To any of the adult ARML team leaders, or anyone who knows stuff about ARML, can you explain to me the process of forming an ARML team?  How complicated is it?  What are the expenses? etc...

Thanks

You might also ask that question of Setofallsets, who was (is?) in the process of forming an ARML team for parts of VA that don't get to go to TJ.

By the way, I'd be interested in hearing how that's turning out.

Different regions organize their ARML teams in different ways. In some states, like my own, the ARML team is chosen, organized and paid for a math teachers organization. Here the ARML Team is a subset of the SC All-State Math Team, which is chosen by the SC Council of Teachers of Mathematics via a series of tests. Once the members are selected for the team, all costs of the trip are paid for by the SCCTM.

In other regions, teams are somewhat more self-organized. Either by schools or by students themselves or by ARML alumni in the region. In some regions the students have to pay a portion of the cost themselves, and in some other regions the teams have fundraisers or corporate sponsors to pay for the trip.

If you're from Michigan, I believe they already have a team. Here's the website: http://www.albion.edu/math/arml/

I don't know how the team is chosen or organized, but I would bet it has something to do with the results of Michigan Mathematics Prize Competition. Information about that may be on the webpage somewhere.

this year, michigan does not have an arml team - at least its not sponsored by the people as last year. yeah, its sad. ragging, if you form a team, count me in.

Yea... Thats why I was asking Tetrahedron.... I don't think I could get in to the regular ARML team, but it's not even forming this year.  :cry:

sadly,  i would have made it to the mi arml team this year, since im one of the mmpc top 100, and generally any of these people can go to arml. but not this year. anyway, we would compete in iowa, is that correct? hmmm... the cost for a team that students could organize seems to be very high.

Isn't Mr. Rusczyk forming one?  How much does it cost?

We're building a team for the SD area; we still haven't finished researching all the costs.  I think the team fee was $250 in recent years, plus lodging and food (not sure what that is yet).  There's also transportation and (for whoever is running the team) liability insurance to get pricing on.

SD is no longer going with San Francisco bay???

The cost of an ARML team is actually fairly high.  You have (I am giving the amounts in respect to Baltimore's team going to Penn State)

1) registration cost (~550)
2) food cost (usually teams try to pay for the students) (~300)
3) transportation cost (~2k)

which means basically 3000.

setsofallsets of trying to form a none TJ ARML team?? That's fairly cool, because we can share a bus setofallsets (email me).  Recruiting students is quite a difficult task for teams that do not have a history.  Mr. Rusczyk, I don't know what kind of liability insurance you are getting, but NYC or Baltimore have never gotten it, we use waive all your rights and your soul forms instead. ;)

San Diego hasn't been going with SFBA for maybe 12 years. (It was a California team back then.) Usually each student from SFBA has to pay somewhere around $100 to go to ARML, but we don't have to deal with transportation costs. (When it was at UNLV, the cost was $250 per student since we had to get plane tickets.)

I was told this year due to funding issue, Michigan will not have ARML team. However, I was also informed that Dr. Khetan from ICAE (http://www.icae.org) is forming a team from Michigan to participate. If interested, you can contact him to see if you can be part of his team.

As for liability, if those teams are chaperoned by teachers, I would have guessed that they're essentially covered by their unions.  I believe waive your soul forms are of little protection.


How many odd divisors does the number $ 5^6\minus{}1$ have? How many even divisors does it have?

[hide]We have
\[ 5^6 \minus{} 1 \equal{} 5^6 \minus{} 1^6 \equal{} (5^3 \minus{} 1)(5^3 \plus{} 1) \equal{} (5 \minus{} 1)(5^2 \plus{} 5 \cdot 1 \plus{} 1^2)(5 \plus{} 1)(5^2 \minus{} 5 \cdot 1 \plus{} 1^2) \equal{} 4 \cdot 31 \cdot 6 \cdot 21 \equal{} 2^3 \cdot 3^2 \cdot 7 \cdot 31
\]
This is the prime factorisation.

The odd divisors of this number are simply the divisors of $ 3^2 \cdot 7 \cdot 31$ and there are $ 3 \cdot 2 \cdot 2 \equal{} 12$ odd divisors. Since $ 5^6 \minus{} 1$ has $ 4 \cdot 3 \cdot 2 \cdot 2 \equal{} 48$ divisors in total, there must be 48 - 12 = $ 36$ even divisors.[/hide]

$ 5^6\minus{}1\equal{}(5\minus{}1)(5^2\plus{}5\plus{}1)(5\plus{}1)(5^2\minus{}5\plus{}1)\equal{}2^3.3^2.7.31$
generally if $ n\equal{}2^a\prod_{j\equal{}1}^{h}p_j^{c_j}$ where $ p_j$ primes $ >2$ then there are $ \prod_{j\equal{}1}^{h}(c_j\plus{}1)$ odd divisor and  $ a\prod_{j\equal{}1}^{h}(c_j\plus{}1)$ even divisors.

here  there are $ (2\plus{}1)(1\plus{}1)(1\plus{}1)\equal{}12$ odd divisor and  $ 3(2\plus{}1)(1\plus{}1)(1\plus{}1)\equal{}36$ even divisors.

Yes, it's pretty straightforward, given that this is Problem 1 of Day 1.


I have heard that there is a way to make almost invisible text by going like this: [hide= ]message[/hide].  And then there is supposed to be a way to click on it or something and see the "invisible" text, but I can't seem to figure out how.  Is there a way?  And if so, how do you do it?

The slash is the wrong way. It's the forward slash (/), not the backslash (\). Also, the first tag does not have a slash. And you view it by quoting.

There are two types of hide tags. One is [code][hide]"invisible text"[/hide][/code] which turns out to be
[hide]"invisible text"[/hide]

The other type is [code][hide=this is hidden text]"invisible text"[/hide][/code] which is used if you want to include a "title" for your hide tag. It turns out to be
[hide="this is hidden text"]"invisible text"[/hide]


any of u play tactics arena online (http://www.digisonline.com/tactics/play) ??
what set-up do you guys use? (describe or printscrn)

if u guys haven't played it before...it's totally awesome..great game to keep you awake at school...most schools don't know about it so it shouldn't be blocked (well now after telling everyone...maybe they'll start noticing lol)

I've played it, but it got a little boring try [url=http://games.swirve.com/utopia/]Utopia[/url]!  :D


ARML
number theory
prime factorization
AMC

[color=cyan]23! can be written as the product of n consecutive positive integers for three values of n.  Find these three values.[/color]

hmmm
sounds like an ARML power question to me...

[hide]

Well, the obvious first case is 23 consecutive integers (1-23).

Then there's 22 consecutive integers (2-23).

Lastly we see that 2*3*4 happens to equal 24. So the product of the integers (5-24) equals 23!... that's 20 consecutive integers.

[/hide]

[color=cyan]What 3 consecutive integers multiplied together give you 23!?[/color]

uh, damn, this is so like an ARML power question i did

I distinctly remember that the complete solution would require the use of the prime factorization of 23!, or something like that.... hopefully


So, when we have SFFT,  how do we know when we'll have integer solutions?  For example

$ 3ab \minus{} 4b \minus{} 4a \minus{} 12 \equal{} 0$.

That has no integer solutions for EDIT : $ 3 < a < b$ (I think).  (It doesn't factor, but it looks like SFFT).

I think I took the equation mod 4, so $ ab$ has to be $ 0 \mod 4$. From there, I saw positive integer solutions..but like, how do we know that there won't be any integer solutions satisfying the condition?

whoops, I meant $ a>3$.  It relates to one of the other problems on AoPS.

It was the one with $ abc\equal{}4(a\plus{}b\plus{}c)$, I plugged in $ a\equal{}3$, but $ b,c > 3$, and I don't think there were solutions.

What Al is trying to say is that an SFFT manipulation will eventually result in some conditions that the divisors of a certain integer (in this case $ 52$) must satisfy (here, congruence to $ 2 \bmod 3$).


function
real analysis
real analysis unsolved

Suppose $\mu()$ is a Radon measure in $\mathbb{R}^{n}$ on some $\sigma$-algebra $\Sigma$ of subsets of $\mathbb{R}^{n}$. Let $f: \mathbb{R}^{n}\rightarrow \mathbb{R}^{m}$ be continuous and proper (i.e. inverse images of open and compact sets are open and compact, resp.) with $f^{-1}(b)$ Borel for $b$ Borel.

Do we know $f^{-1}(f(A)) \in \Sigma$ if $A \in \Sigma$?

Define $f\colon\mathbb R^{2}\to\mathbb R^{2}$ by $f(x_{1},x_{2})=(x_{1},(|x_{2}|-1)^{+})$. I think this map satisfies all of the above assumptions. Let $\Sigma$ be the $\sigma$-algebra of Lebesgue measurable sets in $\mathbb R^{2}$. Let $E\subset \mathbb R$ be a nonmeasurable set. Then $A=E\times\{0\}\in\Sigma$, but $f^{-1}(f(A))=E\times [-1,1]\notin\Sigma$.

But  $f^{-1}((0,1)\times{0})$ isn't open, so your function isn't continuous.  Sorry.

Basically, I want to know if $\mu(f^{-1})()$ is Radon, but the only way I can do that is to construct a $\sigma$-algebra of Radon measurable sets in $\mathbb{R}^{m}$ from $\Sigma$ in $\mathbb{R}^{n}$, right?

Or can we show $\mu(f^{-1})()$ is Radon without doing so?

$f(x_{1},x_{2})=(x_{1},(|x_{2}|-1)^{+})$ is not continuous? :what?:  Its range is $\mathbb R^{2}$, not $\mathbb R$.

To prove that something is a Radon measure, you can consider its action on some space of continuous functions and apply some kind of a Riesz representation theorem.

OOPs :blush: .  I probably interpreted the question wrong then..haha.


Does anyone know how to prove this:

[color=green]Suppose that R is a ring. Show that -(a+b) = (-a)+(-b) for a, b in R.[/color]

I've no idea how to prove it.

Thanks guys.


Let $ABCD$ be a cyclic quadrilateral with $M$ as the point of intersection of its diagonals and circumcentre $O$.
A line through $M$ and perpendicular to $OM$ cuts $AB$, $CD$ in $X$, $Y$ respectively. Prove That $MX=MY$.

[quote="1=2"]Could you PROVE the [[Butterfly Theorem]]??[/quote]
[url=http://en.wikipedia.org/wiki/Butterfly_theorem]Wikipedia[/url]  is a very useful site. :lol:


The area of a rectangle is 80 sq ft. If the width is increased by 1 foot and the length is decreased by 1 foot, the new rectangle has area 90 sq ft. What is the perimeter of the original rectangle, in feet?

$ \left\{\begin{array}{l}
xy\equal{}80 \\
(x\plus{}1)(y\minus{}1)\equal{}90
\end{array}\right.$

So, $ (x\plus{}1)(y\minus{}1)\equal{}xy\minus{}x\plus{}y\minus{}1\equal{}90 \iff \minus{}x\plus{}y\equal{}11$.
Hence, $ x(11\plus{}x)\equal{}80 \iff x^2\plus{}11x\minus{}80\equal{}(x\plus{}16)(x\minus{}5)\equal{}0$.
Therefore, $ x\equal{}5, \ y\equal{}16$.

Thus, $ 2x\plus{}2y\equal{}2(x\plus{}y)\equal{}2(5\plus{}16)\equal{}2\times21\equal{}\boxed{42 \ \text{feet}}$.


trigonometry
trig identities
Law of Cosines

I would use law of cosines:  first, we have $c^{2}=n^{2}+q^{2}-2\cdot6\cdot3\cos{60}=45-18=27$, so $c=\sqrt{27}=3\sqrt{3}$.  Then, we use law of cosines again:  $\cos{Q}=\frac{n^{2}+c^{2}-q^{2}}{2nc}=\frac{0}{18\sqrt{3}}=0$.  Then $\angle C=\cos^{-1}{0}=90$ or $270$.  If it were 270 the sum of the angles would be too big so its 90
QED

[quote="DiscreetFourierTransform"]Assume NCQ is right:
$\tan 60 = \frac{x}{3}$
Tan 60 always equals $\frac{\sqrt{27}}{3}$, regardless of whether the triangle is right or not.

$3 \cdot \tan 60 = x$
By trig identities,
$3 \cdot \frac{\sqrt{27}}{3}= \sqrt{27}$

And this is X, by pythag. thm:
$x^{2}+3^{2}= 6^{2}$
$x = \sqrt{27}$

Thus NCQ is right. QED.[/quote]
I'm pretty sure you can't really assume what you're trying to prove.

[quote="fireemblem13"]
I'm pretty sure you can't really assume what you're trying to prove.[/quote]
I agree.
You can asusme the opposite of what you are trying to prove, and then prove your assumption wrong...

This is known as circular logic.
[hide="one liner"][size=75]All triangles in this setup are congruent with the SAS and the 30-60-90 works so it is congruent to a 30-60-90[/size][/hide]

[quote="Klebian"]In this kind of diagram:
$\setlength{\unitlength}{.5cm}\begin{picture}(5,4) \put(.5,.5){\line(5,0){5}}\put(.5,.5){\line(0,3){3}}\put(.5,3.5){\line(5,-3){5}}\put(-.1,3.5){N}\put(-.1,0){Q}\put(5.5,0){C}\put(2.75,0){3}\put(3,2.1){6}\end{picture}$

If I know angle C is 60 degrees and I am trying to prove Q is right, is it sufficent to say that because cos(60) = 1/2 and QC/NC = 1/2, Q is right? Or is there more I must explain?

