Find static distributions of n charges (n=4 and n=5) such that each charge is in equilibrium.
  (geometry, lengths, and magnitudes)

do u mean static charges and stable equilibrium then it is not possible see [url=http://en.wikipedia.org/wiki/Earnshaw's_theorem]here[/url]

Probably my question is not clear enough. "static" just wants to mention that the system is not moving. And "equilibrium" means the net  Column's force on each charge is 0.

would like to give a thought on it if i have time if u meant unstable equilibrium for stable equilibrium that's impossible see earnshaw's theorem

take q on three vertices of equilateral triangle.

and -q/sqrt(3) at the centre.

for n=4 we are done...

but its unstable equilibrium...

for n=5,
take q on 4 sides of a square,
-q/sqrt(2) at the centre...

but unstable equilibrium..

i did a small mistake for n=5,

correction:
the central charge should be

-q(1/sqrt(2) + 1/4)

I hope this is the solution you were looking for.


calculus
integration
algebra unsolved
algebra

Find the greatest positive integer $n$ such that the systems of equations
$(x+1)^2+y_1^2 = (x+2)^2+y_2^2 = ... = (x+n)^2+y_n^2$
has an integral solution $(x,y_1,y_2,...,y_n)$

This is actually a Vietnamese problem, I believe. Very nice problem by the way.

Working modulo some small number should help to get some restrictions on n.

for n=3,it's obvious that (x,y1,y2,y3)=(2,4,3,0) is suitable.
if n>=4,let S=(x+1)*(x+1)+y1*y1=.......
we know that m*m=0,4,1(mod8),so:
(1)x=0(mod4),if y1 is odd,then S=2(mod8),but(x+2)*(x+2)=4(mod8),so y2*y2=6(mod8),which comes to a contrariety.if y1=0(mod4),then S=1(mod8),but(x+2)*(x+2)=4(mod8),so y2*y2=5(mod8),which comes to a contrariety.if y1=2(mod4),then S=5(mod8),but(x+4)*(x+4)=0(mod8),so y4*y4=5(mod8),which comes to a contrariety.
(2)x=2(mod4),if y1 is odd,then S=2(mod8),but(x+2)*(x+2)=0(mod8),so y2*y2=2(mod8),which comes to a contrariety.if y1=0(mod4),then S=1(mod8),but(x+4)*(x+4)=4(mod8),so y4*y4=5(mod8),which comes to a contrariety.if y1=2(mod4),then S=5(mod8),but(x+2)*(x+2)=0(mod8),so y2*y2=5(mod8),which comes to a contrariety.
(3)x=1(mod4),if y1 is odd,then S=5(mod8),but(x+3)*(x+3)=0(mod8),so y3*y3=5(mod8),which comes to a contrariety.if y1=0(mod4),then S=4(mod8),but(x+4)*(x+4)=1(mod8),so y4*y4=3(mod8),which comes to a contrariety.if y1=2(mod4),then S=0(mod8),but(x+2)*(x+2)=1(mod8),so y2*y2=7(mod8),which comes to a contrariety.

(4)x=3(mod4),if y1 is odd,then S=1(mod8),but(x+3)*(x+3)=4(mod8),so y3*y3=7(mod8),which comes to a contrariety.if y1=0(mod4),then S=0(mod8),but(x+2)*(x+2)=1(mod8),so y2*y2=7(mod8),which comes to a contrariety.if y1=2(mod4),then S=4(mod8),but(x+2)*(x+2)=1(mod8),so y2*y2=3(mod8),which comes to a contrariety.
so,n=3 is the greatest :arrow:


[hide]
$x+z=2y$
$\frac{x+z}{y}=n$
$n=2$
${n}\subseteq{\mathbb{Z}}$[/hide]

nvm my proof sucks.

[quote="deej21"]Getting started is not a place for proofs, even though this one indeed is really easy.
[hide]
$\frac{x+z}2=y$
[/hide][/quote]This proof problem is very elementary and so it is OK. About proofs, this is as far as it can go in Getting Started as set by MCrawford (admin) in the guidelines.

[quote="Inspired By Nature"]I knew it was too easy  :) . I'll post something harder next time.[/quote]No, I don't think you understood what I was saying. If you want to post harder next time, you should post them in Intermediate Topics forum. Only "extremely elementary" proofs are allowed in Getting Started, like the one above.

$x+z=2x+2=2(x+1)=2y$

if you mean that
$x<y<z$

As x, y, and z are consecutive integers, $x = y - 1$, $z = y + 1$. We are trying to prove $\frac{x+z}{y}$ is an integer. Use substitution to get $y - 1 + y + 1$ or $2y$ as the numerator. Since $\frac{2y}{y}$ results in 2 and 2 is an integer, x+z is divisible by y.
$Q.E.D$

No, you have to specifiy that x<y<z. Otherwise, x=7, y=6, z=8, and 15 is not divisible by 6.

Edit: Oops, didn't see DGS's post.

By definition:
$x = y-1$
$z = y+1$
Then:
$(y-1) + (y+1) = 2y$
$\frac{2y}{y} = 2$
As $2$ is an integer, $x+z$ is divisible by $y$ unless $y = 0$.
QED

Without loss of generality, let us assume $x<y<z$ We can rewrite y and z as $x+1$ and $x+2$, respectively. Thus $x+z=2x+2$ and $y=x+1$. Therefore, $\frac{2x+2}{x+1}=2,$ and $y|x+2$.
$\boxed{QED}$

[quote="surge"]$x|y+z$.
Can someone explain what this notation denotes (or link me to mathworld)?[/quote]It means "x divides y+z." So for something like $3|12$, it means 3 divides 12.

[quote="surge"]"Divides", as in, divides evenly? So, the result of $x|y+z$ will always be a subset of
$\mathbb{Z}$.

thanks.[/quote]

Well, technically, though one would usually use the word element instead of subset if it's a single number...

[quote="cincodemayo5590"][quote="José"][quote="LordoftheMorons"]QED[/quote]

What's QED?????[/quote]

quod e·rat de·mon·stran·dum /kwäd Šert ‹demnŠsträndm/ (abbr.: QED)
* used to convey that a fact or situation demonstrates the truth of one's theory or claim, esp. to mark the conclusion of a formal proof.

(yay copy/paste from dictionary :D )--- baaically it comes at the end of a proof and means that you've proven what you were trying to prove.[/quote]

Got it, thanks!

I know some people like to use a black square or a hollow sqaure at the end of proofs instead of QED.

If it's an informal proof you can end it almost how you want...like The End, or as I once did, and the question lived happily ever after... Darij Grinberg has said the German team sometimes end their IMO proofs with PWNED.

[quote="Centy"]I know some people like to use a black square or a hollow sqaure at the end of proofs instead of QED.

If it's an informal proof you can end it almost how you want...like The End, or as I once did, and the question lived happily ever after... Darij Grinberg has said the German team sometimes end their IMO proofs with [b]PWNED.[/b][/quote]
 :rotfl:  :rotfl:  :rotfl: ..... that's HILARIOUS! I think I might start doing that, hope they don't mind  :D

[quote="mathnerd314"]Some people write WWWWW, "which was what we wanted". That notation, however, is generally avoided by the AAAAA, the "American Association for Avoiding Annoying Alliteration".

Other people think QED stands for "quite entertainingly deceptive".[/quote]

Your joke is flawed. It's only alliteration for initial [b]constanant[/b] sounds  ;)

[quote="Inspired By Nature"]x, y, & z are consecutive integers. prove that x+z is divisible by y.[/quote]

[hide]
x=y-1
z=y+1
x+z=y-1+y+1=2y.
x+z is divisible by y.

[/hide]

Just put "This what we wanted to prove, so the problem is finished" instead of inventing new endings.......

[quote="10000th User"][quote="mna851"]It's trivial to generalize this problem as well.  Let $\ x,y,z$ be three consecutive terms of an increasing arithmetic sequence.  The sum of the first and third terms will be divisible by the second.[/quote]That's good, now we have a new challenge (for beginners please):

Write a one-line proof that the sum of the first and third terms is divisible by the second term of an arithmetic sequence. :)[/quote]
[hide]$\frac{a+(a+2n)}{a+n}=\frac{2(a+n)}{a+n}=2$...does this work[/hide]

[hide]We have terms $x$ , $y$ and $z$. We know that $x=n$ , $y=n+1$ and $z=n+2$ . If we add $x$ and $z$ , we get $x+z=2n+2$ . If we divide by $y$ , we get $(2n+2)/(n+1)=2$ .[/hide]

[quote="Centy"]Darij Grinberg has said the German team sometimes end their IMO proofs with PWNED.[/quote]
I'm in the team and end proofs with "q.e.tee." everytime (PWNED is just used by darij).  :D

x+z=2y+1-1
x+z=2y
(x+z)/y=2, which is an integer

[hide]Let the common difference be d.  Assuming that x<y<z, then this means that y=x+d, and z=x+2d.

So x+z=2x+2d, and y=x+d.  And this implies that x+z is a multiple of y.  Hence we have proved what we need to prove.

QED[/hide]

now prove it by geometry! and I don't understand why everyone post the same answer again and again...? waste of time  :(

[quote="Centy"]http://www.mathlinks.ro/Forum/viewtopic.php?t=84559

A correct solution written with elements of L33tsp33k. After reading that I was tempted to end one of my proofs with $\mathsf{OWNAGE!}$.[/quote]

Sounds like something that could fit into the Games & Fun Factory Board: A marathon of proofs written in l33tsp34k.

Back on topic:
How about a proof without words?


Two different natural numbers are selected from the set $ \{1, 2, 3, \ldots , 6\}$. What is the probability that the greatest common factor of these two numbers is one? Express your answer as a common fraction.

We can count the ones that have a common factor other than 1. $ \{2,4\},\{2,6\},\{4,6\},\{3,6\}.$ The total number of pairs is $ \binom{6}{2}\equal{}15.$ The probability is thus $ 1\minus{}\frac{4}{15}\equal{}\boxed{\frac{11}{15}}.$


algebra
polynomial
quadratics
irrational number
complex numbers

find the minimum value for $ n$ (an integer) such that the given polynomial below can be factored in-to 2 polynomials with integer coefficients.
\[ p(x) \equal{} x^4 \minus{} 2(n \plus{} 2009)x^2 \plus{} (n \minus{} 2009)^2
\]

[hide]Here's my approach:
We see that p(x) is a quadratic function. Because its leading coefficient is 1, we easily see that p(x) can be factorized into integer coefficient if (n+2009)^2-(n-2009)^2=k^2, which means:4n.2009=k^2 (1). Therefore, we easily see that n=0 is the smallest integer that satisfy (1). Try to plug n=0 into p(x), we have:x^4-2.2009x^2+2009^2=(x^2-2009)^2=(x^2-2009)(x^2-2009)
Remark: p(x) can be factorized into two distinct polynomial with integer coefficients :wink: . In that case, n=2009 is the smallest integer(since 2009 is a prime). :)[/hide]

[quote="ghjk"]Because its leading coefficient is 1, we easily see that p(x) can be factorized into integer coefficient if (n+2009)^2-(n-2009)^2=k^2, which means:4n.2009=k^2 (1). [/quote]
Why is this so?  By the way, I think you meant $ \deg f(x^2)\equal{}2$.

[quote="mathwizarddude"][quote="ghjk"]Because its leading coefficient is 1, we easily see that p(x) can be factorized into integer coefficient if (n+2009)^2-(n-2009)^2=k^2, which means:4n.2009=k^2 (1). [/quote]
Why is this so?  By the way, I think you meant $ \deg f(x^2) \equal{} 2$.[/quote]
Yes, that's what I meant.
Considering that the condition of having an integer solutions of a degf(x^2)=2 is the delta must be a perfect square(In this case, it doesn't need to have a positive integer solutions since we can factorize it into: (x^2-a)(x^2-b) with some positive integers a and b if delta is a perfect square). If delta is not a perfect square, we can easily infer that two polynomials q(x) and g(x)(q(x).g(x)=f(x)) can't have an integer coefficient because q(x) and g(x) is in this form: (x^2-irrational number).(x^2-irrational number)(we can replace - by +) and the irrational number is, in fact, a coefficient of x^0, which is not an integer.

[quote="mathwizarddude"][quote="ghjk"]Considering that the condition of having an integer solutions of a degf(x^2)=2 is the delta must be a perfect square(In this case, it doesn't need to have a positive integer solutions since we can factorize it into: (x^2-a)(x^2-b) with some positive integers a and b if delta is a perfect square).[/quote]What do you mean by delta?[/quote]
I meant delta=(n+2009)^2-(n-2009)^2=4.2009.n. And I should state it more clearly that because the leading coefficient is 1, we will have two integer roots if and only if our "delta" is a perfect squared number. But do we count the complex numbers as a polynomial with integer coefficient? If it is counted, then n can be arbitrary value. So, I assume we just count polynomial that has only rational roots.

Oh, I see; you are taking the discriminant of the quadratic formula.


inequalities
trigonometry
geometry
angle bisector
geometry unsolved

In triangle $ ABC$, let the angle bisector of $ A$ meet $ BC$ at $ A'$. The projections of $ A'$ on $ AB,AC$ are $ P,Q$, respectively. Let $ a'$ be the length of $ PQ$. Define $ b',c'$ similarly. Let $ a,b,c$ be the sidelengths of the triangle $ ABC$.

Prove that $ (a\plus{}b\plus{}c)^3\geq 64a'b'c'$

$ \left\|\begin{array}{c}
a'\equal{}PQ\equal{}AA'\sin A\equal{}\frac {2bc\cos\frac A2\sin A}{b\plus{}c}\equal{}\frac {4S\cos\frac A2}{b\plus{}c}\\\\
\prod\cos \frac A2\equal{}\frac {p}{4R}\ \ ;\ \ \prod (b\plus{}c)\ge 8abc\ \ ;\ \ R\ge 2r\end{array}\right\|$ $ \implies$ $ a'b'c'\equal{}\frac {16p^4r^3}{R\prod (b\plus{}c)}$ .

