geometry
college
MATHCOUNTS
analytic geometry
function
videos
blogs

Congratulations to Evan O'Dorney, the champion of the 2007 Scripps National Spelling Bee  :D

Also, he said he didn't like the spelling bee. That is pretty awesome@@

Damn, that's some pretty impressive diversity.

[quote="brady00n"]Wow, he's only 11 years old.[/quote]

You're looking at a page from two years ago.

o'dorney lives in the bay area where i live
a local news report said hes skipping high school to go to college to learn mathematics and physics..............

his accomplishments show that hes really amazing and yeah he said he prefers math to spelling.....dude thats pretty wicked

He said he is doing calc and is 13...sound suspiciously like one of us.

When he made the "math is good, music is good, spelling in memorization" quote, I was like  That's EXACTLY what I would have said!
I think somewhere in the program said he was in the Berkeley Math Circle.

[quote="Treething"]I intensely dislike the types of people who win spelling bees. Not because they're pretentious, but because they devote insane amounts of time to godd*mn memorization of words that will have absolutely no effect on life.[/quote]

For an 8th grader, I think as long as they're not playing Warcraft all day, then almost anything they do is commendable ...

Warcraft > your soul

Right, back on topic...

I remember there was this contest called Word Power. I also remember it actually pertaining to life, because even if you don't know how to spell a word, if you have a vague idea on what it means, you won't make a fool out of yourself at one time or another.

I think I got like 4th place in NJ, and this girl I knew got first (or maybe it was second) in the nation.

Lucky, hardworking, whatever. At least she'll get something other than money out of winning, unlike MathCounts winners and spelling bee winners.

[quote="Valentin Vornicu"][quote="billzhao"]For an 8th grader, I think as long as they're not playing Warcraft all day, then almost anything they do is commendable ...[/quote]What's wrong with Warcraft?  :mad: :P[/quote]

I did not just see this. From now on, if someone "catches" me playing Warcraft instead of doing math or soemthing....I'm quoting you.

Anyway, congratz to Evan :P. Does he have an AoPS username?

yeah dude spelling bee is kinda like what but he himself noted that it is kinda like what so dude he gets some props there like that is pretty sweet

There's a lot more to memorization to spelling if you don't want to rely on luck. Knowledge of the common roots of words in English, (Latin, Greek, German, etc.) is always immensely helpful especially since they give you the etymology.

I've never participated in a spelling bee though, they look lame.

[quote="K81o7"][quote="sonny"]...hes...[/quote]

You know, I'm gonna hunt the guy down at MOP and confirm that this is indeed not a word, because I've started to see it a lot.  That, and "definately", as well as "seperate".  As for those of you who have noticed my extensive use of the word "mroe", that is just my inability to coordinate my fingers when I'm typing very quickly.[/quote]
Well it is possible that he's using Mozilla, which at some points brings up the find function when you use the apostrophe key.

Although I definitely have to agree that it really annoys me when people misspell definitely.

[quote="K81o7"][quote="sonny"]...hes...[/quote]

You know, I'm gonna hunt the guy down at MOP and confirm that this is indeed not a word, because I've started to see it a lot.  That, and "definately", as well as "seperate".  As for those of you who have noticed my extensive use of the word "mroe", that is just my inability to coordinate my fingers when I'm typing very quickly.[/quote]

I was typing too fast. No, seriously I am not usually so bad. oh yeah and ask him how to spell your name at MOP

I can say 2 things. He's really nice, and he's really talented.
For example, at the NSB last Friday, he was improvising sonatas, playing concertos he had written at home, and playing really difficult pieces on the piano. Also, he had found a pattern in the carpet (it was something with stars and dots) and came up with a function for it in only 5 seconds. Then, he noticed another pattern in a different carpet and came up with a function for that.

Wow. What a fa-mous person. He'll grow up to be famous.

[quote="vamathletes"] Also, he had found a pattern in the carpet (it was something with stars and dots) and came up with a function for it in only 5 seconds. Then, he noticed another pattern in a different carpet and came up with a function for that.[/quote]

Yes, he had a dozen kids looking on and learning what he was explaining.

He's getting some bad press at digg.com. :( I won't link you to the discussion page but it's in the "Videos" section up at the top.

Here's an interview:
http://www.ifilm.com/video/2862093

[quote="Elemennop"]oh dude that video had me rofling

where'd the reporter find a sandwich? She pulls it out of nowhere!

and that kid is too funny to watch...the 15 second pauses and one word answers were hilarious.

oh and the assburger's comment under the video had me laughing too. Comedy gold, that video is.[/quote]

I haven't seen the video and wont bother to watch it. My daughter was a semifinalist. We were told what would happen if your child was in the finals. After the morning semifinals, the finalists went to the White House. Then it was back to the hotel, where they (and their parents) were sequestered all day. Press interviews were set up during that time. If you had to go anywhere, you went with a guard. They also told what would happen if your child was the winner. The bee ended about 10pm. Then there were some press conferences. Chances are, Evan didn't get to bed until midnight. He had to be up before 6am, to catch a plane for NY (and some morning show).

I don't know about Evan, but a lot of the kids didn't really get a good night sleep the two nights before. The poor kid was exhausted.

I have been reading a lot of online blogs about the spelling bee. I am amazed at how easily people can insult the kids who competed. These are intelligent and hard working kids.

BTW, the term is Aspergers. I spotted several kids who probably were on the spectrum, but not nearly as many as I see when I go to math events.

also huh "not as many as at math events" what I was not actually aware that there were socially dysfunctional people at math events

I mean hmm

I guess I wouldn't meet them

since they were socially dysfunctional

so eh maybe you are right

also I don't actually know what Aspergers is and am just guessing based on [i]context clues[/i]

You should look up what Asperger's is, but oh well those SAT reading skills must have been pretty helpful... but yeah I don't think there are very many socially dysfunctional people at math thingies? Like definitely at MOP there aren't too many... unless by socially dysfunctional you mean awesome... and then yeah what usually there might only be like 2 at a math competition? unless it is really really big? so since positive integers are discrete and all that doesn't leave much for the spelling bee...

That woman who interviewed him is as of this moment my most-hated person ever. "You're making this really hard on me, Evan, my interview team wasn't able to come up with interesting and discussion-stimulating questions and so I'm reduced to asking you about your favorite sandwich and how to spell a word that I can't pronounce or spell myself. This is clearly your fault."

The woman was asking normal, basic starter questions that usually has no problem getting people to jump into a comfortable topic that they can actually discuss. It's the kid's fault that he replied with one word answers after 15 second pauses.

Locked... The original topic was already not quite AMC forum relevant, and it's just spam now.


geometry
rectangle
topology
real analysis
real analysis unsolved

Let us call a rectangle $ \mathcal R$ in $ \mathbb R^2$ be nice if $ \mathcal R$ has sides parallel to the axes. Prove that for any open set in $ \mathbb R^2$ is a countable union of nice rectangles.  :)

Could you please try all nice rectangles whose centers and length and width are rational numbers ?There is a similar theorem for balls in analysis books.

Statements like 'sides parallel to the axes' have no meaning in topology  :wink:

I think what you want to say is that the product topology on R gives the standard topology on R2.

[quote="kenn4000"]im tempted to say that it suffices to prove for the unit open ball?[/quote]

um, any open set in $ \mathbb{R}^2$ has countably many connected components, so it suffices to prove it for connected open sets, and each connected open set is the union of a countable number of open balls, so ...

(i guess the first reduction isn't even necessary)

well this really just equivalent to the problem of showing that

$ A_n \equal{} \{ (j,k)\in (Z^\plus{})^2 : j^2\plus{}k^2<2^n \}$

has the property $ \frac{|A_n|}{2^{n\plus{}1}} \to \frac{\pi}{4}$


modular arithmetic
number theory unsolved
number theory

[quote="mathpk"]Find the $ min(|25^{m}-7^{n}-3^{n}|)$ where $ m$ and $ n$ are positive integers.[/quote]

We have $ 25^{m}=1\pmod{12}$ $ \forall m$
If $ n$ is odd, we have $ 7^{n}=7\pmod{12}$ and $ 3^{n}=3\pmod{12}$ and so $ 7^{n}+3^{n}=10\pmod{12}$
If $ n$ is even, we have $ 7^{n}=1\pmod{12}$ and $ 3^{n}=9\pmod{12}$ and so $ 7^{n}+3^{n}=10\pmod{12}$
So  $ 25^{m}-7^{n}-3^{n}=3\pmod{12}$
And  $ -(25^{m}-7^{n}-3^{n})=9\pmod{12}$

and so $ 25^{m}-7^{n}-3^{n}$ can only be $ 3,-9, 15,-21, 27, \ldots$ and $ |25^{m}-7^{n}-3^{n}|$ can only be $ 3, 9, 15, 21, 27, \ldots$
Since $ m=n=1$ gives $ |25^{m}-7^{n}-3^{n}|=15$, we just have to check the two cases :
Case 1 : $ 25^{m}-7^{n}-3^{n}=3$
Case 2 : $ 25^{m}-7^{n}-3^{n}=-9$

But case 1 can never be true since LHS is always $ 7\pmod{8}$ and RHS is $ 3\pmod{8}$
And case 2 can never be true since LHS is in $ \{0,2,3\}\pmod{5}$ and RHS is $ 1\pmod{5}$

So $ min(|25^{m}-7^{n}-3^{n}|)=15$ and is reached for example with $ m=n=1$


function
MATHCOUNTS
algebra
domain
geometric series
partial fractions
binomial theorem

I just write them out. They wouldn't ask questions with 4 or 5 dice.

When I say I write them out, I don't mean I write all 3 dice out. I write one die out, and the sum of the other two dice out. For example, in your problem, I would say the possible ways to get 10 are (4,6)(5,5)(6,4)(7,3)(8,2)(9,1) where the first number is the sum of the first two dice and the last number is the sum of the last dice. And just remember, for 2 dice you can get the number x in 6-|x-7| ways for 2:le:x:le:12.

[color=cyan]The most systematic way is to start by subtracting the largest possibly number on the most number of dice, and then slowly decrease them -- in other words, work with lists of numbers that are in decreacing order.  So, for making 20 with 5 dice, say, start with
6, 6, 6, 1, 1
and then slowly shift pips from the alrger dice to the smaller.

This approach is identical to the "How manhy ways can you make $1 with coins?" type of problem.  It is also basically what Mystic suggested.[/color]

well, when the problems get complicated enough, it would be advantageous to use generating functions, no?

[color=#20DFFF]To be honest, I know almost nothing about generating functions.  But, being that this is the Mathcounts forum, I think it's safe to say that systematic counting will be the way to go.[/color]

Well he never specified whether or not the question was from MATHCOUNTS...oh well.

Can anyone explain to me what "generating function" is? Shouldn't we use combinatrics? Are they the same thing?

I think, (note the think, I'm not very sure at all) that generating function means that given a set of integers, a sequence for example, you generate a function that gives you the nth number of the set when pluging n in

I know of two kinds of generating functions. One is of the form  :Sigma: a_i x^i which is the basic generating function, and another is of the form :Sigma: a_i x^i/i!, which is the exponential generating function. In both cases, the a_i stand for the ith coefficient, which stand for whatever we're trying to find. (In the case JBL just posted 3 posts ago, it would be the number of ways to make up i cents out of the basic coins). This makes things simpler because we can use sums of infinite geometric series and partial fractions to manipulate these coefficients. Sigh, I suppose I'm not explaining this very well tho...

I'd say generating functions is the way to go for these sorts of problems if the numbers aren't very small.

Generating functions can be really really complicated though.. Anyone want to give a nice post on them?

Basically a generating function is like a clothesline: you hang stuff up on it. They key is not the value of the function for different values in domain, but the coefficient of x^k. For example, consider a set of n coins. There are nCk ways to choose k heads. Our generting function would be f(x)=nC1*x + nC2*x^2 +nC3*x^3 + ... + nCn*x^n. Here we can simplify (by binomial theorem) to f(x)=(1+x)^n. We say that the number of ways to choose k head coins from n coinsis the coefficient of x^k in the function f(x)=(1+x)^n. This example is trivial, but its like a window into the world of generating functions. They are used in particular in partitions theory, card-dealing problems, various combinatorics stuff (like finding a formula for Fibonacci numbers), etc....

You stole that clothesline analogy from generatingfunctionology, didn't you?

Mmm, must...take...Intermediate Counting...

[hide]Reply: plagiarism!  [/hide]

i did steal it from generatingfunctionology. but its a pretty good one, so why not use it?  :D

how would you go about evaluating the generating function?  For example in our dice problem, would you expand (x^1+x^2...x^6)^3, or is therea more clever way to do this.  I know we can change it to ([x(x^6-1)]/(x-1))^3, but how does this hlep us?  How do we evaluate it from there?


calculus
integration
logarithms
function
real analysis
real analysis unsolved

Compute
\[ \int_1^e\frac {dx}{\sqrt {x^2\ln x \plus{} (x\ln x)^2}}.\]

I am sorry, the lower integration limit is 1.

$ \equal{}\int_0^1 \frac{1}{\sqrt{t\plus{}t^2}} \, dt\equal{}2 \text{ArcSinh}[1]$

In the same way as basil mohammed did .

put $ t \equal{} \ln x$

hence the integral reduces to $ \equal{}\int_{0}^{1}\frac{1}{\sqrt{t\plus{}t^{2}}}\, dt\equal{} \ln (  t\plus{}1/2 \plus{} \sqrt{ (t \plus{} 1/2 )^2  \minus{} (1/2)^2 }   )$ from 0 to 1

= $ \ln {3 \minus{} 2 \sqrt {2}  }$

Pardon me if my calculations are incorrect .


analytic geometry
function
vector
advanced fields
advanced fields unsolved

Consider on $\mathbb{R}^{2}$ the connection defined by $\Gamma_{11}^{1}=x^{1}$, $\Gamma_{12}^{1}=1$, $\Gamma_{22}^{2}=2x^{2}$ and the other Christoffel symbols vanishing. (Here $x^{2}$ means the coordinate function of the chart $(x^{1},x^{2})$.) Let $\gamma: [0,1]\to \mathbb{R}^{2}$ be the path $\gamma(t)=(t,0)$. Compute the parallel transport along $\gamma$ of the vector $(\partial_{2})_{0}\in T_{0}\mathbb{R}^{2}$.

