to replace all the denominators with the max of $ a \plus{} b$,$ b \plus{} c$,$ c \plus{} a$

and then everything falls in place[/hide][/quote]
Umm... u can't do the WLOG part because the ineq is not symmetric but cyclic.
And maybe I don't know the details, but it appears that u can't increase the denominator. Because some of the numerators can be negative.
And Arvind, pls don't ignore anurag's request.
I've a solution, but that isn't "Very, very easy" as it appears to you. So there's a better solution? pls post it totally. Pls don't leave hints only.

Sorry akashnil i was not boasting or something

I just thought that my "wrong method" as pointed out by you was correct.

Yes the solution is not "very easy".(very sorry again :oops: )

But u can expand out the whole left hand side to $ a(a\minus{}c)^2$+$ b(b\minus{}a)^2$+$ c(c\minus{}b)^2$

which is greater than or equal to zero.

Please forgive me again akashnil :(

I am very sad.

Hey don't worry dude. I didn't mean you were boasting.
Anyway here is my method which is slightly different than straightforward expanding:
[hide]\[ \sum \frac{a\plus{}b\minus{}2c}{b\plus{}c}\equal{}\sum a\left(\frac{1}{b\plus{}c}\plus{}\frac{1}{a\plus{}b}\minus{}\frac{2}{c\plus{}a}\right)\]
\[ \equal{}\sum a\left(\frac{(a\plus{}b)(c\plus{}a)\plus{}(c\plus{}a)(b\plus{}c)\minus{}2(b\plus{}c)(a\plus{}b)}{\prod (a\plus{}b)}\right)\equal{}\]
\[ \frac{\sum a[(a^2\plus{}\sum ab)\plus{}(c^2\plus{}\sum ab)\minus{}2(b^2\plus{}\sum ab)]}{\prod (a\plus{}b)}\equal{}\frac{\sum a(a^2\plus{}c^2\minus{}2b^2)}{\prod (a\plus{}b)}\]
\[ \equal{}\frac{\sum (a^3\plus{}c^2a\minus{}2ab^2)}{\prod (a\plus{}b)}\equal{}\frac{\sum (a^3\plus{}c^2a\minus{}2ca^2)}{\prod (a\plus{}b)}\equal{}\frac{\sum a(a\minus{}c)^2}{\prod (a\plus{}b)}\ge 0\][/hide]

well this solution is pretty neat... Akashnil good job... :)

@aravind... dude please enlighten us with that lemma....!!!

I don't the know the exact link da.
Let me find out.

Can't this be done with A.M.>=GM?
I tried but failed. Maybe coz its past midnight though :p
The equ. reduces to us requiring to prove a+b+c[u]>[/u]
[u]a^2b+b^2c+c^2a[/u]
              abc
IF we take a+b+c=k, we get max value of RHS numerator as 3k^6/6^6. Now if we take max value of abc^-1 somehow and use it to solve the question?  :ninja:
P.S. I warned you I'm sleepy!

replace b+c,a+b,c+a with x,y,z resp.
x-y=c-a,x-z=b-a,y-z=b-c.
the given inequality changes to:
{(y-z)+(y-x)}/x+{(x-z)+(x-y)}/z+{(z-y)+(z-x)}/y>=0

by expanding this,i think we should get....please help....

[hide]$ the\ given\ inequality\ can\ be\ rearranged\ as\ following \\
\frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{a \plus{} c} \plus{} \frac {a \plus{} c}{a \plus{} b} \ge 2(\frac {a}{a \plus{} c} \plus{} \frac {b}{a \plus{} b} \plus{} \frac {c}{b \plus{} c})\ \ \ ................(5) \\
from\ AM \minus{} GM\ we\ have \\
\frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{a \plus{} c} \plus{} \frac {a \plus{} c}{a \plus{} b} \ge 3\ \ \ ............(3) \\
now\ we\ need\ to\ prove \\
\frac {a}{a \plus{} c} \plus{} \frac {b}{a \plus{} b} \plus{} \frac {c}{b \plus{} c} \le \frac {3}{2}\ \ \ .............(4) \\
we\ know \\
\frac {a}{a \plus{} c} \plus{} \frac {b}{a \plus{} b} \plus{} \frac {c}{b \plus{} c} \equal{} \frac {a}{a \plus{} c} \plus{} \frac {b}{a \plus{} b} \plus{} \frac {c}{b \plus{} c}\ \ \ ...............(1) \\
and \\
\frac {a}{a \plus{} c}\plus{} \frac {b}{a \plus{} b}\plus{} \frac {c}{b \plus{} c} \le \frac {c}{a \plus{} c}\plus{} \frac {a}{a \plus{} b}\plus{} \frac {b}{b \plus{} c}\ \ \ ..............(2)(this\ follows\ from\ rearrangement\ inequality) \\
adding\ (1)\ and\ (2 ) \\
we\ get\ \frac {a}{a \plus{} c} \plus{} \frac {b}{a \plus{} b} \plus{} \frac {c}{b \plus{} c} \le \frac {3}{2} \\
from\ (3)\ and\ (4)\ we\ get\ (5)\ which\ is\ nothing\ but \\
\frac {a \plus{} b \minus{} 2c}{b \plus{} c} \plus{} \frac {b \plus{} c \minus{} 2a}{a \plus{} c} \plus{} \frac {a \plus{} c \minus{} 2b}{a \plus{} b} \ge 0$[/hide] please correct if wrong

[quote="rahul1993"]$ now\ we\ need\ to\ prove \\
\frac {a}{a \plus{} c} \plus{} \frac {b}{a \plus{} b} \plus{} \frac {c}{b \plus{} c} \le \frac {3}{2}\ \ \ .............(4)$[/quote]
Unfortunately this isn't true. Try $ a \equal{} 1,b \equal{} 2,c \equal{} 3$ so you must have done something wrong while proving it. If you are still interested in what went wrong:
[quote="rahul1993"]$ \frac {a}{a \plus{} c} \plus{} \frac {b}{a \plus{} b} \plus{} \frac {c}{b \plus{} c} \le \frac {c}{a \plus{} c} \plus{} \frac {a}{a \plus{} b} \plus{} \frac {b}{b \plus{} c}\ \ \ ..............(2) \\
(this\ follows\ from\ rearrangement\ inequality)$[/quote]
I don't know how you used rearrangement here. Try the same counterexample $ a \equal{} 1,b \equal{} 2,c \equal{} 3$.


algebra
trigonometry
function
inequalities proposed
Inequality

[quote="rogue"]Let $a,b,c>1$ be such that $a+b+c=abc.$ Prove the inequality
$(a^2-1)(b^2-1)(c^2-1)\leq 8.$[/quote]
This is true for all $a\geq0,$ $b\geq0,$ $c\geq0,$ such that $a+b+c=abc.$ ;)

[quote="Lovasz"]Trigonometric substitution works.
Let $\tan A=x, ...$.[/quote]
How does it work? :?

Jensen on the function $f(x)=\ln(x^2-1)$ over $(1,\infty)$ works pretty well here  :)
Like arqady said , it is still true if $a,b,c$ are non-negative since if one of $a^2-1,b^2-1,c^2-1$ is negative then LHS < 0 , RHS >0 which is true .So we only need to consider the case for all of them postive .

[quote="shyong"]Jensen on the function $f(x)=\ln(x^2-1)$ over $(1,\infty)$ works pretty well here  :) [/quote]
How? :?
Jensen on the function $f(x)=\ln(x^2-1)$ over $(1,\infty)$ leads to $(a^2-1)(b^2-1)(c^2-1)\le \left(\left(\frac{a+b+c}{3}\right)^2-1\right)^3$
and  $abc=a+b+c$ implies $\frac{a+b+c}{3}\ge\sqrt{3}$ and $\left(\left(\frac{a+b+c}{3}\right)^2-1\right)^3\ge8 \ldots$

[quote="shyong"]Like arqady said , it is still true if $a,b,c$ are non-negative since if one of $a^2-1,b^2-1,c^2-1$ is negative then LHS < 0 , RHS >0 which is true. So we only need to consider the case for all of them postive .[/quote] It remains to consider the case with two of $a^2-1,b^2-1,c^2-1$ negative which is impossible since $a^2-1<0,$ $b^2-1<0$ imply $ab<1$ and $abc<c<a+b+c.$

Let $ x,y,z >0$ such that $x+y+z=xyz.$ [url=https://artofproblemsolving.com/community/c6h299899p1671345]Prove that[/url]
\[ (x-1)(y-1)(z-1) \le 6 \sqrt{3} -10\]

$$(x^2-1)(y^2-1)(z^2-1)\leq 8$$

Let $a,b,c>0$ and  $a+b+c=abc.$ Prove that
$$(a^2-1)(b^2-1)(c^2-1)\leq 8$$
[url=https://artofproblemsolving.com/community/c6h2951337p26425368]Generalization [/url]
[url=https://artofproblemsolving.com/community/c6h148966p842701]h[/url]


$ \\a,b,c,d>0$ then prove that
$ \\a^4\plus{}b^4\plus{}c^4\plus{}d^4\plus{}2abcd\geq\ 2(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)$

$ a,b,c,d>0$ then prove that
$ \\a^4\plus{}b^4\plus{}c^4\plus{}d^4\plus{}2abcd\geq\ a^2b^2\plus{}a^2c^2\plus{}a^2d^2\plus{}b^2c^2\plus{}b^2d^2\plus{}c^2d^2$

[quote="boroo12"]$ \\
a,b,c,d > 0$ then prove that
$ \\
a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \plus{} 2abcd\geq\ 2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)$[/quote]

Let $ a<b\plus{}c,b<c\plus{}a,c<a\plus{}b,d\rightarrow 0$,
$ a^4\plus{}b^4\plus{}c^4 \ge 2(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)$
$ \Leftrightarrow 0\ge (a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)>0$,It is conflict.

[quote="boroo12"]$ a,b,c,d > 0$ then prove that
$ \\
a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \plus{} 2abcd\geq\ a^2b^2 \plus{} a^2c^2 \plus{} a^2d^2 \plus{} b^2c^2 \plus{} b^2d^2 \plus{} c^2d^2$[/quote]
See here: http://www.mathlinks.ro/viewtopic.php?t=239742 :)


Let $ a$, $ b$, $ c$, be positive real numbers.  Prove that $ \left(1\plus{}\frac{a}{b}\right)\left(1\plus{}\frac{b}{c}\right)\left(1\plus{}\frac{c}{a}\right)\geq 2\left(1\plus{}\frac{a\plus{}b\plus{}c}{\sqrt[3]{abc}}\right)$.

Thanks in advance!

[hide="Hidden by mod"]Mulitply both sides by abc, expand and simplify, we have
$ a^2b \plus{} a^2c \plus{} b^2a \plus{} b^2c \plus{} c^2a \plus{} c^2b \ge 2a^{5/3}b^{2/3}c^{2/3} \plus{} 2a^{2/3}b^{5/3}c^{2/3} \plus{} 2a^{2/3}b^{2/3}c^{5/3}$
Which is true by Muirhead.[/hide]

[quote="dgreenb801"]Which is true by Muirhead.[/quote]

What's Muirhead, and how does ity apply to this problem?

Thanks!

[url=http://en.wikipedia.org/wiki/Muirhead_inequality]The Wikipedia for Muirhead[/url]

Let $ \mathbf{u} \equal{} (u_1, u_2, \ldots u_n)$ and $ \mathbf{v} \equal{} (v_1, v_2, \ldots v_n)$ be two ordered lists of non-negatives such that:

$ \blacktriangleright \; \sum_{i \equal{} 1}^{k} u_i \geq \sum_{i \equal{} 1}^{k} v_i \; , \; \; \; \forall k < n$

$ \blacktriangleright \; \sum_{i \equal{} 1}^n u_i \geq \sum_{i \equal{} 1}^n v_i$

We say, under these conditions, that $ \mathbf{u}$ [i]majorizes[/i] $ \mathbf{v}$

[b]Then[/b], for any list of non-negative numbers $ \mathbf{x} \equal{} (x_1, x_2, \ldots x_n)$, the following inequality holds true:

$ \sum_{sym} x_1 ^{u_1} x_2 ^{u_2} \ldots x_n ^{u_n} \geq \sum_{sym} x_1 ^{v_1} x_2 ^{v_2} \ldots x_n ^{v_n}$

[u]Or, in closed form:[/u] $ \sum_{sym} \prod_{j \equal{} 1}^n x_j ^{u_j} \geq \sum_{sym} \prod_{j \equal{} 1}^n x_j ^{v_j}$

[b]Application:[/b]

Here, $ n \equal{} 3$, and our lists of exponents are $ \mathbf{u} \equal{} ( 2,1,0 )$ and $ \mathbf{v} \equal{} ( 5/3, \; 2/3, \; 2/3 )$. The first list majorizes the second because:
$ 2 \geq 5/3$
$ 2 \plus{} 1 \geq 5/3 \plus{} 2/3$
$ 2 \plus{} 1 \plus{} 0 \equal{} 5/3 \plus{} 2/3 \plus{} 2/3$

So then, we can use this inequality on the list of bases $ \mathbf{x} \equal{} (a,b,c)$ as follows:

$ \sum_{sym} a^2 b^1 c^0 \geq \sum_{sym} a^{5/3} b^{2/3} c^{2/3}$

There are 6 terms on each side, but because the coefficients on $ b$ and $ c$ in the expression of the symmetric sum on the right are equal, it condenses to 3 terms with coefficients of $ 2$. [b]dgreenb801[/b] claims that the resulting inequality rearranges into your inequality.