$ \frac {(a\plus{}b\plus{}c)^3}{64a'b'c'}\equal{}\frac {R}{128pr^3}\cdot\prod(b\plus{}c)\ge$ $ \frac {R\cdot abc}{16pr^3}\equal{}$ $ \frac {R\cdot Rpr}{4pr^3}\equal{}$ $ \left(\frac {R}{2r}\right)^2\ge 1$ .


calculus
derivative
function
inequalities
complex analysis
limit
algebra

let $U$ be a connex bounded open subset of $\mathbb{C}$, $f: U\to U$ an anaytic function and $a \in U$ a fixed point of $f$. Let's note $f_n(z)=f(f(..f(z)))$. Prove that:
1/ $|f'(a)| \leq 1$
2/$|f'(a)| = 1$ if, and only if, $f$ is bijective
3/if $|f'(a)|<1$, the sequence $(f_n)$ converges towards a constant function, uniformly on any compact.

Proof:  Since $U$ is convex, it is simply connected, by Riemann's theorem, it is biholomorphic to the unit disk $\Delta$. Let the biholomorphism from $U$ to $\Delta$ be $h$ and $F=hfh^{-1}$. $F$ is a holomorphism from $\Delta$ into itself fixing some point $b=h(a)$. By the classical Schwarz Lemma we can conclude 1) and 2).
But the following proof doesn't require $U$ is convex.
1) Using Cauchy's inequality, for any holomorphic function $g$ from $U$ to $U$, noticing that $U$ is bounded, we can conclude that $|g'(a)|\leq M$ where $M$ is independent of $g$. Then for $f^{(n)}=f\circ f\circ\cdots\circ f$, we have $|f'(a)|^n\leq M$, so $|f'(a)|\leq 1$. Note: $a$ is fixed.
2) If $f$ is bijective, $f^{-1}$ must also be holomorphic, considering $f$ and $f^{-1}$ leads to $|f'(a)|=1$.

Indeed, the other hand of 2) is also true without the assumption that $U$ is convex, but the proof I know is somewhat long.

3) First prove that there is some neighborhood $V$ of $a$ on which $f^{(n)}$ converges locally uniformly to the constant $a$. Then collect all such $V\subset U$ and use Zorn's Lemma to argue that the maximal element is acturally $U$.
There may be some difficulties. But in this proof you will find that the convexity of $U$ is useless.

sorry, I made a typo : $U$ is only supposed to be open bounded and connected. (not convex).

A:  Find an open neighborhood $B$ of $a$ such that for any holomorphism $h$ from $D$ into itself we have $h(B)\subseteq B$. For this you can take $B$ as an open ball centered at $a$ under the Poincare matric.
B:  Take an increasing subsequence ${n_k}$ of positive integers such that $\lim_{k\to\infty}f'(a)^{n_k}=1$. Then you can prove that $f_{n_k}$ converges locally uniformly to identity on $B$, by uniqueness, $f_{n_k}$ converges locally uniformly to identity on $D$. Here we use the fact that ${f_{n_k}}$ is a normalty. Hence $f$ is injective.
Consequently, $f^{-1}$ is holomorphic.
The following lemma will be usefull in proving that $f$ is surjective.
Lemma: Let $M$, $N$ be two domains in $\mathbb{C}$. $f_k$ are holomorphic open maps from $M$ into $N$ for $k\in\mathbb{N}$ such that $f_k$ converges locally uniformly to $f$. Assume that the Jacobians of $f_k$ don't vanish anywhere and that the Jacobian of $f$ doesn't vanish identically. Then for every $a$, there exists $n_0$ such that $f(a)\in f_n(M)$ for all $n$ bigger than $n_0$.

Proof of Lemma: By Hurwitz theorem, the Jacobian of $f$ doesn't vanish anywhere. so $f$ is locally injective. Take a relatively compact neighborhood $U$ of $a$ such that on $\overline U$ $f(x)=f(a)$ implies $x=a$.
And take $B$ a connected open neighborhood of $f(a)$. Now we conclude by contradiction.
We may assume that $f(a)\notin f_n(\overline U)$ by taking subsequence.
Let $I_n=f_n(U)\cap B$, $J_n=(N-\overline{f_n(U)})\cap B$, $K_n=\partial{f_n(U)}\cap B$.
You can prove that, for $n$ big enough, $I_n$ and $J_n$ are open and nonempty but $K_n$ is empty.
However, $B=I_n\cap J_n\cap K_n$, a contradiction to the connectness of $B$.

In the proof of 3) we will also use Poincare metric for it is such a good metric.

It is so tough that there might be some errors. Do you know some simpler proof?

I have an easy and understandable proof for the first part, since the cyqwhu's proof i couldn't get it because there are a lot of stuff that i havent seen yet:

$Proof$ : Its not hard to see that $f '_{n}(a)=(f '(a))^{n}$ So let $g_{n}(z)=\frac{f_{n}(z)-a}{z-a}$ , if $z\ne a$ and $g_{n}(a)=f '_{n}(a)$ (remember $a$ is a fixed point of $f$ , hence $f_{n}(a)=a$ )  , then $g_{n}$ is a holomorphic function on $U$.    Let $b \in \partial U$ .   Define \[\varepsilon = \min_{z\in \partial U}|z-a|\] , which exists because of $U$ is compact.   Hence

\[\overline{\lim_{z\to b}}|g_{n}(z)|=\overline{\lim_{z\to b}}\frac{|f_{n}(z)-a|}{|z-a|}\leq \frac{2M}{\varepsilon}\] , where $M= \sup_{z\in U}|f(z)|$ ($U$ is bounded).

Therefore, by The Maximum Principle , $|g_{n}(z)| \leq \frac{2M}{\varepsilon}$ for each $z \in U$.   In particular if $z=a$ then $|f'_{n}(a)| \leq \frac{2M}{\varepsilon}$ which is equivalent to $|f'(a)| \leq (\frac{2M}{\varepsilon})^{\frac{1}{n}}$.     Since $M$ and $\varepsilon$ don't depend on $n$, then if $n \to \infty$ you can conclude what you want.


calculus
integration
algebra
polynomial
real analysis
real analysis unsolved

If $ \int_0^{2\pi } {f\left( x \right)dx} \equal{} 0$ then $ \int_0^{2\pi } {f\left( nx \right)P_{n \minus{} 1} \left( x \right)dx} \equal{} 0$ for all polynomial $ P_{n \minus{} 1}(x)$ whose degree $ \leqq n \minus{} 1$.


Haha, I'm wondering here.  I guess I like alternative rock...but I also have some scary hippie music (Joni Mitchell.... um, have you guys heard of the horrible cover of "They paved paradise and put up a parking lot"? well..she did that first!  And did it a whole lot better in the process)  and some doesn't-make-sense music, like Tori Amos.  Um, has anyone here figured out her music?  But at least you can count on her for some bangin' piano playing.  (She went to Julliard, but she was kicked out for "unique style")  Haha, I like AFI (aha!), Linkin' Park, Slipknot, Blink 182, Ani DiFranco (:-D), Deftones, Dropkick Murphys, Selena (Tu conoces..."Baila, baila esta cumbia!"), System of a Down, Tool, and Yeah Yeah Yeahs (They don't love you like I love you...MAAHHHHHAHHHHHAHHHHAHHHAHHHPs, WAIT, they don't love you like I love you, that's the only song I know from them. yup).  And then,  when I'm ghetto, I like Lil' Kim and Luda (Stop, drop, KABOOM...:-D)  Now that I officially scared everyone, and make less sense than the infamous Tori Amos, I'll stop.  Anyway..

What's (or what're) favorite music and bands?

(I had a whole bunch of choices, but apparently, there is a limit on how many options you can put on a poll.  Grr.)

Take one off and put an other, and examples of your "other choices" in parenthenses (i cant spell, whatever). As for fave bands/singers/whatnot- right now I am totally digging Michael Tolcher. He's a new guy coming up, and he's totally hot (musically, you guys, musically). He's out of the Southeast (Georgia I think, anyways, a friend from there told me about him), and he plays happy music- its not all angsty, but its not stupid. You can listen to (a lot) of his stuff on his website, which is, of course!, http://www.michaeltolcher.com If that doesn't make you an instant fan, go to one of his concerts. VERY entertaining performer. Quite funny and down to earth. If you are a fan of Maroon 5, Gavin Degraw, and/or John Mayer, you will definitely like this guy.

Hm.  Well, I did have an "other" option, but I deleted it.  But um, I can't edit the poll now, but hey...just pretend there is an other option.

Tori Amos, Chantal Kreviazuk, and Coldplay, in that order. These people are nothing short of brilliant.

Oh wait, I forgot... Da Vinci's Notebook! (and Weird Al to a lesser degree)

you've forget "trance and techno" in the vote tab. am l the one who like that music (DJ dean, ...) ?

Er, could you define "alternative rock?"  I'm unfamiliar with the term.

At any rate, the music to which I listen primarily consists of music from the 50s through the early 80s.  For the most part, I will generally listen to anything written before I was born (I'm 15 now).  I know this choice of music is both ironic and rather eccentric, but I simply can't help myself.  The particularly ironic thing is, whenever I tune the car radio to the Oldies station, my mother is of the opinion that such music is noisy.  The Beach Boys, Stevie Wonder, Elton John, and the Jackson 5 are my favorite artists (I'm sure I've got more, but I can't think of any at the moment).  Like Bob Seger says: "I like that old time rock'n'roll; that kind of music just soothes the soul..."

Alt rock!!!  I love that new song by modest mouse, so cool.

[b][i][u]GUSTER IS THE BEST BAND TO BE EVER IN EXISTENCE!!![/u][/i][/b]

I'm so pumped because they're coming to my city in September; I was crying with happiness.

Other great bands: The Strokes, Switchfoot, Coldplay, Goo Goo Dolls, and anything from DDR (especially Max 300)

I basically listen to any type of music as long as it's not Country.

Well, you are absolutely in the wrong, the best band is by far Queen. But my favourite is obviously Pink Floyd, havent you listen The Wall or Wish You were here?
  I think that every mathematician should do 2 things:
1) Love number theory
2)  Listen to Pink Floyd

Alternative of course.

I like Linkin Park, Switchfoot, and just really all of it.

The new Modest Mouse song (Float On) is really awesome, just like whoever said... (I'm too lazy to go back and look)

I also like Good Charlotte even though they are technically Punk instead of Alternative

well i prefer rock.heavy metal and alternative suit me .my alltime fave i guns n roses (no matter the amt of abusives they use in their songs).and then metallica is good.arcturus (rather unknown in my place) is good (but i believe it's anti-christ so i have stopped listening to it).linkin park is rather cool.aerosmith,robbie williams are also good.and i like saying it,i hate "slut pop"

I second Robinhood3000I like the oldies.  That music is really neat.  My parents listen to it so I have listened to it all my life.

I can't believe no one has even metioned Brittney Spears!!!!!!!!!!!!!!!!!!!

j/k

My two favorite bands/singers are actually Hoobastank and Faith Hill.

Shall I dare to say COUNTRY! I love country (Tim Mcgraw is the hottest dad Ive ever seen!) I like Alt. rock though too. I like modest mouse they are pretty cool. I like jet. They are so bold to sing that song Cold Hard *****. I like the vines they are pretty neat. Hoobastank is the best. I love The Reason. I like some hip hop not much some of Ushers stuff is pretty neat sounding. I like classical Vivaldi winter is the best. Queen is the band of all bands. One could compare The Darkness to Queen. I like Boheimian Rhapsody thats a cool song. I like oldies too.

Somehow my mind was off when I wrote my post...no one's said Rush yet.  Hey, they're credited with creating 'math rock', whatever that means.  But weird terms aside, they are awesome.

For me, right now it's alternative rock and rap.  Lately I've been listening to Zebrahead, Slipknot, D12, and the Beastie Boys though.  I can stand most music except for R&B, country, and classical, and worst of all, rap mixed with R&B.

[quote="MathFiend"]Does anyone kno the song "Save a Horse, Ride a Cowboy" ??? Who is the artist? Where can I find the lyrics? Like everyone is always singing that song, but I have never heard it. lol.[/quote]

Artist:Big & Rich

Title:Save a Horse, Ride a Cowboy

Album: Horse of a Different Color

Genre: Country

Lyrics:

(intro)
DUM-DE-DE-DUM, DE-DE-DUM-DE-DE-DUM,
DE-DAA-DAAAAA
DUM-DE-DE-DUM, DE-DE-DUM-DE-DE-DUM,
DE-DAA-DAA-DAA-DAA-DAAAA!

Well, I walk into the room
Passing out hundred dollar bills
And it kills and it thrills like the horns on my Silverado grill
And I buy the bar a double round of crown
And everybody's getting down
An' this town ain't never gonna be the same.

(Chorus:)
Cause I saddle up my horse
and I ride into the city
I make a lot of noise
Cause the girls
They are so pretty
Riding up and down Broadway
on my old stud Leroy
And the girls say
Save a horse, ride a cowboy.
Everybody says
Save a horse, Ride a cowboy

Well I don't give a dang about nothing
I'm singing and Bling- Blanging
While the girls are drinking
Long necks down!
And I wouldn't trade ol' Leroy
or my Chevrolet for your Escalade
Or your freak parade
I'm the only John Wayne left in this town

And I saddle up my horse
and I ride into the city
I make a lot of noise
Cause the girls
They are so pretty
Riding up and down Broadway
on my old stud Leroy
And the girls say
Save a horse, ride a cowboy.
Everybody says
Save a horse, Ride a cowboy

(Spoken:)
I'm a thourough-bred
that's what she said
in the back of my truck bed
As I was gettin' buzzed on suds
Out on some back country road.
We where flying high
Fining, whine, having ourselves a big and rich time
And I was going, just about as far as she'd let me go.
But her evaluation
of my cowboy reputation
Had me begging for slavation
all night long
So I took her out giggin frogs
Introduced her to my old bird dog
And sang her every Wilie Nelson song I could think of

And we made love
And I saddled up my horse
and I ride into the city
I make a lot of noise
Cause the girls
They are so pretty
Riding up and down Broadway
on my old stud Leroy
And the girls say
Save a horse, ride a cowboy.
Everybody says
Save a horse, Ride a cowboy

What? What?
Save A Horse Ride A Cowboy
Everybody says
Save a Horse Ride a cowboy.