Step 0: There is a formula for parallel transport in terms of Christoffel symbols.  Use it.

Step 1: Plug your information into the formula to get an ODE.

Step 2: Solve the ODE.

If that doesn't help, then which step is the one you're having trouble with?


Euler
number theory unsolved
number theory

Show that every positive integer is a sum of one or more numbers of the form 2^r*3^s where r and s are nonnegative integers and no summand divides another.

A completely different problem in which $ 2^r 3^s$ appears  [i]mysteriously[/i] . Find all positive integers such that $ \phi(n)$ divides $ n$ where $ \phi$ is Euler totient function. I hope you find it amusing.


quadratics
algebra
geometry
perpendicular bisector

1. (#20, 2006) The quadratic equation $ ax^2 \plus{} bx \plus{} c \equal{} 0$ has the roots $ \minus{} 1$ and $ 5$. Find the value of discriminant of this equation.
A) $ \minus{} 5$
B) $ \frac15$
C) $ 4$
D) $ 5$
E) $ 6$

My solution
[hide] They all are - in fact, any positive number can be the discriminant, by "scaling" the coefficients of the quadratic equation.
[/hide]
2. (#32, 2006) Given the two points $ A( \minus{} 2, 0)$ and $ B(4,0)$, write the equation defining the set of points that are equidisitant from $ A$ and the perpendicular bisector of $ AB$.
A) $ y^2 \equal{} 6x \plus{} 3$
B) $ y^2 \equal{} 6x \minus{} 3$
C) $ y^2 \equal{} \minus{} 6x \plus{} 3$
D) $ y^2 \equal{} \minus{} 6x \plus{} 9$
E) $ y^2 \equal{} 6x \minus{} 9$

My solution
[hide]We can find the vertex to be $ ( \minus{} \frac12, 0)$ and directrix $ x \equal{} 1$, and $ p \equal{} \minus{} \frac32$. Thus our equation is $ x \plus{} \frac12 \equal{} \dfrac{y^2}{ \minus{} 4\cdot \frac32}$, reducing to $ y^2 \equal{} \minus{} 6x \minus{} 3$.
[/hide]
I don't have solutions at hand, so I'm not sure where I went astray...Thanks in advance for any explanation.

I have a copy of the official solutions to the 2006 test, and both of those problems are crossed off and apparently weren't counted towards the official scores (in other words, you are correct on both of them).

Looking at the copies that I have, it appears that these questions may have been thrown out. So your solutions may be correct. But I'm being lazy and not working them myself! :-)

Does anyone know where I might be able to find a copy of last year's GCTM problems and solutions? thanks.

Mr. Fulton, I know you don't approve of tournaments being posted to AoPS, but would it be acceptable if I e-mailed him a copy? If so, Boru, would you please PM me your e-mail address so that I may send you a copy?

I certainly don't have a problem emailing a test between a couple of folks. It's just when it's posted and available to the entire world.

The written test is almost invariably exactly 14 pages (including the cover page). Hmm...

eric do you think you can send me last year's test too?

Okay. Anyone else who wants it, just ask.


From the given condition it can be said that sinx,siny and sinz are all positive.....

From the inequality

Root Mean Square >= AM for +ve numbers we have

sinx^2+siny^2+ sinz^2 >= 4/3 which implies
cosx^2+cosy^2+cosz^2 <= 5/3

Again applying RMS>= AM we have....

cosx + cosy + cosz <= sqrt ( 5)..... :D ....

[quote="Arne"]This one is a little bit tougher than the first one I posted here: :)
Let x, y, z be real numbers such that sin x + sin y + sin z \geq 2.
Show that cos x + cos y + cos z \leq \sqrt5.
[i]Russia 1995[/i][/quote]
Let $x, y, z$ be real numbers such that  $sin x + sin y + sin z \geq {2}.$ Show that $cos x + cos y + cos z \leq \sqrt{5}. $
[i][b]Russia 1995[/b][/i]

[color=darkred][b]PP ([u]Russia 1995[/u]).[/b] Let $x, y, z$ be real numbers such that $\sin x + \sin y + \sin z\ge 2$ . Show that $\cos x +\cos y + \cos z\le\sqrt 5$ .[/color]

[color=darkblue][b][u]Proof[/u].[/b] I"ll use the well-known inequality $\{a,b,c\}\subset \mathbb R\implies |a+b+c|\le \sqrt {3\left(a^2+b^2+c^2\right)}$ . Therefore, $\left(\sum\cos x\right)^2\le$

$ 3\sum\cos^2x=$ $3\left(3-\sum\sin^2x\right)\le$ $3\left[3-\frac 13\cdot \left(\sum\sin x\right)^2\right]\le$ $3\left(3-\frac 43\right)=5\implies$ $\sum\cos x\le \left|\sum\cos x\right|\le\sqrt 5$ .

[u]See [b]PP15[/b][/u] from[/color] [url=http://www.artofproblemsolving.com/blog/81573][color=red][b]here[/b][/color][/url]


Given $ a_{n} > 0$ for all $ n \in \mathbb{N}$ and that $ \sum a_{n}$ diverges, does $ \sum \frac {a_{n}}{1 + na_{n}}$ diverge or converge?

Can I use the limit comparison test with $ \left\{\frac{1}{n}\right\}$ which would give me ${ \lim_{n\rightarrow \infty}\frac {na_{n}}{1 + na_{n}}}$? Is this limit just 1?

No, because $ a_n$ might be less than $ \frac1n$.

$ \frac{a_n}{1\plus{}na_n}$ is comparable to the [b]minimum[/b] of $ a_n$ and $ \frac1n$. That minimum is what you need to work with.
This has been discussed before here, and I think you're missing a condition.

[quote="jmerry"]This has been discussed before here, and I think you're missing a condition.[/quote]
Indeed. The condition that jmerry is referring to is monotonicity. So just what can I do if I deny monotonicity? How about this:

$ a_n\equal{}\begin{cases}\frac1{\sqrt{n}}&n\text{ a perfect square}\\ \frac1{n^2}&\text{otherwise}\end{cases}.$

Once you unscramble the details of this, you'll see that $ \sum a_n\equal{}\infty$ while $ \sum \frac{a_n}{1\plus{}na_n}$ converges.


ABC  triangle , x,y,z>0. Prove that
[(x+y)/z][bc/(s-a)] + [(y+z)/x][ca/(s-b)] + [(z+x)/y][ab/(s-c)] >=8s

still  dont  have  time  to  learn latex...
thanks  anyway!

[quote="nickolas"] still  dont  have  time  to  learn latex...
thanks  anyway![/quote] it would take you at most 10 minutes to learn the simbols which are ussually used in inequalities. Like \geq -> $\geq$, \leq -> $\leq$, \frac {a+b}{c+d+e} $\frac {a+b}{c+d+e}$. :)

Does s denotes the sum of lengths or the semiperimeter?

I found a solution but it is to long, what i did is basically the following:

put s-a=p, s-b=q, s-c=r then p,q,r are positive reals. Now write the inequality in terms of p,q,r,x,y,z and clear denominators expanding everything so you ill get a sum of 24 terms on each side. aply several  times AM-GM using simetry. I think this is standard for this sort of problems, you dont have to think, just expand and aply AM-GM to "kill" terms at each side!

Vetali couldn't you guess that S was the semiperimeter?

:):):)

Nice problem by the way... I solved it!!!

___________________________
"Our days are never coming back... " (Highway Song -> System of a down)


logarithms
modular arithmetic
function
complex numbers
complex analysis
complex analysis unsolved

Where is the mistake?
$ i \pi \equal{}\ln(\minus{}1)\equal{}\ln\left (\frac{1}{\minus{}1}\right )^{\minus{}1}\equal{}\minus{}\ln(\minus{}1)\equal{}\minus{}i \pi$

In using the functional equation of $ \ln$ without any justifying.

Or in other words, what we think of as the exponent and logarithm rules (when everything lives in the positive reals) no longer hold true over the complex numbers without some adjustments to take account of the fact that the argument (and so the logarithm) is multi-valued.

Another way to put it is that that "proof" does reasonably well show that
\[ i\pi\equiv\minus{}i\pi\pmod{2\pi i}\]

Put in yet another way, there is another branch that you have to move onto.

[quote="klismann"]log a=x

->  a>0[/quote]

No, you can perfectly  well define a logarithm on nonzero complex numbers. It's just a
bit more complicated an issue than defining the real logarithm, since the exponential
function is not injective on the complex plane.

Yes, you have to take into account the multiple branches of the logarithm function, and stick some $ 2\pi ik\plus{}\theta$ for $ k\in\mathbb{Z}$ in there.


let $ (a_{n})$ a sequence such that:
$ a_{n\minus{}1}\equal{}\frac{n^2}{a_{n}}\plus{}\frac{a^n}{n^2}\plus{}2$   and $ a_{1}\equal{}1$

prove that [$ a_{n}$]$ \equal{}n$

this doesn't make sense. that should be $ a_{n \plus{} 1} \equal{} ...$, or else the problem wouldn't make sense. and even then the conclusion is wrong because we get $ a_{2} \equal{} 4$. so you might wanna check the problem statement again, wherever you got it from.

i do know of a problem that looks very similar, though...

the sequence $ (a_n)_{n\geq 1}$ is defined by $ a_1 \equal{} 1, a_{n \plus{} 1} \equal{} \frac {a_n}{n} \plus{} \frac {n}{a_n}$ for $ n \geq 1$. prove that $ \lfloor a_n^2 \rfloor \equal{} n$ for all $ n\geq 4$.


Find the closed-form of

$ \sum_{k\equal{}1}^n \frac{F_k}  {5^k}$

where $ F_n$ is the nth Finonacci number that satisfies $ F_{n\plus{}2} \equal{} F_{n} \plus{} F_{n\plus{}1} \ (n \in N)$ and $ F_1 \equal{} F_2 \equal{} 1$.

It's the sum of two geometric series, which doesn't seem to belong in the Olympiad section.


geometry
3D geometry
algebra
polynomial
number theory proposed
number theory

1. Prove that every multiple of 3 can be written as a sum of four cubes (positive or negative).

2. A polynomial $ P(x)$ of degree 2005 satisfies $ P(k)\equal{}2^{k}$ for $ k\equal{}0,1,\cdots 2003$. Find $ P(2004)$.

[quote="mathwizarddude"]1. Prove that every multiple of 3 can be written as a sum of four cubes (positive or negative).[/quote]

$ 3(2p)\equal{}(p\plus{}1)^3\plus{}(p\minus{}1)^3\minus{}p^3\minus{}p^3$

$ 3(2p\plus{}1)\equal{}p^3\plus{}(2p\minus{}5)^3\minus{}(p\minus{}4)^3\minus{}(2p\minus{}4)^3$


geometry
trigonometry
inequalities
triangle inequality
geometry unsolved

Given a circle and a point $A$ inside the circle, but not at its center. Find points $B$, $C$, $D$ on the circle which maximise the area of the quadrilateral $ABCD$.

My solution might be wrong, because it's really easy, and brazil mo are usually hard.
Anyways, here it is:
Let the centre of circle be $O$, $OA=d$, radius of circle is $R$. Denote $|MNPK|$ the area of a polygon $MNPK$
Then:
$|ABCD|\le |AOB|+|BOC|+|DOC|+|AOD|$
$=\frac{1}{2}(AO(OD)sin(AOD)+AO(OB)sin(AOB)+CO(OB)sin(COB)+DO(OC)sin(COD))$
$=\frac{1}{2}(dRsin(AOD)+dRsin(AOB)+R^{2}sin(COB)+R^{2}sin(COD))$
$=\frac{1}{2}(R(d(sin(AOD)+sin(AOB))+R(sin(COD)+(COB))))$
$\le \frac{1}{2}(R(d*2+R*2))$, which is the maximum value of area of $ABCD$. This value is achieved when $sin(AOD)=sin(AOB)=sin(COD)=sin(COB)=1$, ie when $\angle{AOD}=\angle{AOB}=\angle{COD}=\angle{COB}=90^{o}$
so $C$ is he intersection of $AO$ with the circle so that $CO>AO$; $BD$ are endpoints of diameter of circle such that $BD\perp AO$

As a Brazilian, I'm flattered you found Brazil MO hard, but I don't think it's [b]that[/b] hard. There are some medium-to-hard problems indeed, but there are some really easy ones. Like the one you just solved (your solution is correct! But can you make it without trigonometry?).

I first did Brazil MO 2004 and found some probelms hard(like number 6); then I expected Brazil MO 2000 to be same difficulty but this was a really easy probelm, I got it in about 5 min
I will try to find a geometry proof
EDIT: found one, don't have time  right now, I will post tomorrow

Ok here is the geometric solution.Let te centre of the circle be $O$ .Let $BD$ intersect $AD$ at $M$, denote $|MNPK|$ area of polygon $MNPK$, let $h_{1}$ be height in $\triangle{DAB}$ on $BD$, $h_{2}$ is height in $\triangle{BCD}$ on $BD$
Then $|ABCD|=|BAD|+|BCD|= \frac{1}{2}(BD*h_{1}+BD*h_{2})$
$=\frac{1}{2}BD(h_{1}+h_{2})$
We have $h_{1}\le AM, h_{2}\le CM$, hence $h_{1}+h_{2}\le AC$, by triangle inequality, $AC\le AO+OC$, so $AC$ is maximized if it passes through $O$, $H_{1}+h+2$ is maximized if $BD\perp AC$
Also, by triangle inequalty, $BD\le BO+OD$, so $BD$ is maximized if it passes through $O$
Threfore the area is maximized if $AC$ and $BD$ intersect at $O$ and $BD\perp AO$, and so $C$ is the intersection of $AO$ with the circle such that $CO>AO$


AMC
AIME
USA(J)MO
USAJMO
USAMO
AMC 12
probability

I usually get around 140 on the AMC 10, but only about 100 on the AMC 12 :oops: and a 2-4 on the AIME. At this point, should I be concentrating more on getting my AMC 12 score up, or my AIME?

