limit
algebra
polynomial
calculus
trigonometry
geometric sequence
calculus computations

[quote="perfect_radio"]I don't understand one thing: Why did you use L'Hospital? It is fairly useless here, I think.[/quote]

perfect_radio, I don't think using L'hospital rule is disable here. But applying it for this problem, you've made problem become complex, minsoens... :)

[quote="perfect_radio"]Why did you use L'Hospital?[/quote]
it was the first thing that came to mind; i don't know any good proof methods; im a noob like that~
um did i get it right though?

EDIT: ~~100th~~ :lol:

[quote="minsoens"]it was the first thing that came to mind; i don't know any good proof methods; im a noob like that~
um did i get it right though?[/quote]

The solution is quite simple, [hide="hint"] just expanding the products... [/hide]

For a): I think that $\frac{a_1+a_2+\cdots+a_n}{n}.$

oh right. so you didn't have to use calculus after all :)
but the $\lim$ was what made me think of calculus immediately..

Find:
$c)$ $\lim_{n\to\infty} \sqrt[n]{\frac{1}{n!}}$
$d)$ $\lim_{n\to\infty} \sqrt{3\sqrt{5\sqrt{3\sqrt{5\sqrt{\cdots}}}}}$
                             $\sqrt$ appears $n$ times.

[hide="c"] For any $a$, $n!>a^n$ for large enough $n$. Therefore
\[ \lim_{n\rightarrow\infty}\sqrt[n]{\frac{1}{n!}}<\lim_{n\rightarrow\infty}\sqrt[n]{\frac{1}{a^n}}=\lim_{n\rightarrow\infty}\frac{1}{a}=\frac{1}{a}  \]
And since $\sqrt[n]{\frac{1}{n!}}>0$, the limit must be $0$.[/hide]
[hide="d"] Define $a_0=1$ and $a_{n+1}=\sqrt{3\sqrt{5a_n}}$. All $a_n<5$, this can be proven by induction. If $a_n<5$, then $a_{n+1}=\sqrt{3\sqrt{5a_n}}<\sqrt{3\sqrt{25}}=\sqrt{15}<5$. This is an increasing sequence, which can also be proven by induction. Suppose $a_n>a_{n-1}$. Then $a_{n+1}=\sqrt{3\sqrt{5a_n}}>\sqrt{3\sqrt{5a_{n-1}}}=a_n$. This proves that the limit exists. Let $l$ be the limit. Then $l=\sqrt{3\sqrt{5l}}$, or $l=\sqrt[3]{45}$.[/hide]

[quote="perfect_radio"][quote="Lovasz"]$a)$ $\lim_{n\to\infty}(\sqrt[n]{(x+a_1)(x+a_2)\cdots(x+a_n)}-x)$.
[/quote]

First of all, I think that it is  $\lim_{x \to \infty}(\sqrt[n]{(x+a_1)(x+a_2)\cdots(x+a_n)}-x) = \ell$.

Let $P(x) = (x+a_1)\ldots(x+a_n)$.

Use the well-known formula for $a^n - b^n$ to get

$\ell = \lim_{x \to \infty} \frac{P - x^n}{\sqrt[n]{P^{n-1}} + x \sqrt[n]{P^{n-2}} + \ldots + x^{n-1}}$

$= \lim_{x \to \infty} \frac{(a_1+\ldots+a_n) x^{n-1} + O(x^{n-2})}{n \cdot \textrm{something of degree }n-1}$

$= \frac{a_1+a_2+ \ldots + a_n}{n}$,

which is the same as Kunny's answer.[/quote]

Yes, perfect_radio.

to jmadsen: in question $d)$, why don't you try geometric progression? I'll post the solution soon.

to ...: try the questions $e),f)g)$... :)


MATHCOUNTS probability geometry rectangle ratio inequalities

1.
Darts are thrown randomly at a board showing three concentric circles of radii 2, 3, and 4.  If a dart hits inside one of the circles, what is the probability, expressed as a common fraction, that it is in the interior of the circle of radius 3 but not in the interior of the circle of radius 2?

2.
Three tennis balls are tightly sealed in a right cylindrical can, so that the balls cannot move inside the can.  If the radius of each ball is 4 cm, find the number of cubic centimeters in the volume of the space within the cylinder, not taken up by the three tennis balls.  Express your answer in terms if $\pi$.

3.
A custom made window is a single pane of glass consisting of a semicircular sheet of glass on top of the narrow end of a rectangular sheet of glass.  The rectangle has dimensions of 4 feet by 3.5 feet.  How many feet of weather-stripping will be needed to go around the glass?  Express your answer to the nearest integer.

4.
The numbers from 1 through 8 are separated into two sets, A and B.  The numbers in A are multiplied together to get $a$.  The numbers in B are multiplied together to get $b$.  The larger of the two numbers $a$ and $b$ is written down.  What is the smallest possible number that can be written down using this procedure?

5.
Eric's Florist Shop sells roses for 20 dollars per dozen, daises for 5 dollars per dozen, and chrysanthemums for 10 dollars per dozen.  At these rates how much would an arrangement of 9 roses, 6 daises, and 18 chrysanthemums cost in dollars?

6.
Given the equation $\displaystyle \frac{1}{m} + \frac{1}{n} = \frac{2}{15}$, where $m$ and $n$ are positive integers, find the value of $m+n$ such that $m>n$ and $m$ is not a multiple of $n$.

7.
Use the pattern below to express $100^2$ in the form $a^2+b^2-c^2$.  Then find $a+b+c$.
\begin{eqnarray*}
12^2 = 8^2 + 9^2 - 1^2\\
14^2 = 10^2 + 10^2 - 2^2\\
16^2 = 12^2 + 11^2 - 3^2\\
18^2 = 14^2 + 12^2 - 4^2
\end{eqnarray*}

8.
The ratio of the length of a rectangle to its width is the same as that of the diagonal to the length.  If the width is 2, how many units are in the length of the diagonal?

9.
Suppose a number $x$ satisfies $x^2-4x-1=0$.  How much greater is $x$ than its reciprocal?

10.
What is the difference between the largest and the smallest values of $x$ satisfying the following system of inequalities?
Express your answer as a mixed number.
\begin{eqnarray*}
2x-6 \le 6\\
4x+3 \ge 13
\end{eqnarray*}

This was the first post 9 years and 4 days ago.

*cough* hwl I believe McCrawford works for AOPS and is trying to promote discussion on our dead forum

Anyways, my answer for number 1: (too lazy to make solution)

[hide]5/9?[/hide]


trigonometry
logarithms
function
symmetry
real analysis
real analysis solved

What's the explicit bijection enters $]0,1[$ et $\mathbb{R}$?  $x \mapsto 10^n x$ ??  :blush:

How about $x\mapsto \tan \left(\pi \cdot \left(x-\frac 12\right)\right)$?

Indeed! I have shame of what I wrote  :? Thank you

There are plenty of nice answers to this.

$f(x)=\ln\left(\frac{x}{1-x}\right)$ is another one.

And that's just restricting ourselves to smooth functions with smooth inverses and certain symmetry properties.


search
number theory theorems
number theory

What is Bertrand's postulate???can anyone help me!!!
thanks

I suggest you search GOOGLE. That will be better than anyone telling you on this forum !!

Eric W. Weisstein et al. "Bertrand's Postulate." From MathWorld--A Wolfram Web Resource.[url=http://mathworld.wolfram.com/BertrandsPostulate.html]Bertrand's Postulate (form Mathworld)[/url]

For $n \geq 1$, there is always a prime number $p$ with $n \leq p \leq 2n$.

The proof is quite tricky, and a beautiful (and elementary) one can be found in [i]Proofs from THE BOOKS[/i].

Heh, I always thought this 'rule' (theorem?) was Chebycheff's (or w/e is the spelling  :rotfl: )

Also My problem was it, because I knwo it as Chebycheff theorem. :oops:
But thanks for all help me.

What about Sylvester's Theorem:
There is a prime factor $p$ of $\binom{n}{k}$,$p>k$ when $n\geq2k$
It's stronger.I wonder where's the proof.

[quote="shencaili"]What about Sylvester's Theorem:
There is a prime factor $ p$ of $ \binom{n}{k}$,$ p > k$ when $ n\geq2k$
It's stronger.I wonder where's the proof.[/quote]

I want to see also.


trigonometry
absolute value
geometry solved
geometry

Evaluate  of

 cos5.cos10.cos15....cos85=?

Sure, here are two hints:

(a) Let
\[\omega = \cos \left( \frac{2 \pi}{n} \right) + i \sin \left( \frac{2 \pi}{n} \right).\]
Show that
\[x^{n - 1} + x^{n - 2} + \dots + x + 1 = (x - \omega)(x - \omega^2) \dots (x - \omega^{n - 1}).\]

(b) Set $x = 1$.  Take the absolute value of both sides to prove that
\[\sin \left( \frac{\pi}{n} \right) \sin \left( \frac{2 \pi}{n} \right) \dots \sin \left( \frac{(n - 1) \pi}{n} \right) = \frac{n}{2^{n - 1}}.\]


quadratics
vector
linear algebra
matrix
linear algebra unsolved

Let $ P^ \plus{} ,P^ \minus{} ,I,Q \in R^{n\times n}$, $ K\in R^{n\times 1}$, $ M\in R^{1 \times n}$, and assume that
$ Q$ is positive definite, $ P^ \minus{}$ is positive semidefinite whence $ (MP^ \minus{} M^T \plus{} Q)^{ \minus{} 1}$ exists (where $ T$ denotes transpose).

In what sense does $ K \equal{} P^ \minus{} M^T(MP^ \minus{} M^T \plus{} Q)^{ \minus{} 1}$ minimize the quadratic expression

$ P^ \plus{} : \equal{} (I \minus{} KM)P^ \minus{} (I \minus{} KM)^T \plus{} KQK^T$, over $ K$?
Is this minimization of $ P^ \plus{}$ over all vectors $ K$ with respect to the usual ordering for positive semidefinite matrices $ A\le B$ iff $ B \minus{} A$ is positive semidefinite?

Next consider the extension $ P^ \plus{} : \equal{} (I \minus{} KAM)P^ \minus{} (I \minus{} KAM)^T \plus{} KAQA^TK^T$, where $ A\in R^{n\times 1}$, $ K\in R^{n\times n}$ and $ K$ diagonal, where all other matrices are as above.
What is the minimum over $ K$ (with respect to the previous ordering or something )??

Any help will be deeply appreciated.

A small correction,

$ Q > 0$, a scalar, not a matrix!!

Please help me with this.


function
advanced fields
advanced fields unsolved

Denote by C([0, 1]) the set of continuous real-valued functions on [0,1]. Let A be the set of continuously differentiable functions on [0, 1] such that f'(1/4)=f'(3/4) for every f in A. Is A dense in C([0, 1])?

Consider $ C^1([0,1])$ as a subspace of $ C([0,1])$. The linear functional $ T(f) : \equal{} f'(1/4) \minus{} f'(3/4)$ on $ C^1([0,1])$ is [i]not[/i] continuous (it's easy to construct a sequence $ f_n \in C^1$ s.t. $ f_n'(1/4) \equal{} 1, f_n'(3/4) \equal{} 0$, but $ f_n\to 0$ uniformly, for example).
Thus $ A \equal{} \ker(T)$ is dense in $ C^1([0,1])$. Since $ C^1([0,1])$ is dense in $ C([0,1])$, we have that $ A$ is dense in $ C([0,1])$.