[size=59](diagram thanks to I_like_pie's latexing skillz)[/size][/quote]

wow... I_like_pie has crazy texing skillz

yes.  im pretty sure thats all you need to say

you want to use SAS similarity with another 30-60-90 triangle

Are you trying to prove for the general 30-60-90 triangle?


Let $ a,b,c > 0$ prove that \[ \sum_{cycl}\frac{b\minus{}c}{b\plus{}c} \ge 4\sum_{cycl}\frac{a\minus{}c}{2a\plus{}b\plus{}c}\]

[quote="shfdfzhjj"][quote="apollo"]Let $ a,b,c > 0$ prove that
\[ \sum_{cycl}\frac {b \minus{} c}{b \plus{} c} \ge 4\sum_{cycl}\frac {a \minus{} c}{2a \plus{} b \plus{} c}
\]
[/quote]

I think it is wrong.[/quote]
Why  :?:

try $ a\equal{}b$
the following is true :
$ \sum_{cyc}\frac{b\minus{}c}{b\plus{}c}\geq 8\sum_{cyc}\frac{a\minus{}c}{a\plus{}2b\plus{}c}$ but is easy.

[quote="Albanian Eagle"]try $ a \equal{} b$
the following is true :
$ \sum_{cyc}\frac {b \minus{} c}{b \plus{} c}\geq 8\sum_{cyc}\frac {a \minus{} c}{a \plus{} 2b \plus{} c}$ but is easy.[/quote]

Oh,yeah
I sloved it. :)

[quote="Sunkern_sunflora"]Is it MV that you used shfdfzhjj to solve the problem?[/quote]

What do you mean? :maybe:

[quote="shfdfzhjj"]What do you mean? :maybe: [/quote]
Mixed variables! :wink:

[quote="Albanian Eagle"]
the following is true :
$ \sum_{cyc}\frac {b \minus{} c}{b \plus{} c}\geq 8\sum_{cyc}\frac {a \minus{} c}{a \plus{} 2b \plus{} c}$ but is easy.[/quote]
I think it's wrong. Try $ a\equal{}3,$ $ b\equal{}1$ and $ c\equal{}2.$ :wink:

[quote="Albanian Eagle"]sorry. i meant $ a\geq b\geq c$. It is obvious right?[/quote]
Now it's right!


Hi, all --

The HCSSiM alumn database is now available, so if you're an alumn or former staff member, go and either check on your data or enter it fresh.  (And if you know any alumns or former staff members who aren't on AoPS, which includes some of the HS teachers of AoPS members for example, please let them know about this.  I think we have non-bouncing email addresses for roughly half of the HCSSiM alumns.)  In particular, we don't have the 2005 info auto-entered, so the several of you AoPSers who are also HCSSiM 2005 should go enter your data (except for MC alderbank who already did).

The place to go is http://www.hcssim.org/alumns.htm which has a link to the database.  The whole thing is password-protected for privacy from spiders (if you have trouble getting in, let me know) as is each individual record.  Questions?  Just ask.

Great idea! Thanks for setting this up, s-m! Now I can stalk all of my fellow HCSSiM alums... :D


inequalities
ratio
inequalities unsolved

Prove the inequality if $ a,b,c\ge 0$
$ \frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{ab\plus{}bc\plus{}ac}\plus{}\frac{8abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\ge 2$

[quote="RAGHAV GROVER"]Prove the inequality if $ a,b,c\ge 0$
$ \frac {a^{2} \plus{} b^{2} \plus{} c^{2}}{ab \plus{} bc \plus{} ac} \plus{} \frac {8abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\ge 2$[/quote]

It's easy by SOS  :lol:  or "S;S" can slove it  :)

$ \frac{(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2}{2(ab\plus{}ac\plus{}bc)}\minus{}\frac{a(b\minus{}c)^2\plus{}b(c\minus{}a)^2\plus{}c(a\minus{}b)^2}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\ge 0$

$ \sum_{cyc} (\frac{1}{2(ab\plus{}ac\plus{}bc)}\minus{}\frac{c}{(a\plus{}b)(b\plus{}c)(c\plus{}a)})(a\minus{}b)^2\ge 0$

$ \sum_{cyc}(a^2b\plus{}a^2c\plus{}b^2a\plus{}b^2c\minus{}ac^2\minus{}bc^2)(a\minus{}b)^2\ge 0$

WLOG $ a\ge b\ge c$

Let $ x\equal{}a^2b\plus{}a^2c$, $ y\equal{}b^2a\plus{}b^2c$, $ z\equal{}c^2a\plus{}c^2b$. (note that $ x$ is clearly the largest).

Then we have to show $ \sum_{cyc}(x\plus{}y\minus{}z)(a\minus{}b)^2\ge 0$

This is true since $ S_b,S_b\plus{}S_c,S_b\plus{}S_a$ are all positive, so we are done by S.O.S.

I think, using uvw theorem, we can prove it like this:
Let $ 3u\equal{}a\plus{}b\plus{}c, 3v^2\equal{}ab\plus{}bc\plus{}ca; w^3\equal{}abc$
The inequality$ \implies$
\[ \frac{3u^2\minus{}2v^2}{v^2}\plus{}\frac{8w^3}{9uv^2\minus{}w^3}\geq 2\]
\[ \iff \frac{u^2}{v^2}\plus{}\frac{24uv^2}{9uv^2\minus{}w^3}\geq 12\]
Now, fix $ u,v^2\implies$ it is enough to prove when $ w^3$ is minimized. So we only need to consider the cases i)two variables are equal ii) any of them=0
i) If $ a\equal{}b$, we introduce a new ratio $ t\equal{}\frac{c}{a}$ then the inequality implies
\[ \frac{2\plus{}t^2}{1\plus{}2t}\plus{}\frac{4t}{t^2\plus{}2t\plus{}1}\geq 2\] which is true from the trivial inequality (it gives $ (t^2\minus{}t)^2\geq0$)
ii) WLOG $ a\equal{}0$. Then the inequality becomes: $ \frac{b^2\plus{}c^2}{bc}\geq2$, perfectly true.

This completes our proof.
Arqady, am I correct?  :blush:


Let $ f(x) \equal{} a_0 \plus{} a_1x \plus{} a_2x^2 \plus{} a_3x^3 \plus{} a_4x^4$ where $ a_0$, $ a_1$, $ a_2$, $ a_3$ and $ a_4$ are constants with $ a_4 \neq 0$. When divided by $ x \minus{} 2003$, $ x\minus{} 2004$, $ x \minus{} 2005$, $ x \minus{} 2006$, and $ x \minus{} 2007$, respectively, $ f(x)$ leaves a remainder of $ 24$, $ \minus{}6$, $ 4$, $ \minus{}6$, and $ 24$. Find the value of $ f(2008)$.

[hide="Solution"] In other words, $ f(2003) \equal{} 24$, etc.  $ g(x) \equal{} f(x \plus{} 2005)$ is an even polynomial, hence $ g(x) \equal{} a_4 x^4 \plus{} a_2 x^2 \plus{} 4$.  Then $ g(1) \equal{} a_4 \plus{} a_2 \plus{} 4 \equal{} \minus{}6 \implies a_4 \plus{} a_2 \equal{} \minus{}10$ and $ g(2) \equal{} 16 a_4 \plus{} 4 a_2 \plus{} 4 \equal{} 24 \implies 4 a_4 \plus{} a_2 \equal{} 5$ which gives $ a_4 \equal{} 5, a_2 \equal{} \minus{}15$.  Then

$ f(2008) \equal{} g(3) \equal{} 5 \cdot 3^4 \minus{} 15 \cdot 3^2 \plus{} 4 \equal{} \boxed{ 274 }$. [/hide]

[quote="t0rajir0u"]$ g(x) \equal{} f(x \plus{} 2005)$ is an even polynomial, hence $ g(x) \equal{} a_4 x^4 \plus{} a_2 x^2 \plus{} 4$. [/quote]
How does the above follow?

[hide="Well..."] $ g(1) \equal{} g(\minus{}1) \implies a_1 \plus{} a_3 \equal{} 0$
$ g(2) \equal{} g(\minus{}2) \implies 2a_1 \plus{} 8a_3 \equal{} 0$.

Together, we get $ a_1 \equal{} a_3 \equal{} 0$.  In other words, five points uniquely determine a quartic, and we can give an even quartic that works, so it must be the only one. [/hide]

I see.. thanks! I had no idea about how to start this one


Let a,b,c>0
Prove that
$ a^3(b\minus{}c)^2\plus{}b^3(c\minus{}a)^2\plus{}c^3(a\minus{}b)^2 \geq (a\minus{}b)(b\minus{}c)(c\minus{}a)(ab\plus{}bc\plus{}ca)$

Thanks  :)

Inequality is not true.

  Contraexample  $ b\equal{}1,  c\equal{}2, a \to\infty$


group theory
abstract algebra
graph theory
superior algebra
superior algebra theorems

First of all, is it there a theorem that states that if $\sigma,\sigma'\in S_{n}$, then the pair $\left(\sigma,\sigma'\right)$ generates $S_{n}$ if $\textrm{order}\left(\sigma,\sigma'\right)=n!$?  Also, I can see why that might be useful in terms of graph theory, but could someone show me a proof for this?

Thank you.

Is $\mathrm{order}\left(\sigma,\sigma'\right)$ the number of permutations generated by $\sigma,\sigma'$¿
Then its really not hard: the generated elements form a subgroup of $S_{n}$, and since it has the same order than the group $S_{n}$, it is $S_{n}$ itself.


limit
function
logarithms
trigonometry
LaTeX
calculus
calculus computations

let $ f_n(x)\equal{}\frac{1\minus{}\cos(x)\cos(2x)...\cos(2^nx)}{x^2}$
and $ u_n\equal{}\lim_{x\rightarrow 0}f_n(x)$
1) prove that $ u_{n\plus{}1}\equal{}u_n\plus{}2 . 4^n$
2)find $ u_n$ in function of $ n$.

Define:

$ \ln(g_n(x)) \equal{} \sum^{n}_{i \equal{} 0}\ln\left(\cos\left(2^i x\right)\right)$

then we have:

$ u_n \equal{} \lim_{x \rightarrow 0}\frac {1 \minus{} g_n(x)}{x^2}$

Now:

$ \ln(g_n(0)) \equal{} \sum^{n}_{i \equal{} 0}\ln(1) \equal{} 0$ so we can use L'Hopital's rule and get:

$ u_n \equal{} \lim_{x \rightarrow 0}\frac { \minus{} g_n(x) \sum^{n}_{i \equal{} 0}2^i \tan\left(2^i x\right)}{2x}$

Now, we know that $ g_n(0) \equal{} 1$, but the sum equals $ 0$ for $ x \equal{} 0$, so we can use L'Hospital's once more and get:

$ u_n \equal{} \minus{} \frac {1}{2}\lim_{x \rightarrow 0}\left(g_n(x) \left(\sum^{n}_{i \equal{} 0} \minus{} 2^i \tan\left(2^i x\right)\right)^2 \minus{} g_n(x)\sum^{n}_{i \equal{} 0}2^{2i} \sec^2\left(2^i x\right)\right)$

Now, the sum with $ \tan$ again equals $ 0$, but the other one doesn't, so we get:

$ u_n \equal{} \frac {1}{2}\sum^{n}_{i \equal{} 0}2^{2i} \equal{} \frac {1}{6}\left(4^{n \plus{} 1} \minus{} 1\right)$

1) is simply wrong. Take $ n \equal{} 0$, the limit is well know and equals $ u_1\equal{}\frac {1}{2}$ and for $ n \equal{} 1$ we get $ u_2 \equal{} \frac {5}{2}$, but:

$ \frac {5}{2}\neq \frac {1}{2} \plus{} 2.4$

Whoa!

Why are you using periods "$ .$" to indicate multiplication?

Use something like "$ *$" or "$ \cdot$" or use parentheses, please!

LaTeX suggestions: \cdot for $ \cdot$ (much, much nicer looking that *), \sin and \cos for the trig fns.

[color=darkblue][u][b]Proof.[/u][/b] For $ n\in N^*$ , $ f_{n + 1}(x) = \frac {1 - \cos x\cos 2x\cos 4x\ldots\cos2^nx\cos 2^{n + 1}x}{x^2} =$

$ \frac {1 - \cos 2^{n + 1}x + \cos 2^{n + 1}x\left(1 - \cos x\cos 2x\cos 4x\ldots\cos2^nx\right)}{x^2}\implies$

$ f_{n + 1}(x) = \frac {1 - \cos 2^{n + 1}x}{x^2} + \cos 2^{n + 1}x\cdot f_n(x)\ \ (*)$ . Observe that

$ \lim_{x\to 0}\frac {1 - \cos 2^{n + 1}x}{x^2} = \lim_{x\to 0}\frac {1 - \cos 2^{n + 1}x}{\left(2^{n + 1}x\right)^2}\cdot 4^{n + 1} = \frac 12\cdot 4^{n + 1} = 2\cdot 4^n$ .