Considering you're still in middle school (I think?) and can still take the AMC 10, I would study the AIME. The AMC 12 skills should follow. If you do this you'll have a good chance at making the USAJMO. The exception to this is if you think you have a very good chance at getting about a 220 index through the AMC12. Then I would study both the AMC 12 and AIME so you can make the USAMO.

I remember reading this on the USAJMO announcement but can't find where it is right now, so could anyone tell me if this is true or not (in general)?

AMC 10+AIME index>210=USAJMO
AMC 12+AIME index>210=USAMO

That's basically true, the one exception is if you get an 11+ (i'm not sure) AIME after the AMC 10. Then you get a spot in the USAMO.

You still should start Olympiad soon because USAJMO is actually pretty close in USAMO (2 problems in common)

are they actually in common or just similar level?

@topofmath if she gets 2-4 on the AIME, I doubt she can do USAMO problems.

Heck, I get a 5 or 6 average and I've only been able to do like 2 current USAMO problems from the past 10 years.

By correct, do you mean like a 7/7 on the proof or that you've been able to complete the proof?

Also, they are common, which I think is weird because USAJMOers could focus on the proofs to make Red MOP.

I'm at the same level as isabella and I'm taking the AMC12 problem solving series and then the AIME problem solving series. While I'm doing these, I'll be doing AIME's so I'm hoping I can get my average AIME score to about 8-10.

Is this a good idea or should I go straight to the AIME problem solving course?

If you are considering making USAMO, then you have to score relatively high on both AIME and AMC12, so I suggest that you take the AIME course and just do a bunch of practice AMC12's...

The only problem is that I think I would struggle a little bit at the beggining if I took the AIME course right away.

At what point should I start doing USAMO problems?

Also, is it a good idea for me to take Intermediate Counting and Probability before the AIME Problem Solving Series, or does it not matter? Over the next few months I'm looking at something like this:

- Finish Interm Alg
- Finish reviewing Volume 1
- Volume 2
- AIME problem solving series
- Intermediate Counting and Probability
- Precalculus

My next few months look very similar except I'm not doing pre-calc (because I'm doing it in school), and I'm taking the AMC 12 problem solving series. I'm also doing a book called First Steps for Math Olympians.

[quote="AIME15USAMO"] I'm also doing a book called First Steps for Math Olympians.[/quote]

Actually, I've been thinking of getting that - is it any good? Will it help raise my AMC 12 and AIME scores? Anyone else have any feedback about the book?

I like the book. It starts out each section by teaching a topic, then has 3 examples. Then it has 10 problems from past AMC's using the topic which you just learned in increasing difficulty.

I would recommend it if you're looking to increase your AMC 12 and AIME scores. The book starts out at a simple level (arithmetic ratios) and goes to topics like prime decomposition, statistics, etc.

First Step for Math Olympians is a good book, but not really necessary. It's great for an overvew in problem solving; there are 18 chapters covering a wide range of math with nice problems. But if you have all the AMC problems, the book is not really necessary because all the problems in the book come from AMC questions.

The final word is: If you're not worried about the money, go ahead and buy it.  :)  You won't go wrong with it.

That's true, but they do a good job of explaining the topics before you do them. They are also listed in increasing difficulty which is helpful.


Solve equation:
$ 8(\log_2 x)^3\plus{}12(\log_2 x)^2\plus{}46 \log_2 x \minus{}48\sqrt{2} x\plus{}69\equal{}0$

Just observing gives one of the solutions is $ x \equal{} \sqrt{2}$.


analytic geometry
graphing lines
slope
geometry
circumcircle

my school got the stuff some time ago.  2-3 weeks maybe?

you happen to know if anyone from Acton-Boxborough was in the top 20?

To make this thread more exciting, here's a problem:

There is a triangle ABC.
AB = n-1
BC = n
AC = n+1

A is at the origin, C is at (n+1,0), and B is in the first quadrant.

Find the limit of the slope of Euler's line as n->infinity.
(Euler's line passes through the circumcenter, centroid, and orthocenter of any triangle)

[quote="namstap"]hey guys,

did any of you take the massachusetts level 2 olympiad this year?  i figured we would've found our our results by now.  I kindof want to know seeing as its my last week of school.

anyone know anything?[/quote]

What is that?


induction
calculus
calculus computations

Given any recursive formula for a sequence, I think its easy to show if the sequence converges or diverges once you know whether its increasing or decreasing.(What do u think?)

But the problem is its not easy(for me) to conclude whether a sequence is increasing or decreasing. ( I want a solid proof)

for example

can u find which of these sequences is increasing or decreasing(any advice on general for dealing with recursive sequences?)?

 $1) a_{n+1} = 0.5|a_{n}| + 1$  given   $a_1 = -1$

$2) a_1 = 1/4$      and     $a_{n+1} = 2a_{n}(1-a_{n})$

if you want to know whether a sequence is increasing or decreasing just try and find out the sign of a_n+1- a_n. or in some cases the quotient of the two.for instance in the 2nd qn u gave, first try and prove a_n< 0.5 for all n using induction and use this to find the sign of a_n+1- a_n.


Find the current needed for getting 10000kg/day of $ \ce{Fe}$ from molten $ \ce{Fe2O3}$. 1 mole of electrons has electric charge of 96500C.

The current needed is
$ (3 \ e\minus{})\times(10000 \frac{kg}{day})\times( \frac{1day}{86400sec})\times(\frac{1mol}{0.056kg}) \times (\frac{96500 \ C}{1 mol\times e^\minus{}})$


SOS is not necessary here, just prove directly the stronger one that
\[(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \ge \frac{3(a+b+c)^{2}}{ab+bc+ca}.\]
(very easy)

Alternatively, we can also prove a stronger one that
\[(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \ge 5+\frac{4(a^{2}+b^{2}+c^{2})}{ab+bc+ca}.\]

This is done by SOS,

[quote="carlosbr"]
Let $a,b,c$ be positive real numbers,
such that: $a+b+c \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$\\

Prove that:
\[a+b+c \geq \frac{3}{a+b+c}+\frac{2}{abc}. \]
[/quote]

This problem was in the RMO Shortlist in 2006.

I have a (nice) solution .Let $a+b+c=s$ , $ab+bc+ca=q$ and $abc=p$ .
Then by the condition we have
$sp\geq q\geq \sqrt{3ps}$  so $ps\geq 3$ (1) and so by AM-GM $s\geq 3$ (2)
Now we will consider two cases .
1st : If $p\leq 1$ then the inequality is equivalent to
$s\cdot sp\geq 3p+2s$ .Using (1) it suffices to prove that $3s\geq 3p+2s$
or $s\geq 3p$ . But by AM-GM $s\geq 3\sqrt[3]{p}$ so it suffices to prove $p\leq 1$ which holds .

2nd case :$p\geq 1$ so it suffieces to prove that $s\geq \frac{3}{s}+2$
or equivalently $(s-3)(s+1)\geq 0$ which holds due to (2) so QED  :wink:

[quote]Alternatively, we can also prove a stronger one that
$(a+b+c)(\frac{1}{a}+\frac{1}{b+\frac{1}{c})\ge 5+\frac{4(a^{2}+b^{2}+c^{2})}{ab+bc+ca}$
[/quote]
Okie.... S.O.S killed it  :P
We have:
$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\ge5+\frac{4(a^{2}+b^{2}+c^{2})}{ab+bc+ca}$
We will ues :
$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-9=\sum \frac{(a-b)^{2}}{ab}$
$\frac{a^{2}+b^{2}+c^{2})}{ab+bc+ca}-1=\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{ab+bc+ca}$
By expanding expression, above inequality equivalent:
$\sum (a-b)^{2}(\frac{1}{ab}-\frac{1}{ab+bc+ca})\ge 0$
It is true because we have $ab\le ab+bc+ca$ or
$\frac{1}{ab}-\frac{1}{ab+bc+ca}$  :oops:
This is done by S.O.S method.

[quote="hungkhtn"]I think you make mistake in writing the inequality to square form, Hoang. Try it again,

[hide="@hoang"]@Hoang. Why I can't see you online these days?[/hide][/quote]
Ohh... I sorry because I make mistake in writing the inequality to square form. I tried it and I saw. It should be:
$\sum (a-b)^{2}(\frac{1}{ab}-\frac{4}{ab+bc+ca}\ge 0$
huhmm... I will try more  :blush:

I think it should be

\[\sum (a-b)^{2}\left(\frac{1}{ab}-\frac{2}{ab+bc+ca}\right).\]

[quote="carlosbr"][size=150][color=blue]TST PERU - IMO 2007 [Day 1- Q02][/color][/size]

Let $a,b,c$ be positive real numbers,
such that: $a+b+c \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$\\

Prove that:
\[a+b+c \geq \frac{3}{a+b+c}+\frac{2}{abc}. \]
Spanish Communities[/quote]
Hung and Hoang! It's not very complicated as both of you. :wink:
Note that from $a+b+c \ge 1/a+1/b+1/c$, we have $a+b+c \ge 3$, the inequality is equivalent to
\[(a+b+c)^{2}\ge 3+2\left( \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right) \]
By AM-GM inequality, we have
\[2\left( \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right) \le \frac{2}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2}\le \frac{2}{3}(a+b+c)^{2}\]
It suffices to prove that
\[(a+b+c)^{2}\ge 3+\frac{2}{3}(a+b+c)^{2}\]
or equivalently
\[a+b+c \ge 3 \]
which is obvious. :)

Dear Canm

The initial is really easy. I did prove it in the first post. I just show a stronger inequality as Hoang are trying to prove.

[color=blue]hix ... I sorry because I was very stupid  :blush:
I have made a mistake in use identity  :oops:
We have:
$\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}=\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{2(ab+bc+ca)}$
Now I will prove it again
By expanding expression we have the inequality to square form:

$S_{a}(a-b)^{2}+S_{b}(b-c)^{2}+S_{c}(c-a)^{2}\ge 0$

To concretize:

$S_{a}=\frac{1}{ab}-\frac{2}{ab+bc+ca}\\ S_{b}=\frac{1}{bc}-\frac{2}{ab+bc+ca}\\ S_{c}=\frac{1}{ca}-\frac{2}{ab+bc+ca}$

WLOG we assume: $a\ge b\ge c$ can check it easily:  $S_{b},S_{c}\ge 0$. Then we will prove $S_{b}+S_{a}\ge 0$. In fact:

$\ S_{b}+S_{a}=\frac{1}{ab}+\frac{1}{bc}-\frac{4}{ab+bc+ca}\ge \frac{4}{ab+bc}-\frac{4}{ab+bc+ca}\ge 0$

It is true because we have $ab+bc\le ab+bc+ca$ This is done by S.O.S method  :wink:
I hope above solution has not make mistake  :blush:
Thank you very much, hungkhtn. Cẩn's solution is very nice :D[/color]

Yes, it is exactly what I want, Hoang. Similar to me.

[quote="toanhocmuonmau"][quote="carlosbr"][size=150][color=blue]TST PERU - IMO 2007 [Day 1- Q02][/color][/size]

Let $ a,b,c$ be positive real numbers,
such that: $ a+b+c \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$\\

Prove that:
\[ a+b+c \geq \frac{3}{a+b+c}+\frac{2}{abc}. \]
Spanish Communities[/quote]
Hung and Hoang! It's not very complicated as both of you. :wink:
Note that from $ a+b+c \ge 1/a+1/b+1/c$, we have $ a+b+c \ge 3$, the inequality is equivalent to
\[ (a+b+c)^{2}\ge 3+2\left( \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right) \]
By AM-GM inequality, we have
\[ 2\left( \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right) \le \frac{2}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2}\le \frac{2}{3}(a+b+c)^{2}\]
It suffices to prove that
\[ (a+b+c)^{2}\ge 3+\frac{2}{3}(a+b+c)^{2}\]
or equivalently
\[ a+b+c \ge 3 \]
which is obvious. :)[/quote]

excuse... i dont understand how you use AM-GM  :(

I think he is using the inequality

$ \sqrt{\frac{xy\plus{}yz\plus{}zx}{3}}\leq\frac{x\plus{}y\plus{}z}{3}$,

applied to numbers $ \frac{1}a,\ \frac{1}b,\ \frac{1}c$.

It seems obvious if we break it up into $ abc \ge 1$ and $ abc \le 1$.
If $ abc \ge 1$ then by AM-GM $ a\plus{}b\plus{}c \ge 3$ and it is obvious.
If $ abc \le 1$ then we can immediately substitute the given.
We need to show
$ \frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac {1}{c} \ge \frac{3}{a\plus{}b\plus{}c} \plus{} \frac{2}{abc}$
$ (a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca) \ge 2(a\plus{}b\plus{}c) \plus{}3abc$
Again, using our given,
$ (\frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac {1}{c})(ab\plus{}bc\plus{}ca) \ge 2(a\plus{}b\plus{}c)\plus{}3abc$
$ a^2b^2\plus{}b^2c^2\plus{}c^2a^2 \ge 3a^2b^2c^2$
Which is obvious since $ abc \le 1$.
Is this correct?

[quote="littletush"]let us suppose
$a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
then it's equivalent to
$\sum_{sym}a^3b^2\ge 2\sum_{cyc}a^2b^2c$
trivial by Muirhead[/quote]
what's more,it is because $a-\frac{1}{a}$ etc are mjonotonous that we can assume $\sum a=\sum\frac{1}{a}$.