Let $ S \equal{} \frac{b^n\minus{}1}{b\minus{}1} \equal{} b^{n\minus{}1}\plus{}b^{n\minus{}2}\plus{}...\plus{}b\plus{}1$
Since $ 2\mid b$, $ S\equiv b^2\plus{}b\plus{}1 (mod 8)$
Since S is a perfect square and it's odd, $ S\equiv 1 (mod 8)$.
So $ b^2\plus{}b\plus{}1 \equiv 1 (mod 8)$, i.e. $ 8\mid b(b\plus{}1)$.
As $ (b,b\plus{}1) \equal{} 1$, $ 8\mid b$ or $ 8\mid (b\plus{}1)$, but since b is even, the latter is impossible, thus b is divisible by 8.

[quote=STARS]Let $ b$ be an even positive integer for which there exists a natural number n such
that $ n>1$ and $ \frac{b^n\minus{}1}{b\minus{}1}$ is a perfect square. Prove that $ b$ is divisible by 8.[/quote]

Let $b=2k$ for some natural number $k$
Now  as $n>1$ so let's check for $n=2$
Then $2k+1=x^2$ now as $x^2 =(0, 1,4)mod 8$  for every natural number $x$ then $2k+1=x^2$ is only possible if $8$ divides $2k$ hence for $n=2$ :) $8$ will divide $b$
Now for every $n\geq 3$ :) 8 divides $(2k)^n$
Then $\frac{(2k)^n - 1}{2k-1} = x^2$ then  $\frac{(2k)^n - 1+x^2}{2x^2}=k$ now clearly $(2k)^n - 1+x^2$ must be even it is only possible when $x^2=1mod 8$  and as $n\geq 3$ then 8 divides $(2k)^n$ hence at least $4$ divides $k$ then as $b=2k$ so $8$ divides $b$ $\blacksquare$

[quote=parmenides51][quote=ItzARubix]what is Estonian Math Comp?[/quote] there is always google
http://www.math.olympiaadid.ut.ee/eng/html/index.php[/quote]

Oh TIA


Show that the equation $x^2 +y^5 =z^3$ has infinitely many solutions in integers $x, y, z$ for which $xyz \neq 0$.

[quote="Robert Gerbicz"]Let $ n$ a positive integer and $ x=10*{n}^{15},\; y=3*{n}^{6},\; z=7*{n}^{10}$  then $ x,y,z$ is a solution of the equation.[/quote]

[quote="Peter"]Show that the equation $ x^2 \plus{} y^5 \equal{} z^3$ has infinitely many solutions in integers $ x, y, z$ for which $ xyz \neq 0$.[/quote]
Other solution:
[b]Solution 1[/b]
From $ 2^n \plus{} 2^n \equal{} 2^{n \plus{} 1}$
Chose $ z \equal{} 2^{\frac {n \plus{} 1}{3}}$
           $ y \equal{} 2^{\frac {n}{5}}$
         $ x \equal{} 2^{\frac {n}{2}}$
Chose $ 10|n,3|n \plus{} 1,\Rightarrow n\equiv 20 (\mod 30)$
Then it has solution.
In the same way we can prove that if $ \gcd(a,b) \equal{} 1$ then the equation:
$ x^a \plus{} y^b \equal{} z^c$ has integer solution and $ xyz > 0$
 [b]Solution 2[/b] (From Zetax)
We use the lemma : Every integer can represent as sum of a cube and a perfect square.
The equation has equivalent
$ ( \minus{} y)^5 \equal{} x^2 \plus{} ( \minus{} z)^3$ (1)
Chose $ |y|$ not is a cube or a perfect square.
We prove that if $ x,y$ satisfy (1) then it is different from 0.
 [u] Case 1[/u]$ x \equal{} 0$ then $ y^5 \equal{} z^3\Rightarrow y \equal{} m^3$ (tradition!)
 [u]Case 2 [/u] $ z \equal{} 0$ then $ ( \minus{} y)^5 \equal{} z^2,\Rightarrow y \equal{} \minus{} m^2$ (tradition!)
So it has solution satisfy $ xyz$ is different from 0.


Several thieves found a package containing some dollar bills. They
decided to share the money. They found that when they tried to
give $ \$3$ to each thug, they ran out of money and one thug received
nothing. When each thug took $ \$2$, they had $ \$1$ left over. What is
the least possible sum of the number of dollars and the number of
thieves?

[quote="isabella2296"]The answer is 5.

We know the number of dollars has to be divisible by three, and cannot be even.

The smallest number of dollars that work is 3.

By guess and check, we can determine that the number of thugs is 2.

3+2=5, so that is your answer.[/quote]

false. if there are 2 thugs splitting 3 dollars, and they each take 2, there will be one thug with only 1 dollar. so 5 isnt the right answer

the correct answer, therefore, is the next odd multiple of 3, 9

using the formula 2x+1=9, we find there are 4 thugs

9+4=$ \boxed{94}$

[color=white][size=9]Its 13. did you really think i'd do that?[/size][/color]

[quote="Brut3Forc3"]Yeah!
(Quote him...)[/quote]

finally someone reads it!

Let x be the total amount of money and t be the number of thieves.

The first clue gives us 3(t-1)=x.
The second clue gives us 2t+1=x.
Then we set 3(t-1) and 2t+1 equal:

3t-3=2t+1
t=4

There are 4 thieves, so they stole $ \$9$ (Which is nothing if youre going to steal money).
4+9=13

Wow. This was asked quite a while ago.

But thank you for posting an entirely new solution.


let $m \neq n$ be two real numbers such that $m^{2}=n+2$ and $n^{2}=m+2$. find the value of $4mn-m^{3}-n^{3}$.

Hint

$m^{2}=n+2\ \cdots [1]$
$n^{2}=m+2\ \cdots [2]$

$[1]-[2]\Longrightarrow m+n=?$ and $[1]+[2]\Longrightarrow m^{2}+n^{2}=?$, yielding $4mn-m^{3}-n^{3}=4mn-(m+n)(???)$.
How can be found the value of $mn$?

i have no idea where to start either. but this was a question from singapore math olympiad junior section 2006.

*going to take one in a few mins time :X*

[quote="pigley"]let m≠n be two real numbers such that m²=n+2 and n²=m+2. find the value of 4mn-m³-n³.[/quote]

[hide] Subtract the two equations to get, $m^{2}-n^{2}=n-m$, then by difference of two squares $(m-n)(m+n)=-(m-n)$ so $m+n=-1$
Then add the two equations to get $m^{2}+n^{2}=m+n+4=-1+4=3$

Now take what we are trying to find, $4mn-(m^{3}+n^{3})$. So we factor by sum of two cubes, which yields $4mn-(m+n)(m^{2}-mn+n^{2})=4mn-(-1)(3-mn)$

now we need to find $mn$. Since $(m+n)^{2}=m^{2}+2mn+n^{2}=1$, we can substitute to get that $3+2mn=1$, which means that $mn=-1$

So, substitute this back into $4mn-(-1)(3-mn)=3mn+3=\boxed{0}$[/hide]

That's right. You can also find the value of $m^{3}+n^{3}$ by $[1]*m+[2]*n$ in my post.


I'm feeling pretty bad since I didn't get to compete.  I got fifth in IN, so close to getting a big trophy.

I think that I would have made it in any other state with the 40 that I got at state.  Say your state and if I would have made it and what place I would have come in at state

OH - first written, not by much at all though :)

The fourth guy from IN got 54th at Nats, so...

Yeah I know, I was there.  I know them, I just want to see how I might have done because me and him have our on anf off days (we practice together) sometimes he beats me, sometimes I beat him, he beat me at state though

You're pretty sour about not making it to Nats aren't you?

It's okay. Nationals is cool, but it's not everything. In the long run, Mathcounts really doesn't count for much, if anything at all. Just say to yourself "Darn it, I'll do better next year" and you'll be fine.

mods: delete this topic.  I wanted a poll but didn't know how when I first did the topic

Seriously, you gotta chill a little. It sounds like you know more about the competition than the competitors themselves, and you know tons of stuff about your teammates personal lives. Also, it's weird how you were the one who said Indiana should've won even though you're not on their nationals team. You're getting too obsessive over not making it. I'll repeat, it's not the end of the world. You beat everybody in your state competition except for 4 people. That's pretty damn good.

So what if you would've made it to nationals in another state or beaten other people at nationals? What happened happens, and there's no way to change it. You can complain and whine all you want, but that's not going to do anything. So keep complaining if you want, but just remember, it's doing squat for you.

I'll try to control myself. THanks for the advice though.
(please don't be so negative to me  :? )

i know how u feel (although i didnt miss as closely as u did-i even won a dvd player in the raffle but i was still  :( ). i barely missed CD last year and im good at it but anyways.. like yifman said dont get down on yourself. Mathcounts is just ---- Mathcounts
its fun, yea, but you shouldn't get obsessiv over it. Mathcounts is a great competition but it isnt the only one out there. There are tons of high-school competititions that are just as prestigiuos or even more prestigious. If you are in 8th grad eu have 4 years to prove yourself. If youre 7th grade u can make nats next year. Don't complain about what happened w/ should've could've would'ves. You cant change it. Just focusing on working harder in the future. and i dont think yifman was trying to b e mean. he was just giving you the truth.


Right now my references come out looking like this "(name, year)" and I'd like "name(year)".  I use the '\cite' command.  \citeA, \citeN, and \citet don't work.  I'm using Texshop on a mac.  Do you think it has to do with the style files I'm using?

Thanks.

tim

You will need to give a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url] as there's too little to go on. \citeA, \citeN are for the Chicago bibliography syle. The  [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=natbib]natbib package[/url] provides the\citet command to give name (year).


In one step it is possible to exchange a number $ a$ drown on a blackboard with $ 1\minus{}a$ or $ \frac{a}{2}$. Prove that for any given $ a$ it is possible to obtain on a blackboard some number $ b$ which differs form $ \frac{1}{1997}$ not more that on $ 2^{\minus{}1997}$.

This one has been posted recently...

Pierre.

Here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=181196

Pierre.


Its kind of sad how slow everyone reacted to the situation..even the media didnt cover much untill very recent.

[quote="bleumoose"]a. yes they did, maybe not where you live, but where i live, there has been hurricane katrina coverage all over the front page since tuesday (storm hit on monday).
b. and even if there hasn't been much, that would be because the city is 20 ft underwater, theres a huge threat of disease, and people are getting SHOT AT. its not exactly reporter-friendly.[/quote]

The Iraqi War is not very reporter friendly first of all, but people still go there because its their job. I think by slow reaction, Gobraves meant that it took forever for the US troops to come they came like 4 to 5 days after the storm which was partly Bush's fault.

yeah...it sucks that new orleans is below sea level and the dikes arent strong....

How can I believe that a country can rescue people on the other end of the world, bring them freedom and democracy, but unable to save its own people?
Hail, selfless USA.

Wow, great point.

It seems like America never expected the devastation to be so immense. They are saving people right now. I think they are using buses to transport people from the superdome to the astrodome, and that's filled, so they are transporting them to other places.

People are getting rescued in both countries. That's a good thing.