For $ x\to 0$ in the relation $ (*)$ results that $ u_{n + 1} = u_n + 2\cdot 4^n$ a.s.o.

[u][b]An easy extension.[/u][/b] Consider the sequence $ a_n$ , $ n\in N^*$ and a function $ f\ : \ (-\epsilon ,\epsilon )\ \rightarrow\ \mathcal R$ ,

where $ \epsilon >0$ and for which $ \lim_{x\to 0}f(x)=1$ , $ \lim_{x\to 0}\frac {1-f(x)}{x^p}=l\in \mathcal R$ . Denote

$ L_n=\lim_{x\to 0}\frac {1-f(a_1x)f(a_2x)\ldots f(a_nx)}{x^p}$ . Prove that $ (\forall )\ n\in\mathcal N^*$ , $ L_{n+1}=L_n+l\cdot a^p_{n+1}$ .[/color]

like  this problem was posted here.
when $ \alpha$ tend to $ 0$ then we have :
$ \cos \alpha \approx 1 \minus{} \frac {\alpha^{2}}{2}$
or
$ \cos 2^{j}x \approx 1 \minus{} 2^{2j \minus{} 1}x^{2}$
Hence :
$ u_{n} \equal{} \frac {1 \minus{} \prod_{j \equal{} 0}^{n} (1 \minus{} 2^{2j \minus{} 1}x^{2})}{x^{2}}|_{x \equal{} 0} \equal{} \sum_{j \equal{} 0}^{n}2^{2j \minus{} 1}$ you are done

here is a closed formula :
$ \frac{1\minus{}\frac{sin(2^{n\plus{}1}x)}{2^{n\plus{}1}sinx}}{x^2}$


The digits 1, 2, 3, 4 and 9 are each used once to form the smallest possible
even five-digit number. The digit in the tens place is

a. 1         b. 2         c. 3        d. 4 e.     9

We need to minimize the first digit, then the second,etc. We must also make the number even. Hence, the smallest even number is $ 12394$, and the tens digit is $ 9$ or E.


Well, seeing that nobody has posted a mock AIME for this year, I wonder if anybody has created one?

I sorta have one in progress, but it will take time.

EDIT: Oh, and if you haven't seen, I think Altheman posted at least one or two.

Well...it is not really close to competition season so less people are likely to take a mock test. Also school starts in about a week for me so I don't really have the time to compose a test [it gets really time consuming...]. Also it is very likely that all my problems will be going to a certain program...

I am working on problems for my math meet in a month, and there are a couple problems I'm chucking because they're too hard. I'll try to work on one in October.


Let $k,m,$ and $n$ be natural numbers such that $m+k+1$ is a prime greater than $n+1$. Let $c_{s}=s(s+1).$ Prove that the product \[(c_{m+1}-c_{k})(c_{m+2}-c_{k})\cdots (c_{m+n}-c_{k})\] is divisible by the product $c_{1}c_{2}\cdots c_{n}$.

Could you explain  me how you reduced
 $ \frac{\Pi_{i \equal{}1}^{n}(m\plus{}i\minus{}k) \;\Pi_{i\equal{}1}^{n}(m\plus{}k\plus{}i\plus{}1)}{n! (n\plus{}1)!}$
to:
$ \;^{(m\plus{}n\minus{}k)}C_{n}\;\cdot \; \left( \frac{1}{m\plus{}k\plus{}1}\right) \;\cdot \;^{(m\plus{}k\plus{}n\plus{}1)}C_{ n\plus{}1}$
Thank you.

Yeah, that reduction is wrong indeed.

I also think an extra condition $ m>k$ or $ m\plus{}n<k$ wouldould be necessary, otherwise we have to show $ 0|c_1\cdots c_n$, which is obviously false.

Write $ m \plus{} k \plus{} 1 \equal{}: p$ (a prime) $ > n \plus{} 1$.  Note that
$ c_{1}c_{2} \cdots  c_{n} \equal{} (1 \cdot 2) (2 \cdot 3) \cdots  (n \cdot (n \plus{} 1)) \equal{} n! (n \plus{} 1)! \equal{} (n \plus{} 1)(n!)^2$

whilst
$ \begin{array}{cl}
         & (c_{m \plus{} 1} \minus{} c_{k}) (c_{m \plus{} 2} \minus{} c_{k})\cdots (c_{m \plus{} n} \minus{} c_{k})\\
  \mbox{ \equal{}} & \left((m \plus{} 1)(m \plus{} 2) \minus{} k(k \plus{} 1)\right)\left((m \plus{} 2)(m \plus{} 3) \minus{} k(k \plus{} 1)\right)\cdots \left((m \plus{} n)(m \plus{} n \plus{} 1) \minus{} k(k \plus{} 1)\right)\\
  \mbox{ \equal{}} & (2\Sigma_{j\equal{}k\plus{}1}^{m\plus{}1}j) (2\Sigma_{j\equal{}k\plus{}1}^{m\plus{}2}j) \cdots (2\Sigma_{j\equal{}k\plus{}1}^{m\plus{}n}j)\\
  \mbox{ \equal{}} & (m \plus{} 1 \minus{} k)(p \plus{} 1) (m \plus{} 2 \minus{} k)(p \plus{} 2)\cdots(m \plus{} n \minus{} k)(p \plus{} n)\\
  \mbox{ \equal{}} & \Pi_{i \equal{} 1}^{n}(m \plus{} i \minus{} k) \Pi_{i \equal{} 1}^{n}(p \plus{} i)\\
  \mbox{ \equal{}} & {{m\minus{}k\plus{}n} \choose {m\minus{}k}}n!{{p\plus{}n} \choose p}n!\\
  \mbox{ \equal{}} & {{m\minus{}k\plus{}n} \choose {m\minus{}k}}{{p\plus{}n} \choose p}(n!)^2\\
\end{array}$

Now I am going to show that $ (n\plus{}1) \left| {{p\plus{}n} \choose p} \right.$ i.e. $ \frac{1}{n\plus{}1} {{p\plus{}n} \choose p} \in \mathbb{Z}$ and we are done.
$ \frac{1}{n\plus{}1} {{p\plus{}n} \choose p} \equal{} \frac{(p\plus{}n)!}{(n\plus{}1)!p!} \equal{} \frac{1}{p}\frac{(p\plus{}n)!}{(n\plus{}1)!(p\minus{}1)!} \equal{} \frac{1}{p}{{p\plus{}n}\choose{p\minus{}1}}$
Note that $ p | (p\plus{}n)!$ (since $ p<p\plus{}n$), but $ p \not| (n\plus{}1)!$ (since $ n\plus{}1<p$) and  $ p \not| (p\minus{}1)!$ (since $ p\minus{}1<p$).
So the prime $ p \left|\frac{(p\plus{}n)!}{(n\plus{}1)!(p\minus{}1)!} \right.$ $ \blacklozenge$

Phill and Peter,

  The reduction is not wrong. If $ m\plus{}1 \leq k \leq m\plus{}n$, then the result is trivially true as $ c_1c_2...c_n|0$. If $ k < m\plus{}1$, then follow the proof from the second post.

[quote]$ = (m+1-k)(m+k+2)\; (m+2-k)(m+k+3)\;\cdots\;(m+n-k)(m+k+n+1) $[/quote]

or

[quote]$ = (m+1-k)(p+1) (m+2-k)(p+2)\cdots(m+n-k)(p+n)$[/quote]

This I can't see how can be obtained. Could anybody explain the details?


geometry
geometric transformation
rotation
3D geometry
projective geometry
analytic geometry

What is the volume of the solid formed by rotating the triangle whose vertices are $(0,0),\ (5,1),\ (0,3)$ about the $y{\rm -axis}$?  Express your answer in terms of $\pi$.

[hide]
My way: Separate it into two cones, one with height 1, one with height two. They both have radius 5. So now we have 1/3pi*5^2*(2+1), which is indeed 25 pi. But, what is strange to me is that the volume is the same as it would be with one cone, yet the two solids don't abide by Cavaleri's principle...could someone explain this?
[/hide]

You may also like to look at the current thread  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25704]here .[/url]

One of a  less known method to do this is  using  Pappus : (And a new tool in your tool box)

[hide]Essentially You multiply the [area of the triangle]  with [average-value of x-cordinates]  and [$2\pi$]

Eg  $\displaystyle (\frac {5 \times 3} {2})  \times (\frac { 0+5+0} {3}) \times (2 \pi) = 25 \pi$[/hide]

This works only with the type of example rcv posted.

You have three points.

(a,b) (a,c) (d,e)

The volume of the solid when that triangle is rotated around the y-axis is:

 :pi: * (d^3-a^3) * (c-b)
___________________  - :pi: * a^2 * (c-b)

3 * (d-a)


A traditional problem I believe:

Imagine the earth to be a perfect sphere, with a tight steel band around the equator.

The band is cut, and an additional 12" section added.  With a deft flick of the wrist, the band is shaken out until the gap between the Earth and the band is uniform all the way round.

How big is the gap?

A:  Only big enough to shine a light through/under.
B:  Only big enough to slip a piece of paper through/under.
C:  Big enough for a mouse to walk through/under.
D:  Big enough for a penguin to walk through/under.

[hide="Answer"]C; the additional distance added will always be $\frac{1}{2\pi}$ of the distance added to the circumference. In other words, almost $2$ inches are added, which is enough space for a mouse to go through.[/hide]

[hide]If the earth is x feet in circumference then the extra part will be $\frac{x+1}{2\pi}-\frac{x}{2\pi}=\frac1{2\pi}\text{ ft.}\approx1.91\text{ in.}$[/hide]

Good answers, both of you.  NatMC can you hide your solution?  (Also, you didn't actually answer the question, which does not ask for a number, but for what can fit under the gap.)

How am I suposed to know how tall a mouse is?  Answering the questions is not all that you want, a good solution is much better

[quote="JesusFreak197"]Obviously it's not big enough for a penguin to fit through, but way more than big enough to slide a piece of paper under.[/quote]
well, B says ONLY big enough for a paper to fit under/through.

[quote="JesusFreak197"]Obviously it's not big enough for a penguin to fit through, but way more than big enough to slide a piece of paper under.[/quote]
What about those mini penguins and jumbo mice?

lol...just the average penguin and the average mice...lol :rotfl:


Each time I try to edit my Popoviciu proof at http://www.mathlinks.ro/Forum/viewtopic.php?t=19097 the post gets totally f'cked up (spaces disappear, there comes a warning about BBcode inside LaTeX formulas etc). As far as I can say, this is all due to one formula:

[code]$L\left( q_{i}\right) =\alpha q_{i}+\beta \ $[/code]

It was a mistake of mine to put a "\ " (backslash and whitespace) before the dollar sign, but it is not right that it produces such a mess. Obviously, the whitespace is killed by some automatic replacer, and what remains then is a backslash followed upon by a dollar, what is re-interpreted as a dollar sign, what leads to a missing ending dollar.

  darij

You are right that is what is probably generating the bug. I will see what I can do to fix it, however the best idea would be probably to avoid doing that at the end :)


How do you find the number of solutions to:
$x_{1}+x_{2}......x_{r}=n$ if
1. $x_{1},x_{2}........x_{r}$ are all nonnegative integers and $n$ is a positive integer?
2. $x_{1},x_{2}........x_{r}$ are all positive integers and $n$ is a positive integer?

[hide="Answers"]1. ${n+r-1\choose r-1}$

2. ${n-1\choose r-1}$[/hide]

[hide="Outline Proof"]
Put $n$ objects in a row, and have $r-1$ 'walls' you can put between them. $x_{1}$= 'number of objects to the left of the first wall. $x_{2}$= 'number of objects between the first and second wall etc. The number of ways the walls can be arranged is $\binom{n+r-1}{r-1}$

If $x_{i}$ are positive integers rather than non-negative integers, solve the non-negative integer problem for $x_{1}+x_{2}+\dots+x_{r}=n-r$ and add 1 to all the $x$.[/hide]

See [url=http://wangsblog.com/jeffrey/combinatorics.html]here[/url] at the bottom of the page for the "Sorting into bins" argument that is equivalent to these problems only with balls instead of $1$'s.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Let $A$ be an invertible matrix $nxn$ with integer entries such that the set  of ${A^k|k\geq 1}$ is finite. Prove that in this case it has at most $ e^{\sqrt{A nlnn}}$ elements where $A$ is an absolute constant.

Hint: apply [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=229823]another problem by harazi[/url] to the characteristic polynomial of $A$.


group theory
abstract algebra
logarithms
vector
real analysis
real analysis unsolved

Do the two series
$ A\equal{}1\plus{}1\plus{}\frac{1}8\plus{}2^{\minus{}8}\plus{}2^{\minus{}15}\plus{}2^{\minus{}24}\plus{}\cdots$
and
$ B\equal{}2^{\frac{3}4}\left(1\plus{}\frac{1}4\plus{}2^{\minus{}6}\plus{}2^{\minus{}12}\plus{}2^{\minus{}20}\plus{}\cdots\right)$
have equal sum?