[quote="carlosbr"]Let $a,b,c$ be positive real numbers, such that: $a+b+c \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$

Prove that:
\[a+b+c \geq \frac{3}{a+b+c}+\frac{2}{abc}. \][/quote]

$ puttig~r=abc~;~t=a+b+c~;~B=ab+bc+ca~we~have~rt\geq B \\(a+b+c\geq\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Leftrightarrow rt\geq B)~we~well~to~prove~rt^2\geq 3r+2t \\(a+b+c\geq\frac{3}{a+b+c}+\frac{2}{abc}\Leftrightarrow rt^2\geq 3r+2t )\\ \bullet rt\geq B\Rightarrow \left (rt  \right )^2\geq B^2\geq 3rt\Rightarrow rt\geq 3\Rightarrow rt^2\geq 3t\Rightarrow \frac{2}{3}rt^2\geq 2t ~~\fbox1\\\bullet rt\geq B\Rightarrow rt^2\geq tB \geq 9r \Rightarrow \frac{1}{3}rt^2\geq 3r ~~\fbox2 \\ finnally~\fbox1+\fbox2 \Rightarrow rt^2\geq 2t+3r~\blacksquare $

[quote=carlosbr]Let $a,b,c$ be positive real numbers, such that: $a+b+c \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$

Prove that:
\[a+b+c \geq \frac{3}{a+b+c}+\frac{2}{abc}. \][/quote]
http://www.artofproblemsolving.com/community/c6h1184601p5748463
Let $a,b,c>0$  and $a+b+c\geq\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.Prove, that $$a+b+c\geq\frac{3}{abc}$$

[quote=tarzanjunior][quote=hungkhtn]Very easy by SOS method

Notice that
\[\frac{a+b+c}{abc}\le \frac{(a+b+c)^{2}}{abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}=\frac{(a+b+c)^{2}}{ab+bc+ca}.\]
So we only need to prove that
\[(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \ge 3+\frac{2(a+b+c)^{2}}{ab+bc+ca}.\]
It can be proved easily due to SOS method.[/quote]

What is SOS here?[/quote]

SOS is a short form for a powerful inequality bashing/solving technique. :)
More can be found [url="http://artofproblemsolving.com/community/c6h80127p458535"]here[/url]. :)
Hope it helps.

$(a+b+c)(ab+bc+ca)\geq 9abc(AM-GM)$
$\rightarrow$
$(a+b+c)(a^2b^2+b^2c^2+c^2a^2)\geq 3abc(ab+bc+ca)(Cauchy Schwarz)$
$\rightarrow$
$(a+b+c)(ab+bc+ca)^2\geq 3abc(ab+bc+ca)+2abc(a+b+c)^2$
$\rightarrow$
$(a+b+c)(ab+bc+ca)\geq 3abc+2(a+b+c)(Condition)$
$\rightarrow$
$abc(a+b+c)^2\geq 3abc+2(a+b+c)(Condition)$
$\rightarrow$
$a+b+c\geq \frac{3}{a+b+c}+\frac{2}{abc}$ :D

Before taking cases, we will remark some important stuff.
Using AM-HM
$$a+b+c \ge \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{9}{a+b+c} \implies a+b+c \ge 3$$
Now re-write the condition as $abc(a+b+c) \ge ab+bc+ca$ and use $(x+y+z)^2 \ge 3(xy+yz+xz)$ to get
$$(abc(a+b+c))^2 \ge (ab+bc+ca)^2 \ge 3abc(a+b+c) \implies abc \ge \frac{3}{a+b+c} \; \text{and} \; a+b+c \ge \frac{3}{abc}$$
[b]Case 1:[/b] $abc \le 1$
Then apply previous ineqs to get
$$a+b+c \ge \frac{3}{abc} \ge abc+\frac{2}{abc} \ge \frac{3}{a+b+c}+\frac{2}{abc}$$
Equality if $a=b=c=1$
[b]Case 2:[/b] $abc>1$
By AM-GM
$$ab+bc+ca \ge 3(abc)^{\frac{2}{3}}>3$$
We will use this in a smart way
$$a+b+c>\frac{a+b+c}{ab+bc+ca}+\frac{2(a+b+c)}{ab+bc+ca} \ge \frac{3}{a+b+c}+\frac{2}{abc}$$
There is no equality case.
Hence the ineq is proven with equality at $a=b=c=1$, thus we are done :D


calculus
integration
limit
trigonometry
calculus computations

Let $n\geq 2$ be integer. On a plane there are $n+2$ points $O,\ P_{0},\ P_{1},\ \cdots P_{n}$ which satisfy the following conditions as follows.

[1] $\angle{P_{k-1}OP_{k}}=\frac{\pi}{n}\ (1\leq k\leq n),\ \angle{OP_{k-1}P_{k}}=\angle{OP_{0}P_{1}}\ (2\leq k\leq n).$
[2] $\overline{OP_{0}}=1,\ \overline{OP_{1}}=1+\frac{1}{n}.$
Find $\lim_{n\to\infty}\sum_{k=1}^{n}\overline{P_{k-1}P_{k}}.$

I think $n+1$ points $ P_i (i=\overline{0,n})$  must be form the convex polygon? :maybe:
If not, it is very difficult to find that limit. Because, we need a circle circumscribed polygons.

There's nothing difficult to the problem: the lines $OP_{i+1}$ don't necessarily separate $(P_i, P_{i+2})$ and still this doesn't change anything essential to the configuration.

First, note that all $\triangle OP_kP_{k + 1}$ are trivially similar to each other, for $k = 0, 1, \ldots, n - 1$, and that $OP_k = (1 + 1/n)^k$ for all $k$. Thus we may compute

$P_kP_{k+1}^2 = OP_k^2 + OP_{k+1}^2 - 2OP_kOP_{k+1} \cos \frac{\pi}{n}$, by which

$\sum_{k = 0}^{n-1} OP_k \cdot OP_{k+1} = \left( \sum_{k = 0}^{n - 1} \left(1 + \frac{1}{n}\right)^k \right) \sqrt{f(n)}$, with

$f(n) = 1 + \left(1 + \frac{1}{n}\right)^2 - 2\left( 1 + \frac{1}{n} \right) \cos \frac{\pi}{n}$.

Now, by the sum of a geometrical progression, we are left with

$\left( \left( 1 + \frac{1}{n} \right)^n - 1 \right) \sqrt{n^2f(x)}$.

Since $2n^2\left(1 - \cos \frac{\pi}{n} \right) = 4n^2 \sin^2 \frac{\pi}{2n} \to \pi^2$, $2n - 2n\cos \frac{\pi}{n} \to 0$, the desired limit will be

$(e - 1)\sqrt{\pi^2 + 1}$.

EDIT: The solution with "integral calculation" (which should amount to the above) is the following.

Consider the smooth curve $\gamma : [0, 1] \to \mathbb{R}^2$ parametrized by $\gamma(t) = (u(t), v(t))$, with $u(t) = e^t\cos(\pi t)$ and $v(t) = e^t\sin(\pi t)$. Then the value asked for is the limiting value of $|\gamma(0) - \gamma(1/n)| + \cdots + |\gamma((n-1)/n) - \gamma(1)|$, that is, the length of $\gamma$, which can be computed to be

$\int_0^1 \sqrt{(u'(t))^2 + (v'(t))^2} \mathrm{dt}$, that is,

$\int_0^1 \sqrt{(e^t\cos(\pi t) - \pi e^t\sin(\pi t))^2 + (e^t\sin(\pi t) + \pi e^t\cos(\pi t))^2} \mathrm{dt}$, which equals

$\int_0^1 \sqrt{e^{2t} + \pi^2e^{2t}} \mathrm{dt} = e^t \left. \vphantom{\int} \right|_0^1 \sqrt{\pi^2 + 1} = (e - 1)\sqrt{\pi^2 + 1}$, as above.

It doesn't make any difference if re-arrange the points counter closkwise.
Let $\rho(k\pi/n) = \rho_k = \overline{OP_k}$, then $\rho_{k+1}-\rho_k=\Delta\rho_k = \left((1+1/n)-1\right)\rho_k = \frac{\rho_k}{\pi}\frac{\pi}{n}$
This means that $\frac{d\rho}{d\theta} = \frac{\rho}{\pi}$ and so $\rho = e^{\frac{\theta}{\pi}}$ is the curve those points approach to.

From $\rho = e^{\frac{\theta}{\pi}}\quad (\theta\in (0,\pi))$ we see the length

$L = \int_0^{\pi}\sqrt{\rho^2+{\rho}'^2}d\theta = \sqrt{1+\frac{1}{\pi^2}}\int_0^{\pi} e^{\frac{\theta}{\pi}}d\theta$
$= \left(1+\frac{1}{\pi^2}\right)^{1/2} \pi (e-1) = (e-1)\sqrt{\pi^2+1}$


Given a positive integer $ n$, for all positive integers $ a_1, a_2, \cdots, a_n$ that satisfy $ a_1 \equal{} 1$, $ a_{i \plus{} 1} \leq a_i \plus{} 1$, find $ \displaystyle \sum_{i \equal{} 1}^{n} a_1a_2 \cdots a_i$.

Shobber, do you mean this?

Given a positive integer $ n$, find the sum

$ \sum_{\begin{array}{c} a_1,a_2,\cdots ,a_n\text{ positive} \\
\text{integers satisfying} \\
a_1 \equal{} 1\text{ and} \\
a_{i \plus{} 1} \leq a_i \plus{} 1\text{ for all }i\end{array}}\sum_{i \equal{} 1}^{n} a_1a_2 \cdots a_i$.

In this case, the sum equals $ 1$, $ 5$, $ 28$, $ 190$, $ 1594$ for $ n$ being $ 1$, $ 2$, $ 3$, $ 4$, $ 5$, respectively. I can't find this in Sloane's encyclopedia...

  darij

First we can show by induction that for $i\in\{1,...,n\}$, we have $a_i\le i$. Thus $ \displaystyle \sum_{i = 1}^{n} a_1a_2 \cdots a_i\le 1!+2!+...+n!$, with equality if and only if $a_i=i$.

As per request, here is a solution to this problem:

[hide=Solution][b][u]Solution:[/u][/b]

For $a_1=m, a_{i+1}\le a_i+1$, denote $f(m,n)$ as the sum $\sum a_1a_2\cdots a_n$.

Observe that if $a_1=m$, then $a_2$ can be $1,2,\cdots m+1$.

Therefore $f(m,n+1)=m\cdot\sum a_2a_3\cdots a_{n+1}=m(f(1,n)+f(2,n)+\cdots+f(m+1,n))$.

Now we will proceed by induction:

$$f(m,n)=\binom{2n-1}{m+2n-2}\cdot(2n-1)!! \ (*)$$

Induct for $n$:

When $n=1$, it is obvious that $f(m,1)=m$, which satisfies [b](1)[/b].

$$f(m,n+1)=m(f(1,n)+f(2,n)+\cdots+f(m+1,n))=m\cdot(2n-1)!!\cdot\left(\binom{2n-1}{2n-1}+\binom{2n-1}{2n}+\cdots+\binom{2n-1}{m+2n-1}\right)$$

Notice that $$\binom{2n-1}{2n-1}+\binom{2n-1}{2n}+\binom{2n-1}{2n+1}+\cdots+\binom{2n-1}{m+2n-1}$$ $$=\binom{2n}{2n}+\binom{2n-1}{2n}+\binom{2n-1}{2n+1}+\cdots+\binom{2n-1}{m+2n-1}$$ $$=\binom{2n}{2n+1}+\binom{2n-1}{2n+1}+\cdots+\binom{2n-1}{m+2n-1}$$ $$=\cdots=\binom{2n}{m+2n}$$

$$f(m,n+1)=m\cdot(2n-1)!!\cdot\binom{2n}{m+2n}=m\cdot(2n-1)!!\cdot\frac{(m+2n)!}{m!(2n)!}=(2n+1)!!\cdot\frac{2n+1}{2n+m}$$

so $(*)$ holds true at $n+1$.

Thus $(*)$ always holds true, and $f(1,n)=(2n-1)!!$.

Therefore the desired sum is $(2n-1)!!$. [/hide]


function
real analysis
geometry
3D geometry
sphere
calculus
integration

Show that for no value of $ a\in \mathbb{R}$ the function $ f(x)\equal{}|x|^{\minus{}a}, x\neq 0, f(0)\equal{}0$ belongs to $ L^p( \mathbb{R}^n,dm), 1\le p<\infty$ where $ m$ is the Lebesgue measure.

A hint will be highly appreciated.

Hints:

Integrate in (spherical) polar coordinates. $ dx \equal{} r^{n\minus{}1}\,d\sigma\,dr$ where $ d\sigma$ is surface measure on the unit sphere (and the integral of $ d\sigma$ over the whole unit sphere is simply a specific finite number whose exact value can be computed but is immaterial to this problem.)

For $ a p>n,$ focus on the "inner" part of the integral, $ |x|\le 1.$

For $ a p<n,$ focus on the "outer" part of the integral, $ |x|\ge 1.$

For $ a p\equal{}n,$ it diverges both at zero and at infinity.


how to prove it?

$ if:\left\vert f\left( x\right)\minus{}l\right\vert <\frac{\left\vert l\right\vert }{2},then.f\left( x\right) >\frac{\left\vert l\right\vert }{2}$

thank you ,scorpius119, you are kind man

i see my mistake, haha  .  " if: l>o "  haha  ...............

I assume you mean $ |f(x)|>\frac{|l|}{2}$, in which case it follows from
\[ |f(x)|\plus{}|l\minus{}f(x)|\geq |l|\]


1)Find general formula for $2^0+2^1+2^2+....+2^n$

2)Find the 60000000800th term of the sequence 0,3,8,15,24........ and also generalise.

[hide="1"]
use geometric progression
[/hide]

[hide]

there is clearly a sequence..

and hence let the nth term of the sequence be 2^0 + 2^1 + 2^2.... + 2^n

the nth term of the sequence is equal to :

2^(n+1) - 1

[/hide]

[quote="riddler"][hide="2"]
the sequence can progress in different ways cant it?
anyway I think it would go as 0,3,8,15....? Is it like this?
If it is then use arithmatic progressions
[/hide][/quote]

Tell the answer then.......

[quote="robinhe"][hide="#2"]If you add one to each number, then it is a perfect square.  So, the nth number is $n^2-1$.  The 60000000800th term is approxamately $3.6 E21$.[/hide][/quote]

Perfect...i don't know the answer but perfect. :)

Also for the first one you could note that in binary it's 111....111 with however many 1's in between, and that by subtracting 1 from 1000....000 you get 111...111. So $2^{n+1}-1$.


There's a problem something like...

2 x 4 x 6 x 8 x 12

or something like that.

It's 5 integers multiplied and the answers is 12, when it's obviously not.

So is their an explanation for that thing being like that or is it just a messed up question?  :|

When needing to report a problem with a question, review the game after you played (you can click the review button in the lobby).  There are options to report or problem (which sends us an email) or ask on forum (which posts the problem here).