[quote="hwenterprise"]It seems like America never expected the devastation to be so immense.[/quote]I don't want to "play the blame game" as former president Bush said on CNN yesterday, but [b]c'mon[/b]!

There are tons of reports by jurnalists and experts in disaster fields warning about what a 5th degree hurricane can do to New Orleans. They predicted exactly what just happened.

I watched on CNN before Katrina striked Lousiana the director of FEMA and the director of HomeLand Security stating [b]"we are more than prepared for this"[/b], and also stating that [b]"troups are standing by ready to come in with help and relief in case people will need it"[/b].

After [b]4 days[/b] the same directors appear and state that [b]"we didn't know that there are so many people that didn't evacuate"[/b]. I only wonder what did they actually do those 4 days?

[quote="hwenterprise"]They are saving people right now. [/quote]This is the problem. Those people should have been saved a long time ago. Some of them died :( before they could be saved. Some are still desperately asking for help, and help is promised on TV, but it doesn't get there!

Does anyone else belive that this is more of a "human made disaster" rather than a simple "natural disaster"?

[quote="Valentin Vornicu"]Does anyone else belive that this is more of a "human made disaster" rather than a simple "natural disaster"?[/quote]

It is definetly a natural disaster to [i]begin[/i] with, but I think that more of the later part of the storm is man-made. If the victims were saved a while ago, not as much people would've died (what is the death toll now?).

Surely, it was a natural thing that the city was below sea level, but that doesn't have anything to do with quick rescue. That's how a disaster becomes worse than it is. Late efforts to save victims.

"Jumping across the Northern Sea, Mountain Tai under your arm — if you say, ‘I am unable,’ that is really being unable.
Splitting wood for the elderly — if you say, ‘I am unable,’ that is not acting, it is not being unable."
--- Mencius

I've watched a lot on CNN the coverage on Katrina, as I told you even before the hurricane hit the Golf Coast, as well as other media coverage of this from around the world. What I've gathered is that basically there is no death toll yet, as the search and rescue operations are still underway. However it is estimated that the vast majority of people which passed away died not because of the storm itself, but of its repercursions. So all those people could have been saved if the authorities (either local or federal) would have provided the means to evacuate all the people [b]before[/b] Katrina hit New Orleans, or, at least, rigth after it hit.

Let's remember that search and rescue on a mass scale began [b]4[/b] days [b]after[/b] Katrina hit ...

I agree with Valentin, but about this:

[quote="Valentin Vornicu"]Does anyone else belive that this is more of a "human made disaster" rather than a simple "natural disaster"?[/quote]

I don't agree, because you have to give them [i]some[/i] credit. If they didn't do anything, then it would be even worse. They didn't cause the disaster, the storm did. They did have the power to do more, but they didn't. That does NOT mean that they caused the disaster, just that they could have helped more. A  LOT more, actually. "The Bush administration is SO good! They took away Suddam Hussein and are helping Iraq!" Okay, that's nice...but it must be really embarassing to have this kind of disaster in your own country and claim to be helping the rest of the world. The priority should be the country, not other countries!!

Just what [i]I[/i] thought.

[quote="Valentin Vornicu"]So all those people could have been saved if the authorities (either local or federal) would have provided the means to evacuate all the people [b]before[/b] Katrina hit New Orleans, or, at least, rigth after it hit. [/quote]

[quote="Ben Stein"]If able-bodied people refuse to obey a mandatory evacuation order for a city, have they not assumed the risk that ill effects will happen to them?

When the city government simply ignores its own sick and hospitalized and elderly people in its evacuation order, is Mr. Bush to blame for that?[/quote]

[quote="State of Louisiana disaster plan, page 13, paragraph 5, January 2000"]The primary means of hurricane evacuation will be personal vehicles. School and municipal buses, government-owned vehicles and vehicles provided by volunteer agencies may be used to provide transportation for individuals who lack transportation and require assistance in evacuating.[/quote]

[url=http://maps.google.com/maps?q=New+Orleans,+LA&ll=29.968598,-90.089189&spn=0.006110,0.009958&t=e&hl=en]Would you like to see where those buses were?[/url]
Notice the date that picture was taken. Now zoom in... There are about 150 buses there, which figures to about... NINE THOUSAND PEOPLE.

[quote="shreveport times"]But 27 percent of the city's households don't own a car -- a problem that doesn't appear to have been addressed in advance planning models.

The "untested" emergency response system was supposed to be a system of public buses that would transport people out of the city in the event of a disaster.

Instead, buses took people to the Superdome, which was supposed to be a staging point from which to transport people out of the state.

But nearly a week after the hurricane had hit ground, thousands of people were still waiting for buses to take them out of the city. By then, the Superdome had become uninhabitable as the air conditioning cut off, running water stopped and toilets backed up.[/quote]

All I'm saying is, be careful who you blame for things. And please remember that, "right after" the hurricane hit, there was no physical way to evacuate large amounts of people from the city.
And perhaps you didn't know this, but FL, LA, and MS were declared natural disaster areas before Katrina hit shore. AL was declared a natural disaster area on Tuesday morning (or Monday afternoon, I forget). Is that slow?

On a (sort of) related topic, I wonder how many people here have New Orleans kids going to their school now. We have one probably, and another maybe. And I wonder if anyone was planning on applying to Loyola or Tulane... and had a change of plans?

A lot of the displaced kids from colleges are being offered potential spots at various universities around the coutry, tuition-free. Check out some of the http://www.collegeconfidential.com posts on the subject, especially for colleges like Tulane.

[quote="noneoftheabove"]I have been thinking and i am not sure if it is smart to rebuild at the same location because it is in a bowl and the fact that it will erode in about 50 years.[/quote]I belive the Dutch have been living like this for hundreds of years and they are doing pretty well. Maybe Mr. Bush should ask them how they do it?  :)

By no means am I suggesting my views agree with beliefs of rapper Kanye West, but what I am pointing out or rather find amusing, is that someone who is bound and famous for something completely different, can have such a strong, uncensored voice on political and social issues, through his influence of music and un-scipted outbursts, as the one on NBC.  :lol:

[quote="GoBraves"][quote="bleumoose"][quote="white_horse_king88"]West was out of line, but then you think about it and say: Why on the heavenly planet earth would marines with assault rifles be doing patrolling the streets in armored tanks instead of spending time in a search and rescue? Who cares if they loot a little? There are lives at stake!!! The government was terrible in priorities...[/quote]

Gosh, I dunno... maybe it's because gang members have pretty much taken over the city and are terrorizing people and shooting rescue workers... do you think that could maybe be it?[/quote]

No I dont think that maybe it, they took over the city so they could get food, and I dont blame them. Im sure that they would rather get rescued by the marines than rather stay there and continue to roam the place with guns and shoot people who are trying to rescue them.[/quote]

[i]You[/i] don't think they would do that.  But you don't have their mentality.  The people left were/are the poor, lower class people for the most part.  They just let people loose from jail.

There were people looting stores for luxury items.  For example, one person hoarded shoes he got from a shoe store.  He had like 20 pairs.  Taking one pair is justifiable, but 20?  People were definitely looting, not just taking what they [i]need[/i].  Not everyone was this way, but there were enough to make a huge difference.

and does it REALLY make logical sense that someone who would "rather be rescued" would be roaming around with (stolen) guns, shooting at rescue workers? come ON.

Google's tipping in on the blame game here: http://www.google.com/search?hl=en&q=failure&btnG=Google+Search  :lol: and Yahoo and MSN are seconding the motion at http://search.yahoo.com/search?p=failure&sm=Yahoo%21+Search&fr=FP-tab-web-t&toggle=1&cop=&ei=UTF-8 and http://beta.search.msn.com/results.aspx?q=failure&FORM=QBHP

The Yahoo one didn't work, and the MSN one seemed to backfire.  This whole failure thing is really old; I remember searching for "[url=http://www.whitehouse.gov/president/gwbbio.html]miserable failure[/url]" something like two years ago.

[quote="joml88"]
There were people looting stores for luxury items.  For example, one person hoarded shoes he got from a shoe store.  He had like 20 pairs.  Taking one pair is justifiable, but 20?  People were definitely looting, not just taking what they [i]need[/i].  Not everyone was this way, but there were enough to make a huge difference.[/quote]

White Horse King wasn't saying they were looting for survival, he was just saying that the government shouldn't be expending resources trying to stop these people while there are lives at stake. I agree, they need to get their priorities straight.

They tried that.  But the city was in a state of basically anarchy!  Look, they went to New Orleans.  They started rescuing people.  But then they experienced a massive amount of resistance from people remaining in the city.  It went to the point that the government had to do something.  So they clamped down.

You had people shooting at rescuers.  People were not cooperating with the rescuers.  They were not able to get their job done.  Therefore you send in reinforcements to reduce the resistance.

I really don't think that this part of the recovery process was what should be criticized.  There are certainly many other things that you could critique the Federal and Local governments for doing/not doing.

Yes people were looting and shooting and stuff. It doesn't matter. There were people who wanted to be rescued, and they should have been rescued first. If there were people who didn't want to be saved, to hell with them. Don't try and police them until all the others are safe.

[quote="bleumoose"]Like joml said, that makes absolutely [i]no sense.[/i] Here, let me give you just an example of what was going on. There was this hospital full of people that needed to be rescued (patients that couldn't get out of town for some reason or another). So rescue workers were going to rescue them... but they had to hold off, because OTHER PEOPLE were shooting at them. What do you suggest the army do?... They did what the National Guard's JOB is to do- they went down there and policed WHILE rescue workers were doing their job. Do you not understand that?[/quote]

Why the hell would other people shoot at the rescue workers when they are trying to rescue them??!! Either you are really stupid, or the people from New Orleans are really stupid, and Im sure that no human is that stupid to say no when they are in 15 feet of contaminated water when they are being asked to be rescued and instead shoot them.

[quote="GoBraves"][quote="bleumoose"]Like joml said, that makes absolutely [i]no sense.[/i] Here, let me give you just an example of what was going on. There was this hospital full of people that needed to be rescued (patients that couldn't get out of town for some reason or another). So rescue workers were going to rescue them... but they had to hold off, because OTHER PEOPLE were shooting at them. What do you suggest the army do?... They did what the National Guard's JOB is to do- they went down there and policed WHILE rescue workers were doing their job. Do you not understand that?[/quote]

Why the - would other people shoot at the rescue workers when they are - trying to rescue them??!! Either you are really stupid, or the people from New Orleans are really stupid, and Im sure that no human is that stupid to say no when they are in 15 feet of contaminated water when they are being asked to be rescued and instead shoot them.[/quote]

Do you live in a bubble?  This isn't some fairy tale world where everyone lives rationally and does what they are supposed to for society to succeed.  Go watch the news.  Do a search for New Orleans (and find something like [url=http://www.cnn.com/2005/WEATHER/09/01/katrina.impact/]this[/url], for example).  Please stop talking about stuff that you don't even have a clue about.