The sum of each series is approximately $ 2.1289$.

They don't have equal sums.

$ A\minus{}B\equal{}0.0000055766988496...$

OK, I should have noticed that- I added up enough terms to get that.

This result means that the sequence I was interested in has even/odd alternation in its asymptotic behavior.

What was the sequence you were interested in?

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=163158]here[/url]. It's the number of subgroups of $ (\mathbb{Z}/2\mathbb{Z})^{n}$ or its logarithm; the series here are part of the "constant" term. Working out the exact count is a nice exercise- it's a sum over sizes, and you want to think of the group as a vector space.


Are you doing it?

If so, what school are you from? What grade are you and what team are you on?

Sigh...I feel out of my element. Time to switch to the MA forum...

[quote="akalra1"]?

wat is gnhml?[/quote]
Greater New Haven mathematics league

i do the "math leauge" at skool

we go all over fairfield county i think

[quote="worthawholebean"]I'm doing JV and Varsity too.[/quote]

Since when are you doing JV?

i have to say that Hopkins has a lot of extra curricular activites.. :)

Choate Varsity A/B, depending on whether I'm having an ok day....
think I'm reviving an old topic. oh well.
I do JV too. perfect so far.


logarithms
integration
calculus
calculus computations

wassap ?  :D

can you show

$ \sum_{k\equal{}1}^\infty\;\dfrac{(\minus{}1)^k}{k\cdot 2^k\cdot \dbinom {2k}{k}}\;\equal{}\;\boxed{\dfrac{1}{3}\ln\left(\dfrac{1}{2}\right)}$

We have that

$ \int\limits_{0}^{\infty}\frac{x^{k}}{(1+x)^{2k+1}}dx=\frac{1}{k\binom{2k}{k}}$.

It follows, if $ S$ denotes your sum, that

$ S=\sum\limits_{k=1}^{\infty}\int\limits_{0}^{\infty}\frac{x^{k}}{(2x+1)^{2k+1}}\left(\frac{-1}{2}\right)^{k}dx$

$ =\int\limits_{0}^{\infty}\frac{1}{x+1}\sum\limits_{k=1}^{\infty}\left(\frac{-x}{2(x+1)^{2}}\right)^{k}dx$

$ =\int\limits_{0}^{\infty}\frac{-x}{(1+x)(2x^{2}+5x+2)}dx.$

The last integral can be easily evaluated by different techniques, and it equals $ \frac{1}{3}\ln \frac{1}{2}$.


Is there a problem here you need solved? If you want help on just generally solving linear equations, it better to ask about that, then to post all your problems.

[quote="ackter78692"]im sorry i did not know  that u cant post homework i really need help im failing math thats y[/quote]
Please schedule time with your teacher before school, during lunch, during a free period, or after school. If these times are not viable options, you may wish to discuss with your parents hiring a private tutor.


analytic geometry
graphing lines
slope
geometry
perpendicular bisector

[b]Determine whether the point W(2,-1) lies on the perpendicular bisector of line segment UV, endpoints U(3,5) and V(-3,-1). Explain and justify your answer.[/b]

I cant figure it out, if someone could, it would be a pleasure!!!!!

The perpendicular bisector will pass through the midpoint of UV and have a slope that is the negative reciprocal to that of UV.  The slope of UV=1 so the slope of the perpendicular bisector is -1.  We can also determine the midpoint of UV because the x coordinate will be the average of the x coordinates of U and V and the y coordinate will be the average of the y coordinates of U and V.  This means the midpoint is (0,2).  We now know the slope and y-intercept so we know the equation is $y=-x+2$.  To see if W is on this line we simply input the coordinates to see if they satisfy the equation.
$(-1)=-(2)+2$ does not hold true so W does not lie on the perpendicular bisector.

hi i am bad in english, cna u show me in steps please


probability
combinatorics unsolved
combinatorics

This problem isn't really combinatorical, but the probabilty theory is very close to the combinatorics... There is a problem (the solution should use limit probability statements):

One medical company invented some pills that should accelerate the growth of hair. The company wanted to try the pills on people - and so 2 groups of volunteers (each had 100 persons) were chosen to participate in the test. Persons in 1st group didn't take the pills and the experts noticed that the average monthly increment of the length of their hair was 3,5 cm, person in the 2nd group took the pills and the experts found out that the average monthly icrement was 3,9 cm. It's known that [color=blue] the variance (denoted Var(X) [/color] for random variable X) of monthly increment of the length of hair is 0,5 cm^2. Was it only a luck or the pills are really effective? What is the probability that by using the pills the hair will grow faster?

I would like to know, how to use the limit statements from the probability theory (as Central Limit Statement, the Rule of Great Numbers...) on this problem. I know that it isn't so difficult, but this is new for me and I'm not able to applicate it yet.

Thank you very much for help.

You can find the solution of this problem at http://www.planetamoldova.com


find the value of $sin165 \times sin345$

We can find $\sin{165^\circ}$ and $\sin{345^\circ}$ quickly by using sum/difference formula, half angle formula, etc. (For instance, $\sin{165^\circ}=\sin{15^\circ}=\sin{\frac{30^\circ}{2}}$)

Alternately, we can use sum and product property to directly find the value of the expression.

so what is the answer?
(i know it is easy.)

Note that
$2\sin{a}\sin{b}=\cos(a-b)-\cos(a+b)$

Let $a=165^\circ$, $b=345^\circ$. Should be obvious from here.

Edit: fixed signs.
Edit2: actually fixed the signs this time,  :P

Shouldn't the above relationship be the following:

$2\sin(a) \sin(b) = \cos(a - b) - \cos(a + b)$

Yeah, whoops, I meant to fix the signs before but somehow I didn't...

[quote="frt"]We can find $\sin{165^\circ}$ and $\sin{345^\circ}$ quickly by using sum/difference formula, half angle formula, etc. (For instance, $\sin{165^\circ}=\sin{15^\circ}=\frac{1}{2}\sin{30^\circ}$)
[/quote]
I believe you wantrd to type $\sin{15^\circ}=\sin{\frac{1}{2}30^\circ}$

Yeah, you're right :blush:

(Edited now)


A merchant set the selling price of an article 40% above what it cost him.  Then, seeing that it did not sell, he decided to reduce the figure by 15%.  The final price was $8.65.  What did the article cost the merchant?

[hide](1.4x)*0.85 = 8.65

1.19x = 8.65

x = 7.27

I'm not sure if I got the answer right becuase I got a repeating decimal as my answer, but I plugged it into the problem, and it worked.[/hide]

$7.27 gets $8.6513 as the paid price...

Maybe he split a coin into thirteen pieces?


What fraction, in simplest radical form, of a sphere is inside of a tetrahedron that has all 4 vertices on the outside of the sphere?

Please provide an elegant solution.

Also, if possible, find a formula for the area of the tetrahedron given the radius of the sphere.

Do you mean the volume of the tetrahedron, or the surface area?

The radius of the circumsphere is $ \sqrt{\frac{3}{8}}a$

Where $ a$ is the side length.

The volume of a tetrahedron with side length $ a$ is $ \frac{\sqrt{2}}{12}a^3$

The volume of the circumsphere is $ \frac{4}{3}\pi\left(\sqrt{\frac{3}{8}}a\right)^3\equal{}\frac{4}{3}\pi\left(\frac{3\sqrt{3}}{8\sqrt{8}}a^3\right)\equal{}\frac{\sqrt{3}}{4\sqrt{2}}\pi a^3$

Because the entire tetrahedron is contained in the sphere, the fraction is

$ \frac{\frac{\sqrt{2}}{12}a^3}{\frac{\sqrt{3}}{4\sqrt{2}}\pi a^3}\equal{}\frac{4}{3\sqrt{3}\pi}$

or $ \frac{4\sqrt{3}}{9\pi}$


[quote="Valentin Vornicu"]The set of positive integers is partitionated in subsets with infinite elements each. The question (in each of the following cases) is if there exists a subset in the partition such that any positive integer has a multiple in this subset.

a) Prove that if the number of subsets in the partition is finite the answer is yes.

b) Prove that if the number of subsets in the partition is infinite, then the answer can be no (for a certain partition).[/quote]

a) suppose au contraire that the subset $X_i$ in the partition lacks the multiples of $n_i$ for every subset. In that case. where is the number $n_1n_2...n_k$?

b) The following partition will do. $X_n = \{ n^k, k = 1,2,.../ n \neq a^m, m>1 \}$

Daniel

a) Some set gets infinte numbers of the form $n!$.  :D


[hide="HINT for 3 and 4"]
$\ln e^{n}= n$
[/hide]

[hide]1)
After splitting up the first term we get:
$125\cdot5^{2x}+374\cdot 5^{x}-3=0$
Factor:
$(125\cdot 5^{x}-1)\cdot(5^{x}+3)=0$
$125\cdot 5^{x}-1=0 \rightarrow 5^{x}=\frac{1}{125}\rightarrow \boxed{x=-3}$
$5^{x}+3 \rightarrow 5^{x}=-3$ There is no real value for x in this equation.

2)
Take base $e$ of both sides:
$e^{ln(x-2e)}=e^{ln(3/x)+2}$
$x-2\dot e= \frac{3}{x}\cdot e^{2}$
Multiply both sides by $x$ and set the equation equal to 0.
$x^{2}-2e\cdot x-3e^{2}=0$
$(x-3e)\cdot (x+e) = 0$
$x-3e=0 \rightarrow \boxed{x=3e}$
$x+e=0 \rightarrow-e$ cannot be the answer because when you plug it back into your original equation you cannot get the natural log of a negative.[/hide]

Thanks for the help, can u plz help me with other two?

[hide="3"]
I'm assuming you mean $e$ as in the constant, not a variable.
It's a rule of logs (I think) that $\log_{x}{x^{n}}=n$
$\ln$ is just $\log_{e}$
so the fraction becomes $\frac{4-3}{10}=\frac{1}{10}$
[/hide]
[hide="4"]
Same assumption as #3.
Applying the rule from #3 also, we get:
$3+\frac{1}{29}-(-2.5)+\frac{7}{6}$.

I'll leave you to simplify that.
[/hide]

CircleSquared,  that is the correct use of $\log$ rules.
[quote="AstroPhys"]$\ln e^{n}= n$[/quote]

yeah i meant e as a constant.. once again, thanks for the help!


In the figure, AB is an extensible massless string at the ends of which two equal masses m are attached.Another body of mass m is imparted a velocity  making an angle of  with AB. AB= .Find the velocity of A,B &C & the cm of A&B after the impact of B and C.Also find the angular velocity of the system about the cm after the impact of B and C.
(the whole arrangement is placed in a horizontal plane)

I will post my full answer afterwards.
I want to see if my start is right.
My thinking is even after the collision the two balls connected by thread will be connected.
Is that right?

yup...

and is that a typing error -- "extensible string"?
or inextensible

it CAN be extensible,but that wud inc the level :)

The problem clearly states an [b]EXTENSIBLE MASSLESS STRING[/b], its not a typo. :D though that increases the level of the problem
[hide=" Hint"]: An inextensible string wud behave like a rigid rod but this is not the case here.
AB=$ l$ implies that the string is taut and $ l$ is its natural length. Its obvious that there will be some stretching of the string after impact
$ <ABC\equal{}30^o$
[/hide]

PS:I am working on the problem , its hard but not unattainable and i am almost complete .......wud be posting solution very soon......all suggestions and solutions are utmost welcome.

could u plz check my solution for the case in which the string is inextensible.

I think that when the ball with with velocity v hits the second ball an elastic collision takes place and since the balls have the same mass the balls just exchange their velocities.

Could anybody reply whether this part is right

Thx,
Befuddlers

im not sure..
but i guess ur right.. for an instant the velocities will be exchanged..
however.. the ball will then go thru immediate deceleration due to the string...


I was just wondering how much space this website can handle since there are so many posts and other things.

-interesting_move

Good question.  Hope we don't find out the hard way. :shock:

Seriously, the whole forum at this point is just a few megs.  We should have plenty more room...

It's MySQL database (I think this site uses it), and it doesn't take up much space.

Oh, that's good!  Yeah...posts don't take up as much room as I previously thought they would since they're mostly text.

-interesting_move


algebra
polynomial
function
real analysis
binomial theorem

Find the coefficient of x^5 in the expansion of (1-12x^5)^21 * (1+x)^10.

How can I start with this problem?

Thanks.

The first thing I would do is take a deep breath and relax, cuz when I first saw this problem I thought it was harder than it actually is (assuming I did it right of course  :P)

Then consider each polynomial seperately.  How many ways can one term in the first polynomial times another term in the second polynomial yield the variable "x^5"?  (also, do you know the binomial theorem?)

Generating function time!

This is related to something called the 'convolution' of two functions.

If we have two generating functions $f(x)$ and $g(x)$ with coefficients $a_i$, $b_i$, then the coefficient of $c_k$ in $f(x)g(x)$ is given by:

$\sum_{i+j=k} a_ib_j$ or, equivalently

$\sum^n_{i=0} a^{n-1}b^i=a_nb_0+a_{n-1}b_1+\ldots+a_0b_n$ (assuming your sequences start at zero).  Can you see how this might be applicable?