Well, i clicked the report a bad problem or something button, but yesterday when i was playing the game again the same question appeared in another game.  So i reported it again, but i haven't noticed if it's gone or not cuz i only played it few more times after i reported it the second time, so i'm not sure if it got modefied or not.... but yea


James, by himself, can paint 4 rooms in 10 hours. If he hires June to help, they can do the same job together in 6 hours. If he lets June work alone, how long will it take her to paint 4 rooms?

[hide="hint"]How many rooms does James paint per hour?[/hide]

[hide="another hint"]Let the number of rooms per hour that June can paint be $j$.  What equation can you write, in terms of $j$?[/hide]

Your questions:

How many rooms does James paint per hour?

James can paint 4 rooms in 10 hours.

So, 10/4 = 2.5 rooms, right?

Together they paint the room in 6 hours.

What equation?

This is where I got stuck.

Look at each hour.  James paints $\frac{4}{10}=\frac25$ of a room in 1 hour.  June paints $j$ rooms in one hour.  Thus, in one hour, they paint a total of $\frac25+j$ rooms.

Yes, because he paints 4 rooms in 10 hours, so divide by 10 to get the number of rooms painted in 1 hour.

Can you write the equations now?

Sorry...I tried.  Can you write the equation for me?

[hide]If james paints $\frac25$ of a room in one hour, and June paints $j$ rooms in one hour, then together, they paint $\frac25+j$ rooms in one hour.  We know that this is $\frac46$ rooms per hour.

Now, you can solve for $j$  :) [/hide]

[quote="b-flat"][hide]If james paints $\frac25$ of a room in one hour, and June paints $j$ rooms in one hour, then together, they paint $\frac25+j$ rooms in one hour.  We know that this is $\frac{6}4$ rooms per hour.

Now, you can solve for $j$  :) [/hide][/quote]
I think you mean that you want $\frac{4}6$ rooms per hour, because you want 4 rooms in 6 hours  :P


geometry
3D geometry
sphere
Pythagorean Theorem

Mandelbrot December 1990 #7 Round Two

A sphere with radius $3$ is inscribed in a frustrum of slant height $10$. What is the volume of the frustrum? (A [i]frustrum[/i] is the non-conical region formed by cutting a cone with a plane parallel to its base)

Let $s = 10$ be the slant height, $\rho = 3$ the radius of the inscribed sphere, R > r the radii of the frustrum bases and H the height of cone before cutting it into the frustrum, perpendicular to the base, and $h = H - 2 \rho$ the height of the cut-off cone. Cut the frustrum with a plane through its rotational axis. The result is an isosceles tangential trapezoid. Hence, 2R + 2r = 2s, R + r = s = 10. By Pythagorean theorem, $(R - r)^2 + (2\rho)^2 = s^2,$ $R - r = \sqrt{s^2 - 4 \rho^2} = \sqrt{100 - 36} = 8$. Solving, R = 9, r = 1. From similar right angle triangles, the cone height is given by $\frac{H}{R} = \frac h r = \frac{H - 2\rho}{r},$ $H = \frac{2 \rho R}{R - r}$ and $h = \frac{2 \rho r}{R - r}$. The frustrum volume is

$V = \frac{\pi R^2 H}{3} - \frac{\pi r^2 h}{3} = \frac{2 \pi \rho}{3} \cdot \frac{R^3 - r^3}{R - r} = \frac{2 \pi \rho}{3}\ (R^2 + Rr + r^2) = 182 \pi$


hello !
prove

 if an equation: $ ax^2\plus{}bx\plus{}c\equal{}0$ admits at least a solution then :$ a\plus{}b\plus{}c \leq \frac{9}{4} \text{max}(a,b,c)$

Problem statement is not true. Counterexample: $ a\equal{}20$, $ b\equal{}21$, $ c\equal{}1$.

$ 20x^2 \plus{} 21x \plus{} 1 \equal{} (20x \plus{} 1)(x \plus{} 1)$

but $ 20 \plus{} 21 \plus{} 1 \equal{} 42$ and $ 9/4\cdot\max \{ 20,21,1 \} \equal{} 37.8$.

Oops, I divided by 5... :blush:
Sorry, I think I have a proof:

[hide]Looking at discriminant, $ b^{2} \ge 4ac$. So assume $ a\ge b,c$ ($ c\ge a,b$ would be similar). So it suffices to show $ b \plus{} c \le 5a/4$.

If $ a < 0$, then $ b \plus{} c\le 2a < 5a/4$.

If $ a\ge 0$,
If $ 2a \plus{} b \ge 0$, $ 2a \plus{} b\le 3a \implies (2a \plus{} b)^2 \le 9a^2 \implies 4ab \plus{} b^2\le 5a^2 \implies b \plus{} (b^2/4a)\le 5a/4$ . Since $ c\le b^2/4a$, we're done.
If $ 2a \plus{} b < 0$, $ b < \minus{} 2a$ so $ b \plus{} c < c \minus{} 2a\le \minus{} a < 5a/4$ because $ a\ge 0$.

Now we show that if $ b\ge a,c$ then $ a \plus{} c\le 5b/4$.

If $ a < 0$, then $ a \plus{} c < c\le b < 5b/4$, unless $ b<0$. But if $ b<0$, then $ a\plus{}c\le 2b<5b/4$. Similarly for $ c < 0$. So we can assume $ a,b,c\ge 0$.

If $ b > 2\cdot \max \{ a,c \}$, then $ a \plus{} c < b < 5b/4$.

Now suppose $ b\le 2\max \{a,c\}$. WLOG $ a\ge c$ (we can do something similar if $ c\ge a$). So $ b\le 2a$. Then $ b \minus{} 5a/2\le 3a/2$ and $ \minus{} 3a/2\le b \minus{} 5a/2$, so $ (b \minus{} 5a/2)^2\le 9a^2/4$. Multiply out to get $ b^2 \minus{} 5ab \plus{} 4a^2\le 0 \implies b^2/4a \plus{} a\le 5b/4$ because $ a\ge 0$. Since $ b^2/4a \ge c$ ($ a\ge 0$), we're done.[/hide]

EDIT: Minor edit
EDIT 2: Missed a case.

I really hope I didn't miss any cases/miss any signs. I wonder if there is a nicer proof (like one that actually derives it).

I'm pretty sure that your proof cannot be correct because the problem isn't correct. (Just think, we can flip the sign of $ a$, $ b$, $ c$, and the inequality switches).


A thin horizontal disc of radius $ R $ = 10 cm is located within a cylindrical cavity filled with oil whose viscosity $ \eta $ = 0.08 P . The clearance between the disc and the horizontal planes of the cavity is equal to $ h$ = 1 mm . Find the power develpoed by the viscous forces acting on the disc when it rotates with the angular velocity $ \omega $ = 60 rad\s . the end effects are to be neglected.

[hide]Preferably, give the answer and soln. in terms of variables.[/hide]

hello again???? Anyone please atleast try!!!!


[quote="Ligouras"]Let $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that

$ \sum[\frac {a^2(a^2 \plus{} a \minus{} c)}{b^2}]\geq\frac {(a \plus{} b \plus{} c)^2(a^2 \plus{} b^2 \plus{} c^2)}{3}$[/quote]

`The inequality does not hold.`

a counter-example

$ a\equal{}\frac{58}{121},b\equal{}\frac{1}{60},c\equal{}2$
$ \sum[\frac {a^2(a^2 \plus{} a \minus{} c)}{b^2}]\minus{}\frac {(a \plus{} b \plus{} c)^2(a^2 \plus{} b^2 \plus{} c^2)}{3}\equal{}\minus{}\frac{27262494857856217234741}{28036495358593920000}$

[quote="hedeng123"][quote="Ligouras"]Let $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that

$ \sum[\frac {a^2(a^2 \plus{} a \minus{} c)}{b^2}]\geq\frac {(a \plus{} b \plus{} c)^2(a^2 \plus{} b^2 \plus{} c^2)}{3}$[/quote]

`The inequality does not hold.`

a counter-example

$ a \equal{} \frac {58}{121},b \equal{} \frac {1}{60},c \equal{} 2$
$ \sum[\frac {a^2(a^2 \plus{} a \minus{} c)}{b^2}] \minus{} \frac {(a \plus{} b \plus{} c)^2(a^2 \plus{} b^2 \plus{} c^2)}{3} \equal{} \minus{} \frac {27262494857856217234741}{28036495358593920000}$[/quote]

Sorry my Friends:

Let $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that

$ \sum[\frac {a^2(a^2 \plus{} a \minus{} b)}{b^2}]\geq\frac {(a \plus{} b \plus{} c)^2(a^2 \plus{} b^2 \plus{} c^2)^2}{3}$


ratio
combinatorics unsolved
combinatorics

A space city consists of 99 space stations. Every two stations are connected by a space highway. All highways are one-way except for 99 which are two way. A group is defined as a set of four stations such that we can travel from any one to any other of the four along the highways. Determine the maximum number of groups in a space city.

It's a bit confusing. If they are allowed to travel on any number of highways to get from one station to another, then (if we label the stations 1-99) have the 2-way highways between (1,2), (2,3), (3,4), ... (98,99), and the last one wherever. Now anyone can get from any station to any other station. So there are ${99\choose 4}$ groups.

If they arent allowed to do this, we want to make the biggest subgraph that is a complete graph. I can't explain why, but it makes sense. Suppose this number which I am too lazy to compute is $n$. Then there are ${n\choose4}$ groups. But $n-99$ will be left over. We connect them from one station to each of the stations in our complete subgraph, as many times as we can. After the first 3, each additional one will yield a group. So we add $n-102$.  The final result is ${n\choose4}+n-102$. $n$ is actually the largest number of the form ${k\choose2}$ that is less than or equal to 99.So this is completely doable, I'm just lazy.

As a quick explanation to why this works, we want to maximize the ratio of additional groups made to additional edges added. And complete subgraphs I believe do this.


Let be $ a,b,c>0$ such that $ a\plus{}b\plus{}c\ge ab\plus{}bc\plus{}ca$ . Prove that :

i)$ a\plus{}b\plus{}c\ge a\sqrt{bc}\plus{}b\sqrt{ca}\plus{}c\sqrt{ab}$

ii)$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\ge 3$

iii)$ a^2\plus{}b^2\plus{}c^2\ge 3a^2b^2c^2$

[quote="alex2008"]Let be $ a,b,c > 0$ such that $ a \plus{} b \plus{} c\ge ab \plus{} bc \plus{} ca$ . Prove that :

i)$ a \plus{} b \plus{} c\ge a\sqrt {bc} \plus{} b\sqrt {ca} \plus{} c\sqrt {ab}$

ii)$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\ge 3$

iii)$ a^2 \plus{} b^2 \plus{} c^2\ge 3a^2b^2c^2$[/quote]

[hide="1"]
We have to prove that:
$ ab\plus{}bc\plus{}ca \ge \sqrt{abc}(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})$
But $ \sum(\sqrt{ab})^2 \ge \sqrt{abc}(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})$
[/hide]

[hide="hint 2"]
$ (ab\plus{}bc\plus{}ca)^2 \ge 3abc(a\plus{}b\plus{}c)$

[/hide]

[hide="3"]
$ 3(a^2\plus{}b^2\plus{}c^2) \ge (a\plus{}b\plus{}c)^2 \ge (ab\plus{}bc\plus{}ca)^2 \ge 9a^2b^2c^2$
[/hide]


On each Friday last January, the temperature was a record low temperature for the date. On Friday, January 30, the mercury dropped to 5°F, which was 7 degrees higher than the low temperature on the Friday before. The low on January 9 was 2 degrees lower than the low temperature on January 16. On the first Friday of the month, the low temperature was 3 degrees lower than that on January 23. The low temperature on the Friday in the middle of the month was 6 degrees higher than the low temperature on the first Friday.

What was the low temperature reading for each Friday of the month?

First Friday(january 2nd):-5
Second Friday (january 9th): -1
Third Friday (january 16th): 1
Fourth Friday (january 23rd): -2
Fifthe Friday (january 30th): 5

[quote="colts18"]how did you get that?[/quote]
i did simple arithmetic as NoSoupForYou has done.


calculus
integration
logarithms
induction
limit
function
calculus computations

Evaluate $ \int_{0}^{1} (\ln x)^{n}\ dx$ where $ n$ is a positive integer.

$ I_{n}\equal{}\int (lnx)^{n}dx\equal{}x(lnx)^{n}\minus{}n\int (lnx)^{n\minus{}1}dx$
by induction we can prove $ I_{n}\equal{}x\sum_{i\equal{}0}^{n}(\minus{}1)^{i}A_{n}^{i}(lnx)^{n\minus{}i}\plus{}C$

$ \int (\ln x)^{n}\ dx \equal{} x (\ln x)^{n} \minus{} \int xn(\ln x)^{n \minus{} 1}\frac {1}{x}\ dx \equal{} \minus{} n\int(\ln x)^{n \minus{} 1}\ dx$.

Now $ \int \ln x\ dx \equal{} x \ln x \minus{} \int x\cdot \frac {1}{x}\ dx$.

Putting in the limits, this has the value $ \minus{} 1$.

Therefore repeatedly using the reduction formula gives $ \int_{0}^{1} (\ln x)^{n}\ dx \equal{} ( \minus{} n)( \minus{} (n \minus{} 1))...( \minus{} 3)( \minus{} 2)( \minus{} 1) \equal{} ( \minus{} 1)^{n}n!$.

Another approach that can be taken is to make the substitution $ \ln x \equal{} \minus{} z$. This gives us:

$ x \equal{} e^{ \minus{} z}$

$ dx \equal{} \minus{} e^{ \minus{} z} \,dz$

$ \int ^0 _\infty ( \minus{} z)^n \left( \minus{} e ^{ \minus{} z}\right)\,dz \equal{} ( \minus{} 1)^n \int ^\infty_0 \frac {z^n}{e^z} \,dz \equal{} ( \minus{} 1)^n n!$

And how about:

$ \int^1_0 x^m \ln ^n x \,dx \equal{} ?$

Why can $ \ln x$  be defined in $ 0\leq x\leq 1?$

Since $ \lim_{x\to 0^\plus{}}x^m(\ln x)^n\equal{}0$ for $ m>0,$ such integrals may be considered proper. (The integrand extends to a continuous function on $ [0,1].$)

If $ \minus{}1<m\le0,$ we do have an integrand that tends to (plus or minus) infinity as $ x\to0^\plus{},$ but slowly enough for the improper integral to converge.