I don't really get Muirhead...It seems a bit complicated.  Is there an alternative method?  My solution book uses AM-GM, but I don't get how we can use that here.

Thanks!

You can convert Muirhead into multiple applications of AM-GM, in the above case this is,

By AM-GM,

$ \frac {1}{6} \sum_{sym} (a^2b \plus{} a^2c \plus{} ab^2 \plus{} ac^2 \plus{} a^2b \plus{} a^2c) \ge \sum_{sym} a^{5/3}b^{2/3}c^{2/3}$

[quote="JavaMan"]$ \frac {1}{6} \sum_{sym} (\color{red}{\mathbf{a^2b}}$ $ \plus{} a^2c \plus{} ab^2 \plus{} ac^2 \plus{} a^2b \plus{} a^2c) \ge \sum_{sym} a^{5/3}b^{2/3}c^{2/3}$[/quote]

The plus sign that was previously in there is a typo  :wink:

Thank you, I have edited my post accordingly.  :)


ARML trigonometry geometry USAMTS email AMC USA(J)MO

Prizes are available. Does anyone who got 100 know what the prizes are?

-360 Problems
-ARML 1995-2003
-Code Book
-Mathematica
-Countdown
-CPIG
-Music of the Primes
-Book whose name escapes me at the moment

I feel like there were more...

I was about to ask it myself! those things listed - those were the prizes? Does anyone who scored above 40 have an "official list" they could please share?

The other two I forgot (thanks to not_trig for pointing this out to me) were Journey through Genius and The Elegant Universe.

The Elegant Universe is a great book.  It's a physics book, which starts off with traditional physics (Newton), explains Einstein's theories of relativity (very cool) and then goes on to talk about string theory.  Its a great book for anyone who likes physics - I got it last year.

Any idea if mathematica will be left saturday?  And by CPIG, do you mean challenging problems in geometry?  I have it but haven't gone through it yet.

Yes, I think when CatalystOfNostalgia says CPIG it means Challenging Problems in Geometry. Also I think there will be Mathematica left on Saturday, I remember last year I was silver and I was able to get Mathematica

Thanks ln.  I was bronze last year (73) and all I remember is I had 5 choices - Elegant Universe, CPIG, the code book, and two others.  It seems there are a lot more choices than that from the start, so its nice to know that silver does get all the choices.  Not that I think it will make a huge difference for the books, but I've heard Mathematica goes fast - wasn't even a choice for me last year.

Is the Mathematica a full license or is it time limited?  Thanks.

If I remember correctly, it is a full version, but you can only install it on one computer (legally).

for the Mathematica software, do people get duplicates just for that one primary reason? (that you can only install it on one computer)

which prizes are highly recommended for a Bronze winner?  I heard that Mathematica will not be available for bronze winners... =/ (I hope to get Mathematica)

and do you think i will be able to select my prizes if I stay past midnight at monday?

I don't know, in my mind, it would seem too selfish to get multiple copies of Mathematica.

I remember last year as a silver winner, I actually got on at like around 11 PM (mountain time) the day before the prizes were supposed to be released and I was able to choose my prizes.

As of Apr 30 12:07am:

Please choose 3 of the prizes below and click Submit.

	The Code Book by Simon Singh - A sweeping history of the subject of encryption, with technical and mathematical explanations and fascinating personalities.
	The Music of the Primes by du Sautoy - The extraordinary history behind one of mathematics great unsolved mysteries, the Riemann Hypothesis.
	360 Problems for Mathematical Contests by Andreescu/Andrica - Contains 360 mostly very challenging Olympiad problems grouped by subject. Contains solutions. 280 pages. Difficulty: 9/10.
	Challenging Problems in Geometry by Salkind/Posamentier - Interesting geometry problems of varying difficulty and topic. ~240 pages. Difficulty 5-8/10.
	ARML & ARML Power Contests 1995-2003 - A 400-page collection of ARML tests and Power Contests from 1995-2003. See http://www.arml.com for more information about ARML. Difficulty: 5-8
	The Elegant Universe by Brian Greene - A layman's guide to fascinating physics concepts such as string theory, relativity, and quantum physics.
	Count Down - National Book Award finalist Steve Olson chronicles the 2001 United States IMO team. Book signed by the author.
	Mathematica for Students - Powerful mathematical programming tool. Students choosing this prize agree to allow the USAMTS to release their address and email to Mathematica. (MAKE SURE YOUR ADDRESS AND EMAIL ON THE PROFILE ARE CORRECT!!) If you use an operating system besides Windows XP and choose this prize, email your username and operating system to usamts@usamts.org.
	The USSR Problem Book - A collection of challenging problems drawn from past competitions in the USSR.
	Journey Through Genius by William Dunham - An exploration of math history and some fascinating of mathematics through examination of groundbreaking (in their time) theorems.[/img]

[quote="SamE"]Aren't these the same prizes as last year? The only two I didn't have were Music of the Primes and Journey through Genius, so I got a second Count Down for my mom.

Not that I'm complaining, but it [i]would [/i]be nice to see some new books up there.[/quote]

Yeah, I agree. There were only two prizes on the list that I didn't already have, and I had to choose 3  :wink:  Maybe when the USAMTS coordinators get a chance, they can look into getting a newer variety.

Is the student edition of mathematica for six months, a year, or as long as you're a student?

Mine just came in - Music of the Primes. It looks good.

T-shirt's pretty cool too.

Uh, has anyone else not received prizes?

Email usamts@usamts.org if you have not received it by the end of next week.

I haven't...I'll wait until friday until sending an email, like they said.

I received mine today in central Connecticut. For those who were supposed to get 360 Problems, you may not get it; I got Music of the Primes and Countdown instead, which both look very good.

I got mine today. I like the idea on the T-shirt, lol.

The last of them should be showing up this week.  If you haven't received your by one week from today, email us at usamts@usamts.org.

What should I do if I haven't received Mathematica yet...

If you haven't received Mathematica, and you chose it as one of your prizes, then email usamts@usamts.org with all your identifying info.


limit
integration
trigonometry
inequalities
calculus
function
real analysis

Let's try writing the sum as

$S_{n}=\sum_{i=1}^{n^{2}}\frac1n\cdot\frac1{1+\left(\frac{i}{n}\right)^{2}}$

Low look at this as a Riemann sum (for a decreasing function). We have the inequality

$\int_{0}^{n}\frac{dx}{1+x^{2}}-\frac1n<S_{n}< \int_{0}^{n}\frac{dx}{1+x^{2}}$

(The left hand member of this inequality comes from switching from a right-endpoint Riemann sum to a left-endpoint sum.)

Now take the limit as $n\to \infty.$ Both ends of the chain of inequalites converge to

$\int_{0}^{\infty}\frac{dx}{1+x^{2}}=\frac{\pi}2.$

As a question of algebraic tactics, it's frequently useful to have homogeneous expressions - like $n^{2}+i^{2},$ for instance. violetcraze, you did yourself no favor when you substituted $k$ for $n^{2}$ because that broke the homogeneity and made it harder to see whether you were getting the powers of $k$ right.

You could see that 0 is wrong by noticing that the first $n$ terms of the sum are at least $1/(2n)$ each. Since the Riemann sum becomes an integral when $k\to\infty$, you can't keep $k$ fixed in the process.

[quote="Kent Merryfield"]Let's try writing the sum as

$S_{n}=\sum_{i=1}^{n^{2}}\frac1n\cdot\frac1{1+\left(\frac{i}{n}\right)^{2}}$

Low look at this as a Riemann sum (for a decreasing function). We have the inequality

$\int_{0}^{n}\frac{dx}{1+x^{2}}-\frac1n<S_{n}< \int_{0}^{n}\frac{dx}{1+x^{2}}$

(The left hand member of this inequality comes from switching from a right-endpoint Riemann sum to a left-endpoint sum.)

[/quote]

I still don't get the left hand side of the inequality. How do u know that it's lesser than $S_{n}$?   :huh:

$S_{n}+\frac1n$ is greater than the left-endpoint Riemann sum for the integral $\int_{0}^{n}\frac{dx}{1+x^{2}}$, which in turn is greater than the integral itself. Write down the left-endpoint sum and notice that all of its terms, except for the first one, are in $S_{n}$. And the first term is $1/n$.

Because it's a decreasing function, the integral is less than the left-endpoint Riemann sum. That is,

$\int_{0}^{n}\frac{dx}{1+x^{2}}<\sum_{i=0}^{n^{2}-1}\frac1n\cdot\frac1{1+\left(\frac{i}{n}\right)^{2}}$

In turn,

$\sum_{i=0}^{n^{2}-1}\frac1n\cdot\frac1{1+\left(\frac{i}{n}\right)^{2}}=S_{n}+\frac1n-\frac1n\cdot \frac1{1+\left(\frac{n-1}{n}\right)^{2}}<S_{n}+\frac1n$

Now move the $\frac1n$ to the other side of the inequality.


[quote="linzk3"]logb(x) + logx(b) = 8

would
(logb(x))^2 + (logx(b)^2
be equal to 32 or 64?[/quote]

wait I dont understand what your question is asking with all those parenthesis.  If its what I think its asking, its 62..

Im interpreting it as:
$\log_{b}{x}+\log_{x}{b}=8$

What is $(\log_{b}{x})^{2}+(\log_{x}{b})^{2}?$
If that is the question, then I think it would be$8^{2}-2=62$

[quote="linzk3"]logb(x) + logx(b) = 8

would
(logb(x))^2 + (logx(b)^2
be equal to 32 or 64?[/quote]
Please type questions in $\LaTeX$.

If $\log_{b}x+\log_{x}b=8$, then $\left(\log_{b}x\right)^{2}+\left(\log_{x}b\right)^{2}=32$ or $\left(\log_{b}x\right)^{2}+\left(\log_{x}b\right)^{2}=64$?

Yea..so if the question was like that then the answer would be 62..because its just $a^{2}+1/(a^{2})$, and since $a+1/a=8$, $a^{2}+1/(a^{2})=64-2=62$

[quote="linzk3"]logb(x) + logx(b) = 8

would
(logb(x))^2 + (logx(b)^2
be equal to 32 or 64?[/quote]

Its neither.


floor function
FTW
number theory
relatively prime

[quote="GameBot"]Seventy-two 50-cent pieces are lined side-by-side on a table. Joe replaces every second 50-cent piece with a quarter and leaves. Jane replaces every third remaining 50-cent piece with a dime and leaves. Jim replaces every fourth remaining 50-cent piece with a quarter and leaves. What is the number of dollars in the amount of money left on the table? Express your answer to the nearest cent.[/quote]

There are $ 36\plus{}24\plus{}18\minus{}12\minus{}6\minus{}9\plus{}3\equal{}54$ quarters, and thusly 18 half-dollars.

So $ 9\plus{}13.5\equal{}22.50$ dollars.

BOGTRO,

It's not just quarters you're dealing with....there's also dimes.

I seem to be getting $19.20 as an answer though. I'll look in to it.
____________________
~pi man  :blush:  :roll:

Note that $72=2^3\times3^2$. Therefore, the number of half-dollars remaining is equal to the number of positive integers less than $72$, which are relatively prime to $72$. That is equal to $\phi(72)=72(1-\frac12)(1-\frac13)=24$. The number of dimes is equivalent to $\lfloor\frac{72}{3}\rfloor-\lfloor\frac{72}{6}\rfloor=12$. Thus, the nummber of quarters is $72-24-12=36$. The total amount of money is then $36(.25)+24(.5)+12(.1)=9+12+1.2=\boxed{\$22.20}$.

I worked this out many different ways and I don't agree with the answer that is given for this problem.  They got \$19.50.  I think the answer is \$22.20.  I even acted it out by writing all 72 50-cent pieces in.  Another way to work it is to start out with 72 times 50 which is 3600 cents.  Then, subtract off 25 times 36 which is 900.  This takes into account the 1st move of changing every 2nd coin for a quarter.  Finally I divided 72 by 6 since every 3rd coin that has not already been exchanged with a quarter is then replaced by a dime.  That leaves 12 coins replaced with a dime.  40 cents times 12 is 480.  That leaves 3600-900-480 cents which equals \$22.20.

[color=blue]Careful with dollar signs! Edited and fixed. Dollar signs should be proceeded with a "\".

- Your local awesome moderator[/color]

The FTW question is fixed now. It's supposed to read:

Seventy-two 50-cent pieces are lined side-by-side on a table. Joe replaces every second 50-cent piece with a quarter and leaves. Jane replaces every third remaining 50-cent piece with a dime and leaves. Jim replaces every fourth remaining 50-cent piece with a $\text{nickel}$ and leaves. What is the number of dollars in the amount of money left on the table? Express your answer to the nearest cent.