[quote="paladin8"][quote="goodyfresh741"]This is WAY easier than it looks, stuff like convolution isn't even necessary.

[hide]The answer is zero, because from the first polynomial you get coefficent of 1 as just 1, of x^5 as 12*-21=-252, from second polynomial, coeff. of 1 is 1, of x^5 is 10!/(5!^2)=252, add the coefficients, get 252-252=0 as the coefficient of x^5 in the product.  All you do to find those coefficients is use the binomial theorem with exponents of 21 and of 10[/hide][/quote]

That's exactly what I did. But the problem asks for what you get when you multiply the polynomials, not add.[/quote]

Ummm, I'm pretty sure I didn't add, but I multiplied. . .is my answer wrong?  I didn't really multiply in a sense, in that I only added in the coefficients that actually contributed to what was asked for, rather than multiplying out the whole thing (which would be crazy).

Wow, that's interesting. I added them and got 0 also (that's why I thought you did the same thing as me). Sorry, you were right now that I actually do it out.

[quote="blahblahblah"][quote="goodyfresh741"]This is WAY easier than it looks, stuff like convolution isn't even necessary.

[hide]The answer is zero, because from the first polynomial you get coefficent of 1 as just 1, of x^5 as 12*-21=-252, from second polynomial, coeff. of 1 is 1, of x^5 is 10!/(5!^2)=252, add the coefficients, get 252-252=0 as the coefficient of x^5 in the product.  All you do to find those coefficients is use the binomial theorem with exponents of 21 and of 10[/hide][/quote]

Yes, but any time you multiply two polynomials together and collect coefficients with subscripts adding to the same number, you're taking the convolution (or Cauchy product) whether you realize it or not.[/quote]

Point taken, I guess I really was using the convolution operation, just without really thinking about it as such.


[quote="can_hang2007"]Let $ a,b,c$ be real numbers such that
$ ab(a \plus{} b) \plus{} bc(b \plus{} c) \plus{} ca(c \plus{} a) \equal{} 6.$
Prove that
$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \plus{} abc(a \plus{} b \plus{} c) \ge \frac {18}{a^2 \plus{} b^2 \plus{} c^2}.$[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2,$ where $ v^2\in\mathbb R$ and $ abc \equal{} w^3.$
Hence, your inequality is equivalent to $ w^6 \plus{} P(u,v^2)w^3 \plus{} Q(u,v^2)\leq0.$
But $ w^6 \plus{} P(u,v^2)w^3 \plus{} Q(u,v^2)$ gets a maximal value, when $ w^3$ gets an extremal value,
which happens when two numbers from $ \{a,b,c\}$ are equal.
Id est, after homogenization we can assume $ b \equal{} c \equal{} 1,$ which gives $ (a \minus{} 1)^2a^2\geq0.$ Done!
By the way, your inequality is equivalent to
$ \sum_{cyc}a^2b^2(a\minus{}b)^2\geq0.$

[quote="can_hang2007"]Let $ a,b,c$ be real numbers such that
$ ab(a \plus{} b) \plus{} bc(b \plus{} c) \plus{} ca(c \plus{} a) \equal{} 6.$
Prove that
$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \plus{} abc(a \plus{} b \plus{} c) \ge \frac {18}{a^2 \plus{} b^2 \plus{} c^2}.$[/quote]
We have
$ 2[a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \plus{} abc(a \plus{} b \plus{} c)]\equal{}\sum{b^2(a\plus{}c)^2}$
By using Cauchy-Schwarz we get
$ 2[a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \plus{} abc(a \plus{} b \plus{} c)](a^2 \plus{} b^2 \plus{} c^2)$
$ \equal{}(\sum{b^2})(\sum{b^2(a\plus{}c)^2})\ge (\sum{b^2(a\plus{}c)})^2\equal{}36$

[quote="leedt26"][quote="can_hang2007"]Let $ a,b,c$ be real numbers such that
$ ab(a \plus{} b) \plus{} bc(b \plus{} c) \plus{} ca(c \plus{} a) \equal{} 6.$
Prove that
$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \plus{} abc(a \plus{} b \plus{} c) \ge \frac {18}{a^2 \plus{} b^2 \plus{} c^2}.$[/quote]
We have
$ 2[a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \plus{} abc(a \plus{} b \plus{} c)] \equal{} \sum{b^2(a \plus{} c)^2}$
By using Cauchy-Schwarz we get
$ 2[a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \plus{} abc(a \plus{} b \plus{} c)](a^2 \plus{} b^2 \plus{} c^2)$
$ \equal{} (\sum{b^2})(\sum{b^2(a \plus{} c)^2})\ge (\sum{b^2(a \plus{} c)})^2 \equal{} 36$[/quote]
Very nice, leedt26. In fact, my solution is the same as you, too. I am glad that we got the same ideas in two times (the previous problem and this one)  :lol: .

By the way, I post my proof here: http://canhang2007.wordpress.com/2009/11/10/inequality-61-v-q-b-can/

[quote="arqady"]By the way, your inequality is equivalent to
$ \sum_{cyc}ab(a^2 \minus{} b^2)^2\geq0.$[/quote]
Arqady, I think it must be
$ \sum a^2b^2(a\minus{}b)^2 \ge 0.$

[quote="can_hang2007"][quote="arqady"]By the way, your inequality is equivalent to
$ \sum_{cyc}ab(a^2 \minus{} b^2)^2\geq0.$[/quote]
Arqady, I think it must be
$ \sum a^2b^2(a \minus{} b)^2 \ge 0.$[/quote]
Yes, of cause. I have fixed it.


[quote="Virgil Nicula"][color=darkred]In the triangle $ABC$ define the point $D\in (BC)$ for which the inradius of the triangles $ABD$ and $ACD$ have a same length.  Analogously define the points $E\in (CA)$ and $F\in (AB)$. Prove that the lines $AD$, $BE$, $CF$ are concurrently.[/color][/quote]

Are you sure? If I am not mistaken, this is one of the most popular examples of three non-concurrent cevians in a triangle; in fact, the lines AD, BE, CF are concurrent if and only if triangle ABC is isosceles (cf. Paul Yiu, [i]The congruent-incircle cevians of a triangle[/i], Missouri J. Math. Sci., 15 (2003) p. 21-32[/i])...

  Darij

Maybe, [i]Darij [/i]! I received this problem today from the well-known Romanian teacher [i]Sorin Radulescu[/i]. I will communicate him what you said. Thanks. To-morrow I will give you the answer after I will say him this fact.

Darij is right, as usual, together with Paul.

    This is a fine example of Yettians, non-concurrent Cevians.

    They are pretty close (to concurrency), 'but no cigar'.

    Virgil is right, too, by placing this problem in the 4-stars range.

    One more star and it will have enough alcohol  to be drinkable

    as a (strong!) cognac distilled  somewhere in my favorite country

    in Eastern Europe.

    T.Y.

    M.T.

I think the teacher S.R. knew this fact. He put me to the test and he was successful because I lost at least two hours with this problem. I am sorry, I didn't know he made a joke (with good liking). Soon I will "avenge" similarly. Help me, maybe you are a same difficult problem for give him and you send me by private message !. Thanks.


inequalities
function
inequalities unsolved

Find all $ n > 1$ such that the inequality \[ \sum_{i\equal{}1}^nx_i^2\ge x_n\sum_{i\equal{}1}^{n\minus{}1}x_i\] holds for all real numbers $ x_1$, $ x_2$, $ \ldots$, $ x_n$.

let $ a \equal{} \sum_{i \equal{} 1}^{n \minus{} 1}x_i^2$ and $ b \equal{} \sum_{i \equal{} 1}^{n \minus{} 1}x_i$..

consider the function $ f(x) \equal{} x^2 \minus{} bx \plus{} a$... we have that its discriminant equals $ \Delta \equal{} b^2 \minus{} 4a$, and we have that $ b^2 \minus{} (n \minus{} 1)a\leq 0$, so for $ n\leq 5$ we have that $ \Delta\leq 0$, which implies that $ f(x)\geq 0$ for all $ x$...

for $ n > 5$, take $ x_1 \equal{} x_2 \equal{} ... \equal{} x_{n \minus{} 1} \equal{} 1$ and $ x_n \equal{} \sqrt {n \minus{} 1}$ to see that the inequality is false, and we're done  :D


Is it possible to induce a current with just heating?
i.e. if we have metal rods (conducting/non-conducting rods) and a heat source, can we make current and determine the direction and amount of current (if possible in relation to the heat energy)?

yes we can
for that all we gottcha do is connect 2 different metals like Bi-Sb and heat one end and cool the other
we have peltier effect,seeback effect etc in it

For more, see [url=http://www.dself.dsl.pipex.com/MUSEUM/POWER/thermoelectric/thermoelectric.htm]this[/url].

.


[hide="my method"]
Just see the pattern:
$ a_{n \plus{} 1} \equal{} \sqrt {\frac {1 \plus{} a_n}{2}}$
so put $ a_1 \equal{} \cos{\alpha}$
$ \alpha \equal{}\frac{\pi}{4}$

 then $ a_2 \equal{} \cos{\frac {\alpha}{2}}$

 and $ a_3 \equal{} \cos{\frac {\alpha}{2^2}}$

and so on.

So generally $ a_r \equal{} \cos{\frac {\alpha}{2^{r \minus{} 1}}}$
 Then $ \cos{x}.\cos{2x}.....\cos{2^{n \minus{} 1}x} \equal{} \frac {\sin{2^{n}.x}}{2^{n}.\sin{x}}$
Then apply $ lim_{n\rightarrow \infty}$ to get your answer.
[/hide]

this is an ama sum n a kk sir test sum if i rem right n rock was the one to solve it in the test too  :D


symmetry
probability
expected value
combinatorics unsolved
combinatorics

For a permutation $P = (p_1, p_2, ... , p_n)$ of $(1, 2, ... , n)$ define $X(P)$ as the number of $j$ such that $p_i < p_j$ for every $i < j$. What is the expected value of $X(P)$ if each permutation is equally likely?

Let $A_j$ be the event that $p_1$, $p_2$, $\ldots$, $p_{j-1}$ are all less than $p_j$.  That event is equivalent to $p_j$ being the maximum of $p_1$, $p_2$, $\ldots$, $p_j$.  By symmetry, the probability of that event is $\frac{1}{j}$.

By linearity of expecation, the expected value of $X$ is
\[ E(X) = \sum_{j=1}^n \Pr(A_j) = \sum_{j=1}^n \frac{1}{j} \, .  \]
In other words, the answer is the $n$th harmonic number.


trigonometry function calculus derivative

Hi

1. What family relationship are Adam and Bethany? Adam is Chuck's son. Dianne is married to Chuck. Bethany is Fala's neice and Gabriella's granddaughter. Hailey is Fawn's sister, Isaiah's daughter and Bethany's aunt. Gabriella and Isaiah have 3 children, all daughters, and all named so far.

(a)Cousins
(b)Niece-Uncle
(c)Nephew-Aunt
(d)Brother-sister
(e)Wife-husband

2. Three ships are situated as follows : A is 225 miles west of C and B, due south of C bears S 25 degrees 10' E from A. Which of the following are true statements.

(a)B is 579 miles from A
(b)B is 479 miles from C
(c)The bearing of A from B is N25 degrees 10' W
(d)Both b and c
(e)All of the above

These are 25 and 26 of the Elon 2004 High School Math Contest. I don't know how to do 25 and have no idea what 26 means o-o.

Please, no spam, and please provide an explanation.

Okay either Fala is a typo for Fawn and the answer is (d) or Fala is married to Hailey and the answer could be either (d) or (a).

I suspect the commas on the second one are messed up, which is why it is confusing. Perhaps it should read

[quote]A is 225 miles west of C, and B, due south of C, bears S 25 degrees 10' E from A. Which of the following are true statements.[/quote]

What is with all of the non-math terminology. Basically you have a right triangle with right angle at C, AC=225, and CAB=151/6 degrees. Does that help?

Yea what's with all the problems in contests not really having to deal with math lately?

[quote="MysticTerminator"]eh, I'm fine with both of the above problems since they can be solved with purely mathematical principles[/quote]

Well, for the second one, what does bearing S 25 degrees 10 minutes mean....

GAH

[quote="Teki-Teki"]60 second = 1 minute
60 minute = 1 degree
360 degree = 1 circle

basically just more awkward ways of measuring exact angles.  I really think that we should have 60 degrees to a circle and forget about the seconds and minutes, as that way the trig functions of any integer degree measure are expressible in terms of integers with just addition, multiplication, and square roots.[/quote]

Actually, I was asking about the 'bearing' part.

[quote="Teki-Teki"]I really think that we should have 60 degrees to a circle and forget about the seconds and minutes, as that way the trig functions of any integer degree measure are expressible in terms of integers with just addition, multiplication, and square roots.[/quote]

OKAY. The derivative of sin(x) is 30cos(x)/π. And while you're at it, $ \sin{x}\equal{}\frac{e^{30xi/\pi}\minus{}e^{\minus{}30xi/\pi}}{2}$. Hooray for math reform!

But yeah, the degrees/minutes/seconds system is hilariously inefficient. Even more so than "mixed numbers".