Let $ a,b,c$ be non negative real numbers. Prove that
$ \frac {a(b\plus{}c)^2}{b^3\plus{}abc\plus{}c^3}\plus{}\frac {b(c\plus{}a)^2}{c^3\plus{}abc\plus{}a^3}\plus{}\frac {c(a\plus{}b)^2}{a^3\plus{}abc\plus{}b^3}\ge2$

Equality holds for $ a\equal{}b,c\equal{}0$ up to permutation!  :)

[quote="dduclam"]Let $ a,b,c$ be non negative real numbers. Prove that
$ \frac {a(b \plus{} c)^2}{b^3 \plus{} abc \plus{} c^3} \plus{} \frac {b(c \plus{} a)^2}{c^3 \plus{} abc \plus{} a^3} \plus{} \frac {c(a \plus{} b)^2}{a^3 \plus{} abc \plus{} b^3}\ge2$

Equality holds for $ a \equal{} b,c \equal{} 0$ up to permutation!  :)[/quote]
Nice, Lam :). My solution
by Cauchy Schwarz, we have
\[ \sum \frac {a(b \plus{} c)^2}{b^3 \plus{} abc \plus{} c^3} \cdot \sum a^3(b^3 \plus{} abc \plus{} c^3) \ge (\sum a^2(b \plus{} c))^2
\]
It suffices to prove that
\[ (\sum a^2(b \plus{} c))^2 \ge 2\sum a^3(b^3 \plus{} abc \plus{} c^3)
\]
Now, using the fact $ (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \ge 0$, we have done. :)

Thank you,Can  :)  The following resulst are true too

$ \frac {a^2(b \plus{} c)}{b^3 \plus{} abc \plus{} c^3} \plus{} \frac {b^2(c \plus{} a)}{c^3 \plus{} abc \plus{} a^3} \plus{} \frac {c^2(a \plus{} b)}{a^3 \plus{} abc \plus{} b^3}\ge2$

Equality holds for $ a\equal{}b\equal{}c$ or $ a \equal{} b,c \equal{} 0$ or permutations!

and
$ \frac{(a\plus{}b)c}{a^2\plus{}b^2}\plus{}\frac{(b\plus{}c)a}{b^2\plus{}c^2}\plus{}\frac{(c\plus{}a)b}{c^2\plus{}a^2}\ge 2$

Equality holds for $ a \equal{} b,c \equal{} 0$ up to permutation :D

[quote="dduclam"]Thank you,Can  :)  The following resulst are true too

$ \frac {a^2(b \plus{} c)}{b^3 \plus{} abc \plus{} c^3} \plus{} \frac {b^2(c \plus{} a)}{c^3 \plus{} abc \plus{} a^3} \plus{} \frac {c^2(a \plus{} b)}{a^3 \plus{} abc \plus{} b^3}\ge2$

Equality holds for $ a \equal{} b \equal{} c$ or $ a \equal{} b,c \equal{} 0$ or permutations!

and
$ \frac {(a \plus{} b)c}{a^2 \plus{} b^2} \plus{} \frac {(b \plus{} c)a}{b^2 \plus{} c^2} \plus{} \frac {(c \plus{} a)b}{c^2 \plus{} a^2}\ge 2$

Equality holds for $ a \equal{} b,c \equal{} 0$ up to permutation :D[/quote]
The second is known :).
And here is my solution for the first :)
By the Cauchy Schwarz, we have
\[ \sum \frac{a^2(b\plus{}c)}{b^3\plus{}abc\plus{}c^3} \cdot \sum \frac{a^2(b^3\plus{}abc\plus{}c^3)}{b\plus{}c} \ge (\sum a^2)^2\]
It suffices to show that
\[ (\sum a^2)^2 \ge 2\sum \frac{a^2(b^3\plus{}abc\plus{}c^3)}{b\plus{}c}\equal{}2\sum a^2(b^2\minus{}bc\plus{}c^2)\plus{}2abc\sum\frac{a^2}{b\plus{}c}\]
\[ \Leftrightarrow \sum a^4 \plus{}abc\sum a\minus{}2\sum a^2b^2 \ge abc(2\sum \frac{a^2}{b\plus{}c} \minus{}\sum a)\]
\[ \Leftrightarrow \sum a^4 \plus{}abc\sum a\minus{}2\sum a^2b^2 \ge abc(\sum a)(\sum \frac{(a\minus{}b)^2}{(a\plus{}c)(b\plus{}c)})\]
Since
\[ abc(\sum a)(\sum \frac{(a\minus{}b)^2}{(a\plus{}c)(b\plus{}c)}) \le abc(\sum a)(\sum \frac{(a\minus{}b)^2}{c(a\plus{}b\plus{}c)}) \equal{}\sum ab(a\minus{}b)^2\]
It suffices to show that
\[ \sum a^4 \plus{}abc\sum a\minus{}2\sum a^2b^2 \ge  \ge \sum ab(a\minus{}b)^2\]
that is
\[ \sum a^2(a\minus{}b)(a\minus{}c) \ge 0\]
It is Schur's Inequality.
We have done.
Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}b,c\equal{}0$ or any cyclic permutations. :)

Very nice, Lâm. :)

Thanks,Can! My proof is similar yours  :)


Please accept with no obligation, implied or implicit, our best wishes for an environmentally conscious, socially responsible, low-stress, non-addictive, gender-neutral celebration of the winter solstice holiday, practiced within the most enjoyable traditions of the religious persuasion of your choice, or secular practices of your choice, with respect for the religious/secular persuasion and/or traditions of others, or their choice not to practice religious or secular traditions at all. We also wish you a fiscally successful, personally fulfilling and medically uncomplicated recognition of the onset of the generally accepted calendar year 2006, but not without due respect for the calendars of choice of other cultures whose contributions to society have helped make America great. Not to imply that America is necessarily greater than any other country nor is it the only America in the Western Hemisphere. And without regard to the race, creed, color, age, physical ability, religious faith or sexual preference of the wishee. By accepting these greetings you are accepting these terms. This greeting is subject to clarification or withdrawal. It is freely transferable with no alteration to the original greeting. It implies no promise by the wisher to actually implement any of the wishes for herself or himself or others, and is void where prohibited by law and is revocable at the sole discretion of the wisher. This wish is warranted to perform as expected within the usual application of good tidings for a period of one year or until the issuance of a subsequent holiday greeting, whichever comes first, and warranty is limited to replacement of this wish or issuance of a new wish at the sole discretion of the wisher.  :)

[quote="log0"]Translation: Happy Holidays?[/quote]
And a darn politically correct translation at that.

I use the French Revolutionary calendar.  Happy beginning of Nivose, unless you choose not to use that form of calendar.

Can't we solve this controversy by not using holiday greetings in general?  We can do pretty well without them during the rest of the year.

Same to you.
Did you think that up? How much free time do you have?

[quote="Bictor717"]Did you think that up? How much free time do you have?[/quote]
no. i found it on another forum and thought i'd share.

We should all do like in the first season of south park : a non offending christmas play, with Philip glass and a big shark...

Well, if it was really non-offending, then it wouldn't be a christmas play. :P

Reminds me of a cartoon showing a banner that read: Holiday Concert (above it had the words Christmas Concert with a big X over it). The guy below says: "We will now present Handel's 'Hallelujah!'"....


group theory
abstract algebra
linear algebra
matrix
superior algebra
superior algebra unsolved

Let $F$ be a field.
a)Prove that if $G$ is a subgroup of $GL_n(F)$ containing $SL_n(F)$ then $G$ is normal in $GL_n(F)$
b)$n\geq 3$.Prove that if $G$ is normal in $GL_n(F)$ then either $SL_n(F) \subseteq G$ or $G$ consists of scalar matrices.

For a) If $G$ is a group and $H\leq G$ such that $G'\leq H$, then $H\triangleleft G$.

maybe i'm missing something but as far as i understand you are implying that $(GL_n(F))' \subseteq SL_n(F)$ ? Could you explain this moment a little bit more-i haven't known this fact- it ease to prove the inverse
inclusion $SL_n(F) \subseteq (GL_n(F))'$ because every matrix in $SL_n(F)$ is representable as a commutator of matrices from $SL_n(F)$

$SL_n$ is obviously a normal subgroup of $GL_n$, and $GL_n/SL_n$ is abelian ($\det$ induces an isomorphism of $GL_n/SL_n$ onto $F^*$), so the commutator subgroup of $GL_n$ must be contained in $SL_n$.

Oh-i understand it=it was stupid question
Meantime, can't we just say that every matrix in $(GL_n(F))'$ has the determinant 1 and thus it belongs to $SL_n(F)$ ?


geometry
rectangle
inequalities
trigonometry
national olympiad

Here are the most interesting (on my opinion) problems from our
national contest '05.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34896]9.3.[/url] If $x,y,z$ are positive and $x+y+z=1$, prove that
\[\sqrt{xy(1-z)}+\sqrt{yz(1-x)}+\sqrt{zx(1-y)}\leq\sqrt{\frac{2}{3}}.\]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34897]10.2.[/url] If the middle of positive numbers $a,b,c$ is smaller than geometric mean of
the smallest and the largest, prove that
\[abc(a^3+b^3+c^3)\geq a^3b^3+b^3c^3+c^3a^3.\]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34899]10.4.[/url] Every cell of $30\times30$ chessboard is colored black or white and there
are 450 cells of each color. In one move you can cut the table into $3\times1$
rectangles and form a new table $30\times30$ using these rectangles. Prove that it
is always possible to put all the white cells in the upper half-table and all
the black cells in the lower half-table. What is the minimal number of moves that we need
in general?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34898]11.2.[/url] Prove that in an acute triangle the following inequality holds:
\[\sqrt{\sin\alpha\sin\beta\cos\gamma}+\sqrt{\sin\beta\sin\gamma\cos\alpha}+
\sqrt{\sin\gamma\sin\alpha\cos\gamma}\leq\frac{3\sqrt6}{4}.\]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34902]11.3.[/url] Find all composite numbers $n$ satisfying the following property:
if $1=d_1<d_2<\dots<d_k<d_{k+1}=n$ are all divisors of $n$, then $d_1+d_2$ and
$d_1+d_2+\dots+d_k$ are also divisors of $n$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34901]11.4.[/url] Prove that for any $n$ there exists a graph with all cycles length greater
than $2n+3$ and whose vertices cannot be labeled by numbers from 1 to $2n+1$ such that
neighbours differ by 1 modulo $2n+1$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34900]12.4.[/url] 100 prisoners were brought to prison. The guard said them:
"I'll give you one night to talk and after that you will have no
way to communicate. Sometimes I will bring one of you into the
room with a lamp. At the beginning the lamp is off. Leaving the
room you can leave the lamp on or off, as you wish. If some of you
tells me everybody has been in the room, and he is right I'll keep
bringing you into the room until someone else tells me you've all
been in the room. If he is right too I will release you all. If
someone is wrong I will throw you all to crocodiles. And if
everybody says nothing I'll keep bringing you into the room, and
for nobody his visit will be his last visit. "
Can the prisoners find the strategy that will rescue them?


calculus
integration
trigonometry
logarithms
real analysis
real analysis unsolved

2. $ \int_0^{\frac{\pi}{2}} \frac{x\sin x}{1\plus{}\cos ^ 2 x}\ dx\equal{}\left[\minus{}x\tan ^ {\minus{}1}(\cos x)\right]_0^{\frac{\pi}{2}}\plus{}\int_0^{\frac{\pi}{2}} \tan ^{\minus{}1} (\cos x)\ dx$

$ \equal{}\int_0^{\frac{\pi}{2}} \tan ^{\minus{}1} (\cos x)\ dx$.

[u]Comment[/u]

$ \int_0^{\pi} \frac{x\sin x}{1\plus{}\cos ^ 2 x}\ dx$ is more easy than your integral. Japanese high school students could be solved without the knowledge of $ \tan ^{\minus{}1} x$.

3. Let $ x \equal{} \minus{} t$, $ \int_{ \minus{} \frac {\pi}{2}}^{0} \frac {\ln \cos x}{1 \plus{} e^{x \plus{} \sin x}}\ dx \equal{} \int_0^{\frac {\pi}{2}} \frac {\ln \cos t}{1 \plus{} e^{ \minus{} (t \plus{} \sin t)}}\ dt$. Since $ \frac {1}{1 \plus{} e^{x \plus{} \sin x}} \plus{} \frac {1}{1 \plus{} e^{ \minus{} (x \plus{} \sin x)}} \equal{} 1$, Split out the interval of the integral, we have

$ \int_{ \minus{} \frac {\pi}{2}}^{\frac {\pi}{2}} \frac {\ln \cos x}{1 \plus{} e^{x \plus{} \sin x}}\ dx \equal{} \int_0^{\frac {\pi}{2}} \ln \cos x\ dx$.

For solution of 1. - see - http://www.mathlinks.ro/Forum/viewtopic.php?t=168932


calculus
integration
number theory theorems
number theory

Let $m$ be an [b]odd[/b] natural number. Prove that

$1.3.5 ... m = m!/(2^{[m/2]}[m/2]!)$

where square brackets means the integral part (e.g. $[\pi]=3$) .

Sincerely Yours,
Murat Aygen

$2^{[m/2]}[m/2]!=2^{\frac{m-1}{2}}\cdot(\frac{m-1}{2})!$
which equals the product of all the even numbers up to and including $m-1$, and then the result is immediate.

Thanks. Then use this identity to investigate the convergence of the series whose general term is $1-n\log((2n+1)/(2n-1))$. (A Course of MODERN ANALYSIS, E.T. Whittaker, G.N. Watson, p.38).