[i][size=9]Info courtesy of Daps Lyrics[/size][/i]

BoA tehehe...beat of the angel (my angel :-P )

lol...anyways...I like azn music better than american music for some reason *shrugs*

There`s a similar topic on The Round Table..

I voted for alternative rock.

I like many kinds of music, rock (alternative/college/progressive/..), reggae, pop, metal, electronic, new-age, punk..

My favorite artists/bands are: Pink Floyd, Radiohead, R.E.M., Pearl Jam, Guns n` Roses, Interpol, The Shins, The Velvet Underground, PJ Harvey, Sonic Youth, Dire Straits, Pixies, The Doors, Scorpions, The Cranberries, Suede, Oasis, Coldplay, Stratovarius, Dream Theater, Hsker D, Bad Religion, Ramones, A-Ha, Simon & Garfunkel, Bob Marley, Peter Tosh,  Manu Chao, Air, Vangelis, Enigma, Enya, David Arkenstone,  Massive Attack, Belle & Sebastian, Erykah Badu..

I also like soundtracks of Playstation games. Nobuo Uematsu`s (Final Fantasy 7, 8, 9, 4..) and Yasunori Mitsuda`s (Chrono Cross is my favourite soundtrack)

These days I like to listen Goo Goo Dolls, Mike Oldfield, Charlatans UK, Grandaddy..

Currently I`m listening R.E.M.`s new album "Around the Sun", np (now playing) Wanderlust.

Good Charlotte's new album, Chronicles of Life and Death, hits stores on Tuesday, October 5th.  I've been listening to it on MTV"s The Leak today, and it's pretty darned goodly.  I'll pick it up f0 sh0.

Eminem's new album, Encore, comes out on November 16th, along with Destiny Child's new one.

Then Greenday's album just came out.  Also, Jimmy Eat World's new album, Futures, comes out October 19th.  "Pain" is their first awesome single.

So many good choices!

All hail the gods of rock that are Led Zeppelin...

...and the lesser gods and goddesses that are Queen, Pink Floyd, Joan Jett and the Black Hearts, and Cheap Trick.

Anyone ever heard of Emerson Lake and Palmer?
Synthesizers and such. Great music.

Oh and speaking of bands like Led Zeppelin, Queen, Pink floyd...
surely you've heard of [url=http://www2.memlane.com/jmilner/stairwaybackwards.htm]this[/url] urban legend...

EDIT
Don't take that the wrong way... I have nothing against those bands. Those are true classics, and I like them as much as you. I just thought it was interesting that you mentioned the exact same bands that were on the stairwaybackwards website. Right now I'm actually trying to find out if they hid any subliminal messages in the Strong Bad Sings CD.  :?

[quote=Roblox]bump $$ $$[/quote]

YOU DID NOT INCLUDE ELECTRONIC!!!

[quote=RYang2]wow, this is oldest fun factory thread[/quote]

no

The oldest FF thread is titled "I won't probably be active for a week or so because ..." by Valentin Vornicu on a math olympiad.

Also I just bumped this thread(ahahaha)

"RT-2PM2 Topol-M Silo-based intercontinental ballistic missile hits streak"


Solve this
$\left|{x-8}\right|^{5}+\left|{x-9}\right|^{5}= 1$ :P

Since both terms are non-negative, we must have $|x-8|\leqslant 1\land |x-9|\leqslant 1$, which after expanding gives $8\leqslant x\leqslant 9$. Put $x=8+y$ where $0\leqslant y\leqslant 1$. Then $|x-9|=|y-1|=1-y$, hence:

\begin{eqnarray*}&& |x-8|^{5}+|x-9|^{5}=1\\ &\iff& y^{5}+(1-y)^{5}=0\\ &\iff& y^{5}+1-5y+10y^{2}-10y^{3}+5y^{4}-y^{5}=1\\ &\iff& 5y(y^{3}-2y^{2}+2y-1)=0\\ &\iff& y\big[(y-1)(y^{2}+y+1)-2y(y-1)\big]=0\\ &\iff& y(y-1)(y^{2}-y+1)=0\end{eqnarray*}

Since $y^{2}-y+1=0$ has no real solutions, the only possible values for $y$ are $0$ and $1$, giving $x\in\{8,9\}$


If a,b,c>0 prove if the following inequality hold true

                   a/(b+c)+b/(c+a)+c/(a+b)>=(a+b+c)/2 :3sqrt:(abc).
  If this is not true prove the contrary. I hope you do understand what i have written.

Do you mean:
If $a,b,c>0$,
\[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{(a+b+c)}{2\sqrt[3]{abc}}.\]

Expanding the symsums gives something that I don't believe is true.

And indeed, my guess is correct; take $a=1.1,b=1,c=0.9$

yes that's it.But can anyone give me a proof?

It's not true, I provided a counterexample above!


Inequality
polynomial
algebra
n-variable inequality

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

[quote="orl"]Prove the following inequality:
\[\prod^k_{i=1} x_i \cdot \sum^k_{i=1} x^{n-1}_i \leq \sum^k_{i=1}
x^{n+k-1}_i,\] where $x_i > 0,$ $k \in \mathbb{N}, n \in
\mathbb{N}.$[/quote]

This is a direct result of Muirhead's inequality; $(n+k-1,0,0,...0)$ majorizes $(n-1,1,1,...1)$. Alternatively, it follows from the AM-GM by the symmetric sum of $nx_1^{n+k-1}+x_2^{n+k-1}+...+x_k^{n+k-1} \geq (n+k-1)x_1^{n}x_2...x_k$.

Or we can use Chebishev Inequality:

for sequences $ \left\{a_{n}\right\}$ and $ \left\{b_{n}\right\}$ that both are increasing or both decreasing, and real numbers $ \mu_{i}\;\;(1\leq i\leq n)$ we have:

$ \boxed{\sum^{n}_{i \equal{} 1}\mu_{i}a_{i}\sum^{n}_{i \equal{} 1}\mu_{i}b_{i}\leq\sum^{n}_{i \equal{} 1}a_{i}b_{i}\sum^{n}_{i \equal{} 1}\mu_{i}}$

now we have:

$ \sum^{k}_{i \equal{} 1}(1\cdot x^{k}_{i})\sum^{k}_{i \equal{} 1}(1\cdot x^{n\minus{}1}_{i})\leq\sum^{k}_{i \equal{} 1}(x^{k}_{i}\cdot x^{n\minus{}1}_{i})\sum^{k}_{i \equal{} 1}1$

$ \Longrightarrow\frac{1}{k}\sum^{k}_{i \equal{} 1}x^{k}_{i}\cdot\sum^{k}_{i \equal{} 1}x^{n\minus{}1}_{i}\leq\sum^{k}_{i \equal{} 1}x^{n\plus{}k\minus{}1}_{i}$

by AM-GM : $ \prod^{k}_{i \equal{} 1}x_{i}\leq\frac{1}{k}\sum^{k}_{i \equal{} 1}x^{k}_{i}$

and the desired inequality is obtained!


limit
real analysis
real analysis unsolved

Hi, what's your solution to this problem: Show that, by applying Taylor Theorem,

\[ \lim_{k\rightarrow \infty}k(e\minus{}\sum_{n\minus{}0}^{k\minus{}1}\frac{1}{n!})\equal{}0.\]

I'd like to hear about your solution.

Thank you.

you  just need to write the definition of $ e$ and note that $ \sum_{j \equal{} k}^{\infty}\frac{1}{j!} \leq \frac {const \equal{} e}{k!}$ and $ \lim\frac {k}{k!} \equal{} 0$


geometry
ratio
incenter
trigonometry
geometry solved

I'm a bit confused. By ^ACB do you mean angle A of the triangle ABC? I'm asking because in Romania we have the habit of writing angle A as <BAC (the letter of the vertex of the angle being in the middle of the expression, not the left side, as in ^ACB, where A is the first letter).

Even though the convention I'm used to suggests to take ^ACB as angle C of ABC, I'm sure you are referring to angle A because the figure is symmetric with respect to B and C, so there would be no reason for which we can find angle C and not angle B, but then we would know all the angles, and I don't think that's what the problem wants.

Another thing: I think the angle is 60. This is very similar to the first problem from the Chinese Olympiad 2003. In fact, it's a part of that problem. I know I've solved it before and i seem to remember the angle A being 60 deg. All we have to do is prove it :D

Here's the main idea:

We have S(AB2C2)=S(ABC) iff AC2*AB=AB2*AC. Let's assume AB<AC (it's not that mportant, but this is how I drew the picture :)). It's easy to see that the condition above is equivalent to BB2 || CC2 <=> AC2/AB=AC/AB2. Now play with these ratios a bit. I can't work it all out right now, but theorems like Menelaus and the transversal theorem work here.

We can calculate the ratio AC/AB2 by applying Menelaus in the triangle ABC with transversal line C1B2, and then we can find AC2/AB by applying the same thm in the triangle ABC with transversal line B1C2. Just see what you get.

[quote="A1lqdSchool"]Let ABC be the triangle . (K) is the incircle of ABC.
  let B1,C1 respectively be the midpoints of AC,AB. C1K cuts AC at B2
B1K cuts AB at C2. Suppose S(AB2C2)=S(ABC). find ^ACB[/quote]

As Grobber correctly remarked, the angle which has to be found should be < CAB, not < ACB.

I have not much time now for writing solutions for MathLinks, but as for the above problem, I was asked some days ago by a Hyacinthian to solve it, so I had written down a solution, and now I am just copying this solution (the notations are slightly changed: in my solution, the incenter of triangle ABC is called I rather than K):

[b]Extended Problem Statement:[/b] In the following, let [XYZ] denote the area of an arbitrary triangle XYZ.

Let I be the incenter of a triangle ABC. Let $B_1$ and $C_1$ be the midpoints of the sides CA and AB, respectively. Let $B_2$ be the point of intersection of the lines $IC_1$ and CA, and let $C_2$ be the point of intersection of the lines $IB_1$ and AB. Prove that:

[b](a)[/b] We have $[AB_2C_2]=[ABC]$ if and only if A = 60°.
[b](b)[/b] We have $[AB_2C_2]=\frac{[ABC]}{2}$ if and only if A = 90°.
[b](c)[/b] We have $[AB_2C_2]=\frac{[ABC]}{3}$ if and only if A = 120°.

[b]Solution:[/b] All three claims [b](a)[/b], [b](b)[/b], [b](c)[/b] directly follow from the following theorem:

[b]Theorem 1.[/b] For every triangle ABC, we have

$\frac{[AB_2C_2]}{[ABC]}=\frac{1}{4\sin^2\frac{A}{2}}$.

[i]Proof.[/i] Let the B-excircle of triangle ABC touch its side CA at the point Y, and let the C-excircle touch its side AB at the point Z. Then, we have a pretty known lemma:

[b]Lemma 2.[/b] We have $B_1I \parallel BY$.

See below a proof of this lemma.

Of course, $B_1I \parallel BY$ can be rewritten in the form $C_2B_1 \parallel BY$, and after Thales this yields

$\frac{AB_1}{AC_2}=\frac{AY}{AB}$.

Since the point Y is the point of tangency of the B-excircle of triangle ABC with the side CA, we have AY = s - c, where $s=\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC. Also, AB = c. Thus,

$\frac{AY}{AB}=\frac{s-c}{c}$.

Hence,

$\frac{AB_1}{AC_2}=\frac{s-c}{c}$.

Since the point $B_1$ is the midpoint of the segment CA, we have $AC=2\cdot AB_1$, and thus

$\frac{AC}{AC_2}=2\cdot\frac{s-c}{c}$.

Similarly,

$\frac{AB}{AB_2}=2\cdot\frac{s-b}{b}$.

But after a well-known area formula for triangles,

$[ABC]=\frac{AC\cdot AB\cdot \sin\measuredangle CAB}{2}$

and

$[AB_2C_2]=\frac{AC_2\cdot AB_2\cdot \sin\measuredangle C_2AB_2}{2}$.

Since $\measuredangle CAB = \measuredangle C_2AB_2$, we thus get

$\frac{[ABC]}{[AB_2C_2]}=\frac{AC\cdot AB}{AC_2\cdot AB_2}=\frac{AC}{AC_2}\cdot\frac{AB}{AB_2}$
$=(2\cdot\frac{s-c}{c})\cdot(2\cdot\frac{s-b}{b}) = 4\cdot\frac{(s-b)(s-c)}{bc}$.

But a well-known trigonometric formula (half-angle law) tells us that

$\frac{(s-b)(s-c)}{bc}=\sin^2\frac{A}{2}$;

thus,

$\frac{[ABC]}{[AB_2C_2]}=4\sin^2\frac{A}{2}$,

and

$\frac{[AB_2C_2]}{[ABC]}=\frac{1}{4\sin^2\frac{A}{2}}$,

completing the proof of Theorem 1.