And for those people wondering why the U.S. denied help from other countries..Let me tell you this. I can guarantee that under no circumstances will Bush and his administration accept any sort of help from a country because they don't want to be in a position where they owe somebody. It would hurt the "ego" that they have built up. The "greatest" country in the world, trillions of dollars in debt, hit by one of the worst hurricanes in history, and too good to accept help. sigh.......[/quote]

Humans are naturally irrational. I also saw that bit of news, in which the helicopters carrying patients were being fired at. That's just tragic. I'm sorry the national guard wasn't there, but it's apparent that some people don't want to be rescued. However, the fact is, the government is expending resources to secure locations in parts of New Orleans that have been overrun by anarchy - the lootings and shootings. It's all true. However, if those people don't want to be rescued, the government shouldn't interfere. Whereas, all those people at the Covention Center and the Superdome are in need to rescue - the government isn't spending enough resources there. And let me tell you, those places are also hubs of anarchy. I agree that some mediation by the military is required. But the government needs to spread resources wisely. I see tanks patrollying vacant streets, or tanks patrolling streets filled with looters; but I don't understand why the government bothers. The people there don't want rescue, so the troops are putting themselves in danger to control a bit of unlawful behavior while neglecting to bring control to other portions of New Orleans, where thousands are gathered in wait of transport.


When I complie from latex (using winedit) to DVI or PDF, I always generate two extra blank lead pages in the file. Can anyone suggest how I might fix this?

Thank you.

Remove the lines that say \newpage from your TeX file? You might also get a blank page if you have \maketitle or \tableofcontents without \title, \author, \section, or \subsection.

Can you post the TeX source (or at least part of it) so that we can see what the problem is?


[quote="warut_suk"]Let A,B,C be an acute triangle and D,E,F be the feet of the altitudes from A,B,C respectively. Let P,Q,R be the feet of the perpendiculars from A,B,C to EF,FD,DE respectively. Prove that the lines AP,BQ,CR are concurrent.[/quote]
They are concurrent at circumcenter of ABC :D

[hide]By cyclic quads, $ \angle EDC \equal{} \angle A$, and thus, $ \angle DCR \equal{} 90 \minus{} \angle A$, so so $ \sin \angle DCR \equal{} \cos A$. However, Similarly, $ \cos QBD \equal{} \cos A$. We can repeat this argument for the other angles of this triangle, and use trig ceva to finish this off.

Let $ O$ be the point of concurrency. We see that $ \angle COB \equal{} 2 \angle C$ (since $ \angle OCB \equal{} \angle OBC \equal{} 90 \minus{} \angle C$. This holds cyclically, so $ O$ is the circumcenter of $ ABC$, as DangChienbn has mentioned. [/hide]

Dear Mathlinkers,
you can see
http://jl.ayme.pagesperso-orange.fr/,  vol. 3  Cinq theoremes de von Nagel p. 21-22.

Sincerely
Jean-Louis


Solve the equation in natural numbers:
$x^3-p^3=z^2$, with $p$-prime, and $3,p \not | $ $ z$.

Thank you very much, but I can't finish it. Could you please write down these cases? :?

of course $gcd(k,l)=1$,$gcd(x,p)=1$
case 1$gcd(l-k^2,l+k^2)=1$
case 1.1
$3=l-k^2$
$xp=l+k^2$
case1.2
$p=l-k^2$
$3x=l+k^2$
case1.3
$3p=l-k^2$
$x=l+k^2$
case1.4
$x=l-k^2$
$3p=l+k^2$
...
it isn't smart solution i think now you can see next cases

but you haven't showed that there cannot be situation that $(x,l-k^2)>1$ and $(x,l+k^2)>1$


geometry
rectangle
combinatorics proposed
combinatorics

A $ 100 \times 100$ square floor consisting of $ 10000$ squares is to be tiled by rectangular $ 1 \times 3$ tiles, fitting exactly over three squares of the floor.
$ (a)$ If a $ 2 \times 2$ square is removed from the center of the floor, prove that the rest of the floor can be tiled with the available tiles.
$ (b)$ If, instead, a $ 2 \times 2$ square is removed from the corner, prove that such a tiling is not possble.

For part (a), we divide the board in $ 4$ rectangles $ 49 \times 51$, as shown in the diagram below. Since $ 3$ divides $ 51$, these rectangles can be easily tiled.

For part (b), we label all 10000 squares on the board like this:

A B C A B C A B C A ...
B C A B C A B C A B ...
C A B C A B C A B C ...
A B C A B C A B C A ...

No matter how a tile is placed, it always covers one "A" square, one "B" square and one "C" square.
Initially, there are 3334 squares "A", 3333 squares "B" and 3333 squares "C".

If a board can be tiled, it must have the same number of squares of each type.
However this is not the case when we remove a $ 2 \times 2$ square from the up-left or down-right a corner, since those squares have just one A square.
To prove that the other corners don't work either, we use a slightly different labelling:

A B C A B C A B C A ...
C A B C A B C A B C ...
B C A B C A B C A B ...
A B C A B C A B C A ...

Hence, in this case it's not possible to tile the board.


superior algebra
superior algebra unsolved

there is no group G such that G`=[S=4].(hint:S4 is permutation groups and  G` is commutator of G). :blush: [/hide]


Solve if $ a,b,c,d,e \in \mathbb {N}$

and

$ \frac{1}{a}\ \plus{} \frac{1}{b}\ \plus{} \frac{1}{c}\ \plus{} \frac{1}{d}\ \plus{} \frac{1}{e}\ \equal{} \frac{7}{3}$

Without loss of generality, assume $ a\ge b\ge c\ge d\ge e$. Then the fractions are in reverse order. Let the sum on LHS be $ A$. Now $ A\le\frac5e$, so $ \frac73\le\frac5e$, so $ e\le 2$. Now consider cases $ e\equal{}1,e\equal{}2$, and then in the same way narrow down $ a,b,c,d$


induction
inequalities
number theory solved
number theory

Let k>2 integer prove that

p(k+1) + p(k+2) <= p(1).p(2)....p(k)

where the p(i) are primes numbers such that

2 = p(1)<p(2)<....<p(k)<p(k+1)<p(k+2)

Maybe induction on n ?

Suppose p(n + 1) + p(n + 2) < p(1) p(2) ... p(n) for 3 \leq n \leq k.
We'll prove that p(n + 1) + p(n + 2) < p(1) p(2) ... p(n) for n = k + 1.

We have p(k + 1) + p(k + 2) < p(1) p(2) ... p(k).
We want to prove that p(k + 2) + p(k + 3) < p(1) p(2) ... p(k) p(k + 1).
Now we have p(k + 2) + p(k + 3) < 3 p(k + 2) by Bertrand's postulate.
Set p(1) p(2) ... p(k) = P.
Then it suffices to prove that 3 p(k + 2) < P p(k + 1).
But we have p(k + 2) < P - p(k + 1) by the induction hypothesis.
So 3 p(k + 2) < 3P - 3 p(k + 1) < 3P < P p(k + 1) since p(k + 1) > 3.

This inequality is pretty weak, isn't it ?

This famous statement have been proved before Bertran's postulate was.

The original proof is following.

If n>=3 then m=p1*p2*p3*...*pn>p1*p2*p3>6.
Every number >6 is sum of two nonunit coprime numbers: m=a+b, a,b>1 (prove it!).

(a,b)=1 ==> (m,a)=(m,b)=1 ==> if p is prime and p|a, q is prime and q|b then p,q are not in {p1,...,pn} and m=a+b>=p+q>=p{n+1}+p{n+2}


function
inequalities
floor function
ceiling function
combinatorics unsolved
combinatorics

[b]Problem 3.[/b] Let $n\geq 3$ is given natural number, and $M$ is the set of the first $n$ primes. For any nonempty subset $X$ of $M$ with $P(X)$ denote the product of its elements. Let $N$ be a set of the kind $\ds\frac{P(A)}{P(B)}$, $A\subset M, B\subset M, A\cap B=\emptyset$ such that the product of any 7 elements of $N$ is integer. What is the maximal number of elements of $N$?

[i]Alexandar Ivanov[/i]

For each prime $p$, let $N_{p}$ be the number of elements of $M$ with $p$ dividing the denominator, and let $Z_{p}$ be the number of elements of $M$ for which $p$ dvides neither the numerator nor the denominator. If $|N_{p}| \geq 1$, then we must have $|N_{p}|+|Z_{p}| < 7$, and

$-|N_{p}|+(7-|N_{p}|-|Z_{p}| \geq 0$
$2 |N_{p}|+|Z_{p}| \leq 7$
$|N_{p}|+|Z_{p}| \leq 6$.

So if we let $k$ be the number of primes that appear in the denominator of any of the elements of $M$, for each of the $k$ primes  there are at most 6 elements for which the prime does not divide the numerator;  hence, letting $f(n)$ be the maximum cardinality of $M$ for a specific value of $n$,

$f(n) \leq 6k+2^{n-k}, 0 \leq k \leq n$.

Note that the above is a convex function of $k$, so the function is maximized when $k = 0$ or $k = n$.  This gives $f(n) \leq 2^{n}$ and $f(n) \leq 6n$ respectively;  for $n \geq 5$, $6n \leq 2^{n}$, so $f(n) \leq 2^{n}$.  But we can achieve $2^{n}$ elements by simply letting $M$ be all the integer products of the $n$ primes.  Ao $f(n) = 2^{n}$ for $n \geq 5$.

The cases $n = 3$ and $n = 4$ require a more detailed bound.  Summing the inequalities $2 |N_{p}|+|Z_{p}| \leq 7$ over all primes $p$ with $|N_{p}| \geq 1$, we get

$2 \sum_{p \in S}|N_{p}|+\sum_{p \in S}|Z_{p}| \leq 7k$ (1)

where $S$ is the set of primes that appear in the denominators of the elements of $M$. Now, for any element $m$ of $M$, $m$ is of the form

$m = \prod_{i=1}^{a}p_{i}^{1}\prod_{j=1}^{b}q_{j}^{0}\prod_{k=1}^{c}r_{k}^{-1}$,

and for each prime $q_{j}$, $m$ lies in $Z_{q_{j}}$, and for each prime $r_{k}$, $m$ lies in $Z_{r_{k}}$.  So $m$ contributes $b+2c$ to the left and side of equation (1).  Set $g(m) = 2b+c$.  There are $2^{n-k}$ elements with $g(m) = 0$, $2^{n-k}k$ elements with $g(m) = 1$, and $2^{n-k}(k+\binom{k}{2})$ elements with $g(m) = 2$.

For $n = 4$, we have that $6k+2{n-k}\leq 16$ for $0 \leq k \leq 2$.  Let $h(i)$ be the sum of the smallest $i$ values of $g(m)$.  For $k = 3$, we have $2, 6, 12$ elements with $g(m) = 0, 1, 2$, so $h(16) = 2(0)+6(1)+8(2) = 22 > 21$.  So $M$ can have most 15 elements in this case.  For $k = 4$, we have $1, 4, 10$ elements with $g(m) = 0, 1, 2$, so $h(17) = 1(0)+4(1)+10(2)+2(3) = 30 > 28$.  So we can have at most 16 elements, and $f(4) = 16$.

Now for $n = 3$.  For $k = 1$, we have $h(8) = 4(0)+4(1) = 8 > 7$.  For $k = 2$, we have $h(12) = 2(0)+4(1)+6(2) = 16 > 14$.  For $k = 3$, we have $h(13) = 1(0)+3(1)+6(2)+3(3) = 24 > 21$.  So the number of elements cannot be more that 12.  We can achieve this with ${M = \{2, 3, 5, 6, 10, 15, 30, 15/2, 10/3, 6/5, 3/2, 2/3}$.