@mathmamma, gaussintraining

You forgot the last step of exchanging 50-cent pieces with nickels. This last step results in 18 50-cent pieces, 36 quarters, 12 dimes, and 6 nickels, for a total of 19.50 dollars.


trigonometry
function
vector
integration
real analysis
real analysis unsolved

For which real numbers $a$ is the sequence $sin(1^a)+sin(2^a)+...+sin(n^a)$ bounded?

For all $\alpha<-1$, since $\sin\frac{1}{n^{\alpha}}<\frac{1}{n^{\alpha}}$, the sequence is convergence.
For all $\alpha>0$, the sequence is divergence, this is because we can choose ${n_p}$ such that $n_p^{\alpha}\in (2k_{n_p}\pi+\frac{\pi}{4},2k_{n_p}\pi+\frac{3\pi}{4})$.
I'm not quite sure whether it converges in $[-1,0]$.
Anyone got ideas?

there it diverges...
there are many ways to show it: in $[0,1]$ function $\sin$ is increasing, so the sequence is decreasing, and tending to $0$. if it converges, $n\sin(n^a)$ should converge (abel's criteria?), so it can't work.
or, by confrontation (which is more or less the same) with $\sum \frac{1}{n^a}$.

the problem actually comes out for $a>1$... just apparently innocent :)

That's right, but the question was about [b]boundedness[/b], not about convergence. So, the answer is $a<-1$ or $a=1$.
It is not hard to see that in both above cases the sequence is bounded. The real problem is to show that it is unbounded for all other $a$, especially, for $a>1$.  ;)

[quote="fishln"]I have searched this post for quite a long time. I got a solution to these kind of problems: :D

Consider $f(x)$ instead of $f(\frac{1}{n})$ will make problems much easier. In this case, compare $\sin(x^{\alpha})$ to $x^{\alpha}$ one can easily reach the conclusion :idea:  :P[/quote]
Sorry,I can't see what you mean... :(

[quote="fishln"]Since $\sin(x^{\alpha})$ ~ $x^{\alpha}$ around $0$, we can consider $\displaystyle\sum_nx^{\alpha}$ instead of considering $\displaystyle\sum_n\sin(x^{\alpha})$.[/quote]
but It is only to show $-1 \le a<0$ is unbounded.what about other case?

Seems  like it's time for a (sketch of a) solution. :)

1) It is easy to show that for $a<-1$ and $a=1$ the sums are bounded.

2) If $-1\le a\le 0$, the series diverges to $+\infty$

3) Assume now that $0<a<1$. Since $n^a$ is an increasing sequence and $(n+1)^a-n^a\to 0$, the sequence $n^a$ is dense modulo $2\pi$. Fix $M>0$. Choose a very big $n$ such that $n^a\approx \frac\pi 2\mod 2\pi$ and notice that $\sin n^a, \sin (n+1)^a,\dots,\sin(n+M)^a$ are close to $1$ if $n$ is big enough. Therefore, the difference between the $n+M$-th and the $n$-th element of our sequence is almost $M$. From here it follows immediately that the sequence is unbounded.

4) Assume now that $a>1$ is not an integer. Again, fix $M>0$ and consider $k=0,1,\dots,M$. Note that
\[ (n+k)^a=n^a+c_1 n^{a-1}k+c_2 n^{a-2}k^2+\dots+c_{[a]}n^{a-[a]}k^{[a]}+o(1)\qquad\text{as } n\to\infty\, \]
with some positive $c_1,\dots,c_{[a]}$ depending on $a$ only. We are going to use the theorem that says that, for any non-zero $c_0,\dots,c_{[a]}$, the vectors $(c_0 n^a,\dots, c_{[a]}n^{a-[a]})$ are dense in $[0,2\pi]^{[a]+1}$ modulo $2\pi$ Thus, we can choose arbitrarily large $n$ such that $n^a\approx \frac\pi 2$, $c_1n^{a-1},\dots,c_{[a]}n^{a-[a]}\approx 0$ modulo $2\pi$. The end of the argument is the same as in case 3.

5) Assume now that $a>1$ is an integer. We have
\[ (n+k)^a=n^a+c_1 n^{a-1}k+c_2 n^{a-2}k^2+\dots+c_{a-1}n k^{a-1}+k^a \]
with some integer $c_j>0$. The theorem we need now is the one that says that the vectors $(c_0n^a,\dots,c_{a-1}n)$ are dense in $[0,2\pi]^a$ modulo $2\pi$ if no $c_j$ is a rational multiple of $\pi$. Let $t\in[0,2\pi]$.  Fix large $M>0$. Choose $n$ such that $n^a\approx 0, c_1n^{a-1}\approx t, c_2 n^{a-2},\dots,c_{a-1}n\approx 0$ modulo $2\pi$. Then the difference between the $(n+M)$-th and $n-1$-st sum is approximately $D(t)=\sum_{k=0}^M \sin(k^a+kt)$. Note that $\int_0^{2\pi}D(t)^2\,dt=\pi M$, so we can choose $t$ in such a way that $|D(t)|>\sqrt{M/2}$. The rest should be clear.


How did you do on Gauss? 150 on Gauss 8, so I finished Middle School with all perfects in one competition. Yay.

hey someone wanta post the question? im really bored

Gauss was sooooo long ago.

I want to see the questions too.  :)

[quote="nikeballa96"]it was <amc 8 (REALLY EASY). like 492314 people got perfects where i took it.[/quote]

That's impossible, considering that less than 492314 people wrote the contest.
And the results come out that quickly? Damn, I wish Cayley came out that fast...

well, no, our teacher grades them unoficially.


Prove that sin(x) + tan(x) >= 2x when 0 <= x <  :pi: /2.

That doesn't make sense. Plug in 60 for x.

I think x was supposed to be in radians, between 0 and :pi:/2.

Ooh that reminds me of a question we had at Math Bowl 2003:

Let x= :pi: /2003. Put the following in increasing order: sin(x), tan(x), x.

Sorry, yes, x should have been in radians. Fixed it.

[hide]f(x)=sin(x)+tan(x)

f''(x) = -sin(x)+2*tan(x) + 2*tan(x)^3 >= 0 (tan(x)>=sin(x)>=0)

By Taylor-Young:

f(x) = 2*x + x^3/3! f''(c) with in c in (0,Pi/2)

so f(x)-2x >= 0.

I'm not sure it's simpler that just studying f(x)-2x in fact.  [/hide]

belanos: could you clearify that last statement?

Hmm, belenos, that looks correct (I think, I'm not sure steady on my calculus), but I think it can be done without calculus as well.

Here's another similar problem from a local math contest:

Which is closer to 1: 2sin(29) or 2sin(31) (in degrees)?

Of course, this question is very easy with calculus, but there's a nice solution that doesn't use calculus that my father came up with when I showed him this problem.

Hmm, we can use the fact that the sine curve is concave for acute angles.  Is there a better way?

There's a ``better'' way (i.e. one that doesn't even use pseudo-calculus).

Here's the non-calc way to do it.

[hide]2sin(29) = 2sin(30-1) = 2sin30cos-1 - 2sin1cos30 = cos1 -  :rt3: sin1.

similarly, we show that 2sin(31) = cos1 + :rt3: sin1.

Clearly, 2sin(31) is closer, because cos 1 is only a little less than one, and adding a very small value will make it closer to 1.[/hide]

Yeah, my way is similar except without any rt3s or anything:

[hide]

The average of 2sin29 and 2sin31 = sin29 + sin31 = 2cos1 sin30 = cos 1 < 1. So 2 sin 31 must be the one closer to 1.

[/hide]


linear algebra
matrix
superior algebra
superior algebra solved

Let A be in M_2(C).
 For n in N\{0}, x_n=det(A^n+I_2).
Prove that, if x_1=x_2=1,then x_n is in {1,4}

A is similar to a triangular superior
(a c)
(0 b)
then x(n)=(1+a^n)(1+b^n)
the conditions x(1)=x(2)=1 give
(1+a)(1+b)=(1+a^2)(1+b^2)=1

a+b+ab=0
a^2 +b^2 +a^2b^2=0

so a=0 or b=0 or b=1/a

If a=0 => b=0 => x(n)=1
If b= 0 => a=0 => x(n)=1
If b=1/a we have a^2+a+1=0 so a=j or a=j^2 where j=exp(2i \pi /3)

x(n)=(1+j^n)(1+j^(2n))=2+2cos(2n \pi /3)
so x(n) \in {1;4}

I think the place of this problem should be in
College playground-> Superior Algebra

I see you moved this problem in the solved section! I have other solution(s), if someone is interested please ask and I will post it/them.

That the problem is in the solved section no means that it is closed....Please post your solution Lagrangia.

Pierre.

Please Lagrangia post your solution, it is always interesting.
Thx

Let B be from M_2(C) so that det(B+I_2)=1. From Hamilton-Cayley we have: B^2-tr(B)B+det(B)I_2=0_2 (1) and
(B+I_2)^2-tr(B+I_2)(B+I_2)+I_2=02 which is equivalent to: B^2+2B+2I2-(2+tr(B))B-(2+tr(B))I_2=O_2 which gives B^2-tr(B)B-tr(B)I_2=0_2(2)
so from (1) and (2) we get that det(B)+tr(B)I_2=O_2 so we have det(B)+tr(B)=0 (3)

lets denote d=det(A) and t=tr(A). because A^2=tA-dI_2 we get that tr(A^2)=tr(tA-dI_2)=t^2-2d  (4).

we apply (3) to (A) and we get that d+t=0.
we apply (3) to  (A^2) and we get that det(A^2)+tr(A^2)=0 and from (4)
we get that d^2+t^2-2d=0

so we get the sistem :
d+t=0
t^2+d^2-2d=0 with the solutions:
a)t=0 and d=0
and
b)t=-1 and d=1

for a) we get that A^2=02 and so A^n=0n for any n>=2 so in this case x_n=det(A^n+I_2)=det(I_2)=1
for b) we get that A^2+A+I_2=0_2 so A^3=I_2 so we get that:
x_(3n)=det(2I_2)=4
x_(3n+1)=1
x_(3n+2)=1

so x_n is in {1,4}.
problem solved!
cheers!

errr.. I have solved this one using eigenvalues too! :D

cheers!


integration
calculus
derivative
limit
function
calculus computations

How can we found the length of the curve  $ f(x) \equal{} \frac {1}{12}(x\minus{}48)\sqrt x$ where $ x\ge0$ and the vertical line x=48.

I tried to use the formula L=$ \int^{48}_{0}\sqrt {1 \plus{}\ [f'(x)]^2}$ $ dx$
But I think that there is a problem where x=0 because the first derivative is:
$ f'(x)\equal{}\frac {(x\minus{}16) \sqrt{x}}{8x}$ and because x is in the denominator cannot take the value 0.

How can solve this issue??
Any ideas??

Thanks anyone in advance.

This is an improper integral, but it is a convergent improper integral.

We obtain $ L\equal{}\lim_{b\to0^\plus{}}\int_b^{48}\sqrt{1\plus{}[f'(x)]^2}\,dx$

We can see that the integrand resembles $ \frac{c}{\sqrt{x}}$ for some constant $ c$ as $ x\to0^\plus{}.$ (For one thing, if $ f'(x)$ is large, then $ \sqrt{1\plus{}[f'(x)]^2}$ simply looks like $ |f'(x)|.$) And a function that behaves like that does have a finite integral.

$ \int_0^1\frac{dx}{\sqrt{x}}\equal{}2\sqrt{x}\,\bigg|_{0^\plus{}}^1\equal{}2$ is finite.

A monotone (or piecewise monotone) function on a closed bounded interval should have a finite arclength.

hello, inserting your derivative in the square root we get $ \sqrt{1\plus{}\left(\frac{(x\minus{}16)\sqrt{x}}{8x}\right)^2}$ simplifying this for $ x>0$ we have $ \frac{16\plus{}x}{8\sqrt{x}}$ and our indefinite integral is
$ \frac{1}{12}\sqrt{x}(48\plus{}x)$, inserting the upper bound $ 48$ and subtracting the lower bound $ 0$ we get $ 32\sqrt{3}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, inserting your derivative in the square root we get $ \sqrt {1 \plus{} \left(\frac {(x \minus{} 16)\sqrt {x}}{8x}\right)^2}$ simplifying this for $ x > 0$ we have $ \frac {16 \plus{} x}{8\sqrt {x}}$ and our indefinite integral is
$ \frac {1}{12}\sqrt {x}(48 \plus{} x)$, inserting the upper bound $ 48$ and subtracting the lower bound $ 0$ we get $ 32\sqrt {3}$.
Sonnhard.[/quote]

This is exactly my question. How can you integrate with lower bound 0 while you said that "simplifying this for $ x > 0$ we have $ \frac {16 \plus{} x}{8\sqrt {x}}$ and our indefinite integral is..."
Is this right???

Thanks

[quote="Kent Merryfield"]This is an improper integral, but it is a convergent improper integral.

We obtain $ L \equal{} \lim_{b\to0^ \plus{} }\int_b^{48}\sqrt {1 \plus{} [f'(x)]^2}\,dx$

We can see that the integrand resembles $ \frac {c}{\sqrt {x}}$ for some constant $ c$ as $ x\to0^ \plus{} .$ (For one thing, if $ f'(x)$ is large, then $ \sqrt {1 \plus{} [f'(x)]^2}$ simply looks like $ |f'(x)|.$) And a function that behaves like that does have a finite integral.