Remains to prove Lemma 2. This is actually a fundamental fact, but it also can be easily derived from the theorem that the Nagel point N, the centroid G and the incenter I of triangle ABC are collinear, and that NG : GI = 2 (with directed segments). In fact, since the point $B_1$ is the midpoint of the segment CA, the line $BB_1$ is a median of triangle ABC, and hence, it passes through the centroid G of the triangle ABC and is divided there in the ratio $BG:GB_1=2$. Combining this with NG : GI = 2, we get $BG:GB_1=NG:GI$, what, using Thales, leads to $B_1I \parallel BN$. But, being the Nagel point of triangle ABC, the point N must lie on the line BY; hence, the line BN is nothing different but the line BY, and we get $B_1I \parallel BY$, what proves Lemma 2.

Thus, the problem is solved.

Francisco Javier García Capitán kindly sent me a link containing another solution of the initial problem: http://www.animath.fr/UE/audin/exercice.htm problem 4.

  Darij


probability
combinatorics proposed
combinatorics

This is more of a combinatorics oriented than statistics oriented, so I'm posting it here. ;)

If you choose 3 points on a circle. What is the probability that they lie on a semicircle (.

what do u mean not necessarily of the same circle there exista an unique circle passing through three non collinear points...but isn't this more a probability problem so u cud have posted it therr

As I said, this not exactly a probability kind of question. I personally felt it belonged in the olympiad section.

Okay, okay. leave that. If the 3 points are such that the distance between the extremes is greater than $ \PiR$, then the major arc can never form the diameter of any circle.

But then, the minor arc can always be the diameter of any circle. I just gave the sum as in a book. Its a bit confusing actually. I can give the answer if you want.


Prove that if p is a prime of the form 4n+1 where n is a positive integer, then [(2n)!]^2≡-1(mod p)


trigonometry
calculus
calculus computations

if $\alpha\geq\beta\geq\gamma\geq\frac{\pi}{15}$ and $\alpha+\beta+\gamma=\frac{\pi}{2}$ , find the max and min of

$S=\cos\alpha\sin\beta\cos\gamma$

i havent tried yet but it seems that you could just apply Lagrange multipliers, although there might be a "better" solution.


Please reply soon .  If possible with diagram.Thanks

1)How many cubes of length 2cm can be made

 from a cubical wood piece of length 30 cm.

Divide the volume of the big cube by the volume of the small cubes.

So you get $ \frac{30^3}{2^3}\equal{}15^3\equal{}\boxed{3375}$.

You can make 3375 cubes.

thnx a lot for ur quick reply.

 i've sent u friend reqest.

If it was a line of 30 cm that we wanted to break into 2 cm pieces, it's obviously $ \frac{30}{2}$.

If it's a square, it's $ \left (\frac{30}{2} \right )^2$.

So if it's a cube, it's $ \left (\frac{30}{2} \right )^3\equal{}\boxed{3375}$.


limit
logarithms
number theory unsolved
number theory

When I was partitioning integers I found some nice results which I hope maybe verified.
In the usual manner, I had
1.P(n)<P(n-1)+P(n-2) is true for all but finitely many n.
Let F(n) denote fibonacci sequence. Let k be a constant positive integer. Can anyone please prove or disprove thequestion:
Prove that F(n)>P(n+k) holds for all positive integers n but finietly many values.

BOmb

Here's a sketch of a "not-so-combinatorial" proof of your statement (proving these results rigorously requires some tools from analysis)

The infinite product $\prod_{k=1}^\infty {1 \over 1-x^k}$ converges uniformly in the open disk centered at the origin of radius $r$ in the complex plane for any $r<1$.  Therefore the infinite sum $\sum_{n=1}^\infty p(n)x^n$ must also converge on the whole disk as well.  But that implies that $\lim_{n \rightarrow \infty} p(n)^{1 \over n}=1$.   On the other hand, $F(n-k)$ grows exponentially for any fixed $k$.

Could you please elaborate on your proof. Can you please prove it rigorously by analyss, I dont really understand what you've done. Thanks

Bomb

We have $\sum_{k=0}^\infty p(k)x^k=F(x)=:\prod_{k=1}^{\infty} (1-x^k)^{-1}$. It holds for any positive $x$ less then 1. So, $p(k)\le x^{-k}F(x)$, hence $\ln p(n)\le -n\ln x-\sum_{k=1}^\infty \ln(1-x^k)$. Now try to optimize RHS by $x$. Using logarithm's expantion, we have $-\ln(1-x^k)=x^k+x^{2k}/2+x^{3k}/3+\dots$ Now change the order of summands in RHS, you get $-n\ln x+y_1+y_2/2+y_3/3+\dots$, where $y_k=x^k/(1-x^k)=(1-x)^{-1}\cdot z_k$, where $z_k=x^k/(1+x+x^2+\dots+x^{k-1})\le 1/k$. Using all this, we get
\[
\ln p(n)\le -n\ln x+(1+1/4+1/9+\dots)(1-x)^{-1}=-n\ln x+\pi^2/(6(1-x))
\]
For $x=1-c/\sqrt{n}$, we get $\ln p(n)\le (c+\pi^2/(6c)) \sqrt{n}+O(1)$, optimize by $c$, we must take $c=\frac{\pi}{\sqrt{6}}$ and finally

\[
\ln p(n)\le C+\sqrt{2/3}\,\pi \sqrt{n}
\]

the constant $\sqrt{2/3}\,\pi $ is sharp.


One day a restless lion roamed about in his circular cage, of radius 10 meters, along a polygonal path of total length 30 km. Thus, coming to the end of one segment in the path. Prove that, counting both turns to the left and turns to the right (considering each angle to be a postive quantity), he must have turned through a total of at least 2998 radians.

Use a triangle of large perimeter.  This would mean minimal radians, as triangles have
a total angle sum of 180 degrees, or $\pi$ radians.  Say we are dealing with a right triangle inscribed in the circle, as that would give us large perimeter.  Now we only have to deal with the radians calculations.  As <C nears 0, the length of AB nears 0.  At angle 0, we see that he is only pacing back and forth along the diameter.  The diameter's length is 20 meters.  One pace back and forth would give 40 meters.  The translation of 30 kilometers in meters is 30,000 meters.  30,000 meters divided by 40=3,000/4=25*30 equals 750 paces back and forth.  750*$\pi$=2356.  It looks like somebody's wrong here, is it the book?


Triangle $ABC$ is inscribed in the circle $(O)$.Point $M$ (distinct from $B;C$) is arbitrarily chosen on the side $BC$.Let $E;F$ be the centers of the circumcircles of triangles $ABM$ and $ACM$,respectively.The two lines $BE$ snd $CF$ intersect at $D$.Prove that $D$ lies on the circle $(O)$.
Thanks!

[quote="PDatK40SP"]$\angle DBE = 90^{o}-\angle BAE, \angle FCE = 90^{o}-\angle CAE \\ \text{So }\angle BDC = \angle BAE+\angle CAE = \angle BAC,Q.E.D$[/quote]
Why $\angle DBE = 90^{o}-\angle BAE, \angle FCE = 90^{o}-\angle CAE$?

[color=darkblue][b]Nice and very simple problem ![/b]

Denote $\left\{\begin{array}{c}m(\widehat{BAM})=x\\\\ m(\widehat{CAM})=y\\\\ x+y=A\end{array}\right\|$ $\Longrightarrow$ $\left\{\begin{array}{c}m(\widehat{CBD})=m(\widehat{MBE})=90^{\circ}-x\\\\ m(\widehat{BCD})=m(\widehat{MCF})=90^{\circ}-y\end{array}\right\|$ $\Longrightarrow$ $m(\widehat{BDC})=A$ a.s.o.[/color]


[b]$ xy\equal{}12\cdot \sqrt{6}\\
yz\equal{}54\cdot \sqrt{2}\\
zx\equal{}48\cdot \sqrt{3}$[/b]

[hide]the book only addressed the case where each variable is positive, but doesn't the system also work in the case where each variable is just the negative of the positive case?[/hide]

[hide]
$ x^2y^2z^2 \equal{} 186624$
$ xyz \equal{}\pm 432$.
Substituting each of the other equations back in gives the answers.
[/hide]

Maybe your book thought that those solutions were too obvious and trivial to mention?

Of course, you could always divide the equations to solve for variables because none of the variables can equal 0.

They are all positive or all negative, because all products of two variables are positive.

[hide="Hint"]
[hide="Try this first"]
[hide="seriously"]
just multiply the three equations together...then square root both sides, and then solve.
[/hide]
[/hide]
[/hide]

[quote="123456789"][quote="mihail911"][b]$ xy \equal{} 12\cdot \sqrt {6} \\
yz \equal{} 54\cdot \sqrt {2} \\
zx \equal{} 48\cdot \sqrt {3}$[/b]

[hide]the book only addressed the case where each variable is positive, but doesn't the system also work in the case where each variable is just the negative of the positive case?[/hide][/quote]

Which book did you get this problem from?[/quote]
AoPS v2

Really? Which chapter? I don't recall solving it...

ch 13. the one after analytic geometry about algebraic expressions. its a mu alpha theta problem.

Wasn't the original question something like "three faces of a rectangular prism have $ 12\sqrt {6},54\sqrt {2}  48 \sqrt {3}$, find the lengths of the sides of the prism"?

The original question was exactly the same (Chapter 13: Equations and Expressions).

No, it simply said to solve the system of equations.  But if it did discuss a rectangular prism like Temperal said, then $ x$, $ y$, and $ z$ must be positive.  I guess they forgot the negative solutions. :oops:

I see it now. I thought it was a problem... it was exercise 13-7. But...

[quote="mihail911"]its a mu alpha theta problem.[/quote]

In my edition, it doesn't say that.


Don't you think we have discussed a lot about this inequality?

[Moderator edit: Discuss about this inequaltiy on http://www.mathlinks.ro/Forum/viewtopic.php?t=5121 .]

well... yes.. but.. perhaps you should move it to the begginers corner! :)
it may be good for them! :)

sorry again for the duplicate post!

I hope you are joking! This one to the begginers corner? Isn't that too much? Don't forget that 5 is odd and the discution is not trivial in this case. It they were 6 or 8 variables, I would have moved it.


Un triangulo ABC tiene el A=120, llamamos D, E, F al punto de corte de las bisectrices con los lados BC, AC y AB respectivamente. Demuestra que el EDF= 90.

$E$ es excentro de $ABD$
$F$ es excentro de $ADC$

$\measuredangle ADF= \measuredangle FDB =\alpha$

$\measuredangle CDE=\measuredangle EDA=\beta$

$2\alpha+2\beta = 180^\circ \rightarrow \alpha+\beta=90^\circ \rightarrow \measuredangle EDF=90^\circ$


A novel has 527 pages (pages 1 – 527).  How many digits will it take to number all 527 pages from 1 to 527?

[hide]

9 one digits, 90 2 digits, 428 3 digits -- >1473[/hide]


You should have an option, "Rarely" or "Sometimes", because I rarely go play AoPS "For the Win". I only go on so rarely because I haven't experienced much problem solving/MathCounts.

[quote="Cyrazeno"]I played it a few times... but stopped
I was not doing too well... Meh[/quote]
exactly my case.

I go on all the time. My rating proves it. :)

I go on all the time too, but I'm trying to stop. I should be practicing for stuff, but I just get bored.

Maybe you should put this in the FTW forums... :P But every since going into MOEMS it's helped problem solving a bit more.


why is this locked? http://www.mathlinks.ro/Forum/viewtopic.php?t=85011&start=20

maybe because everyone was posting nearly the same thing? :?

Jose, you still haven't read the posting guidelines for the forum, have you?

Each problem should have its own thread in general.

In generaly, people like you  ;)  keep posting problems beyond the scope of the forum within marathon-like threads.  When a problem has its own thread, we can just move it, but in a marathon, that task becomes a moderator's nightmare.

Also, it's easier to search for or document problems with their own threads.


trigonometry
inequalities
geometry
trig identities
Law of Sines

If in a triangle ABC $2m(B)=m(A)+m(C)$ then $2b\geq a+c$.

By the Law of Sines, we have $a=\left (\frac{sin(A)}{sin(B)}\right )b$ and $c=\left (\frac{sin(C)}{sin(B)}\right )b$.  Plugging these into the inequality yields

$sin(A)+sin(C)\le 2sin(B)$.

Since $2m\angle B=m\angle A+m\angle C$, we have $m\angle B=60$.  Thus the above inequality turns into

$sin(A)+sin(C)\le\sqrt{3}$.

Now since $m\angle A+m\angle C=120$, let $A=60-\theta$ and $C=60+\theta$.  Then after the use of the $sin(\alpha+\beta)$ identity, we have

$\sqrt{3}cos(\theta)\le\sqrt{3}$

which is definitely true.


calculus
integration
function
factorial
calculus open

Could somebody help me to do the following integral?

$\int_0^{2\pi}{exp(acos(z))dz}$

a is a real constant For me it would be enough expressing it as a known function (e.g. Bessell). I tried to do it with Mapple but it wasn't able either. Is it possible to find a closed form for it?

Translating into complex terms, this is equal to
$\int_{|z|=1}\exp(a\text{Re }z)\cdot\frac{dz}{iz} =\frac1i\int_{|z|=1}\exp(\frac a2(z+\frac1z))\cdot\frac1z\,dz$.
We need the residue at zero of $\exp(\frac a2(z+\frac1z))\cdot\frac1z$; this is the constant term of the product $(1+\frac a2z+\frac{a^2}{2^2\cdot 2}z^2+\cdots+\frac{a^n}{2^n\cdot n!}z^n+\cdots)(1+\frac a{2z}+\frac{a^2}{8\cdot z^2}+\cdots+\frac{a^n}{2^n\cdot n!\cdot z^n}+\cdots)$, which is
$1+\frac{a^2}{4}+\frac{a^4}{64}+\cdots+\frac{a^{2n}}{2^{2n}\cdot (n!)^2}$.

I'm pretty sure there's no closed form, and I don't know any functions that match it, with that squared factorial in the denominator.

It is at least very easy to approximate accurately.