I installed MiKTeX 2.8 and TeXnicCenter exactly the way the [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Downloads]AoPS Wiki page[/url] says to do it. When I got to step 7 of the TeXnicCenter instructions (configuration), I clicked Close on the tip, Next on the Configuration Wizard, and typed in the address it says to do. I clicked Next, by got an error message. Then, I did it manually and browsed in the folders and found the files it was wanting (latex.exe ...), and clicked next, but I still got the error. Also, the directory I saved it in is the default directory. What am I doing wrong? How can I fix this?

I attached a snapshot of the error. I use Windows XP.

Thanks

EDIT: After trying it in various ways a few more times, I managed to get it to work. Please delete if necessary. Sorry.

Do you know what caused the problem and how you solved it? Your experience may be useful to others who get the same problem.

Yes, I think my problem was that I didn't actually select the folder (bin), I only opened it. I don't know why the copy/paste didn't work though (from the Wiki page).


What is the sum of the least and the greatest three-digit prime numbers that are less than 200?

The biggest prime is 197 and the smallest 2, so we have $ 199$.

@izzy, the question asked about greatest and least 3 digit primes
so the greatest prime is $ 197$ , and the least prime is $ 101$
$ 197\plus{}101\equal{} 298$ F.S.

Actually, 199 is prime, so $ 199\plus{}101\equal{}300$.


How many numbers less than 100 and divisible by 3 are also divisible by 4?

[hide]LCM of 3 and 4 is 12. There are 8 numbers less than 100 that are divisible by 3 and 4.[/hide]

[hide] There are 8 numbers less than 100 and divisible 3 also divisible 4. They are 12,24,36,48,60,72,84,96.[/hide]

[hide="solution"]We are looking for the multiples of 12 that are less than 100. They are 12,24,36,48,60,72,84,96. There are 8 numbers.[/hide]

P.S. If you mean positive and negative numbers, the answer is trivial.

[hide]multiples of 12 then 100/12= abuot 8.  :P [/hide]

[quote="alex__ib"]Imo $0$ is  one of the numbers too  :|[/quote]
[hide="Forgot :blush:"]$9$[/hide]


geometry
perpendicular bisector
Pythagorean Theorem
power of a point

Two intersecting circles have a common chord with length 16 ft. The radii of the circles are 10 ft and 17 ft respectively. Express the distance between the centers of the circles in feet.

[asy]unitsize(6);
path circ1, circ2;
circ1=circle((0,0),10);
circ2=circle((21,0),17);
draw(circ1);
draw(circ2);
pair [] x=intersectionpoints(circ1,circ2);
draw(x[0]--x[1]);
draw((0,0)--(21,0));
draw((0,0)--x[0]);
draw((21,0)--x[0]);
label("$O_1$",(0,0),SW);
label("$O_2$",(21,0),SE);
label("A",x[0],N);
label("B",x[1],S);
label("C",(6,0),SE);[/asy]

$ O_1A=10$ because it is a radius and $ AC=8$ because $ O_1C$ is a perpendicular bisector of the chord, so by the Pythagorean theorem, \[ O_1C=\sqrt{10^2-8^2}\Rightarrow O_1C=6\]
Similarly, $ O_2A=17$ and $ AC=8$, so by the Pythagorean theorem, \[ O_2C=\sqrt{17^2-8^2}\Rightarrow O_2C=15\]
We get
\begin{align*}
O_1O_2 &= O_1C+O_2C \\
&= 6+15 \\
&= \boxed{21} \\
\end{align*}

[hide="alternate solution"]
[asy]unitsize(6);
path circ1, circ2;
circ1=circle((0,0),10);
circ2=circle((21,0),17);
draw(circ1);
draw(circ2);
pair [] x=intersectionpoints(circ1,circ2);
draw(x[0]--x[1]);
draw((0,0)--(21,0));
draw((0,0)--x[0]);
draw((21,0)--x[0]);
label("$O_1$",(0,0),SW);
label("$O_2$",(21,0),SE);
label("A",x[0],N);
label("B",x[1],S);
label("C",(6,0),SE);
pair [] y=intersectionpoints(circ1,(0,0)--(21,0));
label("X",y[0],NE);
pair [] z=intersectionpoints(circ2,(0,0)--(21,0));
label("Y",z[0],NW);[/asy]
Using Power of a Point, we find CY=2 and CX=4, so our answer is 10+17-2-4=21, as we are double counting XY.[/hide]


Find the cosines of all real angles which satisfy the equation

5(Sin2X) +24(1-SinX)=0

By expanding $sinx$ and solving for $sinx$,
$10(sinx)(cosx)+24(1-sinx)=0$
$(sinx)(10cosx-24)+24=0$
$sinx=\frac{-24}{10cos-24}=\frac{12}{12-5cosx}$

By putting $sinx$ in terms of $cosx$ and then factoring,
$\pm\sqrt{1-(cosx)^2}=\frac{12}{12-5cosx}$
$1-(cosx)^2=\frac{144}{144-120cosx+25(cosx)^2}$
$144-120cosx-119(cosx)^2+120(cosx)^3-25(cosx)^4=144$
$(cosx)(5cosx+3)(5(cosx)^2-27cosx+40)=0$

Solve each factor for $cosx$.

The only real solutions are $0$ and $\frac{-3}{5}$.

In an equation like this, look for ways to put the variable parts ($sin(2x)$ and $sin(x)$) in terms of what your solving for (in this case $cos(x)$).  Substitute and solve.  (I actually like to use $u$ or some other variable instead of writing $cos(x)$ all the time.)


Let P(x) be polynomial such that ,P(1)>P(0) ,P(2)>2P(1)-P(0)
   P(3)>3P(2)-3P(1)+P(0)
   P(n+4)>4P(n+3) - 6P(n+2) +4P(n+1)-P(n)
   prove that :P(m)>0 for all m is natural number

First suppose P(x) = x^n

Then it's pretty easy to figure out that you will always have an equality on the last equitation (the first is ok for n>1, the second for n>2, the third for n>3 -- don't worry about these they'll fall right)

put them all on the left side:

P(n+4) - 4P(n+3) + 6P(n+2) - 4P(n+1) + P(n) > 0

as I said this is always zero for P(x) = x^n

So also for n-1, n-2, etc.

So as polynomials are combinations of these, we'll have that when P(x) = a.x^n + b.x^(n-1) + ... + c.x^1
then the thing will result an equality
thus to make it true, we need to add any positive constant to the stuff.

Which makes P(m) always positive.[/i]


I've tried my best to keep up this forum alive, but guess noone really cares about this section..

This forum is still young. Take the patience to it. ;)

Like Mr. Crawford said, you can't make the people to get on the site and enjoy it. You can tell them about it but there's nothing you can really do.

It's upon them.

i'm trying very hard too. but i dunno, i guess a lot of ppl are too advanced for this forum, but it's about just right for me...... how many sixth graders are on this site? i feel like everyone knows more than i do.....

I'm in 7th grade...

Getting Started Section won't be too hard..
It actually helped me alot

Eh, it's going to be alright. People will come eventually. In fact, I'll talk to our Middle school MathCounts team about using the forum to help them when coaches aren't there. People will start coming. Here - try this: Host a competition to kick off the new forum.

[quote="RC-7th"][quote="LuCy4EvA"][quote="solafidefarms"]It's the Young Mathematicains' Talent Search! YMTS! And let's get some problems together.  Lets have it on a weekend, say first in March, and have problems scaled down from school Mathcounts.

Billy[/quote]

wait, do you mean host a contest for the new people or people that's already on AOP?[/quote]

Why not let anyone participate?[/quote]

well, what about the peole that makes the test? and if we are trying to new people to come to this forum, we can't exactly make it suitable for people that's not really into math yet and also for the people that's been here for a while. BUt i"m perfect fine with making two different tests.

Don't worry about it people will start coming.

I personally think this forum has been doing quite well, considering that it's only been around for a few days.  Be patient -- you need to give people time to find it and see how good it is.

[quote]Why do they fear? Nobody will know it's them  In plus this way they can learn [/quote]

They fear it because they feel shy... and I'm pretty sure that they want to post it after they see other people's posts.

well,,

I felt that way too, when I first started

[quote="LuCy4EvA"][quote="Valentin Vornicu"][quote="LuCy4EvA"]just figured that most of my friends have a screen name, but never actually post... like this person I now do the problems but never post anything because they are afraid that it's wrong.... that's why it seems like there's not much people here.[/quote]Why do they fear? Nobody will know it's them  :? In plus this way they can learn  ;)[/quote]

well, I know them... and other people...and beside, I just wrote that to make shinwoo feel better and to say that it's not true that no one comes to this forum... even tho it's true.....[/quote]

that is not true! And i'm not saying this just to make someone feel better....
That's not true Lucy! People just haven't discovered this site yet....
and don't discourage us like this!..... I just got Jennifer to come to this site...she asked me some questions on how to registrate and asked me to help! See, she looked around this site and liked it......

TO EVERYONE: I GOT ONE OF MY FRIENDS TO SIGN UP TODAY!!!!! YAY!! CHEERS!!!!! DON"T GIVE UP!!!

ppl just need time....

I'm a seventh grader and a total IDIOT when it comes to math. Oh well, there's always language arts...
Always multiplying,
Callie Sanders (not my real name)

OMG Don't bump a thread that hasn't been posted in for 5 years!
You have made everyones day horrible now because of that.
STOP!
That is soo rude.

Please do not revive a five year old thread.

[quote="Dsquared"]
You have made everyones day horrible now because of that.
STOP!
That is soo rude.[/quote]

...and so was that.

Oh, so Phire can yell at me in a ff post, but I can't yell at someone that revived a 5 year old thread?
Note ff=fun factory

Hi.

You're not a moderator.

He/she probably doesn't know about the reviving rule.

You're not using the appropriate ways to tell someone to stop. Even if you used polite words, I have PM'ed you personally that you're supposed to remind others of such forum rules by private-messaging and not by continuing to bump the thread.

And yes, that was rude. So was my rant in the Fun Factory post, and I apologize for that, but it was because you were so bad of a spammer that I blew my top watching the forum crumble.


Determine all pair $(m,n)$ of positive integers such that $m$ divides $n^{3}+1$ and $n$ divides $m^{2}+1$.

@Edited by N.T.TUAN.

Anyone?? ... I would like to see the solution too ...

i think i made some progress ... probably obvious to the rest;
use n=m+k
implies m divides 1+ k^3 and (m+k) divides (mk-1)
..dont know how to proceed from here
it has infinitely many solutions tho
of  the form (t^2+1,t^2+t+1)
are u sure it has a fully parametrized set of solutions

cheers

little approach:

All divisors of $m^{2}+1$ are $n_{1}=2$ or $n_{i}\equiv 1 (mod4)$.

case 1: $m$ is odd $\Rightarrow$ one divisor of $m^{2}+1$ is $n_{1}=2$
$n_{1}^{3}+1=9 \Rightarrow m=1,3,9$
missing: all other divisors of $m^{2}+1$

case 2: $m$ is even $\Rightarrow$ all divisors of $m^{2}+1$ are $n_{i}\equiv 1(mod4)$
$\Rightarrow n_{i}^{3}+1\equiv 2(mod4) \Rightarrow m_{1}=2 \Rightarrow n=1,5$
missing: all other $m$ and $n$

All I've got is
- if $(m,n)$ is a solution, then $(m,\frac{m^{2}+1}{n})$ also is a solution,
so now we can assume $n<m$
- if $(m,n)$ is a solution, then $(\frac{n^{3}+1}{m},n)$ also is.

Don't know how to go from here ...


Find the sum of the first 50 terms of the recursive sequence whose $n^{th}$ term is given by $a_n=a_{n-1}+n$ and whose first term is 1.

$A. 22,051$    $B. 22,100$     $C.25,000$     $D.27,250$


Unable to upload file

DEBUG MODE
Line : 259
File : usercp_avatar.php

when I try to change my avatar, I press submit, and it says: no avatar chosen.


Q1 Find number of garlands using 3 flowers of one kind and 12 flowers of other kind

Q2 A parallelogram is cut by two sets of m lines parallel to the sides,then find number of parallelograms formed

Q3 Find total number of six digit numbers $ x_1x_2x_3x_4x_5x_6$having property $ x_1<x_2\leq x_3<x_4<x_5 \leq x_6$

[hide]Ans1=119
Ans2=$ \frac {(m\plus{}2)^2(m\plus{}1)^2}{4}$
Ans3=$ ^{11}C_{6}$ [/hide]

Hi jackie-chan01.

Q2:

Choosing any two of the (now) $ m \plus{} 2$ horizontal parallel lines in $ \binom{m \plus{} 2}{2}$ ways and and two of the $ m \plus{} 2$ oblique parallel lines in $ \binom{m \plus{} 2}{2}$ ways, we get

$ \frac {(m \plus{} 2)^{2}(m\plus{}1)^{2}}{4}$

I've only answered one because I find that this usually spurs someone to answer the others and hopefully the'll do a better job than I've done.

err, why you put the answer at the end of the page while posting it in Unsolved category?
Is it your answer? Check the $ 3^{rd}$ asnwer :D
Hopeful.

For Q1, I have assumed that we can have a garland containing between $ 1$ and $ 15$ flowers.

$ (m, n)$ denotes $ m$ flowers of the first kind (of which there are $ 3$) and $ n$ flowers of the second kind (of which there are $ 12$).

For a garland with $ 1$ flower $ (0, 1)$ or $ (1, 0)$;

For a garland with $ 2$ flowers $ (0, 2), (1, 1), (2, 1)$;

For $ 3$ flowers $ (0, 3), (1, 2), (2, 1), (3, 0)$;

For $ 4$ flowers $ (0, 4), (1, 3), (2, 2), (3, 1)$;

... ... ... ... ....

For $ 14$ flowers $ (2, 12)$ or $ (3, 11)$;

For $ 15$ flowers $ (3, 12)$.

This gives $ 51$ garlands. Either I've done something wrong or I've misunderstood the question.

[quote="AndrewTom"]

Choosing any two of the (now) $ m \plus{} 2$ horizontal parallel lines in $ \binom{m \plus{} 2}{2}$ ways and and two of the $ m \plus{} 2$ oblique parallel lines in $ \binom{m \plus{} 2}{2}$ ways, we get

$ \frac {(m \plus{} 2)^{2}}{4}$

[/quote]
Hi
$ \binom{m \plus{} 2}{2} \equal{} \frac {(m \plus{} 2)(m \plus{} 1)}{2}$, so your answer should be the same as the one given by jackie.