Just to finish things off, obviously $f(1) = 3$;  for $n = 2$, we cannot take all 9 possible elements, but removing $1/6$ is sufficient, so $f(2) = 8$.

Hint. Every prime $p_{k}$ is maximum in 3 denominators. Thus, $|N|\leq 3n$.

Hint. On the competition many good participants got result $\left\lfloor \frac{8n}3 \right\rfloor$ which is [b]NOT[/b] the answer.

Okay, it looks like you are requiring that the numbers are non-integral.   :(

The formula $2 |N_{p}|+|Z_{p}| \leq 7$ that I derived above shows that $|N_{p}| \leq 3$.  Suppose first that not all primes appear in the denominator.  Then $|N| \leq 3n-3$.  I claim this is in fact achievable.  Let $P = \prod_{i = 1}^{k}p_{i}$.  For each $p_{k}$, $k = 1,2, \ldots, n-2$, let

$S_{k}= \{\frac{P}{p_{k}^{2}}, \frac{P}{p_{k}^{2}p_{k+1}}, \frac{P}{p_{k}^{2}p_{k+1}p_{n}}\}$

and

$S_{n-1}= \{\frac{P}{p_{n-1}^{2}}, \frac{P}{p_{n-1}^{2}p_{1}}, \frac{P}{p_{n-1}^{2}p_{1}p_{n}}\}$.

Then $M = \bigcup_{i=1}^{n-1}S_{i}$ satisfies the requirement.

If all primes appear in the denominator, then let $S_{1}, S_{2},$ and $S_{3}$ be the number of primes that appear in $1,2,$ and $3$ denominators respectively, and let $n_{i}= |S_{i}|, 1 \leq i \leq 3$.  If $p$ is in $S_{3}$, then for at least two of the elements with $p$ in the denominator, there is another prime $q$ such that $q$ does not divide the numerator.  On the other hand, there is at most one other element for which $p$ does not divide the numerator. (from $2 |N_{p}|+|Z_{p}| \leq 7$) If $p$ is in $S_{2}$, then there is at least one element for which a prime $q$ does not divide the numerator, and at most 3 elements for which $p$ does not divide the numerator.  If $p$ is in $S_{1}$, there is at most 5 elements for which $p$ does not divide the numerator.
  Thus we have

$n_{3}+3 n_{2}+5 n_{1}\geq 2n_{3}+n_{2}$
$2 n_{2}+5 n_{1}\geq n_{3}$
$3 n_{2}+6 n_{1}\geq n$.

Thus $|N| = 3n_{3}+2n_{2}+n_{1}= 3n-n_{2}-2n_{1}\leq 3n-\frac{n}{3}= \frac{8n}{3}$.

Again, this bound is achievable for $n \geq 4$.  Let $m = \lfloor \frac{2n}{3}\rfloor$.  For $k = 1,2, \ldots, m-1$, let

$S_{k}= \{\frac{P}{p_{k}^{2}}, \frac{P}{p_{k}^{2}p_{k+1}}, \frac{P}{p_{k}^{2}p_{m+\rceil \frac{k}{2}\lceil}}\}$

and

$S_{m}= \{\frac{P}{p_{m}^{2}}, \frac{P}{p_{m}^{2}p_{1}}, \frac{P}{p_{m}^{2}p_{m+\rceil \frac{m}{2}\lceil}}\}$.

For $k = m+1, m+2, \ldots, n-1$,

$S_{k}= \{\frac{P}{p_{k}^{2}}, \frac{P}{p_{k}^{2}p_{k+1}}\}$

and

$S_{n}= \{ \frac{P}{p_{n}^{2}}, \frac{P}{p_{n}^{2}p_{m+1}}\}$.

This satisfies the required condition.

For $n \geq 7$, we have $3n-3 \geq \lfloor \frac{8n}{3}\rfloor$.  So $f(n) = 3n-3$ for $n \geq 7$, and $f(n) =\lfloor \frac{8n}{3}\rfloor$ for $4 \leq n \leq 9$.

For $n=3$, we can let $M = \{3/2, 5/2, 15/2, 2/3, 5/3, 10/3, 6/5 \}$.  This is optimal;  to do better, we must have $n_{3}= 2$ and $n_{2}= 1$.  But for $p, q$ in $S_{3}$, we have $p$ appearing in three denominators, but at most twice in numerators of elements  with $q$ in the denominator. Similarly for $q$.  So the product of the six elements has $p$ and $q$ in the denominator, so any remaining elements must have both $p$ and $q$ in the numerator.  That leaves room for only one other element, $\frac{pq}{r}$.  So seven elements is maximal.


Use complex numbers - isn't it sort of... obvious?

If you want the full proof
[hide]
$P(x) = (x + 1)\left((x^2 + 1)^{\ell}(x - 1)^{\ell} + x^{4\ell - 5}\right)$ where $\ell = k + 1$.

Now let $\omega \neq 1$ be a solution of $\omega^5 = -1$.

Then since $(x^2 + 1)^{\ell}(x - 1)^{\ell} = (x^3 - x^2 + x - 1)^{\ell}$ and $\omega^4 - \omega^3 + \omega^2 - \omega + 1 = 0$, we have

$P(\omega) = (\omega + 1)\left(\omega^{4\ell} + \omega^{4\ell - 5}\right) = (\omega + 1)\omega^{4\ell - 5}(\omega^5 + 1) = 0$

and we are done. [/hide][/hide]


ceiling function
number theory
prime factorization

Is there a quick way to find the sum of the factors of a number (other than finding them all and adding)?

Yes.It's hard to explain, so let me give you an example:
Prime factorize the number, for example 500. This is $2^2*5^3$, so the sum of its factors is $(2^0+2^1+2^2)(5^0+5^1+5^2+5^3)=1092$.
Another example: The prime factorization of 14553 is $3^3*7^2*11^1$, so the sum of its factors is $(3^0+3^1+3^2+3^3)(7^0+7^1+7^2)(11^0+11^1)=27360$.

To add to that, does anybody want to get the formula for the product of the roots?

[hide="hint"] How many factor-pairs will each number have and what will that mean?[/hide]

Um...I am not too good at thinking so....um...does anybody mind telling me the second theorum thingy.....

Isn't there also a formula for the number of factors?

[quote="kyyuanmathcount"]To add to that, does anybody want to get the formula for the product of the roots? [/quote]
[hide="Is this right?..."]For example, 12 has 1,2,3,4,6,12 as divisors. Pair them up, 1 with 12, 2 with 6, 3 with 4. Notice how each pair has a product of 12, there are 3 pairs, so the product of the divisors is $12^3$.

General case: N has the divisors
$d_1,d_2,\cdots d_{n-1},d_n$ Pair them up with the reverse,
$d_n,d_{n-1}\cdots d_2,d_1$

The number of pairs is equal to the number of divisors, and each divisor appears twice.
So $\sqrt{N^{\text{(number of divisors)}}}$ gives the product of the divisors. (???)[/hide]

[quote="pianoforte"][quote="kyyuanmathcount"]To add to that, does anybody want to get the formula for the product of the roots? [/quote]
[hide="Is this right?..."]For example, 12 has 1,2,3,4,6,12 as divisors. Pair them up, 1 with 12, 2 with 6, 3 with 4. Notice how each pair has a product of 12, there are 3 pairs, so the product of the divisors is $12^3$.

General case: N has the divisors
$d_1,d_2,\cdots d_{n-1},d_n$ Pair them up with the reverse,
$d_n,d_{n-1}\cdots d_2,d_1$

There are n pairs, and each divisor appears twice.
So $\sqrt{N^{\text{(number of divisors)}}}$ gives the product of the divisors. (???)[/hide][/quote]

Actually, you need to put the number of divisors over two, because if there are n divisors, there are n/2 pairs that each multiply to the number.

Plus, you can find the number of factors.  Eg: 500 is 5 cubed times 2 squared.  The number of factors is the number of the exponent plus one times all of the other numbers of the exponents plus one.  5 3power+1=4 2 2power+1 is 3 4*3 is 12.

[quote="pianoforte"]Okay, but
[hide]$N^{\frac{number of divisors}{2}}=N^{(number of divisors)*1/2}=\sqrt{N^{number of divisors}}$[/hide]
so it's pretty much the same thing. But dividing by two first is easier, so yeah...[/quote]

Huh?  How is division by 2 equal to a square root?  I don't get it.

$\sqrt{n}=n^{1/2}$.

Generally, a number to the n-root is equal to that number to the power of 1/n.

My bad.  I didn't realize that $N^{\frac{number of divisors}{2}}=N^{(number of divisors)*1/2}=\sqrt{N^{number of divisors}}$ was a power, I thought it was simple multiplication.

For perfect square numbers, the number of factors is always odd. Thus, you cannot

divede by two there. You must add one first!!!!!!!!!!!

for example, take 64 = 2^6 (7 factors).........

You don't have to divide add one with a square number. With four it is $1\times2\times4=8=4^{\frac{3}{2}}$

In general, including squares, it should be $\sqrt{N^{\lceil\frac{\text{number of divisors}}{2}\rceil}}$


How did you get your third equation? This is confusing...

P.S. I don't know how to prove it, either. I graphed the two equations. I was curious as how to prove it algebraically.

Well, $1-x^{3}\ge 1-x$ because $x \le x^{3}$...

Then $x^{3}+y^{3}+1-x^{3}\ge x-y+1-x \Rightarrow y^{3}+1 \ge-y+1$

Aha... that's the problem  :D

EDIT:
Now, after thoroughly *ahem* reading the problem, you think that it only works inside a strangely defined circle. :wink:

That's pretty interesting...

[quote="towersfreak2006"]Are you sure the problem is correct? $x=y=0$ gives \[x^{3}+y^{3}=x-y,\] but \[\left(x-\frac{3}{5}\right)^{2}+y^{2}>\frac{2}{5}.\] [/quote]

actually, $\frac{9}{25}<\frac{10}{25}=\frac{2}{5}$

Yes. If $x=y=0$, then the inequality says:
$\frac{9}{25}\leq \frac{2}{5}$, which is true.

Whoa, for some reason, I thought it was $\frac{9}{5}$.  :blush:

[quote="not_trig"]Can $x^{3}+y^{3}= x-y$?

We know that $x^{3}\le x$,
so \[1+y^{3}\ge 1+y \Rightarrow y^{3}\ge y\] and $y \ge 1$ and $x-y \le 0$...

I'm probably wrong though. Please point out my error :)[/quote]

Isn't x-y>0 because x^3 and y^3 are positive?


calculus
derivative
calculus computations

Using the Substitution method, find the maximum value of

4x-2xy+3y subject to the constraint 4x-y=2

I can do the 1st part:

4x-y=2--->
4x-2=y

substitute this into the original equation:
4x-2x(4x-2)+3(4x-2)

Hope that right so far! but dont know where to go from there? any one could help plz?

thanks ;)

After substitution, you have an ordinary calculus one-variable optimization problem. You should know how to do that.

Your substitution is correct, so just take the derivative which comes out to $20-16x$, and set it equal to zero to obtain $x=\frac{5}{4}\Rightarrow y=3$, thus the maximum value of $4x-2xy+3y$ is $\frac{13}{2}$.