[quote="jmerry"]I don't know any functions that match it, with that squared factorial in the denominator.[/quote]
Ah, but there is such a function - a Bessel function.

This appears to be $J_0(ia).$

Is that the same than $I_0(a)$?

Thanks a lot for your help


1/Find all values of $ \alpha\in N$ satisfying: $ tan\alpha^o\in Q$
2/Prove that: $ tan34^o>\frac{2}{3}$

[quote="Kyoshiro"]1/Find all values of $ \alpha\in N$ satisfying: $ tan\alpha^o\in Q$
2/Prove that: $ tan34^o > \frac {2}{3}$[/quote]

[hide="Solution 1"]
Now $ \tan{45^o} \equal{} 1,\tan{180n^o} \equal{} 0 \in \mathbb{Q}$,and for all $ x\in \mathbb R,\tan(x \plus{} 180^o) \equal{} \minus{} \tan{x},\tan(90^o \minus{} x) \equal{} \frac {1}{\tan{x}}$,thus it's enough by cousiderd for the case $ \alpha\in [1,44]$.

The tangent addition theorem says $ \tan(x \plus{} y) \equal{} \frac {\tan{x} \plus{} tan{y}}{1 \minus{} \tan{x}\tan{y}}$.

by using this repeatedly,we have the next lemma.

Lemma:Let $ n$ be a positive integer,then if $ \tan(\alpha^o)$ is rational,$ \tan(n\alpha^o)$ is also rational.(but it's necessary that all tangent can be defined.)

1)If $ \alpha$ is not dividible by $ 3$,then suppose $ \tan(\alpha^o)$ is a rational.by the lemma,$ \tan(30\alpha^o)$ is also rational,but this absolute value is equal to $ \tan{30^o} \equal{} \frac {1}{\sqrt {3}}$ or $ \tan{60^o} \equal{} \sqrt {3}$ which are irrationals,contradiction!

2)If $ \alpha$ is dividible by $ 3$ but not by $ 9$,then suppose $ \tan(\alpha^o)$ is a rational.by the lemma,$ \tan(10\alpha^o)$ is also rational,but this absolute value is equal to $ \tan{30^o} \equal{} \frac {1}{\sqrt {3}}$ or $ \tan{60^o} \equal{} \sqrt {3}$ which are irrationals,contradiction!

3)If $ \alpha$ is not dividible by $ 9$,i.e.,$ \alpha \equal{} 9^o,18^o,27^o,36^o$.
Suppose $ \tan{9^o}$ or $ \tan{18^o}$ or $ \tan{27^o}$ is rational,then,by the lemma and $ \tan(90^o \minus{} x) \equal{} \frac {1}{\tan{x}}$,we can say $ \tan{36^o}$ is also rational,but $ \tan{36^o} \equal{} \sqrt {5 \minus{} 2\sqrt {5}}$ is irrational,contradiction!

therefore,we see that such $ \alpha\in \mathbb{N}$ is $ \alpha \equal{} 90n \minus{} 45$ or $ 180n$ ($ n\in \mathbb {N}$)[/hide]

[quote="Kyoshiro"]1/Find all values of $ \alpha\in N$ satisfying: $ tan\alpha^o\in Q$
2/Prove that: $ tan34^o > \frac {2}{3}$[/quote]

[hide="Solution 2"]
By the tangent addition theorem we have

$ \tan{34^o}\equal{}\frac{\tan{30^o}\plus{}\tan{4^o}}{1\minus{}\tan{30^o}\tan{4^o}}$.

then,we see that $ \tan{34^o}>\frac{2}{3}$ is equivalent to

$ \tan{4^o}>\frac{24\minus{}13\sqrt{3}}{23}$.

now,
$ \tan{4^o}\equal{}\tan{\frac{\pi}{45}}>\frac{\pi}{45}>\frac{3,14}{45}\equal{}0,0697...>0,0691...\equal{}\frac{24\minus{}13\cdot 1,73}{23}>\frac{24\minus{}13\sqrt{3}}{23}$,

so that is true.[/hide]


algebra
polynomial
function
complex analysis
superior algebra
superior algebra unsolved

Hi there!
It is known that Tchebychev polynomials commute one with other.
Also the family $ P_{n}(x) = x^{n} $ is a commuting family of polynomials.
I was thinking about this and have composed some problems about
commuting polynomials. The first of them are very simple or obvious,
but some of them are interesting, for my opinion.
The coefficients of polynomials can be considered in $ \mathbb{C} $
or in $ \mathbb{R} $, as you wish.
I will consider them in $ \mathbb{R} $, but all the statements are
true also for $ \mathbb{C} $.

1) If a polynomial commutes with linear polynomial $ Q(x) = ax + b $
then it is linear or constant.

From now on all polynomials are monic.
We see that $ ( \mathbb{R}, + ) $ acts on the set of monic polynomials
by $ L_{a}P(x) := P(x + a) - a $, $ a \in \mathbb{R} $. It is easy to see that
if polynomials $ P, Q $ commute, so $ L_{a}P , L_{a}Q $.
We say that polynomials $ P_{1}(x) , P_{2}(x) $ are equivalent if
there exists $ a \in \mathbb{R} $ such that $ P_{2} = L_{a}P_{1} $.

2) If for $ n>1 $ a polynomial $ P(x) $ commutes with a polynomial
$ Q(x) = x^{n} $, then it is of a form $ P(x) = x^{m} $ for some natural $ m $.

3) Prove that for a polynomial $ Q(x) = x^{2} - 2 $ for every degree there exists
a polynomial, which commutes with it.

4) Suppose we have fixed some natural $ n>1$.
Prove that almost every polynomial of degree $ n $ does not commute with any
polynomial of degrees between $ 2 $ and $ n-1 $.

5) Of course every polynomial $ P(x) $ commutes with $ P^{(n)}(x) := P \circ \ldots
\circ P (x) $ ( $ n $ times ), for $ n \geqslant 0 $. If this set contains
all polynomials which commute with $ P(x) $,then we call such a polynomial
poor. For each degree find many pairwise
nonequivalent polynomials which are poor and prove that.

6) Prove that if $ P(x) $ commutes wit $ Q(x) $ and $ Q(x) $ commutes with
$ R(x) $ then $ P(x) $ commutes with $ R(x) $.(It is important that all polynomials are monic)

So we see that property of commuteness is an equivalence relation ( on monic polynomials).
So every polynomial $ P(x) $ is contained in some maximal commuting family of
polynomials. Call a maximal commuting family lucky, if it contains polynomials
in each degree. So we know that there exists lucky family
$ P_{n}(x) = x^{n} $ , $ n \geqslant 1 $. Also we have a "Tchebychev" family $ \{ T_{n}(x) \}$
from problem 3) . Also we have of course for every $ a \in \mathbb{R} $ that
$ \{ L_{a}P_{n} \} $ is a lucky family and also $ \{ L_{a}T_{n} \} $ is a lucky family.

7) Prove that there do not exists other lucky families.

Here is one problem which is not connected to others:
( actually it must be appeared in Analysis section)

8) If an entire function $ f: \mathbb{C} \rightarrow \mathbb{C} $ commutes with
a polynomial (you can take it with complex coefficients) of degree at least
$ 2 $, then $ f $ is a polynomial.

Wow! Nice, Leva! :)

I have a question: in your first problem, don't you need some conditions on $a,b$? What if the polynomial is $P(x)=x^3$ and $Q(x)=-x$? Or maybe you meant to say that $P$ commutes with all linear polynomials? :? I don't think so, because that would make the problem trivial. Actually, it is pretty easy, except for the case $a=-1$, when we might have some trouble like that above.

For the second one:

We may assume that $P$ has no roots equal to $0$ (separate the part $x^m|P$ with maximal $m$ from $P$). We have $P(x^n)=P^n(x)$. After we derive that we get $nx^{n-1}P'(x^n)=nP'(x)P^{n-1}(x)$. This means that $x^{n-1}|P'(x)$. From that we get $x^{n(n-1)}|P'(x^n)$, so $x^{n(n-1)}|P'(x)$. We repeat this to find that $x^m|P'(x)$ for all $m$, meaning that $P'(x)=0$, so $P$ is constant. Since we assumed it was monic, we get $P=1$. When we put back the part $x^m$ we had separated, we get $P(x)=x^m$.

Sorry, grobber!  :blush:
Indeed, I a little bit huried with the 1-st problem.
Indeed, it is not always true, so I apology,
so the corrected problem is :
For given $ a,b $, classify all polynomials which commute with $ ax+b $.

Also in problem 6 I assume that the polynomials $ P,Q,R $ have degrees at least $ 2 $.

Ok, thanks for clarifying that.

For the third proble, the polynomial $Q$ is a member of the commuting family $2T_n\left(\frac x2\right)$, so we can take our polynomials to be precisely these: $U_n=2T_n\left(\frac x2\right)$ (they're also monic, if that's one of the requirements of the problem).

Your questions are very interesting.For more this type of problems you can check the book"Polynomials" by E.J. Barbeau.Its a springer publication & a very nice book.

[quote="Leva1980"]Why nobody wants to solve my problems?  :(  :(  :([/quote]
They are very nice, just too many for one post. I especially like #8. Try posting them one at a time.  ;)

Not so many people on this forum have three degrees in math! ;)

But the problems are elementary, as their solutions
( except the problem 8, which Fedja solved)  :)

Also, Ihave only 2 degrees, I am doing the 3rd now.


calculus
integration
function
trigonometry
Galois Theory
calculus computations

I would like to know about some impossible integrals (to be expressed as finite  sum/product of elementary functions)
for example $\int e^{x^{2}}, \int e^{-x^{2}}, \int\sqrt{a^{2}\sin^{2}x+b^{2}\cos^{2}x}, a>b$.
Can someone help me to find more about this subject? Maybe you can recommend me some books which treat this kind of subject. I would like to know the proof that they are not expressible as elementary functions and maybe some ways to aproximmate them when the integrals are definite.

Thank you.

You can start with Wikipedia and continue with the references mentioned there.

[url=http://en.wikipedia.org/wiki/Differential_Galois_theory]Differential Galois theory[/url]: to prove that something is not expressible as an elementary function.

[url=http://en.wikipedia.org/wiki/Quadrature_%28mathematics%29]Numerical integration[/url]: to approximate definite integrals.

I will refer to another of [b]greengrass[/b]'s [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=617028#617028]posts[/url] -- this one has links to three papers that may interest you. :)

By the way this has been discussed many times, and I think the current thread has the most links out of all the ones I've seen, so I'm going to stick in some terms to make it more searchable:

elementary antiderivative
elliptic integral
closed form

Hopefully this helps people find the thread in the future. ;)

it seems that improper or definite integrals of the second
form you asked can be solved
the solution can be found [url=http://www.joma.org/images/cms_upload/delgado09010344816.pdf]here[/url]
the bibliography that is presented there has solutions to your first
integral as well(in the same conditions as the second one)

The values of improper integrals $\int_{0}^{\infty}e^{-x^{2}}dx$ and $\int_{-\infty}^{\infty}e^{-x^{2}}dx$ are well-known. It's not clear to me what other "definite integrals of the second
form" you had in mind. And the improper integral of $e^{x^{2}}$ is divergent.


calculus
derivative
LaTeX
function
trigonometry
calculus computations

[quote="p4nda"]Find the four second partial derivatives for f(x,y) = y^2e^x + xycosx

I am stuck on the last part... here's what I got so far:

Zx = y^2e^x - ysinx
Zxx = y^2e^x - ycosx
Zxy = ?

Zy = 2ye^x + xcosx
Zyy = 2e^x + xcosx
Zyx = ?

I need help with solving for xy. Both should end up being the same. :huh:[/quote]

Why not?

You've made mistakes both in calculating $ Z_{x}$ and in calculating $ Z_{yy}$.  Try it again, and think carefully about every derivative of every term.

Also, your posts will be much nicer to read if you use LaTeX.  An FAQ on LaTeX in the forum is at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690

Okay, from solving Zy, Zyy, Zyx... I got 2e^xy. But I'm having a difficult time solving for Zx, Zxx, Zxy, even though it ends up the same as Zyx. I need help on finding the derivative "xycosx" as if the y is treated as a constant.

1) There is no partial derivative of that function which is equal to either $ 2e^{xy}$ or $ 2ye^{x}$.

2) Can you calculate $ \frac{d}{dx}\left( x \cdot \cos x\right)$?

3) Seriously, use LaTeX.  Or at least use parentheses.


Correct me if I'm wrong but isn't it n^3

idk how though

[hide]The sum of the first $ n$ odd numbers is $ n^2$, easily proven since $ n^2 \minus{} (n \minus{} 1)^2 \equal{} 2n \minus{} 1$.

So the nth row sums to:
$ T_n ^2 \minus{} T_{n \minus{} 1}^2 \equal{} \left(T_{n \plus{} 1} \plus{} T_{n \plus{} 2} \right) \left( T_{n \plus{} 1} \minus{} T_{n \plus{} 2} \right)$
$ \equal{} \left(\frac {n(n \plus{} 1)}{2} \plus{} \frac {n(n \minus{} 1)}{2} \right) \left( \frac {n(n \plus{} 1)}{2} \minus{} \frac {n(n \minus{} 1)}{2} \right)$
$ \equal{} n^2 \cdot n \equal{} n^3$,

where $ T_n$ is the nth triangular number.[/hide]

You could also prove that the nth row starts with
1+n(n-1)
either by induction or using arithmetic sequences, and since the nth row has n elements that constitute an arithmetic sequence, we can calculate the sum, arriving at the same formula.


diameter = $\pi$ $\longrightarrow$ radius = $\frac{\pi}{2}$
volume = $\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (\frac{\pi}{2})^3 = \frac{\pi^4}{6}$  :)

[quote="236factorial"]What is the volume of a sphere with a diameter of $\pi$?  :D[/quote]

[hide]Volume of sphere $=\frac{4}{3}*\pi*\left(\frac{\pi}{2}\right)^3$

$=\frac{4\pi}{3}*\frac{\pi^3}{8}$

$=\frac{4\pi^4}{24}$

$=\frac{\pi^4}{6}$[/hide]

delete this post.
i accidentaly hit submit instead of preview. :wallbash_red:

[hide]the volume is $\frac{4}{3}\pi \cdot r^3$
in this case the radius is $\frac{\pi}{2}$
so now the volume is$\frac{4}{3}\pi \left(\frac{\pi}{2}\right)^3$
$\frac{\pi}{2}^3$
is $\frac{\pi^3}{8}$
$\frac{\pi^3}{8} \cdot \frac{4}{3} \cdot \pi=\frac{4 \cdot \pi^4}{24}=\boxed{\frac{\pi^4}{6}}$

 :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)
[/hide]

I love to double post too, math92  ;)

Say, Drunken_Math, where did you get the formula for the sphere? $\frac{\pi}{2}$ for the radius?