$ \int_0^1\frac {dx}{\sqrt {x}} \equal{} 2\sqrt {x}\,\bigg|_{0^ \plus{} }^1 \equal{} 2$ is finite.

A monotone (or piecewise monotone) function on a closed bounded interval should have a finite arclength.[/quote]

Do you agree with the Dr Sonnhard Graubner's post??
Because I understand that your post and Dr Sonnhard Graubner's say the same thing. Am I right??

Thanks a lot

[quote="Kent Merryfield"]I was focused on answering the question you asked - as to how this curve can have a finite arclength and how the integral from zero to somewhere should be define. Sonnhard simply treated this as an exercise in computation. That said, there is no conflict between our posts.

Have you studied improper integrals?[/quote]

I do understand what you said about improper integrals.
Thanks a lot!!


AMC AIME USA(J)MO USAMO geometry inequalities college

I have a few questions about AIME and USAMO prep.
My goal for next year is to get some of harder problems in AIME (10-15). What would be a good way to achieve this?? Would practicing with past AIMEs be a good way? ( I did about half of the past AIMEs).

For USAMO , I really don't know where to start. :huh:
I probably could do the first pbm...
How should I prep for USAMO? I heard that The Art and Craft of Problem Solving is a pretty good book for USAMO prep.
What else should I do??

Thanks in advance!

For the AIME, going through AoPS Vol. 2 and practicing all the old AIME's should be enough to get you well past the cutoff.

For the USAMO, I recommend just getting a bunch of problem books (or using the Resources section on this site) and just going through lots of problems. Also, the AoPS Olympiad classes (and WOOT) are good ways to keep on track. Check out old MOP stuff, too (mostly lecture notes).

Thanks for your advise.

But where are those MOP stuff?? :blush:

Thanks.

[quote="paladin8"] Check out old MOP stuff, too (mostly lecture notes).[/quote]

Where can these be obtained?

Here's a collection of Reid Barton's MOP lecture notes: http://web.mit.edu/rwbarton/Public/mop/. They're mostly problems with a little bit of instructional material.

This is a huge packet on Euclidean Geometry by Kiran Kedlaya: http://math.mit.edu/~kedlaya/geometryunbound/geom-080399.pdf

Kedlaya's notes on Inequalities (from AoPS!): http://www.artofproblemsolving.com/Resources/Papers/KedlayaInequalities.pdf

Thanks! I've always wondered where everyone gets all this MOP stuff ;)

I was thinking of ordering the official solutions pamphlets, but the AMC website only seems to have 1992 and 1994...Does anyone know where I can find the rest?

[quote="krustyteklown"]just a note

i think Kiran's notes (both geometry unbound and the inequality one) are quite near the upper end of the olympiad spectrum;

I think for geometry, richard's class here on aops Olympiad Geometry which I am in now, is quite comprehensive

for inequalities, I think a packet by Bjorn Poonen from the Berkeley Math Circle had some less advanced problems and techniques

Here it is: http://mathcircle.berkeley.edu/inequalities.pdf

Edit: Link to Bjorn's packet[/quote]

well they are MOP notes... i would expect them to be rather difficult. but yeah, i agree some of that stuff is probably way beyond a lot of people (myself included).


I see that pi has gotten extremely popular...

Problem is...eventually, how will we differentiate?

Well does it matter? Cause why do you need to differentiate?

Don't trouble yourself over it.  I do have to admit that is phail tho.


I've recently installed LaTeX in my computer and now trying to include diagrams into documents by asymptote. How can I do it? (if I can do it that is) Is there a package that I must include with \usepackage in TeXnicCenter? If so then what is that package? Thank you for your help.

Instructions can be found in the Asymptote section of the AoPSWiki at [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration#Using_Asymptote_in_Latex]Using Asymptote in Latex[/url]

That says to go to "C:\Program Files\texmf\tex\latex\.", but I can't find the texmf folder anywhere.

Ah, that needs changing. texmf is the standard path for LaTeX files on any system but MiKTeX uses as a default something like C:\Program Files\MiKTeX 2.6\tex\latex\ so replace texmf by MiKTeX 2.6.


geometry
function
trigonometry
logarithms
inequalities
email
complex numbers

The Southern California Mathematical Olympiad (SCMO) is an annual mathematical contest sponsored by the Math Zoom Academy.  It’s a 2 round contest.  The first round runs at the end of October, and is administered in Southern California schools and Math Zoom Academy sites in California.  The final round runs at the beginning of December, at a location in Southern California.  The 2009 SCMO Final Round will be held on Sunday, December 6.

The SCMO aims to stimulate the interest of high school students in mathematics and develop their problem solving abilities.  The competition is designed to be a fun and exciting experience.  The SCMO consists of the First Round and the Final Round.  The First Round has two parts.  Part I contains 10 short answer questions.  These questions cover the subject areas of algebra (functions, equations, trigonometry, logarithm, complex numbers, etc.), geometry (plane and solid), number theory, and combinatorics that appear in the high school math curriculum, but with more focus on deeper understanding of concepts and advanced problem solving skills.  The questions range from easy to relatively difficult, to challenge students with various levels of mathematical skills in high school and some advanced students in middle school.  Part II of the First Round contains two questions that require full solutions.  Participants will write complete proofs or solutions with all supporting arguments, and the grading is based on the correctness of the solution as well as the clarity and mathematical rigor.  Besides the areas covered in Part I, the questions in Part II can also involve inequalities, functional equations, recurrences, and various other Math Olympiad topics.

The top performers of the First Round are invited to participate in the Final Round.  The Top 3 participants of the First Round will be in the Grand Challenge Round, where they are given a series of questions and present their solutions to a panel of judges.  Each participant has 15 minutes to present his or her solutions, and then spend 5 minutes to answer questions from the judges.  The ranking of the Top 3 shall be decided based on the correctness of the solutions, presentation skills, and response to judges’ questions.

Besides the Top 3, the next 40 best participants from the First Round are invited to attend the Team Challenge of the Final Round.  The participants shall be divided into 4 teams and given a set of math problems; the team members work together to solve the problems and give answers to the host.  The time limit of the Team Challenge is one hour.  At the end of one hour, the team that answered the most questions is the team winner.

The Final Round is a fun and entertaining event.  It begins with a mathematical talk, then the Team Challenge, the Grand Challenge, and ends with the Award Ceremony.

Admission is free!  Everyone is invited.

The First Round is extended to Nov. 6, based on requests from schools.
There is still time to sign up!  Te location of the First Round test is at your schools or at Math Zoom sites.  If you plan to come to the Math Zoom office to take it, please email info@mathzoom.org to schedule.  The available time is 1pm to 7pm, Monday through Friday.
Location: 5420 Trabuco Road, Suite 180, Irvine, CA 92620.


$\pi = lim \ 2^{n} \sqrt {2_{1}- \sqrt {2_{2}+ \sqrt {2_{3}+ \sqrt { 2_{4} + \dots + \sqrt {2_{n}}}}}}$ (the subscripts are here just to show that the number  of $2s$ under the square roots is $n$).

Note: that's very easy but you need the right idea ;)

Hmm.. I think I have an idea, but, just tell me if it's a good start :) :
$\sqrt 2 = 2 cos \dfrac {\pi}{4}$

Yes, it is. Now use the trigonometric formulas for half angles.

This problem (hard if seen the first time) has a nice generalization :

$x_n = \epsilon_0 \sqrt {2_{1} + \epsilon_1 \sqrt {2_{2}+ \epsilon_2 \sqrt {2_{3}+ \epsilon_3 \sqrt { 2_{4} + \dots + \epsilon_n \sqrt {2_{n}}}}}}$ with $\epsilon_i = \pm 1$

Prove $x_n$ converges (and give the limit in the form of a converging serie).


LaTeX
number theory unsolved
number theory

Let $p_1, p_2, \ldots, p_{25}$ are primes which don’t exceed 2004. Find the largest integer $T$ such that every positive integer $\leq T$ can be expressed as sums of distinct divisors of $(p_1\cdot p_2 \cdot \ldots \cdot p_{25})^{2004}.$

[b]Please post your solutions.[/b] [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]Use $\LaTeX$ please![/url] Please omit jokes/smilies etc. Comments and generalizations are welcome. If you noticed that the comment has been discussed elsewhere, please provide a link. If you don't know the link(s) do NOT post and state that the problems has been discussed many times. If the provided solution is not complete, right or in $\LaTeX$-style I would be happy if you could (re-) post your/the solution in this thread again! At the end of your (re-)written solution post the links to those insufficient solutions as follows:

[hide="Remark"]
- first link, e.g. http://www.mathlinks.ro/Forum/viewtopic.php?p=346917#p346917
- second link
etc.
[/hide]

[b]Happy Problem-Solving![/b] :)

here is my solution:

[b]Lemma[/b]
If the sequence $1=a_1<a_2<...<a_n$ satisfies the condition:  $a_{i+1}\le 1+a_1+...+a_i$  for any $1\le i\le n-1$ then  $a_1+a_2+...+a_n$ is the largest number T such that  every positive integer $\le T$ can be expressed as sums of distict terms of the sequence.

here is my solution:
Denote $M=(p_1p_2...p_{25})^{2004}$
[b]Lemma 1[/b]
If the sequence $1=a_1<a_2<...<a_n$ satisfies the condition:  $a_{i+1}\le 1+a_1+...+a_i$  for any $1\le i\le n-1$ then  $a_1+a_2+...+a_n$ is the largest number $T$ such that  every positive integer $\leq T$ can be expressed as sums of distict terms of the sequence.
proof:  The lemma is very easy.
Lemma 2:
If the sequence $1=a_1<a_2<...<a_n$ satisfies the condition:  $a_{i+1}\le 2a_i$  for any $1\le i\le n-1$ then  $a_1+a_2+...+a_n$ is the largest number $T$ such that  every positive integer $\leq T$ can be expressed as sums of distict terms of the sequence
proof: use lumma 1.

Let $1=n_1<n_2<...<n_N$ are all divisors of $M$
If $min\{p_1;p_2;...;p_{25}\}$=2. We will show that $n_{i+1}\le 2n_i$ for any $1\le i\le N$.
Suppose $p_1=2$. and $n_i=2^{t}p_2^{t_2}...p_{25}^{t_{25}}$.
Case 1: if $t< 2003$ we have  $n_i<2n_i=2^{t+1}p_2^{t_2}...p_{25}^{t_{25}}$ $\Rightarrow 2n_i|M$ $\Rightarrow n_{i+1}\le 2n_i$
Case 2: if $t=2004$. we have $min\{t_2;t_3;...;t_{25}\}\le 2003$. Suppose $t_2\le 2003$.
Let x is the positive integer such that $2^x<p_2<2^{x+1}$ ( Always exist x)
we have  $n_i<2^{2004-x}p_2^{t_2+1}...p_{25}^{t_{25}}<2^{2005}p_2^{t_2}...p_{25}^{t_{25}}$
and $2^{2004-x}p_2^{t_2+1}...p_{25}^{t_{25}}| M \Rightarrow$ we have $n_{i+1}\le 2n_i$.
Use lemma 2: We have $T=n_1+n_2+...+n_N=\frac{(p_1^{2005}-1)...(p_{25}^{2005}-1) }{(p_1-1)(p_2-1)...(p_{25}-1)}$.

if $min\{p_1;..;p_{25}\}\ge 3$ then $M=1$.

.

Actually the rest part is simple just use induction on the number of primes.


Find the values of $ a$ such that  the equation

$ x^4\plus{}ax^3\plus{}bx^2\plus{}ax\plus{}1\equal{}0$

has 3 different real roots.