Let sequence  $ {u_{n}}$ with $ u_{n} \equal{} 1 \plus{} 1 \plus{} \frac {1}{2!} \plus{} \frac {1}{3!} \plus{} \frac {1}{4!} \plus{} .... \plus{} \frac {1}{n!}$
Prove: Lim $ u_{n} \equal{} e$  with e=lim $ (1 \plus{} \frac {1}{n})^n$

It is easier to start from the other definition of $ e$.  Consider the binomial expansion of $ \left( 1 \plus{} \frac{1}{n} \right)^n$ and try to relate it to a sum of reciprocal factorials.

[quote="t0rajir0u"]It is easier to start from the other definition of $ e$.  Consider the binomial expansion of $ \left( 1 \plus{} \frac {1}{n} \right)^n$ and try to relate it to a sum of reciprocal factorials.[/quote]
Sorry, I don't understand :(

You have the expansion $ \left( 1 \plus{} \frac{1}{n} \right)^n \equal{} \sum_{k\equal{}0}^{n} {n \choose k} \frac{1}{n^k} \equal{} \sum_{k\equal{}0}^{n} \frac{1}{k!} \cdot \frac{n(n\minus{}1)...(n\minus{}(k\minus{}1))}{n^k} < \sum_{k\equal{}0}^{n} \frac{1}{k!}$.  This is one bound.  Can you find the other?


If anyones wants to mod this please pm me!
Signups begins
1. lotsofmath
2. ernie
3. cf249
4. Chompy
5. dragon96
6. westiepaw
7. bramblestar
8.
9.
10.
11
12.
13.
14.
15.
16.

I will join and die later than in egg toss 2.

inininniinnininininiinnininiinni

 :ddr:

Oooohhh, I join!!!

1. lotsofmath
2. ernie
3. cf249
4. Chompy
5. dragon96
6. westiepaw
7. bramblestar
8. spaceguy524
9. meewhee009
10. jjfun1
11
12.
13.
14.
15.
16.

1. lotsofmath
2. ernie
3. cf249
4. Chompy
5. dragon96
6. westiepaw
7. bramblestar
8. spaceguy524
9. meewhee009
10. jjfun1
11. EggyLv.999
12. Scampi!
13.
14.
15.
16.

Maybe we should just start now.
The game is about to die.

Let's see now...
[b]Throw at Chompy.[/b]

Chompy's dead :o!
8. spaceguy524
10. jjfun1
11. EggyLv.999

Eggy's Dead :o
8. spaceguy524
10. jjfun1
[size=200]FINAL TWO![/size]

Hehehe...
I knew you were going to throw at each other.
Sooo...
an easier way to determine this...
Pick a number between 1 and 10 :D

Bum, bum, bum...
THE NUMBER WAS 3.
Spaceguy has one the round (?)
Sorry jjfun.

SO NOW THE AWESOME SPACEGUY CHANCE O' LIFEEE!

Mkay.
So...
There are 2 houses in front of you.
A blue one, and a red one.
One has the door of life, where you will exit Egg and return to um... DisneyWorld.
And one has a rabid dog that will pwn you. Hard.
Choose wisely.
What would YOU like to egg?

I would like to egg the [b]red house[/b]

[size=59]I hope I choose the right one.[/size]

Jeez.
Why didn't I pick my lucky number?
And WHY do I always finish second?
Hmm. =[
Can I mod the next game? (If there is one that is)

YOU PICK THE RED HOUSE.
You throw your egg...
AND IT TURNS INTO A DOOR.
You then skip out the door...
and then you see Mickey Mouse...
only a sign of DisneyWorld :).
The End.
Of Egg Toss 3.
If anyone wants to mod the next game be my guest.


Looks like operatorname won't work anymore. I'm sure my LaTeX-code in http://www.artofproblemsolving.com/Forum/viewtopic.php?t=175350 was OK and none has changed it during couple of years but LaTeX does not render the code properly anymore.

Kent has already posted this question in the Support subforum about a month ago. Seems like Valentin thinks it is a minor issue (though I find it rather irritating exactly because of all those scrambled posts like yours). :(


Let $ABCD$ be a tetrahedron. $O$ is point in its interior, so that $OA=OB=OC=OD$. If  $d(O,(ABC))=d(O,(BCD))=d(O,(ABD))=d(O,(ACD))$ then it follows immediatly that ABCD is regular? If yes, i would really appreciate a proof.

it seems that all the areas of the faces of the tetrahedron are equal. try to prove it


Let $ ABC $ be a triangle with area 1 and let $ x $ be a number such that 0<$ x $<1.

Let $ A' $, $ B' $, and $ C' $ be points on $ BC $, $ CA $ and $ AB $ respectively such that:

$ BA' : A'C = CB' : B'A = AC' : C'B = (1-x) : x $

Find the area of triangle $ A'B'C' $ in terms of $ x $.

I have found the area to be $ 1-3x+3x^2 $, with an assumption which i cannot prove...  :( .

The assumption is that $ A'B'C = A'C'B = B'C'A $. Can anyone verify and/or prove me right or wrong??

ah.. there we are.. AMOC extension programme 4 set 2 problem 2..

u are in fact correct.. i remember doing the programme a year or two ago... and remember doing that question too!

Umm.. no, you can't assume they're equal but thats the case that i trialed and errored first when i did it..  Seeing as Mr. Ball generally allows a [i] little [/i] help...

a) try using A=absinC... (this is how i did it)or

b) try comparing the areas of the triangles within ABC (this is how they did it)


i found that eating carrots always helps. but i dont think you can take that into mathcounts

When I get nervous I chew on my corrective rubber band. But I rarely get nervous.

humbug. i did find that i had to go to the bathroom lots of times inbetween rounds of mathcounts. so yah, always a good idea to relieve yourself

ComplexZeta said that drinking water helped him, and it seems to be helping me too on practice tests (though I haven't gotten the chance to try on a real test).

He meant during the test.  You are allowed bottled water, I believe.

For things like AMC's, or tests where you take them at your school, its your school's decision... i dont think theres a hard rule saying you are or are not allowed to drink water.  For example. my school wont let us drink in classrooms, so we cant bring stuff in when we're taking those tests.

We can. In fact, teachers recommend kids to bring waters because dehydration can be a problem even up here.

I wish we could.

Im from MA too, but dehydration's never been a problem i dont think.

Well haven't you experienced a scorching summer day? Weather here can be [b]extreme[/b] and I know that from experience. (I walk between schools, and last year there were like a foot and a half inchs of snow+it was waaaaay below 0 [b]Fahrenheit[/b])
As for summer, there have been cases of dehydration when the drinking fountain was broken. Most times it's just fatigue or loss of appetite which is considered no big deal, so not a lot of people realize that some people are dehydrated.

Oh good.  We agree on something.   8-)   That's always a good thing.

usually before some big event you're always nervous, but once it gets started you calm down a lot.  Im not sure that anything will really help... you'll be nervous no matter what.  But it doesnt really matter since it'll go away during the test and you'll be fine.

I'm not usually all that nervous during math competitions until I finish the test, when I get extra time to think about which problems I probably messed up and which ones I should be checking. I get far more nervous when playing in chess tournaments.

[quote]I'm not usually all that nervous during math competitions [/quote]

Well, it is only logical that a MOPper would be less nervous when taking an AIME than another mathlete just struggling to barely make USAMO...

im not sure, but i think the key is that he said 'during' competitions.  This is probablly true for most people... the real anxiety usually comes before the test.  I remember watching on some show (dateline i think), the empire that Roberto Alomar spit on was returning to umpiring after a year off or something like that.  His heart rate was really really high before the game started, but after the first pitch it went down a lot, cause he was in familiar territory and he was confident that he knew what he was doing.  If you prepared well for a test and are confident in your abilities, the nervousness will no doubt subside after the test has started.

No matter how good you are at anything, im sure you'll still be nervous before a big event... for example, i guarantee you professional athletes are pretty nervous before big games.

Hmm...well, I am nervous before, during, AND after competitions.  8-) But it doesn't matter if Im nervous before or after, my nervousness could only affect me if I freak out during the test.

Yes, the key is ``during.'' If I finish early, I'm nervous.


I have decided to make an FTW tourney (it is different from Mewto's).  The tourney will consist of two rounds, a normal game round, and a coundown round (which the top N make).  The normal games will all be ernie-created, ernie-certified :D, unrated, 15 problem, 15 second time games.  The password for all these games will simply be, "ernie".  Each participant must have played at least 5 games to qualify for the countdown round, which will determine the winners.  At the end of each normal game, you may not leave until I have told you because I need to record each score.  If the normal games are not created by "ernie", the games are not counted on your overall average.The five games must be complete by next Friday, June 20, 2008.  The countdown round will be based on the winners of the normal game round.  To be eliminated from the countdown round, you need to lose two times.  For these to be official, the winning player as well as the losing player must pm me the results (score, time).  Since I am using a  very unexplainable method (maybe not :lol:), I will not explain the countdown round bracket (unlike Mewto, meh.).  Since I am the moderator of this tournament, I will [b]not[/b] be participating.  Have fun!

SIGN UP HERE!

btw, each particpant of [i]ernie's tourney[/i] will recieve a certificate of participating and/or placing in the normal and countdown rounds

wow... i hav a lot of spare time  :D  :D  :D

Count me in. But what is the time? :lol:

ASH-POKEMON SHALL JOIN AND PWN
(I WISH) :!:  :!:  :!:

Uh oh, mewto joined well...  :o

I guess I'll join to give him the satisfaction of pwning me.

btw, normal games r 4 SEEDING ONLY

they do not count in placement

unless u hav the highest average.

*alexhhmun joins after 1 full months of no games.

I will fail.   :thumbup:

so far, i have:

1   mewto55555               2 scores
     visveswara                 0 scores
     murder                       0 scores
2    thdanh90                   2 scores
3    ash-Pokemon              2 scores
4    MathMathMath...         2 scores
      skaldjfhrulez                0 scores
      superllamel                  0 scores
       alexhhmun                  0 scores
dynamo729 0 scores

remember i need 5 scores from each participant by June 20!

[quote="skaldjfhrulez"]meh can i drop out?[/quote]

sure... cuz now its only 22 players!

yay!

an even #!

[quote="(^_^)"]join happy face tourney!!!

 :lol:[/quote]

 :rotfl:

[b]Prelims[/b]

9 Yongyi781 VS 20 bob einstein (GAME A1)
12 isabella2296 VS 17 dynamo729 (GAME A2)
7 Brut3Forc3 VS 22 PWNER (GAME A3)
14 VickAL VS 15 murder (GAME A4)
13 alexhhmun VS 16 permutations (GAME A5)
8 budi713 VS 21 visveswara (GAME A6)
10 thdanh90 VS 19 MathMathMath... (GAME A7)
11 ash-Pokemon VS 18 funboii777 (GAME A8)

[b]Bye Round[/b]

9/20 VS 3 mewto55555 (GAME B1)
12/17 VS 6 lotsofmath (GAME B2)
7/22 VS 14/15 (GAME B3)
13/16 VS 8/21 (GAME B4)
10/19 VS 4 RomanianGenius (GAME B5)
11/18 VS 5 hurdler (GAME B6)

[b]Quarter Finals[/b]

winner of B1 VS winner of B2 (GAME Q1)
winner of B3 VS 1 superllamel (GAME Q2)
winner of B4 VS 2 FantasyLover (GAME Q3)
winner of B5 VS winner of B6 (GAME Q4)

[b]Semi Finals[/b]

winner of Q1 VS winner of Q2 (GAME S1)
winner of Q3 VS winner of Q4 (GAME S2)

[b]Finals[/b]

winner of S1 VS winner of S2

[b]Third Place[/b]

loser of S1 VS loser of S2

LOL FANTASYLOVER AND SUPERLLAMEL ARE ALREADY IN QF BRACKET

 :rotfl:  :rotfl:  :rotfl:


Cards numbered 1 to 9 are arranged at random in a row. In a move, one may choose any block of consecutive cards whose numbers are in ascending or descending order, and switch the block around. For example, 9 1 $\underline{6\ 5\ 3}$ $2\ 7\ 4\ 8$ may be changed to $9 1$ $\underline{3\ 5\ 6}$ $2\ 7\ 4\ 8$. Prove that in at most 12 moves, one can arrange the 9 cards so that their numbers are in ascending or descending order.