[quote="gman"]hey, to the genius who came up with the title: it's '[i]radii[/i]' not '[i]radiuses[/i]'.[/quote]

Save that babble and solve the problem first  ;)


calculus
integration
trigonometry
calculus computations

show that $ {\int_{0}^{\infty}}{\sin{x^2} dx}$ convergents

Substitute $ x^a\equal{}y$ in the integral $ \int_0^\infty \sin x^a dx$ ($ a>1$) and then integrate by parts....


geometry
inequalities
trigonometry
inequalities unsolved

Prove that in each triangle we have the inequalities
1)$\sqrt{bc(p-a)}+\sqrt{ca(p-b)}+\sqrt{ab(p-c)}\leq\ 3R\sqrt p$.
2)$\sum\cos\frac{B}{2}\cos\frac{C}{2}\leq\frac{9}{4}$

at least the first one is easy if i'm not mistaken(the hour is late so it is posible to make silly mistakes).i haven't got the time to post a full solution but i think only the main ideas will be enough.use $R=abc/4S$then substitute $a=(x+y)/2,b=(y+z)/2,c=(x+z)/2$use am-gm for the radical and it will be obvious. ;) if you need more details just say and i will try to post it tomorrow morning


This Topic is just making easy this problem http://www.mathlinks.ro/Forum/viewtopic.php?t=125070


find all natural $x,y$ such that:$2^x-11^y=5$


geometry
3D geometry
function
quadratics
complex numbers
algebra

[quote="Jeremic"]Ok so i was playing around with complex numbers today and i forund something iteresting and soo wierd  :huh:  :huh:  ok

$\sqrt-1=i$
$i^{2}=-1$
So now how does
$(\sqrt-1) (\sqrt-1) =-1$

when you normaly mutiply rotts you multiply the radicands to get teh answer but the answer should be 1 not -1  :|[/quote]
Think about it this way. $\sqrt{x}*\sqrt{x}=\sqrt{x^{2}}=x$

[quote="Jeremic"]Ok so i was playing around with complex numbers today and i forund something iteresting and soo wierd  :huh:  :huh:  ok

$\sqrt-1=i$
$i^{2}=-1$
So now how does
$(\sqrt-1) (\sqrt-1) =-1$

when you normaly mutiply rotts you multiply the radicands to get teh answer but the answer should be 1 not -1  :|[/quote]

Let $x$=$-1$.
Let us start like this:

$\sqrt{x}\cdot\sqrt{x}$
$(\sqrt{x})^{2}$
$\sqrt{x^{2}}$
$=x$
Plug in $-1$ for $x$.

oh wait, this is really weird...

[quote="Jeremic"]that works but i dont understand one thing

if you have $\sqrt-1^{2}=-1$ but if you evaluate that $-1^{2}=1$ and $\sqrt1=1$[/quote]
Well actually, the property $\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}$ only holds if $a,b\geq0$, which is why you can't use it with $\sqrt{-1}$. (That's why it got you an incorrect answer.)

[quote="DanK"][quote="Jeremic"]ohhh ok so i have to use $\sqrt-x=x$ ok thank you for all teh help[/quote]

Where did you get $\sqrt{-x}=x$ from?[/quote]

i think he forgot to add a square...

[quote="anirudh"]$\sqrt{x^{2}}$
$=x$[/quote]
This is invalid, because $\sqrt{x^{2}}=|x|$, although $\sqrt{x}^{2}=x$

[quote="nutz_for2.718281828"]Well, $\sqrt{x}\cdot \sqrt{x}= x$ for all real x. In this case, x=-1, which is a real number.[/quote]

As stated earlier, by nebula42, the case is only true when $\sqrt{x}\in\mathbb{R}$, implying that $x\geq0$.

Also, $\sqrt{x^{2}}=|x|$, so $\sqrt{\imath^{2}}=|\imath|$...

bflat, thats EXACTLY what I said. But he's talking about $\sqrt{x}^{2}$

Yes, I did mess up on the part about $\sqrt{x^{2}}$ instead of $(\sqrt{x})^{2}$, but I still don't think you get the point.

$(\sqrt{x})\in\mathbb{R}$, not what you said.  You only said that $x\in\mathbb{R}$, which causes $(\sqrt{x})\not\in\mathbb{R}$, if $x<0$.  The identity is true only if $x\geq0$.

He's wondering why $(\sqrt{-1})^{2}=-1$ and not 1. For ALL real x, $(\sqrt{x})^{2}=x$ even when x is negative. Or at least the AoPS book Vol. 1 says.

To stop freaking yourself out, you need to be more cautious with the symbol $\sqrt{\cdot}.$

Every positive real number has two square roots. Every nonzero complex number has two square roots (and three cube roots, and four fourth roots, and so on). What does the symbol $\sqrt{\cdot}$ mean? In order for it to define a function, we must have an agreed-upon convention. That convention is this:

For nonnegative real numbers $x,$ the symbol $\sqrt{x}$ denotes the unique nonnegative number $y$ such that $y^{2}=x.$

Now, what about complex numbers? The symbol $\sqrt{-1}$ lies utterly outside the reach of that convention. Trying to "define" $i$ as $i=\sqrt{-1}$ falls woefully short of being a valid mathematical definition. Once we have the complex numbers constructed, we will have that $i$ is [b]a[/b] square root of $-1,$ but that's not the same thing as saying that it is [b]the[/b] square root of $-1.$ There are two complex numbers with equal claim to being the square root of $-1.$ One of them is $i$; the other one is $-i.$

We do need to construct the complex numbers somehow - perhaps as ordered pairs with a defined arithmetic. That's essentially that same as saying that complex numbers are objects that can be written in the form $z=x+iy$ where $x$ and $y$ are real and a central feature of the arithmetic is that $i^{2}=-1.$ It's not enough to say that; you need to grub through the proof that this gives you a [i]field[/i]. That is, we need to show that addition is commutatitve and associative and has an identity and inverses; that multiplication is commutative and associative and that there's an identity and that nonzero numbers have multiplicative inverses, and that multiplication distributes over addition.

Now: suppose $z^{2}=u$ and $w^{2}=u$ where $z,w,$ and $u$ are complex numbers. (That is, $z$ is a square root of $u$ and $w$ is a square root of $u.$) Can we conclude that $z=w?$ No, we can't. This is a quadratic equation and has two roots. The most we can conclude is that either $z=w$ or $z=-w.$

And there's no [i]continuous[/i] way do make up a rule that specifies which of the two possibilities is "the" square root of a complex number.


geometry
3D geometry
algebra
factorization
difference of cubes
special factorizations

[quote="8parks11"]Factor 250x^3-2

i got an answer but I heard this was extremely complicated.
I don't get why...[/quote]

[hide]Notice that this is just 2(125x^3-1).

And remember that a^3-b^3=(a-b)(a^2+ab+b^2)

Goes from there...[/hide]

thats what I got but i heard this was some weird theory and i searched it and got nothing.. ( i heard i had to use that 5^3=125 and stuff)

Look at it like this:
$250x^{3}-2$
Factor out a two:
$(2)(125x^{3}-1)$
Use the method for factoring a difference of cubes:
$2(5x-1)(25x^{2}+5x+1)$
And we are finished.


geometry
circumcircle
incenter
geometry proposed

Let $ (O) $ be the circumcircle of the triangle $ ABC $ . $ (I) $ is the incircle of the triangle $ ABC $ . $ (I) $ tangents $ BC, CA , AB $ , respectively , at $ D , E , F $ . A circle $ (O_1) $ externally tangents $ (I) $ at $ D $ and internally tangents $ (O) $ at $ K $.  A circle $ (O_2) $ externally tangents $ (I) $ at $ E $ and internally tangents $ (O) $ at  $ M $.  A circle $ (O_3) $ externally tangents $ (I) $ at $ F $ and internally tangents $ (O) $ at $ N $ .$ A_1 $ is the midpoint of the arc $ BC $ (not contains A) . $ B_1 $ is the midpoint of the arc $ AC $ (not contains B) . $ C_1 $ is the midpoint of the arc $ BA $ (not contains C) . $ KA_1 $ intersects $ BC $ at $ A_2 $ . $ MB_1 $ intersects $ AC $ at $ B_2 $ . $ NC_1 $ intersects $ BA $ at $ C_2 $ . Prove that : $ A_2 , B_2 , C_2 $ are collinear

I will prove that $A_2, E, F $ are collinear.

From here directly $AD$ is polar of $A_2$, …
(three polars intersect in Gergonne point) hence $A_2,B_2,C_2$ are collinear.

WLOG suppose $b > c$. Then $B$ lies between $A_2$ and $C$

Let $A_3$ be other midpoint of arc $BC$. $A_3A_1$ is diameter.

From previous problems on ML we have $A_3, D, K$ are collinear.

Let $X$ be the midpoint of $BC$. $A_3, X, A_1$ are collinear.

 $A_3K \perp A_2K$ $A_1X \perp BC$. So $A_1XDK$ concyclic.

$A_2B . A_2C = A_2K . A_2A_1= A_2D . A_2X$

$A_2X^2 - \frac {a^2}{4}= A_2X . (A_2X - DM)$

$DM = (b - c)/2$ and $A_2X (b - c)/2 = a^2/4$,

$A_2X  = \frac {a^2}{2(b - c)}$

$A_2B = A_2X - a/2 = \frac{a}{2(b-c)} (a+c-b)$,

$A_2C =A_2X +a/2 = \frac{a}{2(b-c)}(a-c+b)$

$\frac {A_2B}{A_2C} = \frac{p-b}{p-c}$

Applying Menelaos for $ABC$ and points $E,F,A_2$ we get that $E, F,A_2 $ are collinear.

Please look:  ( http://groups.yahoo.com/group/Hyacinthos/message/6208 ). Many thanks to Darij for this proof.
 Now: $AK,BM,CN$ concur at $P$ (see: http://www.mathlinks.ro/Forum/topic-36681.html  ) and $AA_1,BB_1,CC_1$ (bisectors) concur at $Q$ (incenter of $ABC$) in other words $KMN$ and $A_1B_1C_1$ are the circumcevian triangles of the two points  $P$ and $Q$. Now the cut  points $A_2,B_2,C_2$ of $(KA_1,BC),(MB_1,AC),(NC_1,BA)$ are collinear.

Could you please tell me what the circumcevian triangles are ?  :?:


Let $ a, b, c$ be three positive real numbers such that $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\equal{}3$. Find the max value of $ Q\equal{}\frac{ab}{a^3\plus{}b^3}\plus{}\frac{bc}{b^3\plus{}c^3}\plus{}\frac{ca}{c^3\plus{}a^3}$.

Thank.

$ \frac{ab}{a^3 \plus{} b^3} \leq \frac{\frac{1}{a} \plus{} \frac{1}{b}}{4}$ is equivalent to $ 4a^2 b^2 \leq (a\plus{}b)(a^3 \plus{} b^3 )$ which is trivially true.

This yields $ Q \leq \frac{\frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c}}{2} \equal{} \frac{3}{2}$

[quote="BillClin"]Let $ a, b, c$ be three positive real numbers such that $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \equal{} 3$. Find the max value of $ Q \equal{} \frac {ab}{a^3 \plus{} b^3} \plus{} \frac {bc}{b^3 \plus{} c^3} \plus{} \frac {ca}{c^3 \plus{} a^3}$.

Thank.[/quote]
[b]Here is another solution[/b]:
It is very wel- known that $ a^3 \plus{} b^3\ge ab(a \plus{} b)$
$ \Longrightarrow Q \equal{} \frac {ab}{a^3 \plus{} b^3} \plus{} \frac {bc}{b^3 \plus{} c^3} \plus{} \frac {ca}{c^3 \plus{} a^3} \le \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \le \frac {1}{2}(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \equal{} \frac {3}{2} \Longrightarrow max Q \equal{} \frac {3}{2}$.


calculus
derivative
function
real analysis
real analysis solved

let $f$ be a continuous function on the open interval $]a,b[$.
Suppose that for every $x \in ]a,b[$, $lim_{h->0} \frac{f(x+h)+f(x-h)-2f(x)}{h^2}=0$.
Prove that $f=ax+b$

It is a direct consequence from http://www.mathlinks.ro/Forum/viewtopic.php?t=20566 , since we obtain that $f$ is convex and concave in the same time.


[quote="Allnames"]Let
\[ x^2 \plus{} y^2 \plus{} xy \equal{} 1
\]
Denote
\[ f(x,y) \equal{} |(m \plus{} 1)x^2 \plus{} (m \plus{} 2)y^2 \plus{} (m \minus{} 1)xy|
\]
Find the $ Min(Max(f(x,y)))$ there Min means  the minimal value and Max means the maximal one[/quote]
No one gives me a help. Where are you, mathlinkers being good at inequality ?