Let $ y\equal{}x\plus{}\frac 1x$.
Then equation $ y^2\plus{}ay\plus{}b\minus{}2\equal{}0$ had 2 different roots, one of them $ y_1\equal{}\pm 2$, second $ y_2>2$ or $ y_2<\minus{}2$.
It give $ |b\minus{}2|>4$ and $ a\equal{}\pm (2\plus{}\frac{b\minus{}2}{2}).$


[quote="Utkirstudios"]let x,y,z>0  then prove:
\[x^{6}+y^{6}+z^{6}+9(x^{3}y^{3}+y^{3}z^{3}+x^{3}z^{3}) \ge 10xyz(x^{3}+y^{3}+z^{3}) \]
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80197


Let the real numbers $ x_1,x_2,x_3,....x_n$ belong to the interval $ [\minus{}1,1]$ such that the sum of their cubes is zero.Prove that their sum cannot exceed $ \frac n3$

trig kills it
clearly for each $ x_{i}$ there exists an $ \alpha_{i}$ such that $ x_{i} \equal{} \sin \alpha _{i}$
also $ \sum_{i \equal{} 1}^{n} x_{i}^{3} \equal{} 0 \Rightarrow \sum_{i \equal{} 1}^{n} \sin^{3} \alpha_{i} \equal{} 0$
now $ \sin 3\alpha_{i} \equal{} 3 \sin \alpha_{i} \minus{} 4 \sin^{3} \alpha_{i}$
so $ \sum_{i \equal{} 1}^{n}\sin 3\alpha_{i} \equal{} 3\sum_{i \equal{} 1}^{n} \sin \alpha_{i} \minus{} 4\sum_{i \equal{} 1}^{n} \sin^{3} \alpha_{i}$
so $ \sum_{i \equal{} 1}^{n}\sin 3 \alpha_{i} \equal{} 3\sum_{i \equal{} 1}^{n} \sin \alpha_{i}$
this gives $ \sum_{i \equal{} 1}^{n} \sin \alpha_{i} \leq \frac {n}{3}$
so $ \sum_{i \equal{} 1}^{n}x_{i} \leq \frac {n}{3}$
so done  :)


trigonometry
inequalities
geometry proposed
geometry

Suppose $\alpha,\beta,\gamma$ are angles of an acute triangle. Prove the inequality \[\sin 2\alpha+\sin 2\beta+\sin 2\gamma\leq2\left(\frac{\sin \alpha\cos \beta}{\cos(\alpha-\beta)}+\frac{\sin \beta \cos \gamma}{\cos(\beta-\gamma)}+\frac{\sin\gamma\cos\alpha}{\cos(\gamma-\alpha)}\right).\]

bodan are you sure? because when triangle is equilateral, there isn't equality!I think it need const $2$ somewhere!

\[\sin 2\alpha+\sin 2\beta+\sin 2\gamma\leq2\left(\frac{\sin \alpha\cos \beta}{\cos(\alpha-\beta)}+\frac{\sin \beta \cos \gamma}{\cos(\beta-\gamma)}+\frac{\sin\gamma\cos\alpha}{\cos(\gamma-\alpha)}\right).\]  \[\sum_{cyc}\cos \alpha \sin \alpha \leq \sum_{cyc}\frac{\sin \alpha\cos \beta}{\cos (\alpha-\beta)}= \sum_{cyc}\frac{\sin \alpha\cos \beta}{\cos \alpha\cos \beta+\sin \alpha\sin \beta}\] $a=\cos \alpha, b=\cos \beta, c=\cos \gamma$ \[\sum_{cyc}a\sqrt{1-a^{2}}\leq \sum_{cyc}\frac{b\sqrt{1-a^{2}}}{ab+\sqrt{1-a^{2}}\sqrt{1-b^{2}}}\] Let's see what we can do with Cauchy  :wink:


algebra
polynomial
vector
linear algebra
linear algebra unsolved

Let $T$ be a linear operator on a finite-dimensional vector space, and suppose that the characteristic polynomial of $T$ splits. Let $\lambda_1, \lambda_2,...,\lambda_k$ be the distinct eigenvalues of $T$, and for each $i$ let $p_i$ be the order of the largest Jordan block corrsponding to $\lambda_i$ in a Jordan canonical form of $T$. Prove that the minimal polynomial of $T$ is

$
(t-\lambda_1)^{p_1} (t-\lambda_1)^{p_1}\cdot\cdot\cdot (t-\lambda_k)^{p_k}
$

My native language is Spanish, how do you pronounce, in English, the word "eigenvector" ?

Thanx

Do you think this should be in intermediate topics?

:huuh:  :huuh:  :huuh:  :huuh:  :huuh: uuuhhhh....$4$?

I'll move it to a more suitable forum.  And you say it like "eye gen", the first syllable is just like the letter "i" then you say "gen" with a hard g like the g in game.


Find all $ x,y \in Z$ that: $ x^2 \plus{}x\plus{}41\equal{}y^2$

[hide="My solution"]$ x^2\plus{}x\plus{}41\equal{}y^2\Rightarrow x^2\plus{}x\plus{}(41\minus{}y^2)$ so $ x\equal{}\frac{\minus{}1\pm \sqrt{4y^2\minus{}163} }{2}$
so $ 4y^2\minus{}163\equal{}k^2$ for some nonnegative interger $ k$ so $ (2y\minus{}k)(2y\plus{}k)\equal{}163$ $ \because 163$ is prime so $ 2y\minus{}k\equal{}1$ and $ 2y\plus{}k\equal{}163\Rightarrow y\equal{}41,k\equal{}81$

Hence $ (x,y)\equal{}(40,41)$[/hide]

$ x^2 \plus{} x \plus{} 41 \equal{} y^2$

Multiplying by 4 and factoring,

$ (2x\plus{}1)^2 \plus{} 163 \equal{} 4y^2$
$ 163 \equal{} (2y\plus{}2x\plus{}1)(2y\minus{}2x\minus{}1)$
But 163 is prime so one of these is 1 and the other is 163
Solving gives 2 solutions :

$ (x,y)\equal{}(40,41) or (\minus{}41,\minus{}41)$

[quote="gokussj3"]$ x^2 \plus{} x \plus{} 41 \equal{} y^2$

Multiplying by 4 and factoring,

$ (2x \plus{} 1)^2 \plus{} 163 \equal{} 4y^2$
$ 163 \equal{} (2y \plus{} 2x \plus{} 1)(2y \minus{} 2x \minus{} 1)$
But 163 is prime so one of these is 1 and the other is 163
Solving gives 2 solutions :

$ (x,y) \equal{} (40,41) or ( \minus{} 41, \minus{} 41)$[/quote]

The complete solutions are:

$ ( \minus{} 41, \minus{} 41)$
$ ( \minus{} 41, 41)$
$ (40, 41)$
$ (40, \minus{} 41)$

When you substitute into $ x^2 \plus{} x \plus{} 41 \equal{} y^2$ for $ y \equal{} \minus{} 41$, and later for $ y \equal{} 41$,
you'll get two different pairs of values for $ x$.


Two circles, A and B, are externally tangent to each other on the line $ y \equal{} x$ at the origin. The two other tangents of the circles are drawn, and another line that is parallel to the line $ y \equal{} x$ and tangent to circle A is also drawn. Assuming that circle A has radius 666 and circle B has radius 137, what is the area bounded by the three lines drawn that do not include the line $ y \equal{} x$?

Fixing your wording does not mean to add a picture :wink:

[quote="mathcrazed"]There are usually three tangents... cough. The line in between the two circles... and what do you mean by other? You never even specified the first one, unless you mean the line y=x.  But when you say they are externally tangent to each other, it kind of says that A and B lie on the line... Fix your wording[/quote]A and B are externally tangent to each other on the line.
At the origin.
Like the diagram I drew for you guys.

The two circle's point of tangency is the origin.


Hi,
I am interested in learning fluid mechanics, I did a search on the Internet and found out about an MIT course by Prof. Ascher H. Shapiro, which is (actually, was) available online at: http://www.fluidmechanicsvideo.com , unfortunately, this web site is no more available.  So, does any one know where I can find those videos.

Please help.

Thank you in advance.

Check http://ocw.mit.edu

Always a helpful source of MIT courseware

You may just find your videos there


Here it is!  If you have answered the first Relay Race, then you can do this.  If you haven't done the first one then you can't do this. :) Here it is:

1. Angela Wilson processed 300 applications for food stamps during the month of June. During the month of July, she processed 10% fewer applications. How many applications did she process in July?

--------------------------------------------------------------------------------

2. A personnel officer drove from Lake Charles to a conference in Baton Rouge. The total distance for the round trip was 240 miles. The time required to travel one way to Baton Rouge was two hours. Due to heavy traffic during the return trip to Lake Charles, an extra hour was required. How much slower was the personnel officer traveling on the return trip?

--------------------------------------------------------------------------------

3. A state park is budgeted at an amount 9 times the amount budgeted for a nearby city park. If the combined yearly budget of both parks is 1,000,000 dollars, what is the average monthly budget of the city park?

--------------------------------------------------------------------------------

4. The estimated completion time for a particular 100 item test is 3 1/3 hours. Ten applicants actually took the test and completed it in 3 hours. What is the difference between the actual and estimated rate of completion per item?

--------------------------------------------------------------------------------
5.  Newman's Fine Haberdashery store is having its annual spring sale You can buy suspenders and socks for 18 dollars. You can buy socks and a tie for 22 dollars. You can buy a tie and cufflinks for 25 dollars. You can buy a shirt and a belt for 37 dollars. You can buy cufflinks and a shirt for 33 dollars. You can buy a belt and socks for 17 dollars. How much does each article cost?
--------------------------------------------------------------------------------
6.  A crew of 8 people can build a concrete wall in 6 days. Suppose 4 more people had joined the crew at the start. Assume that each person works at the same rate as each of the other people. How many days would it have taken the new crew to build the same wall?
-----------------------------------------------------------------------------------
7. Betty wants to purchase a bicycle but is 23 dollars short. Claire wants to purchase the same bicycle but is 25 dollars short. If they combine their money, they will have just enough to buy the bicycle. What is the cost of the bicycle?
-----------------------------------------------------------------------------------
8. Two dogs run around a circular track 300 feet long. One dog runs at a steady rate of 15 feet per second, the other at a steady rate of 12 feet per second. Suppose they start at the same point and time. What is the least number of seconds that will elapse before they are again together at the starting point?
-------------------------------------------------------------------------------------
9.  A kangaroo chases a rabbit which starts 150 feet ahead of the kangaroo. For every 12-foot leap of the kangaroo, the rabbit makes a 7-foot leap. How many leaps will the kangaroo have to make to catch up to the rabbit?
--------------------------------------------------------------------------------------
10.  Henry was able to buy some 23-cent stamps and some 15-cent stamps for a total of exactly 2.50. How many 15-cent stamps did he buy?
--------------------------------------------------------------------------------------
PM me the answers!

these people got round 1 correct:
 LuCy4EvA
 colts18
 Phelpedo
 bob123
 Hamster1800
 Xantos C. Guin
 nat mc
 gr8sk8r
 chess64
did i get everyone?  if you want to know the answer to Round 1(people who didn't get it) i'll post it here for you.  Just tell me if you want it.

you still don't have my name there...  :mad:

no one has sent me a PM except for colts18.
So far after one week the standings are
1st place: colts18 with a score of 2/10

When are they due?  :?

Edit: For #3, I assume you want the answer rounded to the nearest cent.

They are do June 25.
And #3 you should round to the nearest cent.

There are some new leaders!!!
1st place:  Xantos C. Guin (9/10)
2nd place:  chess64 (8/10)
3rd place  colts18 (2/10)

If you send me the answers i will tell you which ones you get wrong and let you try them one more time. If you get it wrong again you can't try them again.  Does that make sense?

NOTE:  everyone had trouble with #4

[quote="math92"]NOTE:  everyone had trouble with #4[/quote]

I think you're wrong. I'll PM you my solution.

New standings!!!
1st(tied): chess64 and Xantos C. Guin(9/10)
3rd: colts18(2/10)

And i'm not wrong on #4 because when i got these questions they came with answers.

[quote="math92"]Does anyone else want to try this?  Otherwise Xantos C. Guin and chess64 will be the only ones allowed to do Round 3[/quote]

Hey!!! I did this a long time ago....

i will....i just gotta get around to actually doing them

EDIT: pm sent.

EDIT2: mwahahaha I'm the new leader (and permanent lol)

EDIT3: aww...it turns out i'm not the new leader....just a new co-leader.

can i still participate? or is it too late? if it's not too late ill PM you the answer tomorrow... lol this is weird, when i saw that only ppl who did the first round can particpate i felt kinda sad, then found out that i did the first round. So can i stlil participate?

[quote="math92"]OMG!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
You guys did get #4 right!!!!!!!
New standings:
1st place: Xantos C. Guin (10/10)
               bob123(10/10)
               Hamster1800(10/10)
               chess64 (10/10)
5th place: colts18 (2/10)[/quote]

But in the previous 2 ranking posts you said that Xantos had 9 right?!

But i forgot that he had number #4 correct. chess64 also got #4 correct so i changed the scores.

Is there going to be a deadline so we can get on with Round 3?  :P

[quote="Xantos C. Guin"]WOOT!
I win!
 :10:[/quote]

\/\/00t! So do I!  :10: Lol


Find the greatest postive number $ n$ such that $ n^{2}\plus{}2009n$ is perfect square.

Let $ n \equal{} p^2m$ so that $ p,m$ are positive integers and $ m$ is square-free (i.e. do not have a square as its factor).

Then the question becomes $ p^2(p^2m^2 \plus{} 2009m)$ is a square
$ \Leftrightarrow m(p^2m \plus{} 2009)$ is a square. As $ m$ is square-free, $ m|p^2m \plus{} 2009 \Leftrightarrow m|2009$

As $ 2009 \equal{} 7^2 \times 41$ and $ m$ must be square-free, $ m\equal{}287,41,7$ or $ 1$

Case 1:$ m \equal{} 287$, then $ p^2 \plus{} 7 \equal{} x^2$ is a square.$ (x \minus{} p)(x \plus{} p) \equal{} 7$ gives $ p \equal{} 3$ as solution $ \Leftrightarrow n \equal{} 3^2 \times 287$
Case 2:$ m\equal{}41$, then $ p^2\plus{}49\equal{}x^2$ is a square.$ (x\minus{}p)(x\plus{}p)\equal{}49$ gives $ p\equal{}24$ as the largest solution for $ p$ $ \Leftrightarrow n\equal{}24^2 \times 41$ as solution.
Case 3:$ m \equal{} 7$, then similarly $ (x \minus{} p)(x \plus{} p) \equal{} 287$ gives $ p \equal{} 143$ $ \Leftrightarrow n \equal{} 143^2 \times 7$ as solution.
Case 4:$ m \equal{} 1$, then $ (x \minus{} p)(x \plus{} p) \equal{} 2009$ gives $ 1004$ as the largest solution for $ p$ $ \Leftrightarrow n \equal{} 1004^2$

So $ n \equal{} 1004^2$ is the required answer.