Light is about $ 5\cdot 10^{14}$ Hz and $ 4\cdot 10^{\minus{}7}$ m in wavelength. So, if I run around a circle of radius $ 6\cdot 10^{\minus{}8}$ m at an angular acceleration of $ 5\cdot 10^{14}\frac{sec}{sec^2}$, or about $ 2\cdot 10^9\frac{rad}{sec^2}$, that means I am running at the speed of light. Assuming I weigh 70 kg I only need to run at .0004 m/s to accomplish this. Hooray!

[quote="jb05"][quote="Teki-Teki"]I really think that we should have 60 degrees to a circle and forget about the seconds and minutes, as that way the trig functions of any integer degree measure are expressible in terms of integers with just addition, multiplication, and square roots.[/quote]

OKAY. The derivative of sin(x) is 30cos(x)/π. And while you're at it, $ \sin{x} \equal{} \frac {e^{30xi/\pi} \minus{} e^{ \minus{} 30xi/\pi}}{2}$. Hooray for math reform!

But yeah, the degrees/minutes/seconds system is hilariously inefficient. Even more so than "mixed numbers".

Light is about $ 5\cdot 10^{14}$ Hz and $ 4\cdot 10^{ \minus{} 7}$ m in wavelength. So, if I run around a circle of radius $ 6\cdot 10^{ \minus{} 8}$ m at an angular acceleration of $ 5\cdot 10^{14}\frac {sec}{sec^2}$, or about $ 2\cdot 10^9\frac {rad}{sec^2}$, that means I am running at the speed of light. Assuming I weigh 70 kg I only need to run at .0004 m/s to accomplish this. Hooray![/quote]

HUH

ME NO UNDERSTANDY

ALL THAT STUPIDY PHYSICY STUFFF

ALL THAT STUPIDY CALCULASY STUFFF

ALL THAT STUPIDY DEGREEY/MINNUTEY/SECONDY STUFFF

BTW YOU CANNOT RUN AT THE SPEEEEEEEEEEEEEED OF LIGHT B/C THEN YOU WOULD BE STUCK IN A SINGLE TIME FRAAAAAAAAME


logarithms
inequalities
rearrangement inequality
inequalities unsolved

$ 1.)$ Let $ a,b,c \in (0;1)$ ; . with $ 0 < a \leq b \leq c < 1$ .

Compare the numbers :
$ A \equal{} \log_ba \plus{} \log_cb \plus{} \log_ac$ ; and $ B \equal{} \log_ab \plus{} \log_bc \plus{} \log_ca$ .

Could somebody post a complete proof ( maybe with rearrangement inequality ) ?

Let $ x \equal{} \log_b a, y \equal{} \log_c b$, which are both greater than or equal to $ 1$.  Then

$ A \equal{} x \plus{} y \plus{} \frac {1}{xy} \equal{} \frac {x^2 y \plus{} y^2 x \plus{} 1}{xy}$
$ B \equal{} \frac {1}{x} \plus{} \frac {1}{y} \plus{} xy \equal{} \frac {x \plus{} y \plus{} x^2 y^2}{xy}$

and

$ xy(B \minus{} A) \equal{} x^2 y^2 \minus{} x^2 y \minus{} y^2 x \plus{} x \plus{} y \minus{} 1 \equal{} (xy \minus{} 1)(x \minus{} 1)(y \minus{} 1) \ge 0$.


probability
linear algebra
matrix
calculus
derivative
expected value
algebra

Hi,

Is there a closed form solution to solving for $x$ to the following equation?

$p = \sum_{k=0}^{n}\begin{pmatrix}n \\ k \end{pmatrix}k x^{k}(1-x)^{n-k}$

where $p \in [0,n]$.

Approximation to this is also welcome.

This is simply the expected value of a binomial distribution with $n$ trials and probability $x$ of success on any one trial. The answer is that this expected value is $nx.$ There are many, many different possible ways to compute this, some analytical, some using more probabilistic reasoning.

Here's an analytic method:

Let $f(y)=\sum_{k=0}^{n}\binom{n}{k}y^{k}(1-x)^{n-k}.$

Then by the binomial theorem, $f(y)=(y+1-x)^{n}.$

$f'(y)=n(y+1-x)^{n-1}.$

$yf'(y)=ny(y+1-x)^{n-1}=\sum_{k=0}^{n}\binom{n}{k}ky^{k}(1-x)^{n-k}$

Now evaluate both sides when $y=x.$

[quote="Xevarion"]I think the sum is just $nx$ or something ... Write out the binomial coefficient, absorb that extra $k$ into it, then take out an $nx$ from the sum, and you are left with the sum of all the probabilities for a binomial distribution with $n-1$ instead of $n$. I didn't write it out though, I'm not 100% sure. You should check.[/quote]
Instead of being lazy, here's what I meant:

$\sum_{k=1}^{n}\binom{n}{k}kx^{k}(1-x)^{n-k}$
$\sum_{k=1}^{n}\frac{(n!)k}{k!(n-k)!}x^{k}(1-x)^{n-k}$
$\sum_{k=1}^{n}\frac{n(n-1)!}{(k-1)!(n-k)!}xx^{k-1}(1-x)^{n-k}$
$nx\sum_{k=1}^{n}\binom{n-1}{k-1}x^{k-1}(1-x)^{n-k}$
$nx\sum_{j=0}^{n-1}\binom{n-1}{j}x^{j}(1-x)^{n-1-j}$
$nx$

Note that I threw out the first term in teh sum because it's 0. At the end, I used that to change the index of summation to what I wanted. It's just some constant ($nx$) times something familiar: the binomial distribution with the same probability but only $n-1$ trials. We know the sum of the probabilities for all events in the binomial distribution is 1, so the sum is just $nx$.

The second part of the question: Compute

$\sum_{k=0}^{n}\binom{n}{k}k^{2}x^{k}(1-x)^{n-k}-(nx)^{2}$

(In other words, the variance.)

Thank you Kent and Xevarion.

Kent: Do you mean
$\sum_{k=0}^{n}\binom{n}{k}k^{2}x^{k}(1-x)^{n-k}-(nx)^{2}$?

If so, then its variance is 0 (using the differentiation method suggested by you).

Yes, that's what I meant, and I have edited.

But the variance is not zero. It's positive.

Well, we could write that $k^{2}$ as $k(k-1)+k$:
\[\sum_{k=0}^{n}\binom{n}{k}k(k-1) x^{k}(1-x)^{n-k}+\sum_{k=0}^{n}\binom{n}{k}k x^{k}(1-x)^{n-k}-(nx)^{2}. \]
Using Xevarion's trick twice, we get that the [url=http://en.wikipedia.org/wiki/Variance]variance[/url] is probably equal to $nx(1-x)$.


function
vector
calculus
integration
real analysis
real analysis unsolved

[quote="jessa234"]1. i was wondering if you could tell me how i should approach this problem:

Let f:R->R^3 be a differentiable, vector-valued function and
g:R->R be a strictly increasing vector-valued function.  Let h = fog:R->R^3.  Show that the paths traced by f and h are equal and that h'(t) = f'(g(t))g'(t).  h is called the reparameterization of f by g. an additional piece of info is that g is onto - that is, for every y in R, there is an x  such that g(x) = y.
------------------------------------------------------------------

i tried reparameterizing according to arclength.  let s1, s2, s3 = arclength of f, g, and h.
functions are f = f(t), and g = g(b).

s1(t) = [integral]|f'(u)| du
ds1/dt = |f'(t)|

s2(b) = [integral]|g'(u)|du
ds2/db = |g'(b)|

s3(b) = [integral]|f'(u)|du
s3/db = |f'(g(b))|
      = |f'(g(b))*g'(b)|

i have no idea where to go from here.  i don't even know if i started off correctly.  how do i put all that together to show that the path traced by f = path traced by h?  any help would be appreciated.[/quote]

What you posted looks like what you will need to do to show that arclength is well-defined, which is related to this problem.

As for the problem, you need to assume that $g$ is a differentiable function $g:\mathbb{R}\rightarrow\mathbb{R}$. It is odd to call it vector-valued. The first part follows from the fact that g is onto, which follows from the intermediate value theorem. The second is just a statement of the chain rule.


$ABC$ is an acute angled triangle. Taking D,E,F to be the feet of perpendiculars from A, B, C to sides BC, CA, AB respectively, we extend DE past E and DF past F such that they meet the circumcirle of $\triangle ABC$ at points X and Y respectively.

Show that the arcs $AX$ and $AY$ are of the same length.

[quote="JL"]
Show that the arcs $AX$ and $AY$ are of the same length.[/quote]

I don't think so. If I haven't misunderstood the problem, they are not equal starting from the problem conditions.  :?

[hide]Let $H$ be the intersection of the altitudes $BE$ and $CF$. By vertical angles, $\angle EHC\cong \angle FHB$.
Since $\angle BEC\cong \angle CFG$ (both right angles), $\angle ECH\cong \angle FBH$ because the sum of the angles in each triangle ($EHC$ and $FHB$) are both 180 degrees.
Note that $\angle ECH$ is inscribed in arc $AY$ and $\angle FBH$ is inscribed in arc $AX$. The measures of the two angles are equal, so the measures of the arcs are equal.[/hide]

sorry, but if i didn't draw the diagram wrongly, $\angle ECH$ is not inscribed in arc $AY$. the same goes for $AX$.

[quote]I don't think so. If I haven't misunderstood the problem, they are not equal starting from the problem conditions.[/quote]
How do you see that Andreas? I've been having a lot of trouble completing this question. Hope you could clarify. Thanks in advance  :)


induction
number theory solved
number theory

Let M be a set with n elements, let P be the union of the following sets: M and all the subsets of M.
Determine the number of pairs (A,B) if A \in P, B \in P and either A is the same the set as B or a subset of B !

mr: induction, binomial formula
hr: split into partial problems, systematic attempts

Since it's a nice result it may deserve a (short) proof  :)  ?

For each subset B with k elements there are 2^k subsets included in B.

The number is then  \sum C(n,k)2k  = 3n


How do you guys deal with spaces in LaTeX when they need to go in between math mode stuff? For example, I like to put a space between $ \exists$ and $ a \in A$, but if you write [i]\exists a \in A[/i] you get $ \exists a \in A$, i.e.  there exists no space. I prefer $ \exists$ $ a \in A$. I also often encounter situations where I want to put a space between text and math symbols, etc.

So far I've been using the command \; or I've been separating my text with dollar signs. But both of these solutions are kind of annoying (i.e. seemingly inelegant). What do you guys do?

Well I think by writting \exists\,a\in A produces $ \exists\,a\in A.$

It makes sense.

I've used \, and \: and \;, but I always thought they were kind of ugly. So is this what's used in professional papers?

Spaces are always ignored in maths mode as they are only there to help one read the code. But you can add a normal space by using \ (\ followed by a space eg $ x\ y$). \,  \:  \; are easy shortcuts to smaller spaces but of course there is \quad and \qquad for larger spaces.
quad = space of current type size eg 10pt
qquad = 2 quads
\, = 3/18 of a quad
\: = 4/18 of a quad
\; = 5/18 of a quad
\! = -3/18 of a quad

You can have a space of any width you like with \hspace{[i]width[/i]} and
\newcommand{\mysp}{\hspace{1mm}}
would allow you to use \mysp anywhere you wanted a 1em space


The stem-and-leaf plot shows the number of minutes and seconds of one ride on each of the 17 top-rated roller coasters in the world. In the stem-and-leaf plot, 2 20 represents 2 minutes, 20 seconds, which is the same as 140 seconds. What is the median of this data set? Express your answer in seconds.

\begin{tabular}{c|ccccc}
0&28&28&50&&\\
1&00&02&&&\\
2&20&25&35&43&45\\
3&00&00&00&30&36\\
4&00&00&&&\\
\end{tabular}

[quote="GameBot"]The stem-and-leaf plot shows the number of minutes and seconds of one ride on each of the 17 top-rated roller coasters in the world. In the stem-and-leaf plot, 2 20 represents 2 minutes, 20 seconds, which is the same as 140 seconds. What is the median of this data set? Express your answer in seconds.

$ \begin{tabular}{c|ccccc}
0&28&28&50&&\\
1&00&02&&&\\
2&20&25&35&43&45\\
3&00&00&00&30&36\\
4&00&00&&&\\
\end{tabular}$[/quote]

There are 17 numbers, so the 9th number is 2 minutes, 43 seconds, or 163 seconds.


Six married couples attend the marriage enrichment planning meeting.  How many committees of five can be chosen if no committee is to include a husband and wife team?

I think the answer is $ 384$, but the answer key claims it is $ 792$.

My work is this: $ \binom{6}{5}(2 \times 2 \times 2 \times 2 \times 2 \times 2) \equal{} 384$, and I think its right, so why is it 792 (if it is, the key could be wrong)?

Yeah, Im pretty sure the answer is not 792. 792=12C5, which doesnt specify the restrictions.

So think of the problem like this:
A B C D E F.......men
A B C D E F.......women

Where two letters of opposite genders cant be in the same set.

If we considered A B C D E F to be just one person, we have 6C5=6 ways to choose a person.
But they arent the same person. Once we have chosen the 5 letters to be in a group, wlog lets say we chose A B C D E, we actually have the following arrangement:
A B C D E.......men
A B C D E.......women

And there are 2x2x2x2x2 ways to choose whether we want a man or woman.