Se consideră paralelogramul ABCD, cu unghiul D>90. Perpendiculara din D pe AC taie cercul ABC în două puncte. Să se arate că unul dintre acestea este simetricul lui D faţă de AC, iar celălalt este ortocentrul triunghiului ADC.

Caut mai exact o soluţie sintetică...cu un pic de trigonometrie mi-a ieşit.

Sunt si eu curios! Levi, puteti presenta o solutie in acest sens?
 :D

Soluţia - sintetică, ce-i drept - nu mă mulţumeşte 100%. Cu ani în urmă, exista în manualul de clasa a IX-a o "problemă cu pătrat" care suna cam aşa: "în interiorul unui pătrat ABCD se ia un punct M astfel încît unghiurile
MCD şi MDC să fie ambele de măsură 15. Să se arate că triunghiul MAB este echilateral".

   Soluţia cea mai la îndemînă era considerarea unui punct M' astfel încît triunghiul M'AB să fie echilateral, urmată de demonstrarea faptului că M'=M. În ataşament (format .PDF), iată şi o altă soluţie - accesibilă şi elevilor de clasa a VI-a - şi care mie cel puţin îmi place mult mai tare. În spiritul acesteia sper să existe o soluţie şi pentru problema 11342...

[u]Principiul demonstratiei "directei" cu ajutorul "reciprocei" (care deseori este mai simpla sau chiar imediata ca in cazul nostru) prin utilizarea unei functii injective.[/u]

Sa reluam problema din clasa a IX - a care m-a fascinat si pe mine prin anii 1957 (eram in clasa a VIII - a), mai ales cand profesorul nostru ne-a oferit solutia (identica cu cea mentionata de tine, [i]Bpisai[/i]):
" Fie patratul $ABCD$ si un punct $M$ in interiorul acestuia si pe mediatoarea laturii $[AB]$.. Sa se arate: $m(\widehat {MCD})=15^{\circ}\Longleftrightarrow MA=AB$."

[u]Solutie.[/u] Notam mijloacele $M_1$, $M_2$ ale laturilor $AB$, $CD$ respectiv. Sa dam un grad de libertate punctului $M$ prin deplasarea acestuia pe mediatoarea laturii $[AB]$ catre latura $[CD]$. In timp ce punctul $M$ se apropie de latura $[CD]$,  masura unghiului $\widehat {AMB}$ descreste strict de la valoarea $\pi$ cand $M: =M_1\in [AB]$ pana la valoarea $2\arctan\frac 12<\frac{\pi}{3}<\pi$ cand $M: =M_2\in [CD]$. Deci va exista o singura pozitie a punctului $M$ pentru care $m(\widehat {AMB})=\frac{\pi}{3}$. Si acum vom putea folosi reciproca acestei probleme care este evidenta: $MA=AB\Longrightarrow m(\widehat {MCD})=15^{\circ}$.  Functia folosita $f\ : \ [M_1M_2]\rightarrow R,\ M\in [M_1M_2]$, $f(M)=m(\widehat {AMB})$ este injectiva.

[u]Observatie.[/u] Se pot oferi numeroase exemple de probleme de geometrie (si nu numai) cre se rezolva in aceasta maniera. Iata un exemplu semnificativ care este o problema cu adevarat dificila:

" Fie triunghiul $ABC$ pentru care $A=50^{\circ}$ si exista un punct $P$ in interiorul sau astfel incat $PB=PC$, $E\in PB\cap AC$, $F\in PC\cap AB$, $m(\widehat {PEF})=50^{\circ}$, $m(\widehat {PFE})=30^{\circ}$. Sa se determine $m(\widehat {ABE})$."

[u]Rugaminte.[/u] In ipoteza fericita ca ati rezolvat-o (nu neaparat cu principiul mentionat mai sus) postati solutia in continuare.

Eu stiu o solutie prin metoda reducerii la absurd la problema asta cu patratul. Apropo, ea apartine lui Ivan Block. ;)

Dar repede ai aparut, Cezar ! Nici nu am apucat sa termin redactarea corecta a reply-iului ca ai si strecurat un reply. Ce parere ai de acest principiu de rezolvare ? Poti oferi o aplicatie al lui tot in geometrie ? Daca da, te rog sa-l asezi tot aici. Cel care m-a invatat cu adevarat matematica si totodata m-a facut sa o iubesc este acest profesor de liceu de la care de fapt am invatat acest principiu.


Find all functions $f: \mathbb R^{+} \to \mathbb R^{+}$ satisfying
\[ (1\plus{}yf(x))(1\minus{}yf(x\plus{}y))\equal{}1, \quad \forall x,y \in \mathbb R^\plus{}.\]

Let $ g(x)\equal{}xf(x)$, then we get $ g(x\plus{}y)\equal{}\frac{(x\plus{}y)g(x)}{x\plus{}yg(x)}$.
If $ y\equal{}zx$, then $ g(x(1\plus{}z))\equal{}\frac{(1\plus{}z)g(x)}{1\plus{}zg(x)}$ or \[ g(t)\equal{}\frac{t}{t\plus{}g_1(x)}, g_1(x)\equal{}x\frac{1\minus{}g(x)}{g(x)},t\equal{}x(1\plus{}z).\]
It mean $ g_1(x)\equal{}const\equal{}a$. It give $ g(x)\equal{}\frac{x}{x\plus{}a},f(x)\equal{}\frac{1}{x\plus{}a}.$

[quote="∫FaiLurE∮"]Let $ R^ \plus{}$ be the set of all positive real numbers. Find all functions $ f: R^ \plus{} \to R^ \plus{}$ satisfying
$ (1 \plus{} yf(x))(1 \minus{} yf(x \plus{} y)) \equal{} 1$ ,$ \forall x,y \in R^ \plus{}$[/quote]

Set $ g(x) \equal{} \frac {1}{f(x)} \minus{} x$. The given equation is equivalent to $ f(x) \equal{} f(x \plus{} y) \plus{} yf(x)f(x \plus{} y)$ or $ g(x \plus{} y) \equal{} g(x)$ for all $ x,y \in \mathbb{R_ \plus{} }$. It follows that $ g$ is constant, hence $ f(x) \equal{} \frac {1}{x \plus{} k}$ for some arbitrarily chosen $ k \ge 0$.

[quote="FelixD"] ... hence $ f(x) \equal{} \frac {1}{x \plus{} k}$ for some arbitrarily chosen $ k > 0$.[/quote]

To be precise, $ k\equal{}0$ fits too.  :)

[quote="pco"][quote="FelixD"] ... hence $ f(x) \equal{} \frac {1}{x \plus{} k}$ for some arbitrarily chosen $ k > 0$.[/quote]

To be precise, $ k \equal{} 0$ fits too.  :)[/quote]

Of course it does^^... Thanks, pco.


geometry rectangle trigonometry function AMC USA(J)MO USAMO

[quote="Fierytycoon"]Hmm...is it customary for people to establish their own personal end of proof symbols?  :D[/quote]

I much prefer consistency in end of proof symbols as in other matters of mathematical notation. The "Halmos square" (a black filled-in rectangle) is all right, except that I see it used for purposes other than marking the end of a proof, even in math textbooks. I personally would end proofs with Q.E.D. if I were writing a math book myself (and would explain what "Q.E.D." means) but I have nothing against other conventions for showing the end of a proof.

I make my students write "QED" at the end of their proofs.  For a forum like this, using a smiley or such is fine, but it's probably not a good idea to use "invented notation" on a regular basis - it confuses everyone else needlessly.

Richard Feynman talks about this in one of his books - he had invented his own notation for trig functions and log functions which he used for awhile, until he had to waste too much time explaining to others what he was doing.

Sometimes when I'm bored I try to come up with original end-of-proof symbols, but I tend to stick with the Halmos symbol. Sometimes when I finish a proof online, I write $\blacksquare$.

i dont see why you should not make an end of proof symbol. If your proof has been finished, your grader will realize the fact that the its over, and the end symbol is only there for fun, or personality, or whatever. ive heard of people making up end of proof symbols at imo (gabriell carrol drew a gastropod i think) so i dont see what the problem is. a contest is not a book, it is a lot more personal, so i think it makes sense to end with your own symbol.

Neal's into QED a lot... i'll let him tell the rest of the story..

I don't think you should use a personalized end of proof symbol on an olympiad. The graders should not have any idea who is writing the proof since it may influence the score.

I meant for casual work, not for the USAMO or something.

well, if it were casual work, i would just spend 20 mintues drawing a very cool design!

I personally write the last two lines like this:
∴(for those of you who can't see this, it's the "therefore" sign with the three dots.) (statement)
(statement)

For example:
∴ 2x=y
x=y/2

or something like that. The whole point of using the same  mathematical symbols in every country in the world (such as =, numbers, etc.) is to reduce all the work translation is going to take. I mean, just by one country having it's own system of measurement, a whole rocketship or something like that :) fell from the sky. Think about what'll happen if some country described the number 1 as 1 and another country wrote 2 for it.
∴ I think we should stick with the orthodox symbols. Q.E.D. :)

Q.E.D doesn't look that nice in print, but so does the halmos symbol. In my opinion.

Yes, I think. He looks Swedish enough to me! :)
On the same note, tokenadult...*abruptly stops talking* :D

I feel kind of weird after thinking for like 2 month that you're Indian...
It's kind of odd when an avatar suddenly goes from Indian to Swedish... :)

Oh well, better than being mistaken for a girl [b]6[/b], count'em, [b]6[/b] times because of avatar :lol: (no offense to girls, just trying to make a point.)

that avatar that mindspa had was of ramujan (sorry if i misspelled it, especially since i am indian), i thought mindspa was indian also, but then i was corrected a few months ago

Yes, I did recognize his face because his picture was in my textbook.
but still, since avatar represents who you are, I thought he was Indian.

I was also surprised when you told me you were Indian. I thought you were American (or Caucasian) because of your avatar.
So it's kind of like a flip between you two. :)

Well probably no one had any doubt on what nationality I am  :D

lol

tare: my avatar is from the game chrono cross, the main character serge

lol[b]t[/b] :)
it'd be funni[b]er[/b] if you weren't actually Chinese :lol:  :P

I have never heard of that game. My philosophy on games, music, movies, etc.: Watch it, then decide whether you want to pay or not. :twisted:

tare i thought you were a girl for a really long time. especially from the pumpkin with pi carved out, that made me particularly confident.

Um, okay...the toll is up to nine now

[hide]at this rate, I could really fool a lot of people...no one knows that I'm a...[/hide]

besides, the pumpkin was made by a male sculptor, along with some other pumpkins i can not discuss here. It was pretty famous a long time ago when someone spammed it all over the place, but now, no ones knows the legacy


Give $ k \equal{} \frac {n \plus{} 1}{\sqrt {n^2 \minus{} 2n \minus{} 1}}$ with $ n \in N*, n > 2$
and $ x_1,..,x_n$ are $ n$ positive numbers such that $ x_i \in [\frac {1}{k},k]$
Prove that $ \frac {x_1}{x_2 \plus{} x_3} \plus{} \frac {x_2}{x_3 \plus{} x_4} \plus{} ... \plus{} \frac {x_n}{x_1 \plus{} x_2} \geq \frac {n}{2}$


The point $A(4, 0)$ is a vertex of regular hexagon $ABCDEF$, whose side is 8 and whose interior lies completely within Quadrant I.  If vertex $D$ has coordinates $(x, y)$, compute the ordered pair $(x, y)$.

[hide]
In order for all interior of hexagon to be in Q1, B has to be at least on x-axis and that force B(or F) to be on (0,4sqrt3)

D is the one opposite of A, so it is at (12, 8sqrt3)

sorry I cant use LaTex YET.

[/hide]

[hide]the hexagon can only fit if it has one leg along the x axis and one vertex touching the y axis, forming a 30-60-90 right triangle with legs $4, 4\sqrt{3}, 8$.  $D$ is twice the height of the triangle up, and 8 across, so $(4,0)+(8,8\sqrt{3})=\boxed{(12,8\sqrt{3})}$[/hide]

ah, i see geehoon beat me to it :lol:

Okay, you guys are now certifiably ready to succeed at NYSML.  Break a leg!


Okay so whats the deal with people wanting to run?

Somebody mentioned that a mentor might lead runs early in the morning?

I was hoping to do around 35 miles a week. Maybe more if the heat isn't so bad.

I'm up for it. Though I'm rather out of shape; I haven't run in the past week or so due to various reasons.

[quote="mdk"]you guys like running a lot, don't you[/quote]
When whatever God you believe in was handing out brains, you must have thought he said trains and asked for a slow one.

*zing*

LOL...

but a bit harsh.

I like running to the extent that I want to do it consistently. The main reason that I run is to stay in shape. Also my friends are destroying me in summer XC [in terms of training]. So I want to at least try to 'keep up.' Also running feels good...at times...I also want to have a good XC season. I did well as a soph [like top 10 at my school for my age]. But then as a junior I was lazy as hell. Now I want to at least be average. Hence, I want to do some serious running over the summer.

blargh I still don't get why you would want a slow train

like HEY THERE'S A TRAIN I HOPE IT'S PRETTY SLOW

I actually like running and run at least 5 fives a week, otherwise how can I play 30 minutes every game and in the track season win about every race I participate in.

haha, I'm down for early morning runs... just hope it doesn't get too hot. ><

oh trust me. it will be too hot. you'll just have to deal.

Well, I went last year and there was one mentor who brought us over to a cement loop about a third of a mile long and we ran there. It was like from 7:15- 7:45 and then class started 9:00. I'm actually not too sure about that time frame there... It wasn't too hot in the morning... only 98 degrees by the end of the run  :P . And by the end of the three weeks last year, it was just me running... Anyways, I need to run everyday probably since I'm, well, a track/xc devotee... I'm actually considering running after dinner from 7:00-8:30 because I might need at least an hour for running and I'm not the most fond of waking up crazy early. Yup. That's it. I'm not the greatest at math, but that's ok.  :lol:

[quote="PenguinIntegral"][quote="Altheman"]@humDdum: Well, I am motivated to run the whole 3 weeks. So you won't be alone. I am not too amped for running in a circle everyday. I'd like to run like 5 miles a day-that translates into 15 laps. Also cement is not so good for xc training...we need grass or something. Running late at night might work...but I am not sure.