[quote="stephencheng"]Let $ n \equal{} p^2m$ so that $ p,m$ are positive integers and $ m$ is square-free (i.e. do not have a square as its factor).

Then the question becomes $ p^2(p^2m^2 \plus{} 2009m)$ is a square
$ \Leftrightarrow m(p^2m \plus{} 2009)$ is a square. As $ m$ is square-free, $ m|p^2m \plus{} 2009 \Leftrightarrow m|2009$

As $ 2009 \equal{} 7 \times 287$
Case 1:$ m \equal{} 2009$, then the question $ \Leftrightarrow 2009(p^2(2009) \plus{} 2009)$ is a square $ \Leftrightarrow p^2 \plus{} 1$ is a square, impossible.
Case 2:$ m \equal{} 287$, then similarly $ p^2 \plus{} 7 \equal{} x^2$ is square.$ (x \minus{} p)(x \plus{} p) \equal{} 7$ gives $ p \equal{} 3$ as solution $ \Leftrightarrow n \equal{} 3^2 \times 287$
Case 3:$ m \equal{} 7$, then similarly $ (x \minus{} p)(x \plus{} p) \equal{} 287$ gives $ p \equal{} 143$ $ \Leftrightarrow n \equal{} 143^2 \times 7$ as solution.
Case 4:$ m \equal{} 1$, then $ (x \minus{} p)(x \plus{} p) \equal{} 2009$ gives $ 1004$ as the largest solution for $ p$ $ \Leftrightarrow n \equal{} 1004^2$

So $ n \equal{} 1004^2$ is the required answer.[/quote]
$ 2009 \equal{} 7^2 \cdot 41$

We can find all numbers satisfying the required condition.
Hint
\[ n^2 \plus{} 2009n \equal{} p^2
\]

\[ \leftrightarrow (2n)^2 \plus{} 4*2009n \plus{} (2009)^2 \equal{} (2p)^2
\]

\[ \leftrightarrow (2p \minus{} 2n \minus{} 2009)(2p \plus{} 2n \plus{} 2009) \equal{} 2009^2 \equal{} 7^4.41^2
\]
It is easy to complete our solution  :)


Explain why no grouding wire is needed in an electric drill made with a plastic case?

[hide]The grounding wire is to protect against current accidentally delivered to the case. A plastic case is an insulator and cannot carry current.[/hide]


Let $p,q,r$ be three prime numbers and $n$ a positive integer such that
\[ p^n + q^n = r^2. \]
Prove that $n=1$.

If $n=2k$ we have $(p^k)^2+(q^k)^2=r^2$, so $p^k$ or $q^k$ must be even, so $k=1$. In this case we may assume $p=2$ and we get $q^2=(r-2)(r+2)$. It's easy to dismiss this case.

This shows that n must be odd, so we may assume $n\geq 3$. We get $p+q=r,\ \frac{p^n+q^n}{p+q}=r$. This means that $p^2-pq+q^2\leq p+q$, which is false in the given conditions.

This only leaves the case $n=1$.

[quote="grobber"]If $n=2k$ we have $(p^k)^2+(q^k)^2=r^2$, so $p^k$ or $q^k$ must be even, so $k=1$. In this case we may assume $p=2$ and we get $q^2=(r-2)(r+2)$. It's easy to dismiss this case.
[/quote]
I don't think that $k=1$ in this case. I mean $p^k$ can be even, without forcing $k=1$. You obtain the equation $2^{2k} + q^{2k} = r^2$.

If [tex]2^{2k}+q^{2k}=r^2[/tex], then we have [tex]q^{2k}=(r+2^k)(r-2^k)[/tex], which can't be true.

Why can't that be true? You must prove it :D (many students did not prove this).

Let [tex]r+2^k=q^a[/tex], and [tex]r-2^k=q^b[/tex].
If [tex]b\neq0[/tex], we have [tex]q|2^k[/tex], which means [tex]q=2[/tex] and [tex]2|r[/tex], thus we have [tex]r=2[/tex] and [tex]2^{n+1}=4[/tex], which gives [tex]n=1[/tex].
If [tex]b=0[/tex], we have [tex]r=2^k+1[/tex] and then [tex]q^{2k}=2^{k+1}+1[/tex].
In order for [tex]r=2^k+1[/tex] to be prime, [tex]k[/tex] must be a power of 2.
when [tex]k\geq4[/tex], we have [tex]4|k[/tex] and thus [tex]q^{2k}\equiv2+1=3 (mod 5)[/tex], contradiction.
[tex]k=1,2[/tex] gives [tex]q^{2k}=5,9[/tex], which is also a contradiction.
Q. E. D.

I know all that seemed obvious to me when I wrote it, I just don't know why :).

Sorry! :blush:


can anyone explain the quantum theory and  how it is different from normal physics?
please provide links for :
1.theory of relativity
2.quantum mechanics
3.thermodynamics

See [url=http://en.wikipedia.org/wiki/Quantum_mechanics]here[/url] for some basic information on quantum mechanics.


The first one isn't, since you're performing the function on $y$, not $x$.Tthough the second one is, since the square root function is restricted to the nonnegative numbers, then I think that's both a function and it's one-to-one. Correct me if I'm wrong.

Generally, a function is single-valued. Thus, if $f$ is a function, then for all $x$ (in the domain), there does not exist more than one $y=f(x)$. In this case, the "function" can be $y=\sqrt x$ or $y=-\sqrt x$, i.e. for all $x\neq0$, there are two $y$'s. Therefore, $y^{2}=x$ is not a function of $x$.

[quote="Genecks"]Why is this not a function of X?

y^2 = x

I could write it as the following:

$y = \sqrt{x}$

By doing so, y is dependent on X.[/quote]

y is also equal to $-\sqrt{x}$

when you graph both of them, you can see that it is not a function by the vertical line test

you made a common mistake - you took the principal root.  when you see $\sqrt{x}$, you assume that this denotes the positive (principal) root.  if you go into complex numbers, you assume that the argument of $\sqrt{z}$ is $\frac{1}{2}\arg{z}$, but it could also be $\pi+\frac{1}{2}\arg{z}$.

at any rate, the fact that a given root is the principal root can only be assumed if the radical occurs in the original problem statement.  For example $a = 2+\sqrt{2}$ implies that $a \approx 3.414...$. and not $a \approx 0.585...$

since the radical is not given in the original problem statement ($y^{2}= x$) the principal root cannot be assumed and $y = \pm\sqrt{x}$.


topology
advanced fields
advanced fields unsolved

If $ A,B\subset \mathbb R$ and $ A$ is an open set prove that the set $ A\plus{}B$ is also open.
$ A\plus{}B\equal{}\{a\plus{}b|a\in A, b\in B\}$

I was told that the proof was very short and that I should find a way to express $ A\plus{}B$ but I don't see how to do this since you do not know anything about $ B$.

That is what I was thinking but how do you know that $ A\plus{}\{b\}$ is open?

How do you know any set is open?  You choose an arbitrary element and show that there's a ball around it.  How do you know there's a ball around it?  Because you know there's a ball around every element of $ A$ in $ A$, and this set looks exactly like $ A$, but translated.

oh wow i should have seen that.  :blush:  thanks


geometry
vector
complex numbers
geometry unsolved

[quote="Line"]Hi, I dunno how to prove a collinearity.

I've got some exercises that I'd like to learn how solve them.

Here's the first one:

In a $\triangle ABC$, let $D$ and $E$ points on $\overline{BC}$ and $\overline{AB}$ respectively. Let $F$ a point on side $\overline{AC}$ such that $\overline{EF}\parallel \overline{BC}$ and let $G$ a point on side $\overline{BC}$ such that $\overline{EG}\parallel \overline{AD}$. If $M$ and $N$ are middlepoints on $\overline{AD}$ and $\overline{BC}$, prove that $M$, $N$ and middlepoint on $\overline{FG}$ are collinear.

Thanks for your cooperation  :)[/quote]

You can see at  http://www.mathlinks.ro/Forum/viewtopic.php?t=80163

Kostas Vittas.

Here's a proof using complex numbers  :lol:
From Thales Theorem $\rightarrow$ $\frac{AE}{EB}=\frac{DG}{GB}$ and $\frac{AE}{EB}=\frac{AF}{FC}$ ,from this two $\rightarrow$ $\frac{DG}{GB}=\frac{AF}{FC}=\alpha$ , with $\alpha \in R$.
From this we can write $z_{p}=\frac{z_{F}+z_{G}}{2}=\frac{z_{A}+z_{D}+\alpha(z_{B}+z_{C})}{2(1+\alpha)}$
But $z_{M}=\frac{z_{A}+z_{D}}{2}$ and $z_{N}=\frac{z_{B}+z_{C}}{2}$ .
Therefore we have $\frac{z_{P}-z_{N}}{z_{M}-z_{N}}=\frac{1}{1+\alpha}\in R$ QED.

Thanks, but, like I said, I wanna learn 'cause I'm novice.

I don't know complex numbers but thanks for posting anyway.

Someone can give me a proof using basic geometry?

[quote="Line"]...

Someone can give me a proof using basic geometry?[/quote]

I think that in the above link I gave you, there is actually enough basic geometry, to solve your problem.

Kostas Vittas

Yeah, but, derezins uses vectors to solve the problem.

I dunno vectors.

I've been thinkin' if someone can give me another proof.

Dear Line.

You are right about the Derezins's solution, but after him has been posted a solution by my self, which is an elementary proof. Please read it carefully and I will give you the best explanations I can.

Kostas Vittas.

Exactly.

I know there are thousands ways to proof a collinearity.

What I'm lookin' for, is a proof based on angles, or using Menelaus' Theorem.

Something like that.

My apologizes for disturbing.


The sequence $ \{x_n\}$ of real numbers is defined by $ x_1\equal{}\frac{29}{10},x_{n\plus{}1}\equal{}\frac{x_n}{\sqrt{x_n^2\minus{}1}}\plus{}\sqrt{3},\forall n\geq 1$.
Find a real number $ a$ (if it exists) such that $ x_{2k\minus{}1}>a>x_{2k},\forall k\in N$.


Let $ P$ be on the unit circle and $ Q$ is a regular 2007 - gon whose vertices are on the same unit circle. If the distances from $ P$ to all vertices are respectively $ x_1,x_2,x_3,x_4,.......x_{2007}$ then $ \sum_{k \equal{} 1}^{2007}x_k^4 \equal{} ?$

[quote="mffmartinelli"]I think that I've a solution to this one.

Let O the center of the unit circle. By cosine's law on $ \Delta POX_k$ $ (1\le k \le n)$: $ x^2_k = 2 - 2\cos[x + (k - 1)\theta]$, where $ \theta = \dfrac{2\pi}{n}$ and $ x = P\hat{O}X_1$.

We can write: $ \cos[x + (k - 1)\theta] = \dfrac{e^{i[x + (k - 1)\theta]} + e^{ - i[x + (k - 1)\theta]}}{2} = \dfrac{Aw^{k - 1} + \dfrac{1}{Aw^{k - 1}}}{2}$, where $ A = e^{ix}$ and $ w = e^{i\theta}$. Notice that $ w^n = 1$.

So we can notice now:

$ \sum_{k = 1}^{n}x^4_k = \sum_{k = 1}^{n}{[2(1 - \dfrac{Aw^{k - 1}}{2} - \dfrac{1}{2Aw^{k - 1}})]^{2}}$        $ \to$

${ \sum_{k = 1}^{n}x^4_k = 4\sum_{k = 1}^{n}{(1 + \dfrac{A^2w^{2(k - 1)}}{4} + \dfrac{1}{4A^2w^{2(k - 1)}}} - Aw^{k - 1} - \dfrac{1}{Aw^{k - 1}} + \dfrac{1}{2})}$ (*).

We can show that $ \sum_{k = 1}^{n}{w^{q(k - 1)}} = \sum_{k = 1}^{n}{{(w^q)}^{k - 1}} = \dfrac{{(w^q)}^n - 1}{w^q - 1} = \dfrac{{(w^n)}^q - 1}{w^q - 1} = \dfrac{{(1)}^q - 1}{w^q - 1} = 0$ (**), for all $ q \in \mathbb{Z}$ such that $ q \neq mn$ $ (m \in \mathbb{Z})$.

Apllying (**) in (*), we'll have finally: $ \sum_{k = 1}^{n}x^4_k = 4\sum_{k = 1}^{n}\dfrac{3}{2} = 6n$.[/quote] Your solution looks great.


[quote="ch1n353ch3s54a1l"][quote="ragnarok23"]1. Give an answer and a solution.
2. Post another question.
[size=200]Problem The First[/size]
$\text{If e varies directly as g, and inversely as h, and if e}=6\text{ when g}=18\text{ and h}=12,\text{find g when e}=12\text{ and h}=3$[/quote]

[hide]Um..there's already a basics marathon.
I'll answer your question anyways.
Since e varies directly as g, and inversely as h, $\frac{eh}{g}$ is constant. $\frac{6\times12}{18}=\frac{12\times3}{g}$ and $\boxed{g=9}$. [/hide][/quote]
Ahh....yes I see....
I suppose I will post my problems there then.