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

[b]Definitions:[/b]
Let $a_{n}$ be the maximum number of moves required to turn a permutation of $1,2, \dots ,n$ into increasing order.

Let $X$ (given any arrangement) be the number of pairs in decreasing (i.e., the wrong) order.

We note that $0 \leq X \leq^{n}C_{2}$, that the cards are in increasing order iff $X=0$ (and decreasing order iff $X=^{n}C_{2}$), and that given any arrangement not in either of thee two states it is possible to increase or decrease $X$ by 1, by changing around two adjacent cards.

[color=darkred]Lemma 1: $a_{4}=3$[/color]
Here if $X \leq 3$, then we can reduce it to 0 by a sequence of changing around adjacent cards. If $X \geq 4$, we can increase it to 6 in 2 moves, then swap the cards around.

[color=darkred]Lemma 2: $a_{5}=4$[/color]

If $X \leq 4$, then we can reduce it to 0 by changing one pair of cards at a time.
If $X \geq 7$, then we can increase it to 10 similarly and then swap all the cards around
If $X= 5$ and there are three cards in either decreasing or increasing order, then we can swap their order around, at which point $X \leq 3$ or $X \geq 8$. From here we can finish as before, in only three moves.

So we may assume that there are no consecutive three cards in increasing or decreasing order. Therefore the row of cards goes 'up-down-up-down' or 'down-up-down-up'. We can say that it goes 'down-up down-up' without loss of generality because an 'up-down-up-down' order is equivalent to one 'down-up-down-up' with the card 1 interchanged with 5 and 2 with 4.

Let us call the first card $a$, the second $b$, and $c,d,e$ for the third, fourth and fifth cards.
Clearly either $b=1$ or $d=1$, since $a,c > b$ and $c,e > d$. If $b=1$, then we begin by swapping $a$ and $b$ around, from which point we only have four cards to arrange, which takes 3 moves by Lemma 1.
So we assume that $d=1$.
Since $a,c > b$, $b=2$ or $3$.
If $b=3$ then $e=5$ and we have two cases to consider: $43512$ and $53412$. We can check that these are possible in 4 moves.
If $b=2$, then$e=$ one of $3,4,5$. If $e=5$ then we are done by Lemma 1. If $e=4$ or $3$, then there are 4 cases to consider: $32514 52314$ $42513$ $52413$. We can easily check that these are possible in 4 moves.
Hence any arrangement of $1,2,3,4,5$ can be made increasing in 4 moves, that is $a_{5}=4$

[color=darkred]Lemma 3: $a_{n}\leq a_{n-1}+2$[/color]
Suppose we have a permutation of $1,2 \dots n$, and that the last card is $k$. We arrange the other $n-1$ cards into increasing order: this takes $a_{n-1}$ moves at most. We then have $1,2, \dots (k-1),(k+1) \dots n,k$. We can swap $(k+1), \dots n$ around, at which point the arrangement is $1,2, \dots (k-1),n,(n-1) \dots k$. Then we swap the bit from $n$ to $k$ around.
Thus we have arranged $n$ cards into increasing order in $a_{n-1}+2$ moves, proving this Lemma.

We have that $a_{5}=4$, by Lemma 3 $a_{6}\leq 6$, $a_{7}\leq 8$, $a_{8}\leq 10$, $a_{9}\leq 12$. Therefore it takes at most 12 moves to arrange cards 1 to 9 in increasing order.

You cannot bring $n$ to the beginning of the list with one move. Consider $n = 5$ with permutation $2,1,5,3,4$.

/spawn

Define a move to be when we switch the order of a block.

Note that it takes at most 3 moves to get a block of any 4 distinct integers to be strictly increasing or decreasing just by moving blocks with 2 numbers in them. So it takes at most 4 moves to get a block of 4 distinct integers to be strictly increasing. It also takes at most 4 moves to get a block of 4 distinct integers to be strictly decreasing.

Also, note that there will always exist a block with 5 numbers such that 9 is either the first number in the block or the last number in the block. When reading left to right, if 9 is the first number of the block, then we want our entire sequence from 1 to 9 to be strictly decreasing. If 9 is the last number of the block, we want our entire sequence to be increasing. So it takes at most 4 moves to get such blocks of 5 into strictly increasing or decreasing order.

Observe that if it takes a block with $k$ numbers $m$ moves to get it strictly in increasing or decreasing order, it will take $m+2$ moves to get a block of $k+1$ numbers that contains this entire $k$-block into strictly increasing or decreasing order. Without the loss of generality, suppose we have $a_1 , a_2, a_3, a_4, ..., a_k, a_{k+1}$ with the first $k$ terms in increasing order already and it took $m$ moves to get the sequence $a_1, a_2, a_3, ..., a_k$ into strictly increasing order. If $a_{k+1} > a_k$, we're done so suppose there exists an $a_i$ such that $a_{i-1} < a_{k+1} < a_i$ for some $1\le i \le k$ (if $i=1$, then this algorithm will work too). Then make a move on $a_i, a_{i+1}, a_{i+2}, ..., a_k$. Then the sequence will be $a_1, a_2, ... a_{i-2}, a_{i-1}, a_k, a_{k-1}, ..., a_{i+2}, a_{i+1}, a_i, a_{k+1}$. Make another move on the block that contains $a_k, a_{k-1}, ..., a_i, a_{k+1}$. Now our sequence is in strictly increasing order.

Therefore, if it took 4 moves to get our special block with 5 numbers, then it takes 6 moves for 6 numbers, 8 moves for 7 numbers, 10 moves for 8 numbers, and 12 moves for 9 numbers.

BAMO WHAMO COMBO WOMBO

/despawn

We generalize and induct. We claim the answer is $2n - 6$ moves for general $n$ with $n \geq 5$.

For the base case $n = 5$, observe that one of $5$, $1$ is in the positions $1, 2, 4, 5$. Thus we spend at most one move moving one of $5$, $1$ to an outermost number. WLOG we're moving $5$ to the first position. This guarantees that $1$ is in positions $3$ or $4$, since otherwise it would've been in one of the end positions to begin with.

Consider the case where $1$ is in position $3$ to begin with. Notice that if the element in position $2$ is $2$ or $3$, then the number of inversions is at most $3$. Swapping two adjacent elements decreases the number of inversions by $1$, so this gives us $1234$ after at most $3$ moves and then we flip it to win.

Otherwise we have $4$ in position $2$. However, here we begin with $54$, and we can obviously sort $321$ from any sequence of length $3$.

Now consider the case where $1$ is in position $4$ to begin with. The idea now is that we look at reverse inversions so that the sequence tends towards $4321$; then we can just apply what we did for the previous case. Thus the base case is proven.

Let $f(n)$ be the number of moves required to sort a sequence of length $n$. Then, the idea for going from $n - 1$ to $n$ is that we fix the first number, call it $i$. Then, after sorting, we get either \begin{align*}
	i, 1, 2, 3, \cdots, i - 1, i + 1, i + 2, \cdots n,
\end{align*} or we have \begin{align*}
	i, n, \cdots, i + 2, i + 1, i - 1, \cdots, 3, 2, 1.
\end{align*} Each of these costs $f(n - 1) = 2n - 8$ moves, so we just need to sort it to a sorted length $n$ sequence in $2$ moves.

For the first case, just flip the sequence from $1$ to $i - 1$, then flip the sequence from $i, i - 1$, to $1$.

For the second case, do something similar: flip $n$ to $i + 1$, then flip $i$ to $n$. This proves the inductive step, and so we have solved the problem.

Let $x_i$ denote the maximum number of moves it takes to put a permutation of $1,\ldots,i$ in ascending or descending order. We wish to show $x_9\leq 12$.

Lemma: $x_{i+1}\leq x_i+2$
Proof: Consider a row of $i+1$ cards. Let $a$ the card that is last in the row. In $x_i$ or fewer moves, it is possible to put the first $i$ cards in order. WLOG assume they are in ascending order. Consider the (possible empty) set of cards including all cards greater than $a$. Since they are not equal to $a$, they are all in the first $i$, so they have been sorted, so they are in ascending order and form a block of consecutive cards. Reverse this block of cards. Then, all of the cards greater than or equal to $a$ form a block of consecutive cards, since the cards greater than $a$ are at the end of the first $i$ cards and $a$ comes right after. In addition, they are in decreasing order, because the cards greater than $a$ were in ascending order but were reversed, and $a$ comes after them. So, the block of cards greater than or equal to $a$ can be reversed, making all of the cards be in ascending order. So, $i+1$ cards can be sorted in $x_i+2$ moves or fewer.

Lemma: $x_5\leq 4$
Proof: Firstly, we will show $x_3=1$. We know that at least one of $1$ and $3$ is in the first or last position. Assume WLOG $1$ is in the first position. Then the row is either $1, 2, 3$ or $1, 3, 2$. In the first case it's already in the right order and in the second case you can switch $3$ and $2$.

Next, we claim that if there are 4 cards, it can be put in order in 2 or fewer moves if $1$ or $4$ is in the first or last place. Assume $1$ or $4$ is in the first or last place. WLOG, let $1$ be in the first place. Then, the $2, 3,$ and $4$ can be put in decreasing or increasing order in 1 move. Then, it is either $1, 2, 3, 4$ or $1, 4, 3, 2$, and in the second case you can reverse the last $3$. So, it takes at most $2$ moves to sort if $1$ or $4$ is in first or last place.

We will now show that any row of 5 cards can be put in increasing or decreasing order in 4 or fewer moves. Either 1 or 5 is not in the middle spot. Assume WLOG 1 is not in the middle spot. Then, there is a block of 4 consecutive cards where 1 is first or last. So, that block can be sorted in 2 or fewer moves. Then, in a similar way to the previous lemma, the other card can be sorted in in 2 or fewer moves. So, the row of 5 cards can be sorted in 4 or fewer moves. So, $x_5\leq 4$.

Putting these two lemmas together we get that $x_9\leq 12$.

Let $f(n)$ be the least number of moves needed to guarantee being able to make the numbers monotonic.