I'm doing okay in like 85 degrees here.

AND I GOT TOTALLY SOAKED TODAY RUNNING. It was like...hey its raining...I am 2 miles away from home...[/quote]

So one day, I'm like a mile from home, right.

And there is this one street that goes straight to my house.

And on the side of the street I was downwind from, it was garbage night. I still have the shirt I wore, and it still smells.[/quote]

WOW THAT IS PRETTY INTERESTING

I'm a sprinter first and foremost, but I'm doing cross country next season in high school so I'm considering running 3 miles a day... if I do I will SUCK though. Wait so are you saying we should like run in a group? Too many different ages...

Altheman, what grade are you going intO?

Yea...I am not sure if group running is going to fly... I am almost sure that I will be running shirtless everyday [not only to conserve shirts, but because it will be hot]. I agree that the speed differences are hard for a group run. I also generally like to run alone unless it is like 90+ mins.

I will be a senior in high school. I am like average on our team. At the end of xc season last year, my time was like 19:02 for 3 miles [which was worse than my sophomore time of 18:30].

Oh and I was running today in like 90 degree weather. When I was going out, it wasn't so bad because there was a breeze. But if you are running in the heat with the sun beating down on you with no breeze, it is pretty bad. I had to cut my run short by like half a mile and just walk home.

I think it should be broken up into semi-equal sections, so that we do a loop of a landmark or something than come back to campus, so some people can stop, then the rest go and do a loop around another landmark, and come back and let people off, etc.

Also, another stipulation : if we, being manly math runners, run with our shirts off and hot girls chase after us at the sight of our six packs and huge biceps, and are like "Hey I haven't seen you around here, why are you here?", the answer is NOT "math camp".

I can do a mile in 3:20.

That's 3 hours and 20 minutes, half way through I usually set camp and cook a meal.

Why the - is this forum filled with totally unhelpful posts? Like we are actually trying to figure stuff out and then we just get cluttered with spam. Direct spam to the blogs  :D

I have not heard of a mentor doing running. Last year at AMP there were no mentors, only 1-4 students running.

Anyways, I'll run in the mornings.

If we were to run, I'd prefer like 6:30 to 7:00 or something obscenely early if possible...

anyhow just ran two miles, 15:25 =/ so not sure how many people will be that slow.

I really should get a watch to like give myself a wake-up call for my slow ---. Anyway, I think it is pretty obvious that we need to have it so people can just stop. The thing is that there is no shame in stopping if you are overheating. Like usually it is good to push yourself when running, but pushing yourself in heat is the major exception.

I have been reading a bit about running in the heat. I guess the main point is that [b]you need to hydrate well[/b]. That means like drinking like a pint of water before you start. Then drinking some water or w/e every 15-20 mins. Other considerations are: a light sunscreen like SPF 15, avoiding direct sunlight, wearing sunglasses, wearing special running clothing. In my totally unqualified opinion, these last few things are not that important since I run without them and I am fine. To jump on the 'PenguinIntegral' bandwagon, we might wear sunglasses to look awesome.  :lol:

I guess for timing, we should get out pretty early. I might just like start running at like 7:15. I'd like to find a mile loop or something. That way I could just like run 5 of them and be done. Obviously that would get boring, but we will see.

it looks like the first week is going to be nice and cool (low 90s), but I am sure that it will get into the 100's by the time we leave (yay Texas weather)

Actually it looks like it's going to be Seattling (raining) on us for the first week or so as it's been doing almost all summer.Hydration is important, and I would definitely recommend some sort of wicking material to run in. No sunscreen should ever be necessary unless you're albino, and if you're running in the morning, chances are the sun will be staring at you or, more likely, vice versa, so sunglasses will come in handy.

My violin teacher recommends we bring boric acid to kill giant tree roaches.
 :huh:

sweet. That's reassuring. Math is hard enough, let alone math while killing giant bugs.
thanks =)

sorry to interrupt, but that is pretty slow

I'm starting to think that night running will definitely be better than morning running. Definitely.

I e-mailed the UTD xc coach. Provided that he responds, I will do what he suggests.

In my school's xc league, 16 minutes for 3 miles is quite fast for guys, 19 minutes is quite fast for girls. We have four or five guys at 16:20 and three or four girls under 19.

hey, that's like my MC countdown speed :D
three problems in 16 minutes is pretty good xP

maybe I can get someone to sneak us out on wednesdays to run with the greenhill team...zach maybe?

btw im a girl. just thought you needed to know. :whistling: shirts on would be quite nice.

:D mkay, no i did not realize that

co-ed running doesn't tend to fly...as i mentioned earlier...but w/e, we'll see

[quote="Fanatic"]mdk, were you talking to me?

14:29 yesterday... dunno how i got so much faster. first day my knees hurt though, so bad I had to run on the grass for a few stretches of like 100 meters.

Edit: fine I know there are like beasts that do 3 miles in 16 minutes, but
A) I'm a sprinter
B) I'm entering 9th grade[/quote]

yes, I was talking to you, but that's okay, you're still young


My guess [hide="is"]8. Since x-3 has to be 5 if the equation at least makes sense, there has to be 8 there originaly. We can check: (x/2)^(1/2) + 1=(8/2)^(1/2)+1=4^(1/2)+1=3, which is 6/2 as well.[/hide]

EDUCATED guess.

[quote="neelnal"]The are x apples on a tree. Jack takes (x-3)! - 118 apples. Jill then takes half of the remaining. The number of apples remaining is (x/2)^(1/2) + 1.

How many apples were originally on the tree.

Try not to use brute force to solve this. It is about a #12 or #13 on the AMC 12 level question, so not very hard or very easy, but some where in the middle and requires work to do it without simply plugging in numbers and is just for fun. I'll give the answer after people have a jab at it.[/quote]

[hide]The answer has to be at least 8, because $4!$<118, and we can't have negative apples. I quickly see that 5 is the answer, because $5!=120$, $120-118=2$ $8-2=6$, $6/2=3$, $8-2-3=3$,
$8/2=4$, 4^1/2=2, $2+1=3$[/hide]

Does that count as brute force? Sorry :rotfl:

mustafa I was wondering could you calculate 6! as well?
As it comes to 720, which means 602were taken; which is over the original number and as the factorial increases, it increases more.  Also, as the answer cannot be below 5! and the factorial has to be an integer as there is no fraction or negative factorial, the only answer is 5! during the working?
Just a thought.  This may mix up the problem but definitely proves the solution.  This doesn't seem to be that brutal as we are using numbers given to us, not guessing random digits.  Also this and your solution shows workings.

I'm looking forward to the solution you have neelnal.

[quote="neelnal"]The are x apples on a tree. Jack takes (x-3)! - 118 apples. Jill then takes half of the remaining. The number of apples remaining is (x/2)^(1/2) + 1.

How many apples were originally on the tree.

Try not to use brute force to solve this. It is about a #12 or #13 on the AMC 12 level question, so not very hard or very easy, but some where in the middle and requires work to do it without simply plugging in numbers and is just for fun. I'll give the answer after people have a jab at it.[/quote]
[hide="non brute force"]
$\frac{(x-3)!-118}{2}=\sqrt{.5x}+1$
$(x-3)!-118=\sqrt{2x}+2$
$(x-3)!-120-\sqrt{2x}=0$
at which point we go to the graphing calculator :P[/hide]
EDITED SOL. Just realized I made an error

[quote="Farenhajt"]Or we just note that $\sqrt{{x\over 2}}$ must be integer, hence $x$ is of the form $2k^{2}$. For $k=1$ the number is too small, for $k=3$ is already too large ($(x-3)!=15!$), and $k=2$ is just right.

(Boy, does this sound like Goldilocks...)[/quote]

 :rotfl:


Let $a,$ $b$ and $c$ are positive numbers. Prove that
$\sqrt[3]{2(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+9}\geq\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}.$

Cubing both sides with $x^{3}= \frac{a}{b}$, etc, reveals that it is equivalent to
$\sum x^{4}yz+2x^{3}y^{3}+3x^{2}y^{2}z^{2}\geq  \sum 6x^{3}y^{2}z$, where sums are symmetric, i.e. Schur times $xyz$ and Schur on $xy, xz, yz$, so it seems pretty strong; not sure how to actually do the problem yet, of course (I'm playing around with weighted Jensen on $f(t) = \sqrt[3]{t}$?)

[quote="mathpk"][quote="MysticTerminator"]
$\sum x^{4}yz+2x^{3}y^{3}+3x^{2}y^{2}z^{2}\geq \sum 6x^{3}y^{2}z$[/quote]
R.H.S. must be $\sum_{sym}3x^{3}y^{2}z$.[/quote]

No, MysticTerminator is right. Note that LHS has sym sum also, not cyc sum.

[quote="mathpk"]How to do this one arqadi? :D[/quote]
In fact, MysticTerminator has solved this problem and in accuracy by mine way:
$\sqrt[3]{2(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+9}\geq\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\left(\frac{a}{b}+\frac{2a}{c}+3\right)\geq3\cdot\sum_{cyc}\left(\sqrt[3]{\frac{a^{2}}{bc}}+\sqrt[3]{\frac{bc}{a^{2}}}\right)\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(x^{3}-x^{2}y-x^{2}z+xyz)+2\cdot\sum_{cyc}(u^{3}-u^{2}v-u^{2}w+uvw)\geq0,$
where $x=\frac{1}{u}=\sqrt[3]{\frac{a}{b}},$ $y=\frac{1}{v}=\sqrt[3]{\frac{b}{c}}$ and $z=\frac{1}{w}=\sqrt[3]{\frac{c}{a}}.$ :)

I think, the problem for two variables in very famous, it's from Hojoo Note!?!


[quote="zhou_qiang"]The answer is 30.[/quote]

do you care to show the method?

[quote="zhou_qiang"]The answer is 30.[/quote]

Please show why it has to be 30, since I haven't found an answer yet.

[quote="xxxyyyy"]$ PQR$ is a triangle with $ PQ = 33$, $ PR = 21$ and $ QR = m$, an integer.

There are points $ D$, $ E$ on the sides $ PQ$, $ PR$ respectively such that $ PD = DE = ER = n$, an integer.

Find $ m$.[/quote]

Consider the triangle $ PDR$ with $ PD=DE=ER=n$ and $ PR=q$.
Let $ \theta = \angle ERD = EDR$. Then $ \angle DEP = \angle DPE = 2\theta$

[img]10058[/img]
Let $ F\in PD$ such that $ RF=RP$. Then $ \triangle RPF \sim \triangle DPE\Rightarrow \frac{PF}{RP}= \frac{PE}{DP}\Rightarrow \frac{PF}{q}= \frac{q-n}{n}\Rightarrow$

$ PF = q\cdot \frac{q-n}{n}= \frac{q^{2}}{n}-q \ \ (1)$

[img]10059[/img]

Let $ w$ be the circle centered at $ R$ and with radius $ q$

Let $ Q$ on the ray $ PF$ and set $ QP=r$ and $ QR=m$

$ \blacksquare$ [color=blue][b]case 1)[/b] $ Q$ is in the extension of the segment $ PF$[/color]

Then $ QF \cdot QP = QR^{2}-q^{2}\Rightarrow QF = \frac{m^{2}-q^{2}}{r}$

$ PF = PQ-QF \Rightarrow PF = r+\frac{q^{2}-m^{2}}{r}$

$ \blacksquare$ [color=blue][b]case 2)[/b] $ Q$ is in the interior of the segment $ PF$[/color]

Then $ QF \cdot QP =-(QR^{2}-q^{2}) \Rightarrow QF = \frac{q^{2}-m^{2}}{r}$

$ PF = PQ+QF \Rightarrow PF = r+\frac{q^{2}-m^{2}}{r}$
$ \hline$
At any position of $ Q$ on the ray $ PF$ we have $ \boxed{PF = r+\frac{q^{2}-m^{2}}{r}}\ \ (1)$ (even if $ Q\equiv F$)

From $ (1)$ and $ (2)$ we find that

$ \frac{q^{2}}{n}-q = r+\frac{q^{2}-m^{2}}{r}\Rightarrow$

$ q^{2}r-qnr = r^{2}n+(q^{2}-m^{2})n\Rightarrow$
$ q^{2}r = (r^{2}+q^{2}+qr-m^{2})n\Rightarrow \boxed{n = \frac{q^{2}r}{r^{2}+q^{2}+qr-m^{2}}}\ \ (3)$
$ \hline$

Applying to $ (3)$ the data of the problem we get that $ n = \frac{21^{2}\cdot 33}{33^{2}+21^{2}+21\cdot 33-m^{2}}= \frac{14553}{2223-m^{2}}$

So we have to find $ m$ such that $ 2223-m^{2}\mid 14553$

Observing within the possible values of $ m\in\{13,14,...,53\}$ we find that only $ m=30$ gives solution, with corresponding $ n=11$

First use cosine rule. We will get n^2=n^2+(21-n)^2-2(n)(21-n)cos(angleP). Evaluate the expression of cosP in terms of n.Then apply cosine rule again to tiangle PQR, and since m is an integer, we can factor it out.


How did NJ'ers do?

I got 139.5 (amc 10a) and 6 aime.

126/8=206 :mad:

2 perfect scorers on AMC10 A were both from AAST :D

:| 133.5/10 .. kind've disappointed as a senior


Hi,

can someone suggest me a text or a book which introduces to the different topics of inequalities?

Here is it  :wink:

http://www.animath.fr/