So Im getting the answer of $ \binom{6}{5}(2\times 2\times 2\times 2\times 2) \equal{} 192$

But then there is a 99% chance Im wrong, so can anyone confirm this?

I confirm Thunder's solution. 007math, I think you misread the answer key; after all, the last 2 digits are correct, and 7 can be easily mistaken as 1.

Oh, OK.  I get it know.  The test this was taken from was multiple choice and all answer choices were within 200 of 792, so I don't think its a typo.  I guess the writers forgot to think about the restrictions when coming up with the answer.  Anyway, thanks!

That would make sense, because the answer without restrictions is 792. Interesting.


geometry
geometric transformation
reflection
rotation
angle bisector
geometry unsolved

[quote="silouan"]Let a convex quardilateral $ABCD$. The $DA,CB$ intersect at $F$ and the $AB,DC$ intersect at $E$. The bisectors of the angles $DCF,AED$ are perpendicular . Prove that the angle bisectors of these angles are parallel to the angle bisectors of the angles which formed from $AC,BD$[/quote]

I don't think the problem is correct.Vary point $D$ on $EC$ then bisector of the angle formed by $AC,BD$ vary,But angle bisectors of $DCF,AED$ are still constant.

I couldn't find it in IMOSL 1986,by the way.

It is long listed, proposed by Austria.

[u] Babis[/u]

 (Source : IMO compendium- Book)

I made a typo in the problem It was DFC the angle instead DCF .Now try to prove it
as
Let a convex quardilateral $ABCD$. The $DA,CB$ intersect at $F$ and the $AB,DC$ intersect at $E$. The bisectors of the angles $DFC,AED$ are perpendicular . Prove that the angle bisectors of these angles are parallel to the angle bisectors of the angles which formed from $AC,BD$

[color=blue][b]Lemma[/b][/color].[color=darkblue]let $\ell_{1}\ ,\ell_{2}$ be two line that $\angle (\ell_{1},\ell_{2})=\alpha\ ,\ \ell_{1}\cap\ell_{2}=A$ .
now let reflection of shape $\mathcal X$ wrt $\ell_{1}$ be $\mathcal X'$ and reflection of $\mathcal X'$ wrt $\ell_{2}$ be $\mathcal X''$.
then $\mathcal X''$ is rotation of $\mathcal X$ wrt $A$ with angle $2\alpha$.[/color]

in this problem $\alpha=90^{\circ}$ now let $\ell$ be reflection of line $DA$ wrt angle bisector of $AED$ notice that $\angle (\ell,CD)=\angle(AD,AB)$ from the lemma we have $\ell \parallel BC$hence $\angle(BC,BD)=\angle(EA,AD)$ hence quadrilateral $ABCD$ is cyclic,the rest is easy.


integration
calculus
function
real analysis
derivative
calculus computations

A long time ago, Kent Merryfield gave a nice proof for an instance of the following proposition which I have generalized. See if you can prove it:

Let $ f: [0,1] \to R$ be Riemann integrable. Let $ u: [0,1] \to R$ be $ C^1$, non-zero on $ ]0,1]$ and have $ u' > 0$ on $ ]0,1[$. Suppose $ \int\limits_0^1 f \equal{} 0$. Show that there is some $ 0 < c < 1$ for which $ \int\limits_0^c {uf} \equal{} 0$

[b]Lemma)[/b] Suppose $ f, g$ are continuous on $ [0, 1]$ with $ f(0) \equal{} f(1) \equal{} 0$ and $ g(x) > 0$ on $ (0, 1)$. Then $ h(x) \equal{} \int_{0}^{x} [f(x) \minus{} f(t)] g(t) \, dt$ has a zero in $ (0, 1)$.

[hide="Proof."]We may assume $ f$ is not identically zero.

Case 1) Suppose the maximum and minimum of $ f$ are both nonzero. Condition imposed on $ f$ shows that there is $ a, b \in (0, 1)$ such that $ f(a) \equal{} \max f > 0$ and $ f(b) \equal{} \min f < 0$. Then $ h(a) > 0$ and $ h(b) < 0$.

Case 2) Suppose $ f \geq 0$ or $ f \leq 0$ on $ [0, 1]$. By replacing $ f$ to $ \minus{} f$ if needed, we may assume $ f \geq 0$. Then there is $ a \in (0, 1)$ such that $ f(a) \equal{} \max f > 0$. Then $ h(a) > 0$ and $ h(1) < 0$.

Since $ h$ is continuous, Intermediate value theorem is applicable in both cases and we are done.[/hide]

Now let's return to the original problem. Put $ F(x) \equal{} \int_{0}^{x} f(t) \, dt$. Then by integration by parts

$ \int_{0}^{x} f(t) u(t) \, dt \equal{} [F(t)u(t)]_{0}^{x} \minus{} \int_{0}^{x} F(t)u'(t) \, dt \equal{} \int_{0}^{x} [F(x) \minus{} F(t)]u'(t) \, dt.$

Then the rest follows from the Lemma.

But this proof is not satisfactory for me, because it used some advanced version of integration by parts.

[quote="Kent Merryfield"]Can you link to the post that you were working from? I don't remember it.[/quote]

Check  your PM. I've changed the hypotheses slightly so that u is non-zero on [0,1] since someone has pointed out that the proof I have as of now doesn't work when the interval is ]0, 1[.

[quote="sos440"]
But this proof is not satisfactory for me, because it used some advanced version of integration by parts.[/quote]
Well, that's just because SouthStar said integrable and not continuous, so it's not like you can get away from it. The formula holds as long as $ f$ is integrable and $ u$ is absolutely continuous. This is because then $ F$ is also absolutely continuous. (For absolutely continuous functions $ F' \equal{} 0$ almost everywhere does imply $ F$ is constant, and by the lebesgue theorem the derivative of $ F$ is $ f$ a.e.).

It's a nice proof

[quote="Kalle"]For absolutely continuous functions $ F' = 0$ almost everywhere does imply $ F$ is constant, and by the lebesgue theorem the derivative of $ F$ is $ f$ a.e.[/quote]

My proof is nothing but an overkill as long as this problem belongs to just an introductory analysis level. And it explicitly mentions 'Riemann integral', which seems to imply such possibility. That's why I said 'not satisfactory', in the sense of elementariness. [img]http://myhome.shinbiro.com/~sosQED/images/m_kiss.gif[/img]

I suspect if it can be simplified into more elementary one. One of my idea is to change the order of integration. This can be done like this:

\begin{eqnarray*}
\int_{0}^{x} f(t)u(t) \, dt
& = & \int_{0}^{x} \int_{0}^{t} f(t)u'(s) \, dsdt\\
& = & \int_{0}^{x} \int_{s}^{x} f(t)u'(s) \, dtds\\
& = & \int_{0}^{x} [F(x) - F(t)] u'(s) \, ds.
\end{eqnarray*}

But I'm not sure whether it can be justified in the range of Riemann integral theory or not. (Of course, Fubini theorem ensures this...)

[quote="max_tm"]A little big unrelated, but in trying to solve this problem I got something that I feel must certainly be wrong so I'm wondering if somebody can help me out: I showed that $ \int\limits_0^1{uf} \equal{} 0$

[hide]
Since $ u$ is $ C^1$ it is continuous on $ [0,1]$ and so by EVT it attains some max. $ M$ on this interval.
Since $ f$ is integrable (and it's integral is equal to $ 0$), choose a $ \delta$ s.t. for all $ \epsilon > 0$ and partitions $ X$ on $ [0,1]$ where the mesh of $ X$ is less than $ \delta$, $ |S| < \frac {\epsilon}{M}$ where $ S$ is any Riemann sum for $ f$ on $ X$.

Now using the same delta, we get that $ |T| < M|S| < \epsilon$ where $ T$ is any Riemann sum for $ uf$ on $ X$.

I just don't feel like this could be true, so what did I mess up on here?
[/hide][/quote]

I think you're assuming that f is non-negative.

[quote="sos440"][quote="Kalle"]For absolutely continuous functions $ F' = 0$ almost everywhere does imply $ F$ is constant, and by the lebesgue theorem the derivative of $ F$ is $ f$ a.e.[/quote]

My proof is nothing but an overkill as long as this problem belongs to just an introductory analysis level. And it explicitly mentions 'Riemann integral', which seems to imply such possibility. That's why I said 'not satisfactory', in the sense of elementariness. [img]http://myhome.shinbiro.com/~sosQED/images/m_kiss.gif[/img]

I suspect if it can be simplified into more elementary one. One of my idea is to change the order of integration. This can be done like this:

\begin{eqnarray*} \int_{0}^{x} f(t)u(t) \, dt & = & \int_{0}^{x} \int_{0}^{t} f(t)u'(s) \, dsdt \\
& = & \int_{0}^{x} \int_{s}^{x} f(t)u'(s) \, dtds \\
& = & \int_{0}^{x} [F(x) - F(t)] u'(s) \, ds. \end{eqnarray*}

But I'm not sure whether it can be justified in the range of Riemann integral theory or not. (Of course, Fubini theorem ensures this...)[/quote]

My interest in posing this problem was not so much the proof as whether this is a special instance of a stronger proposition. For example can positivity of the derivative be dropped for, say, non-negativity?


topology
group theory
abstract algebra
superior algebra
superior algebra unsolved

I have a exercise from a book that I have to do but I have some questions on it . The following problem :
   The stabilizer of a point x in X consists of those elements g in G for which g(x)=x. Show that the stabilizer of any point is a closed subgroup of G, and that points in the same orbit have conjugate stabilizers. Where G is a topological group acting on topological space X.
    I don't believe affirmation that stabilizer of any point is a closed subgroup of G . It takes me so much time to do it. I don't believe but not assure it's wrong.
    I want to know ideals of you about the problem above. Could you give me an answer ? Thanks !

Hi.

I think the stabilizers are not closed in general. But if the map $G\times X\toX: (g,x)\mapsto g(x)$ is continuous and X is a T_1-space (so all {x} are closed) then this should be true:

Consider the map $\phi_{x}: =g\mapsto g(x)$ which is continouos. The stabilizer is now equal to $\phi_{x}^{-1}(\{x\})$ and therefore closed in G.

Thanks for reply from you !
Of course , If every point {x} is closed then the affirmation is true. Could you  have any example to illustrate for that in the general case isn't closed group? I have an example but be a complex example. I don't like it. I hope you can find for me a more pretty example. Thanks !


floor function
combinatorics unsolved
combinatorics

[quote="math10"]For each positive integer $ x$,such that $ S(x) \equal{} \{[kx]|k \equal{} 1,2,...\}$. Let $ x_1,x_2,x_3$ be three real number greater than $ 1$ such that $ \frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} \frac {1}{x_3} > 1$.Prove that there exist $ i$ and $ j$ with $ 1\leq i < j\leq 3$ such that $ S(x_i) \cap S(x_j)$ has infinitely many elements.

Hint:Using Inclusion -Exclusion Principle.[/quote]
Of course:
$ |S(x_1) \cap S(x_2)| \plus{} |S(x_2) \cap S(x_3)|  \plus{} |S(x_3) \cap S(x_1)| \equal{} |S(x_1)| \plus{} |S(x_2)| \plus{} |S(x_3)| \minus{} |S(x_1) \cup S(x_2) \cup S(x_3)| \plus{} |S(x_1) \cap S(x_2) \cap S(x_3)|$
Denote $ S_n(x) \equal{} \{[kx] \mid k \in \mathbb{N}, [kx] \le n\}$.
Then $ |S_n(x_i)| \ge \left \lfloor \frac{n}{x_i}  \right \rfloor$.
And $ |S_n(x_1) \cup S_n(x_2) \cup S_n(x_3)| \le n$
Since $ |S(x_1) \cap S(x_2) \cap S(x_3)| \ge 0$ we get:
$ \max \{ |S_n(x_i) \cap S_n(x_j)| \} \ge \frac{\left \lfloor \frac{n}{x_1} \right \rfloor \plus{} \left \lfloor \frac{n}{x_2}  \right \rfloor \plus{} \left \lfloor \frac{n}{x_3} \right \rfloor\minus{}n}{3}$
Now $ [x] > x\minus{}1$, so:
$ \max \{ |S_n(x_i) \cap S_n(x_j)| \} \ge n\frac{\frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} \frac {1}{x_3}\minus{}1}{3} \minus{}1$.
Let $ \frac{\frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} \frac {1}{x_3}\minus{}1}{3} \equal{} t > 0$.
Then:
$ \max \{ |S_n(x_i) \cap S_n(x_j)| \} \ge nt\minus{}1 \to \infty$ as  $ n \to \infty$ and we are done.


How do I prove that if you have n+1 integers, no greater than 2n, there will always be at least one integer which is an integral multiple of another?

[hide="Hint"]
Consider the set of the largest odd divisors of each of the $ n \plus{} 1$ numbers. Since all of the numbers $ \le 2n$, how many possible largest odd divisors are there?
[/hide]


The most natural reason is that a moderator considered it as more appropriate for the Pre-Olympiad forum (and I would agree).
But could you please stop writing your problems with that decision into various threads¿ It's annoying. There exist the idea of PM's for such things.