For 3, there are only 4 cases to consider.

Case 1: all numbers are distinct. We choose 6 numbers from 1 to 9. Hence $ C^9_6$

Case 2: $ x_6\equal{}x_5$ and $ x_3\ne x_2$We choose 5 from 1 to 9. Hence $ C^9_5$

Case 3: $ x_3\equal{}x_2$ and $ x_6\ne x_5$We choose 5 from 1 to 9. Hence $ C^9_5$

Case 4: $ x_3\equal{}x_2$ and $ x_6\equal{}x_5$ We choose 4 from 1 to 9. Hence $ C^9_4$

They add to $ 462$.

Can anyone explain why it is equivalent to choosing $ 6$ numbers from $ 11$?  :blush:


Who in here is taking the 2007 Open Exam? This is my first year, and I'm almost done cramming the Campbell book.

its my first year taking the opening exam, and i ll be pretty happy to make it to the semifinals.  do you think cramming on an ap bio prep book (for 2.5 weeks) will help me pass the open exam (havent had ap bio yet, but i felt that most of the prep book is review)?

does anyone know where to get a hold of some practice exams? my teacher is out of town so i can't access any through him.

I can't help you guys with any practice exams, but I can say that an AP Biology preparation book will help, but only marginally. The test covers the Campbell text and, I believe, a couple of other "optional" resources. If you know everything in Campbell, you should do well to make the semifinals. It will then depend on how well you can deal with critical thought problems in various areas of biology.

Hey, does anyone know the approximate cut-off mark for semi's?

I believe last year's cut-off was 26/50 (correct me if I'm wrong), with an upper bound of 45.

cutoff was 26/50, with an upper bound of 41.  There were I think 5-10ish people who scored in the 39-41 range.

I know that one question from last year's test had an error (so they gave everyone full credit for it). But which one was it?
Also, what was the cutoff for 2004?

what are the rules concerning discussion of test questions?

I believe today is the last day of testing. But here in South Jersey everyone's basically on a snow day. So I don't know how that works out.

Even though today was the last day of testing, my teacher told me to not discuss the questions until the results are in (~March). CEE gave some schools on the east coast permission to take it a few days late because of snow issues.

We just got results today.

I qualified with 33/50.

Passed!  38/50

Back to studying for the semifinals :(


function
limit
floor function
calculus
calculus computations

Prove that the function f difined by f(x)=E(x) (xE Q)
                                                  f(x)=x (xE R-Q)
dont have a limit in any point of R

[ E(x)<=x <E(x)+1

For any  $x \in \mathbb{R}$,
let $(a_{n})_{n \in \mathbb{N}}$ be a sequence of irrational numbers, such that $\lim_{n \to \infty}a_{n}= x$,$(b_{n})_{n \in \mathbb{N}}$ be a sequence of strictly increasing rational numbers, such that $\lim_{n \to \infty}b_{n}= x$.

[b] case I:[/b] $x \in \mathbb{N}$
$\lim_{n \to \infty}f(a_{n}) = x \not= x-1 = \lim_{n \to \infty}f(b_{n})$

[b] case II[/b]: $x \in \mathbb{R}\setminus \mathbb{N}$
$\lim_{n \to \infty}f(a_{n}) =x \not= \lfloor x \rfloor = \lim_{n \to \infty}f(b_{n})$

Hence $f$ is not continuous at any $x \in \mathbb{R}$


geometry
3D geometry
modular arithmetic
Additive Number Theory

Show that each integer $n$ can be written as the sum of five perfect cubes (not necessarily positive).

With $4$ cubes we can write every integer multiple of $6$: $(k+1)^3+(k-1)^3+(-k)^3+(-k)^3=6k$.
Since $i^3\equiv i\pmod{6}$ for $i=0..5$, we can thus write all integers as the sum of $5$ cubes.


In the figure, line AB is tangent to the circle with center O, and line AO ~= line OB
(I can't get the = sign to go underneath).

If the radius of the circle is 6, what is the area of triangle AOB?

The circle's radius makes the height of the triangle = 6.

The circles' radius divides the triangle into 2 special right triangles (Angle B = 30°).

The base = √3 * 6 = 4.24

The area of the little triangle = 12.72

The two triangles are identical, so, the area of AOB = 25.44

Am I on the right track?

[quote="The Zuton Force"]Your methodology is correct, however you have a computation error:

$ \sqrt {3} \times 6 \neq \sqrt {3 \times 6}$

The former is what you want, and is equal to about 10.392
The latter is what you used, and is equal to about 4.242

Your answer should be about [b]62.352[/b].

In general, the squareroot of any number greater than 1 is also greater than 1. So, $ \sqrt {3} \times 6 > 1 \times 6 \equal{} 6$, so if you get something like $ 4.242 < 6$, you should take a second look at your calculations --[/quote]

There's also an answer that includes a radical. Is there another way to do it? Can 62.352 be broken down?

$ 6\times 6\sqrt3\equal{}36\sqrt3\approx 62.352.$

6 is the height and 6 root 3 is half the base.

[quote="pi^2/6"]$ 6\times 6\sqrt3 \equal{} 36\sqrt3\approx 62.352.$

6 is the height and 6 root 3 is half the base.[/quote]Thank you thank you!


function
inequalities
limit
triangle inequality
real analysis
real analysis unsolved

Let $ f(z)\equal{}1\minus{}\frac{1}{2^z}\plus{}\frac{1}{3^z}\minus{}\frac{1}{4^z}\plus{}\cdots$.
i)show that  the sum converges uniformly and absolutely on compact sets to an analytic function if $ Re(z)>1$.
ii)Moreover prove that $ f(z)\equal{}(1\minus{}2^{1\minus{}z})\zeta(z)$ for $ Re(z)>1$
iii) show that convergence is conditional when $ z\equal{}x$ is real and $ x\in (0,1)$
iv) more generaly show that the statement is still true for the infinite strip $ 0<Re(z)<1$

it's easy:
i)  just estimate by $ \parallel{}$ use triangle inequality and you get $ < \zeta(x)$ where $ x>1$.
ii)  use the fact that
\[ \sum \frac{1}{(2k)^{z}} \equal{} \frac{1}{2^{z}} \sum \frac{1}{k^{z}}\]
iii) it's enough to show that $ \lim _{n \to \infty} \frac{1}{n^{z}} \equal{} 0$
iv) similary as iii)
v)  what ?

the last part is the interesting part that i wasnt able to get....
the question is to show that the sum shown converges conditionally if $ 0 < Re(z) < 1$
(obviously holds when $ z$ is a real number, by alternating series test, but i cant get it in this case)

I might be wrong but I think KentMerryfield might have posted something similar to this quite a while ago...


How many of the positive integer divisors of 100 are not prime?

The only factors that ARE prime are $ 2$ and $ 5$.  The rest are not.  We count the number of factors fast by using prime factorization, which is $ 2^2 \times 5^2$.  We take the powers, add $ 1$ to each, and multiply them together.  We have here $ (2\plus{}1)(2\plus{}1)\equal{}9$.  We can then do $ 9\minus{}2\equal{}\boxed {7}$ to get our answer.


trigonometry
analytic geometry
conics
parabola
modular arithmetic
complex numbers

1. Sin(pi/5) in ( a + sqrt(b) ) / c form

2. Factor x^4 -2*x^3+7*x-4 into (x^2+bx+c)*(x^2+dx+e) form

[quote="E^(pi*i)=-1"][hide="2"]
$ x^4 \minus{} 2x^3 \plus{} 7x \minus{} 4 \equal{} (x^2 \minus{} 3x \plus{} 4)(x^2 \plus{} x \minus{} 1)$.
[/hide][/quote]
would you explain how you get it ?

[hide]
You didn't specify, but I figured that the coefficients were probably integers.  So, to get the right constant, $ c$ and $ e$ have to be $ 2$ and $ \minus{}2$, $ 4$ and $ \minus{}1$, or $ \minus{}4$ and $ 1$.  The first turns out to be impossible if you try it.  For the second, you get the equations $ b\plus{}d\equal{}\minus{}2$, $ 3\plus{}bc\equal{}0$, and $ 4c\minus{}b\equal{}7$, which are easy to solve to find $ (b,c)\equal{}(\minus{}3,1)$.
 [/hide]

[hide="1"]
$ \frac{\pi}{4}\minus{}\frac{\pi}{20}\equal{}\frac{\pi}{5}$
so use sine of a difference, and once all the terms are out, simplify with half angle twice

yeah it's really messy

[/hide]

another question about definition.
Is r in cylindrical coordinates positive only? Or it can be positive and negative?
Do any one remember the formula for finding distance between the cylindrical coordinates?

[quote="pianoforte"]Dude, how are we going to find $ \sin \pi/20$...[/quote]
Half angle identities?
Twice?

[quote="E^(pi*i)=-1"][hide]
You didn't specify, but I figured that the coefficients were probably integers.  So, to get the right constant, $ c$ and $ e$ have to be $ 2$ and $ \minus{} 2$, $ 4$ and $ \minus{} 1$, or $ \minus{} 4$ and $ 1$.  The first turns out to be impossible if you try it.  For the second, you get the equations $ b \plus{} d \equal{} \minus{} 2$, $ 3 \plus{} bc \equal{} 0$, and $ 4c \minus{} b \equal{} 7$, which are easy to solve to find $ (b,c) \equal{} ( \minus{} 3,1)$.
 [/hide][/quote]
Then if question like this:
How many real solutions does the equation [b]x^4-4*x^3+x^2-2*x-1=0[/b] have?
How do you do it?

Well, we can convert it to complex numbers, and use DeMoivre's Theorem to evaluate. Let pi/5=a
cis(pi)=cis(a)^5=isina^5+5sina^4cosa-i10sina^3cosa^2-10sina^2cosa^3+i5sinacosa^4+cosa^5
0=isina^5-i10sina^3cosa^2+i5sinacosa^4
And uh, I guess you just solve...and get ugly stuff.

More!
1. If n is the smallest positive integer for which there exist positive real numbers a and b > 0 such that  (a+bi)^n=(a-bi)^n, find b/a.
2. What is the shape of the graph of [b]|z| = Im(z)+1[/b] ? ( Im(z) is the imaginary part of z.
3. The Gaussian Integers are the complex numbers a+bi such that a and b are integers. Which of the following is prime over the Gaussian Integers? (Note that, for example, 2+i is prime, although it has the factorixations i(-2i+1) and 1(s+i); multiplication by numbers 1, i, -1, or -i does not make a factorization.)
A 1+3i
B 13
C 2i
D 5+i
E 7

[quote="linj"]More!
1. If n is the smallest positive integer for which there exist positive real numbers a and b > 0 such that  (a+bi)^n=(a-bi)^n, find b/a.
2. What is the shape of the graph of [b]|z| = Im(z)+1[/b] ? ( Im(z) is the imaginary part of z.
3. The Gaussian Integers are the complex numbers a+bi such that a and b are integers. Which of the following is prime over the Gaussian Integers? (Note that, for example, 2+i is prime, although it has the factorizations i(-2i+1) and 1(s+i); multiplication by numbers 1, i, -1, or -i does not make a factorization.)
A 1+3i
B 13
C 2i
D 5+i
E 7[/quote]
Is the answer to number 3 E?

[hide="1"]We have $ (a+bi)^n=(a-bi)^n$.  Let $ r\text{cis}\,\theta=a+bi$, where $ \text{cis}\,\theta = \cos\theta+i\sin\theta$.  Thus $ r\text{cis}\,(-\theta)=a-bi$.  By DeMoivre's, $ r^n\text{cis}\,n\theta=r^n\text{cis}\,(-n\theta)$.

Divide both sides by $ r^n$ to get $ \cos(n\theta)+i\sin(n\theta)=\cos(-n\theta)+i\sin(-n\theta)$.  The real parts cancel out ($ \cos x=\cos -x$).  Divide both sides by $ i$ t get $ \sin(n\theta)=\sin(-n\theta)$.  Since $ \sin -x=\sin x$, we can conclude that $ \sin(n\theta)=0$.  Thus $ (a+bi)^n$ is a real number.

Trying $ n=2$ yields $ \frac{\pi}{2}|\theta$.  However, this will not satisfy $ a,b>0$.
Trying $ n=3$ yields $ \frac{\pi}{3}|\theta$.  This works.  Thus $ \frac{b}{a}=\tan\theta=\boxed{\sqrt{3}}$.
[/hide]

[hide="2"]Let $ z=a+bi$, where $ a,b$ are real.  Thus we have $ \sqrt{a^2+b^2}=b+1$.  Square both sides to get $ a^2+b^2=b^2+2b+1$ or $ a^2=2b+1$, which is a $ \boxed{\text{parabola}}$.[/hide]

[hide="3"]A isn't prime, since $ (1+i)(2+i)=1+3i$.
B isn't prime, since $ (3+2i)(3-2i)=13$.
C isn't prime, since $ (1+i)^2=2i$.
D isn't prime, since $ (1+i)(3-2i)=5+i$.
Assume E isn't prime.  Since $ 7$ is real, then for some integers $ a,b$, $ (a+bi)(a-bi)=a^2+b^2=7$.  However, $ n^2\equiv 0,1,4\pmod 8$, so this is impossible.  Thus the answer is $ \boxed{\text{E}}$.

I found most answers using the fact that $ z_1z_2=z_3\Rightarrow |z_1|\cdot|z_2|=|z_3|$, and by definition of Gaussian integer, $ z_1,z_2,z_3$ are of the form $ \sqrt{n}$, where $ n$ is a positive integer.[/hide]


Find the smallest $ n$ for which the following holds true
$ \frac{1}{\sin 45.\sin 46}\plus{}\frac{1}{\sin 47.\sin 48}\plus{}.....\plus{}\frac{1}{\sin 133.\sin 134}\equal{}\frac{1}{\sin n}$

All the angles are in degrees...

There is no smallest n, you can keep subtracting 360. Perhaps you mean smallest positive n?

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See Post #10 by gotztahbeazn.