The equation $ ax^3\plus{}bx^2\plus{}cx\plus{}d\equal{}0$ has three distinct solutions. How many distinct solutions does the following equation have:

$ 4(ax^3\plus{}bx^2\plus{}cx\plus{}d)(3ax\plus{}b)\equal{}(3ax^2\plus{}2bx\plus{}c)^2?$

Set $ f(x) \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$
We have f'(x)=$ 3ax^2 \plus{} 2bx \plus{} c$
f''(x)=$ 6ax \plus{} 2b$
f'''(x)=$ 6a$
Equation <=>2f(x)f''(x)=(f'(x)$ )^2$
Set $ g(x)$=2f(x)f''(x)-(f'(x)$ )^2$
We have g'(x)=2f(x)f'''(x)=$ 12a^2(x \minus{} \alpha)(x \minus{} \beta)(x \minus{} \gamma)$ with $ \alpha;\beta;\gamma$ is three root of $ f(x);\alpha < \beta < \gamma$
Then
$ g(\alpha)$=-(f'($ \alpha$)$ )^2 < 0$
$ g(\beta)$=-(f'($ \beta$)$ )^2 < 0$
then this equation has 2 root


A triangle with vertices at (1003, 0), (1004, 3), and (1005, 1) in the xy-plane is revolved
all the way around the y-axis. Find the volume of the solid thus obtained.

[quote="kunny"]You can use Pappus-Guirdin's theorem.

[hide="Answer"]$ 5020\pi$[/hide][/quote]

did you spell that right because I cant find it on google. it only comes up with one link which is this one.

is this problem doable without calculus?


In $ \Delta ABC$, $ D,E,F$ are three points on $ AB,BC,CA$ such that, $ BD \equal{} \frac {1}{3}AB$; $ CE \equal{} \frac {1}{3}BC$; and $ AF \equal{} \frac {1}{3}CA$. Join $ A,E; B,F; C,D$. Now if the points of intersections of $ AE,BF$ and $ CD$ be $ P,Q$ and $ R$ then show that $ \Delta PQR$ = $ \frac {1}{7}\Delta ABC$.

hello, can you write down the whole problem exactly please?
Sonnhard.

sorry for my mistake. it was a typing error.
i have edited the problem now.

thank you...

$ A' \equal{} \frac{4}{7} \cdot \frac{2}{3} A \minus{} \left ( \frac{2}{7} A \minus{} \frac{1}{7} \cdot \frac{1}{3} A \right ) \equal{} \frac{1}{7} A$


integration calculus limit geometry rectangle function algebra

Consider:

$ \int_{ - \infty}^{\infty}\frac {Cos(x)}{Cosh(x)}dx = \pi Sech(\pi/2)$

Why can't I just use a sequence of increasingly larger half circular contours:

[img]http://img530.imageshack.us/img530/6/contouragr5.jpg[/img]

Surely in the finite case:

$ i_1 = \int_{ - \pi}^{\pi}\frac {Cos(x)}{Cosh(x)}dx = 2\pi i r_0 - \int_0^{\pi}\frac {Cos(\pi e^{it})}{Cosh(\pi e^{it})}\pi i e^{it}dt$

$ i_2 = \int_{ - 2\pi}^{2\pi}\frac {Cos(x)}{Cosh(x)}dx = 2\pi i (r_0 + r_1) - \int_0^{\pi}\frac {Cos(2\pi e^{it})}{Cosh(2\pi e^{it})}2\pi i e^{it}dt$

$ i_3 = \int_{ - 3\pi}^{3\pi}\frac {Cos(x)}{Cosh(x)}dx = 2\pi i (r_0 + r_1 + r_2) - \int_0^{\pi}\frac {Cos(3\pi e^{it})}{Cosh(3\pi e^{it})}3\pi i e^{it}dt$

where $ r_n$ are the residues at $ (2n + 1)\pi i$ and equal to:

$ r_n = i( - 1)^{n + 1}Cosh\left[\frac {(2n + 1)\pi}{2}\right]$

Now, as I increase n without bounds, both the sum of the residues and the integral over the circular component of the contour go to infinity.  Can I not still conclude:

$ \mathop\lim\limits_{n\to\infty}i_n = \mathop\lim\limits_{n\to\infty}\left\{ - 2\pi\sum_{k = 0}^{n - 1}( - 1)^{k + 1} Cosh(\frac {(2k + 1)\pi}{2}) - \int_0^{\pi} \frac {Cos(n\pi e^{it})}{Cosh(n\pi e^{it})}n\pi i e^{it}dt\right\} \overset{?}{ = } \pi Sech(\pi/2)$

That is, does this limit exists and can it be shown to be equal to $ \pi Sech(\pi/2)$ thus showing that at least in this case, the Residue Theorem holds for a divergent sum of residues.

I've made a little progress with this:  If I use a box contour with the top on $ n\pi$ I can write the integral as:

$ \int_{ - \infty}^{\infty}\frac {Cos(x)}{Cosh(x)}dx = \frac { - 2\pi \mathop\sum\limits_{k = 0}^{n - 1}( - 1)^{k + 1}Cosh\left[\frac {(2k + 1)\pi}{2}\right]}{1 + ( - 1)^{n + 1}Cosh[n\pi]}$

that means of course:

$ - 2\pi \sum_{k = 0}^{n - 1}( - 1)^{k + 1}Cosh\left[\frac {(2k + 1)\pi}{2}\right] = \pi Sech[\pi/2]\left(1 + ( - 1)^{n + 1}Cosh(n\pi)\right)$

for all $ n$ so that I can at least get rid of the sum and write:

$ \mathop\lim\limits_{n\to\infty}\left\{\pi Sech[\pi/2]\left(1 + ( - 1)^{n + 1}Cosh[n\pi]\right) - n\pi i\int_0^{\pi}\frac {Cos(n\pi ie^{it})}{Cosh(n\pi ie^{it})}e^{it}dt\right\}$

Are you trying to evaluate the integral? If so, you can simply consider the rectangle contour with height 2pi.

Hello Soarer,

Thanks.  No, I'm trying to evaluate it with a half-circular contour.  I can do the rectangular contour.  But just suppose that wasn't an option.  He was on vacation or something.  How might I evaluate the resulting limit I posted above?  This brings up a very interesting problem:  How do I use antiderivatives to evaluate:

$ \int_{C}\frac{Cos(z)}{Cosh(z)}dz;\quad C=\{z : z=\pi e^{it}\; 0\leq t \leq \pi\}$

Now, the antiderivative of that integral can be expressed in terms of the Hypergeometric function.  Apparently though the Hypergeometric function is multivalued and $ C$ cuts through two branches so I can't just plug in the upper and lower limits to get at the answer.  I need to split up the domain similar to what we do for $ \int_{C(0;1)}\frac{dz}{z}$.  So what I'd like to know is how to split up the Hypergeometric function like we split up $ Log$ for $ 1/z$ to do that.  That is, I'd like to write it as:

$ \int_{C}\frac{Cos(z)}{Cosh(z)}dz=
 H_1(f(z))]_0^b+ H_2(f(z))]_b^{\pi}\displaystyle$ where $ H_n$ are suitable analytic branches of the Hypergeometric function.

You guys have any ideas?


trigonometry
function
geometry
geometric transformation
homothety
rectangle
number theory

Helping child with math problem
Thanks for any pointers to the formula using variables r & c
I'm not sure if this involves a trig function like the cosine.
I believe the answer to problem 1 is 144 tiles, but I don't see the relationship yet.

A Marching Strip
The King is planning a new rectangular courtyard in the palace which will be laid out using square tiles.  He has chosen some very pretty, but very inexpensive, tiles.  However, his advisor has informed him that some of the tiles will have to be a more expensive type.

You see, visiting dignitaries will always walk along a certain diagonal, from one corner of the courtyard to the opposite corner, so the tiles along the diagonal will get lots of wear and tear.  Therefore, those tiles must be more durable (and hence more expensive).  That is, every tile that contains a segment of the diagonal must be an extra strength, expensive tile. (If a tile touches the diagonal only at a corner it can be a regular tile.)

Then there's an example with a 4 by 6 courtyard (4 down, 6 across).  The diagonal goes from the bottom left hand corner to the top right hand corner.  In this case there are 8 expensive tiles.

Question 1
The king wants to know how many of these special tiles he needs to order.  At the moment, he is planning to have a 63 by 90courtyard.  That is, he will have 63 rows of tiles with 90 tiles in each row.  If he stays with this plan, how many expensive/special tiles will he need?

Question 2
The king keeps changing he mind about the dimensions of the courtyard, so it would be especially helpful if you could find a general formula for him.

Suppose the courtyard had r rows with c tiles in each row.  How many special/expensive tiles will the king need?
(The example that needed 8 special tiles had r =4 and c =6.)

This is definitely the wrong forum (it is a number theory problem but should be in high school basics, for sure).

Connect a diagonal line from one corner to another. We count the number of squares as follows: each time a vertical line is crossed, it follows that the square immediately to the left was traversed. When a horizontal line was crossed, a square immediately below was traversed. (View the line as moving left to right bottom to top, and pick the diagonal to be situated to make such a setup possible).

Now it is clear that we have passed through $m+n-l$ squares, where $l$ is the number of lattice points passed through (since in this case, both a vertical and horizontal line is traversed yielding to an overcount of one per incidence).
How do we calculate $l$? Basic number theory shows that $l=gcd(m,n)$ which follows intuitively (or, since we are in the geometry forum, there is a homothecy taking the large rectangle to a smaller rectangle with diagonal connecting a vertex of the large rectangle to the first lattice point passed through. Then we may take the projection of each passed through lattice point on to the sides of length $m,n$ respectively and notice that $l|m,n$. Furthermore, the diagonal must pass through the maximum number of lattice points possible, so $l$ is the gcd)

Either way, the result is rather trivial and is $m+n-gcd(m,n)$ which generalizes in dimensions.

Thank you for your assistance.  Sorry about placing this post in the wrong forum.

No problem, but you should also try a search of this problem using the search button above, type in something like "gcd diagonal"
if the search button doesn't work use google site search

It was discussed before.


geometry
geometric transformation
superior algebra
superior algebra unsolved

I'm using this to try to prove Cauchy's Theorem using a group action but I'm held-up on one part of defining a set $ X$:

Let $ G$ be a finite group whose order is divisible by a prime $ p$.  Define the set
\[ X \equal{} \left\{\left(g_{1},g_{2},\dots ,g_{p}\right)\in \underbrace{G\times G\times\cdots\times G}_{p\textrm{ times}}\ \Big\vert\ g_{1}g_{2}\cdots g_{p} \equal{} e\right\}.
\]
How many elements does $ X$ have?  If we can't determine this directly, how do we show that $ \left|X\right|\equiv 0\bmod p$?  If each of the $ g_{i}$ were distinct, the order would be $ kp!$ for some $ k\in\mathbb{N}$ but this is clearly not the case since elements like $ \left(e,e,\dots ,e,g_{p \minus{} 1},{g_{p \minus{} 1}}^{ \minus{} 1}\right)\in X$.

Suppose $ g_1,\ldots,g_{p\minus{}1}$ are chosen arbitrarily. How many possibilities for $ g_p$ are there? owk

So there would be one such possible $ g_{p}$? (Namely $ g_{p}\equal{}\left(g_{1}g_{2}\cdots g_{p\minus{}1}\right)^{\minus{}1}$).  Therefore, the size of $ X$ would be $ \left|G\right|^{p\minus{}1}$, right?

Thanks :).

And then consider $ \mathbb{Z}/\mathbb{Z}_p$ acting on $ X$ by cyclic translations, and determine the fixed points.


Still? Is it too easy or hard?
I want some answers about this.


probability Gauss function counting distinguishability factorial

There are 6 Red Marbles 5 Green Marbles and 4 White Marbles in a Cup. Four marbles are taken without replacement.

What is the probability that there will be at least 1 marble of each color?

I have the solution to this, but it makes absoloutely no sense to me. I was wondering if anyone had a solution that isn't tedious (ie multiplying it all out and adding.)

I don't think there is any simple way. Though multiplying it out doesn't seem too tedious - just gotta work out the probabilities of 2,1,1  1,2,1  and  1,1,2.
If you wanted to do the 'total ways of one of each' / 'total ways', the numerator would be simple to work out but the bottom is impossible without multiplying it out, as far as I know. Its sorta the same as 'how many 4 letter words can you make out of mississippi?' which has no formula or anything you have to work out the cases. I always wondered if there was a nicer way, was told there isn't :(

I don't kno if it needs to be that difficult necessarily. A somewhat simpler method would be to calculate:

The probability that you pick 4 marbles of one color (just add it up red, green, white)

Then calculate the probability that you pick only marbles from each pair of colors (red+green, red+white, green+white).