[quote="ZetaX"]The most natural reason is that a moderator considered it as more appropriate for the Pre-Olympiad forum (and I would agree).
[/quote]
Why? Explain please. Thank you.

[quote="Megus"]To make things clear: I moved it, just as ZetaX said, because I considered it too easy for the olympiad section.[/quote]
I don't agree with you. If it too easy, what are roots of moved equation?
I don't ask solution from you. Only roots. Thanks. ;)


geometry
3D geometry
prism
rectangle
Pythagorean Theorem
AMC

i saw a problem that utilized this information on some AIME...i didn't know how to find the distance between two points in the three-dimensional (x, y, z) plane....is it just a continuation of the distance formula sqrt((x2-x1)^2)+((y2-y1)^2)+((z2-z1)^2)?....can i get feedback on how to do this?

Yes, the answer is indeed $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}$.

Think about it as a rectangular prism with sides x,y,z. The diagonal of an x,y rectangle is $\sqrt{x^{2}+y^{2}}$. Combine that diagonal with the z-edge to get $\sqrt{\sqrt{x^{2}+y^{2}}^{2}+z^{2}}=\sqrt{x^{2}+y^{2}+z^{2}}$

Well yeah, I used two right triangles. Here's a picture to help illustrate.

Basically $AB=\sqrt{x^{2}+y^{2}}$ and then $AC=\sqrt{\sqrt{x^{2}+y^{2}}^{2}+z^{2}}=\sqrt{x^{2}+y^{2}+z^{2}}$

And this "extended pythagorean theorem" becomes the "extended distance formula" the same way the regular 2D pythagorean theorem becomes the 2D distance formula.

Distance between $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ is $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}$


I want to show that if A, B are hermitian and commute then I can find P s.t. PAP* and PBP* are both diagonal.
So I know that if v is an eigenvector of A with evalue a then Bv is also an eigenvector of A with the same eigenvalue. Could I maybe play with the basis of evectors of A and prove the P exists?

It's a standard fact around here that comes up a lot here: two diagonalizable matrices that commute are simultaneously diagonalizable.

Here's [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=175612]a thread[/url] with some discussion of the solution. The only real difference is that we restrict ourselves to unitary similarity.

this is used in physics i believe - that commuting hermitian matrices can be simultaneously diagonalized...
yep - i think i know what to do from here... thanks!


Find maximum $n$ such that for every $n$-element subset of
the set $\{1,2,3,......, 75\}$ the pairwise differences of the elements of it are different.

Wait...
Wouldn't it have to be 2?
Otherwise, the set could have 3 consecutive numbers, which would mean two differences of 1...

K81o7 is right as far as i can tell. Although i'd like to see how you got the fibonacci answer.

i was wrong.
can K81o7 explain his solution ?

[quote="spx2"]a lower bound to this number would be the lowest fibonnaci below or equal to 75[/quote]

Hey I want maximum of $n$,not minimum.I dont have any complain about your idea of finding minimum(which is 2 :) ) but  please find maximum of $n$.
I have found that $n$ can be $10$, that is there is a ten element subset of $\{1,2,3,4 ..., 75\}$ which has all pairwise differences of its elements different.
And obviously $n$ can not be $\ge13$ which follows from dirichlets theorem.
So try to prove or disprove that n=11,n=12.

Waiting for solution :juggle:

[quote="spider_boy"][quote="ZetaX"]Well, the way the problem is written above, the maximum is $n=2$ ;)
I think this is how it should be:
[quote="spider_boy"]Find maximum $n$ such that there is a $n$-element subset of
the set $\{1,2,3,......, 75\}$ the pairwise differences of the elements of it are different.[/quote][/quote]
true thank you ZetaX.But dear other members you could warn me earlier,As ZetaX.[/quote]

No one?


Let $ x$ represent the sum of the first 25 positive odd integers. Let $ y$ represent the sum of the next 25 positive odd integers. What is the value of $ y\minus{}x$?

x is 1+3+5+...+49

y is 51+53+55+...+99

nth element of x is 50 fewer than nth element of y.

and there are total 50 elements in both set.

so 50*25=1250

answer : 1250


Hi,

   If the nth term of a sequence is given by the expression 2 times 4 to power n-1.  what is the value of the units digit of the 131 st term in the sequence?

   Thank you!

   Jackie[/quote]

The last digit is periodic. Work out a few terms until you see the period.

This is standard number theory. In particular, it's modular arithmetic.

Its basically 2^ an odd integer.The last digit of such type of nos are 2,8.Well I guess thats hint enough just check when do these nos repeat.Use modular arithmatic.
Good luck...


algebra
polynomial
Euler
Galois Theory
algebra unsolved

Solve this equation system
$ \{  \begin{array}{l}    x^3+y^3 = 2 \\    8x +xy+ y =2   \end{array}  $

Can you give a solution of this equation?

hello, by inserting $ y\equal{}\frac{2\minus{}y}{8\plus{}y}$ in the first equation we have
$ (2\minus{}y)^3\plus{}y^3(8\plus{}y)^3\equal{}2(8\plus{}y)^3$
$ 8\minus{}12y\plus{}6y^2\plus{}511y^2\plus{}192y^4\plus{}24y^5\plus{}y^6\equal{}1024\plus{}384y\plus{}48y^2\plus{}3y^3$
$ y^6\plus{}24y^5\plus{}192y^4\plus{}509y^3\minus{}42y^2\minus{}396y\minus{}1016\equal{}0$.
The equation can by solved with numerical technique.
We have $ y_1 \approx 1.25981378804860843692$ as the real soltution.
Sonnhard.

I think it's not good  :roll:
I want to have a exactly result.

hello, what do you mean with a exactly result? This is a polynomial of degree six and has no solutions in radicals.
Sonnhard.

But it could performed as a sum of radical and ration.I think this solution is not good.I need an other solution.

[quote="tanlsth"]Solve this equation system
$ \{ \begin{array}{l} x^3 + y^3 = 2 \\
8x + xy + y = 2 \end{array} $[/quote]

Where is the problem? My solution is not valid and it result 1 unsolve equation.

We want to know your solution!

@tanlsth: Hi Tan-my good brother, I'm Tuan (L_Euler in VMF). Can you help me to study English to post Topic in this Site? Thank you very much! :)

no, don't post your solution...

maybe: hint?

does the factorization $ (x\plus{}1)(y\plus{}8)\equal{}10$ help?

Dr Sonnhard Graubner is right and tanlsth is wrong.
The polynomial in $ y$ obtained by Sonnhard is correct and its Galois group is $ S_6$. Then this polynomial has no solutions in radicals.


Let $ a,b,c$ be [b]real numbers[/b] such that $ a \plus{} b \plus{} c \equal{} \frac {3}{2}$. Prove that:
\[ 27 \plus{} 44(ab \plus{} bc \plus{} ca)^2\ge 60(ab \plus{} bc \plus{} ca) \plus{} 12abc \plus{} 16(ab^3 \plus{} bc^3 \plus{} ca^3)
\]
I think it is hard and nice. How about you?

[quote="zaizai-hoang"]Nobody give me some comments for this?
Hint:
The inequality is equivalent to:
\[ 4\sum_{cyc} a^4 \plus{} 17\sum_{cyc} a^2b^2 \plus{} 8abc \sum_{cyc} a\ge 4\sum_{cyc}a^3b \plus{} 16\sum_{cyc}ab^3
\]
When does equality hold? Do you think can_hang2007's theorem works here?  :roll:[/quote]
Yes, it works of course. here is it


If $k, l, m$ are positive integers with $\frac{1}{k}+\frac{1}{l}+\frac{1}{m}<1$, ﬁnd the maximum possible value of $\frac{1}{k}+\frac{1}{l}+\frac{1}{m}$.

[hide="This probably will not be rigorous  :P "]
I think it's clear that one of the three integers must be $2$, so let $k=2$. Then, we have that \[\frac{1}{l}+\frac{1}{m}< \frac{1}{2}\implies (l-2)(m-2) > 2 \] so at least one of $l,m$ must be $>4$. Let $m > 4$ be this integer.

Now, if both integers were $>4$ we would have the maximum value as $\frac{1}{2}+\frac{1}{5}+\frac{1}{5}= \frac{9}{10}$, which it is most likely not. So, we have that $l=3$.

The inequality is now $\frac{1}{m}< \frac{1}{6}$, so, to maximize the expression, $m$ must be $7$, which gives the maximum $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}= \boxed{ \frac{ 41}{42}}$ [/hide]

[hide="not sure that this is rigorous..."]Like cincodemayo said,, the maximum for three numbers is 41/42. Let the sum of the first three numbers be equal to it's maximum. Then the fourth number has to be at most 1/43. Then the next number has to be 1+2*3*7*43=1807, and then the next number is 1+2*3*7*43*1807, which is 3263443, and so on. We could define it recursively:

$x_{n}=1+x_{1}x_{2}x_{3}x_{4}\cdots x_{n-1}$

Then I guess that the sum would be:

$\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}=\frac{x_{n+1}-2}{x_{n+1}-1}$

$=\frac{x_{1}x_{2}x_{3}x_{4}\cdots x_{n-1}x_{n}-2}{x_{1}x_{2}x_{3}x_{4}\cdots x_{n-1}x_{n}-1}$[/hide]

This is the first time in the pre-olympiad forum. :oops:

To rigorize cincodemayo's solution, we only need to note that if we don't have a $\frac{1}{2}$ then we must have all three fractions at most $\frac{1}{3}$ and one which is at most $\frac{1}{4}$, so the maximal possible sum is $\frac{1}{3}+\frac{1}{3}+\frac{1}{4}= \frac{11}{12}$.  1=2's solution extends it with a greedy approach, which certainly [i]seems[/i] like it should always give the .  The question is, can you prove to me that there is never a better solution?

Incidentally, [url=http://www.research.att.com/~njas/sequences/A000058]here[/url] is the entry for the sequence 1=2 suggests in the OEIS.

for sake of contradiction, suppose that there exists a triple (a,b,c) with $a\ge b\ge c$ such that $1>\sum\frac{1}{a}>\frac{41}{42}$ we have that $\frac{41}{42}< \sum\frac{1}{a}\le \frac{3}{c}$, hence, $c\le \frac{126}{41}=3.07...$, then cases... $c=3$, we have $\frac{1}{a}+\frac{1}{b}>\frac{9}{14}\implies \frac{2}{b}>\frac{9}{14}\implies b<\frac{28}{9}=3.11...$, but $b\ge c$, so $b=3$, and we have $\frac{1}{a}>\frac{13}{42}\implies a<\frac{42}{13}=3.23...$ but $a\ge b=3$, so $a=3$, giving our sum to be $1$, contradiction

else, c=2, and follow sam's logic [which i have not looked at, but i presume to be correct]

[quote="N.T.TUAN"]You are true! Following problem is difficult than above problem.
Let $n$ be positive integer. If $x_{1},x_{2},...,x_{n}$ are positive integers with $\frac{1}{x_{1}}+\frac{1}{x_{2}}+...+\frac{1}{x_{n}}<1$, ﬁnd the maximum possible value of $\frac{1}{x_{1}}+\frac{1}{x_{2}}+...+\frac{1}{x_{n}}.$[/quote]
[hide="Solution"][url]http://www.mathlinks.ro/Forum/viewtopic.php?p=234089#p234089[/url][/hide]

$ml+mk+kl\ge mlk-1$, $\frac{1}{m}+\frac{1}{l}+\frac{1}{k}\ge 1-\frac{1}{mlk}$, so we have to maximize $mlk$. Suppose $m  \ge l \ge k$. If $3 \le k \le m \le g$ we have contradiction(because $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ exceeds $1$), so suppose $k=2$. Now, if $l\ge 4$ we obtain $\frac{1}{k}+\frac{1}{l}+\frac{1}{m}\le\frac{5}{6}$, so take $l=3$, and of course $m=7$ and we obtain $\frac{41}{42}>\frac{5}{6}$.


trigonometry
algebra
polynomial
function
induction
inequalities
strong induction

Prove that for $n=1,2,3,...$ $\cos\left( n\arccos x \right)$ is a polynomial function of degree $n$.

Ahh, a very nice problem! Strong induction works very well!  :)

[hide]Show that the statement is true for $n=1,2$ and then use the trig identity $\cos{(k+1)y}= \cos{ky}\cos{y}-\sin{ky}\sin{y}$ to complete the proof.

BTW, aren't these Chebyshev polynomials?[/hide]

Additional problem: try to find out some properties of the coefficients of these polynomials (they are not quite random).
For example: up to multiplication with powers of two, they are integers.
Or express the constant term in terms of $n$.

[quote="gotztahbeazn"]hmm but wat do we know about the $\sin$ functions? since the two $\sin$ functions multiplied together in the strong induction could be of $n+1$ degree, and when the $n+1$ degree $\cos$ functions subtract that it could no longer be $n+1$ degree...[/quote]
There is a small trick that you need to use to finish off your induction case:

$\sin{ky}\sin{y}= \frac12{(\cos{(k-1)y}-\cos{(k+1)y})}$

Use this in the expression you obtained earlier. I guess you will find things a lot easier now! :)