This is only 6 calculations which are fairly simple and easy to keep track of.
4 white marbles = 4/15*3/14*2/13*1/12
4 Green marbles = 5/15*4/14*3/13*2/12
4 Red marbles = 6/15*5/14*4/13*3/12

4 Red or Green = 11/15*10/14*9/13*8/12
4 Red or White = 10/15*9/14*8/13*7/12
4 Green or White = 9/15*8/14*7/13*6/12

Of course, at the end, you subtract your answer from 1.

Since there are 4 colors and you must have at least 1 of each color, then some color must be drawn twice. So either you have 2R1G1W, 2G1R1W, or 2W1R1G. In each of these cases there are 4C2*2C1 ways of drawing these combinations.

Now for the first combination, 2R1G1W, the probability of any one of those permutations of colors is (6/15)*(5/14)*(5/13)*(4/12) = a.

Similarly for 2G1R1W, the probability for each way of drawing them is (5/15)*(4/14)*(6/13)*(4/12) = b. And for 2W1R1G it equals (4/15)*(3/14)*(6/13)*(5/13) = c.

So the total probability of any one of these combinations should be (4C2*2C1)*(a+b+c) = 48/91

Gauss, your solution does make sense except for one thing.

How do you know/figure out that there are 4C2*2C1 ways of drawing any of the wanted combinations?

Well I didn't use 4C2 * 2C1 I just used 4!/2, 4! ways of arranging the 4 colours (note non-american spelling :P) over the 2 ways the reds can be swapped. Comes out to the same thing so I suppose 4C2 * 2C1 is just another way of counting it, theres heaps of different ways.

There is a formula for finding the number of four-letter sequences (if you're allowed to rearrange letters) you can make out of Mississippi. At least, I think there is.

Given a word with [i]n[/i] length and [i]k[/i] length of sequences,

let the [i]q()[/i] function make the triangluar number of what is inside the parentheses (the triangular number is (1 + 2 + ... + x), where x is the inside total

let [i]u[/i] = number of sequences

[color=#66CC99][i]q(n-k)q(k) = u[/i][/color]

I think this works.

Does that take into account repeats? As in, are you considering the s's distinguishable or indistinguishable in that formula?

You didn't make a big green smiley face! Congrats! :mrgreen:

Yeah, basically a formula would need to have a variable for each different letter counting how many of those letters there are so we could get rid of repeats. Which pretty much shows that it is impossible to find a simple formula, you need to split it up into cases.

No, it doesn't take into account repeats.

For repeats, the formula is:

[color=#33FF66][code]  q(n-k)q(k)
-------------- = u
    abcd...[/code][/color]

where a, b, c, and d, and any following, are the number of repeats of one specific letter. For instance, Mississippi would have a = 4 (i), b = 4 (s), and c = 2 (p).

I think. :mrgreen:

I'm not sure if that works either! The point is that if you have to make words using <all> the letters, you can divide by a!, b!, c!, ... (Factorials because thats the number of ways to arrange the common letters). However when we are making smaller size words, it depends on which repeats we have in the smaller words, some of these will have no repeats, some will have other combinations of repeats. Either way you need to consider all the cases separately.

How do you change your username/password?


台湾问题一直是中国人心中的“头等大事”，是非常值得讨论的话题。欢迎大陆同胞、台湾同胞、港澳同胞、华人华侨、外国友人前来讨论。
1.言论仅代表个人观点
2.对不同意见要兼容并包，要以理服人，不要进行人身攻击
最新台湾事务新闻：海峡会会长汪道涵先生逝世，未能看到台湾问题解决，深表遗憾。沉痛悼念汪道涵同志！

It had been disgussed several months ago.
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=republic++china&t=20139


Here is a nice problem for beginners:

Into a box are put ten smaller boxes.  Each of these smaller boxes is either left empty or filled with ten still smaller empty boxes.  Of all the boxes, exactly six have other boxes inserted into them.  Of all the boxes, how many remain empty?

From Math League 1981.

[hide]
Number of boxes is 1+10+5*10=61.
61-6= 55 empty boxes
[/hide] :D


Interior numbers begin in the third row of Pascal's Triangle. The sum of the interior numbers in the fourth row is 6. The sum of the interior numbers of the fifth row is 14. What is the sum of the interior numbers of the seventh row?

The sum of the interior numbers of the $ 7$th row is $ 2((7\minus{}1)C1)\plus{}2((7\minus{}1)C2)\plus{}(7\minus{}1)C3$, since $ 6C4$ and $ 6C5$ are the same as $ 6C1$ and $ 6C2$ respectively.

Evaluating,
$ 2((7\minus{}1)C1)\plus{}2((7\minus{}1)C2)\plus{}(7\minus{}1)C3$
$ \equal{}2(6)\plus{}2(15)\plus{}20$
$ \equal{}\fbox{62}$.

Isn't this $ 64\minus{}\binom{7}{0}\minus{}\binom{7}{1}\equal{}\boxed{56}$?...

The sum all of the the numbers in nth pascal row is $ 2^n$ (If the first row is considered row 0), but since this is not the case the formula becomes $ 2^{n\minus{}1}$. Since the exterior numbers are always 1, we get:
\[ 2^6\minus{}2\equal{}\boxed{62}
\]

Two things AIME: first, since the pascal triangle starts with $ 2^0$, you subtract $ 1$, so you do combinations with $ n\minus{}1$ instead of $ n$, and notice that interior numbers are not $ nC0$ and $ nc1$, but $ nC0$ and $ ncn$.


Let a be greatest root of 4x^3 - 2x^2 - 15x + 9,
and b be greatest root of 12x^3 + 6x^2 - 7x + 1.

Show that a^2 + 3b^2 = 4.

[quote="Jean-Louis"]Let $ a$ be the greatest root of $ 4x^{3}\minus{}2x^{2}\minus{}15x\plus{}9 \equal{} 0$ and let $ b$ be the greatest root of $ 12x^{3}\plus{}6x^{2}\minus{}7x\plus{}1 \equal{} 0$. Show that $ a^{2}\plus{}3b^{2}\equal{} 4$.[/quote]
[size=117][color=darkblue][b]Where keep you the dollars ?![/b][/color][/size] Wrote you wrong at least a coefficient ?


When there is an unread topic, I have the little yellow thing to the right of the name. I click that, then it goes to the first post. After a few seconds, it goes to the last post. The problem is, it doesn't say which posts are unread. In the bottom corner, it is white. Is this the same for everyone? Shouldn't it be fixed?

If I click on the last page, then scroll down, it is yellow.

Elaborate...

If you mean the yellow thingy that says there's an unread post, it will stay yellow until you read all the posts. You don't really have to read them if you don't want, it's not like AoPS times you to judge whether or not you read the posts.

No, I mean that when I click "View latest posts", it shows that none of the posts are unread, but some actually are.

For example, if I post something, and then someone posts something while I am, say checking my email, and then I come back and click on the yellow thing beside the topic name, it doesn't show any posts that are unread. I see the new post(s), but it says it is read.

[quote="myyellowducky82"]If the posts are older (about a week) then it will turn white.[/quote]

Actually, just a bit over a day :wink:


Every other question makes sense on the worksheet except # 7. All questions above are needed for someone to answer number 7. which i have, but #7 does not make sense?

I think he meant to post a reply in another thread but started a new topic.


[quote="jli"]A really bad pitcher throws 64 balls in a row.  how many home runs are scored?[/quote]

[hide="the wording says"]
The answer is 0.  4 walks is still not a home run.[/hide]

[hide="what it probably means"]
64 balls in a row means 16 walks.  First 3 get on, 13 left.  After each of those 13 are walked, 13 runs come in.[/hide]

@CircleSquared: I'm not certain, but there are no rules that say that a batter can only go up once per inning.  So basically, the number of players on a team isn't needed for this problem.

Actually, you don't know how many runs came in.  Maybe someone was taking a lead off second base, and the pitcher got him out. :D  :D

No home runs are scored, and you don't know how many runs did come in, it says that the pitcher threw 64 balls, it didn't say what happened before or after that.

How do you know that it doesn't mean balls as in baseballs?

(whole different story...)

Jorian

dude who thehell cares about baseball

Its has less action then golf, the sport sucks.

just do the problem as it is presented instead of alternate versions that are obvioulsy not meant to be.

[quote="alanchou"]Maybe it's a better idea to do the problem and not suggest cases that are almost ridiculous.

@usaha: Think before you write.  That was probably offensive to those who enjoy baseball.[/quote]

just saying..kind of like how the world series ratings were an all time low last year..
wait who won again?

Almost all the posts here (incl mine) are spam. The problem wasn't that great, it should have said like 64 balls, 0 strikes for clarifications, and i think "homeruns" was meant to be used. I remember posting a problem some time ago about the amount of dirt in and x*y*z area. That's a trick question and I feel it's not really needed in the MathCounts forum if the question's not that great (this one's not that hard as where that dirt one would be hard if you multiplied those things). Just some things to think about... (I bet you guys will vote this as spam, but please do read what I wrote...
Jorian

[quote="jli"]A really bad pitcher throws 64 balls in a row.  how many home runs are scored?[/quote]

I don't really understand baseball or football, so I don't know about throwing, pitching, or making home runs.  :huh:


Ring Theory
topology
superior algebra
superior algebra unsolved

if the ring R be a UFD and S $ %Error. "display" is a bad command.
\subseteq$ R is a closed set with the action multiplication.and S has the member unity.then the fraction ring $ %Error. "display" is a bad command.
\ s^{-1}R$,is a UFD.(for the first condition of UFD we should show that every member $ %Error. "display" is a bad command.
\dfrac{a}^{1}$ in     $ %Error. "display" is a bad command.
\ s^{-1}R$    decompose to the irreducible    members.and for the second condition of UFD  we should show that every irreducible member in $ %Error. "display" is a bad command.
\ s^{-1} R$ is prime .)

Suppose $ R$ is a $ UFD$, for any $ a\in R$, there exist unit $ u$ and
irreducible elements (primes) $ p_i$'s such that $ a \equal{} u\prod_i p_i$ . For $ s\in S$, $ \frac {a}{s} \equal{} \frac {u}{s}\prod_i \frac {p_i}{1}$
 $ \frac {u}{s}$ is a unit and $ \frac {p_i}{1}$ is a prime or a unit of $ S^{ \minus{} 1}A$ , for each $ i$. Therefore, every element of $ S^{ \minus{} 1}R$
has a factorization into primes (irreducible elements). Suppose that $ \frac {q_j}{s_j}$'s $ \frac {r_k}{t_k}$'s are irreducible elements (and so are primeswhy?) and $ v, w$ are units in $ S^{ \minus{} 1}R$ satisfying
$ v\prod_{j \equal{} 1}^f \frac {q_j}{s_j} \equal{} w\prod_{k \equal{} 1}^g\frac {r_k}{t_k}$ and $ f \le g$ . We may write $ v \equal{} \frac {v_1}{v_2}$ , $ w \equal{} \frac {w_1}{w_2}$
for $ v_i$'s, $ w_i$'s in $ S$.
It follows that $ t\prod_j q_j \equal{} s\prod_k r_k$ where $ s \equal{} w_1v_2\prod_j s_j$ $ t \equal{} v_1w_2\prod_k t_k$.
Note that all $ q_j$'s and $ r_k$'s are primes in $ R$. In addition, $ (qj)\cap S \equal{} \empty$so $ q_j\not{|} s$ and hence $ q_j$ must divide
$ r_{k_j}$ for some $ k_j$ . Reordering if necessary, we may assume $ q_j$ divides $ r_j$ and hence $ q_j$
and $ r_j$ are associates as $ r_j$ is irreducible. We also have $ f \equal{} g$; otherwise, there would
exist a $ k > f$ such that $ r_k$ divides $ t$. Thus the factorization into irreducible elements
must be unique.


Given a positive rational number q = a/b and an odd positive integer n, find a polynomial, with integer coefficients written in a simple closed form, that has q^(1/n) + q^(-1/n) as a root.

Tchebyshev's polynomials point an answer for us: \[ab(x+\sqrt{x^{2}-4})^{n}+ab(x-\sqrt{x^{2}-4})^{n}-2^{n}(a^{2}+b^{2}).\]


Show that $a_{1}+1\equiv a_{2}+1 \pmod{8}$

I don't get this part of the proof:

$\frac{a_{1}+1-a_{2}-1}{8}= \frac{(a_{1}+1)-(a_{2}+1)}{8}.$

Since the last fraction is an integer, $a_{1}+1\equiv a_{2}+1 \pmod{8}.$

How is the last fraction is an integer?

I think the idea is to show that addition is valid under modulus arithmetic.
He wants proof that we can add any number (like 1) to a side of an already true modulus equation.
I.e. show that if $a_{1}\equiv a_{2}\pmod n$, then $a_{1}+k \equiv a_{2}+k \pmod n$

[quote="jeez123"]yes... sorry, I forgot to include those.

Anyways, I still don't get the proof... how the fraction proved that it is true.[/quote]
[hide] Since $a_{1}\equiv a_{2}\bmod 8$, we have that $\frac{a_{1}-a_{2}}{8}$ is an integer.  Thus, $\frac{a_{1}+1-a_{2}-1}{8}$ is an integer.  Notice, that this is $\frac{a_{1}+1-(a_{2}+1)}{8}$.  So that is an integer.  Thus, we have that $a_{2}+1\equiv a_{2}+1\bmod 8$ [/hide]