Consider a list of $n+1$ numbers then take the first $n$ numbers and make it monotonic in at most $f(n)$ moves. If it is increasing let the sequence be \[1,2,\dots,k-1,k+1,\dots,n,k.\] Then, flip $(k+1,\dots,n)$ and flip $(n,n-1,\dots,k)$ to get $1,2,\dots,n.$ If it is decreasing let the sequence be \[n,n-1,\dots,k+1,k-1,\dots,1,k.\] Then, flip $(k-1,\dots,1)$ and then flip $(1,2,\dots,k)$ to get $n,n-1,\dots,1$ as desired. Thus, $f(n+1)\le f(n)+2.$

Note that $f(3)=1$ by checking the six cases. Thus, if there is a one or four in either of the two ends, we can do it in two moves. Otherwise, we can do it in three. Now, regardless of position of $5$ in the case with $5$ cards, if either one or four is one of the two ends, then $f(5)\le 4$ as desired. Otherwise, either two or five is one of the two ends. This is basically the same thing, so we're done.

ISL Marabot solve

First, there is at least $3$ numbers in one side of $9$ and we can check $6$ cases to see that we can always sort these four numbers in two moves.

Now,let there be $x,a_1,a_2,\dots a_k$ in the row in this order where $a_1,a_2,\dots a_k$ are sorted. Now if this is sorted with $x$ then we are done. Otherwise let, $x$ should lie after $a_{i}$, then we perform,
$(x,a_1,\dots, a_i,a_{i+1},\dots, a_k) $
$\mapsto (x,a_i,\dots, a_1,a_{i+1},\dots, a_k)$
$\mapsto (a_1,\dots, a_i,x,a_{i+1},\dots, a_k)$
So we can sort them in $2$ moves.
And we can do the similar process if $x$ lies after the sorted list.
So we can add the rest $5$ numbers into the sorting list in at most $10$ moves.    $\blacksquare$


Is there a formula that can be used to calculate the number of different ways n items can be chosen from an infinite amount of m different types of items?

If so, does the following method work to solve this type of problem:

At caroll's faerie empire, there are five different types of faeires. Gertrard wanted to eat 29 faeries, and he wants at least one of every type. How many different combinations of faeries can he eat?

I subtracted 5 from 29 b/c one faerie is required from each type, which left me with 24 cherries to choose from 5 different types. So the answer would be the number of ways to choose 24 cherrries from 5 different types. Does this work?

How do you count the number of ways to have 5 different types of 24 cherries?

Well that's what I'm asking. There must be a formula...?

It's the classic sticks and stones.  Divide your 29 cherries into 5 groups, and calculate how many ways there are of arranging the dividers.

Sorry tarquin, didn't know it was on your competition, I'll edit those posts now.

so there's 32 places in which to put 5...um... sticks? so 32c5?

That's almost right.  Although if I have any arithmetic skills it ought to be ${29+4 \choose 4} = {33 \choose 4}$.

Edit--I just saw the condition that he had to have one of each.  sirorange's answer should be right--try to figure out why that is.

Hmm this shouldn't be discussed until after May 4 but whatever.

I haven't received that many anyway.

5 different types. 4 separators. Subtract 5 for at least 1.

Simple/10

[hide="Solution"]
Four dividers to make five groups, 24 cherries, 23 spaces for the dividers. 23C4.
[/hide]

I think a formula would be along the lines of $\binom{n+r-1}{r-1}$, where $n$ is the number of objects selected and $r$ is the number of different objects.

Yeah, but you have to adjust it due to special cases, like this one.

Yeah, sorry tarquin (notice the name-changing) , but just want to clear something up cuz I might not get another chance before leaving for the competition. I already pmed tarquin my answers so this isn't gonna help my score.

there are 29 cherries, which equates to 4 dividers and 33 total slots. You can put dividers in any slot except the first and last one, so subtract 2 for 31 total slots. What am I missing?

EDIT- oh i see, subtracting 2 only accounted for 2 types of *faeries*. So you would need to subtract 5 because there are 5 places dividers can't be?


I'm learning to program (in pascal), and am using the USACO site for interesting questions.

However, I have had problems reading in files.  I can output fine, but have failed to read a single input file.

Any help in doing this would be much appreciated.

in c++

ifstream fin("filename.in);
ofstream fout("filename.out");

fin>>a;
fout<<a<<endl;

this names the files, gets the input, and gives the output.

if you program in java or pascal you'll have to find someone else.

The contest is going on right now, so once you familiarize yourself with the input and the output you probably should do the contest in the next couple days.

I found [url=http://math.uww.edu/~harrisb/courses/cs171/file_io.htm]this[/url] on the internet.  I don't really know pascal, so I'm not sure if this would help or not.

And good luck with the USACO pages.  And I agree with jeffrey, you should tryout the contets.


Let a1 < a2 < ... < an be real numbers. Show that
a1a24 + a2a34 + ... + ana14 \geq a2a14 + a3a24 + ... + a1an4.

Use induction.

The result holds trivially for n = 2.
Now suppose it holds for n-1, then to prove it holds for n it suffices to prove that
a_(n-1)a_n^4 + a_na1^4 - a(n-1)a_1^4 - a_1a_n^4 >= a_na_(n-1)^4 - a_1a_(n-1)^4

That is
a_1^4(a_n - a_(n-1)) + a_n^4(a_(n-1) - a_1) >= a_(n-1)^4(a_n - a_1). (*)

Since a_n - a_1 > 0, the homogeneity allows us to assume that a_n - a_1 = 1.
Thus, (*) follows immediately from Jensen inequality, using the convexity of f(x) = x^.

Pierre.


Dear friends in India!
 Which olympiad books (in english) would you suggest me  that are in sale and could I order them(editor ect). In Greece there are many people  from India (I have a student in my school) and  someone can buy them for me in India.Please send titel and edition.
 Thank you in advance !

 Babis Stergiou - Greece

Is there anything specific u want...actually we all(i suppose) follow foriegn authors

In an old post I had read a messange , where  it was suggested some indian olympiad books (the autor was frim India), but I lost the name. I supposed that in a so big country with great mathematical education and serious olympic presence at IMO , there could be some nice books.
 If you find something , please write.

 babis

[quote="pardesi"]ok i get u...
there is a publication called Prism book house it publishes a book on Functional equations by B.J.Venkatachala and another called Problem Primer to Mathematical Olympiads.
Also there is a book Called Challenges and Thrills of Pre-College Mathematics by C.R.Pransechar,B.J.Venkatahala and K.N. Ranganathan.It is published by New Age international.

I think these are the few books published...actually most here follow Engel and similar famous authors along with these...i will post some other s if i remeber...[/quote]

 Thank you . Today , a friend from Romany sent me these titles:

a)J.N.Kapur, Mathematical Olympiad Problems Book I (1991), Books II, III (1992), Books IV, V, VI (1993), Mathematical Sciences Trust Soc., New Delh
b)J.N.Kapur, Mathematical Olympiad Problems Book I (1991), Books II, III (1992), Books IV, V, VI (1993), Mathematical Sciences Trust Soc., New Delh
c)Mathematics Olympiad Primer, by V. Krishnamurthy, C.R. Pranesachar, K.N. Ranganathan and B.J. Venkatachala
d)Challenge and Thrill of Pre-College Mathematics, by V. Krishnamurthy, C.R. Pranesachar, K.N. Ranganathan and B.J. Venkatachala
e)5. Functional Equations, B J Venkatachala (Prism Books Pvt. Ltd., Bangalore)
Singapore International Mathematical Olympiad-serie
 Of course , some of them are mentioned by you above. Please , if someone can find an internatioanl store in India , which have tha book and can sent them world wide , please write.

 Thank you - Babis(Greece)

i had J N Kapur but somehow, i mysteriously lost it  :rotfl:
actually its there in our school library donated by an olympiad sir n i took a xerox  :)  Its really good with all imo n indian olympiad problems n also some proposed problems i think  :)

Old Math Olympiad handouts from India and elsewhere are the best places to study from (IMHO). With the huge resources that AoPS offers today, books serve very little purpose apart from introducing one to the theory. Of course , sentimental value is there :D


inequalities
calculus
derivative
function
inequalities proposed

Given $ a,b,c,d \ge 0$,and $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}1$.Prove that:
$ a^3\plus{}b^3\plus{}c^3\plus{}d^3\plus{}8(1\minus{}a)(1\minus{}b)(1\minus{}c)(1\minus{}d) \ge 1$
This inequality very nice and hard.The solution in this book long and hard.So,I hope you have a nice solution. :)

Some idea, hope it works.

Let $ f(a,b,c,d)\equal{}a^3\plus{}b^3\plus{}c^3\plus{}d^3\plus{}8(1\minus{}a)(1\minus{}b)(1\minus{}c)(1\minus{}d)$. Then we can prove that, if $ b\plus{}d\le 1$, and if $ b',d'$ are such that $ b\plus{}d\le b'\plus{}d'\le 1$, and $ b^2\plus{}d^2\equal{}b'^2\plus{}d'^2$, then $ f(a,b,c,d)\ge f(a,b',c,d')$.

Suppose $ a\ge b\ge c\ge d$. Then $ b\plus{}d\le 1$. Hence we can move $ b$ and $ d$ and make it nearer, until either $ b'\equal{}d'$ or $ d'>c$ and $ b'\plus{}d'\equal{}1$. If it is the second case, continue again with $ b\equal{}b',d\equal{}c$. Taking limit (or whatever, there must be some way to avoid calculus) we see that we can assume $ b\equal{}c\equal{}d$.

Guess the rest is solvable, I didn't try though..

[quote="Darkseer"]Some idea, hope it works.

Let $ f(a,b,c,d) \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \plus{} 8(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d)$. Then we can prove that, if $ b \plus{} d\le 1$, and if $ b',d'$ are such that $ b \plus{} d\le b' \plus{} d'\le 1$, and $ b^2 \plus{} d^2 \equal{} b'^2 \plus{} d'^2$, then $ f(a,b,c,d)\ge f(a,b',c,d')$.

Suppose $ a\ge b\ge c\ge d$. Then $ b \plus{} d\le 1$. Hence we can move $ b$ and $ d$ and make it nearer, until either $ b' \equal{} d'$ or $ d' > c$ and $ b' \plus{} d' \equal{} 1$. If it is the second case, continue again with $ b \equal{} b',d \equal{} c$. Taking limit (or whatever, there must be some way to avoid calculus) we see that we can assume $ b \equal{} c \equal{} d$.

Guess the rest is solvable, I didn't try though..[/quote]
Thanks :) ,but I think it is wrong :maybe:

[quote="quykhtn-qa1"]Thanks :) ,but I think it is wrong :maybe:[/quote]Bad luck for me?  :|

Still could you just please give a counter-example, or some rough idea about where it will break down?

I hate to show my computations (because it is just differentiation, though I believe there must be a nicer way to do it). But if someone is willing to check it:

[hide]We want to prove that when $ b \plus{} d\le b' \plus{} d'\le 1$ and $ b^2 \plus{} d^2 \equal{} b'^2 \plus{} d'^2$, we have $ f(a,b,c,d)\ge f(a,b',c,d')$, where $ f(a,b,c,d)$ is the LHS of the inequality.

Let $ A \equal{} b \plus{} d$ and $ B \equal{} bd$. We have $ A^2 \minus{} 2B$ fixed. When comparing $ f(a,b,c,d)$ and $ f(a,b',c,d')$. As $ a,c$ is fixed and $ B$ is a function of $ A$, we shall denote $ f$ by $ f(A)$. We will check that $ 2f(A)$ is decreasing for $ A\le 1$. We have \[ 2f(A) \equal{} 2a^3 \plus{} 2c^3 \plus{} 2b^3 \plus{} 2d^3 \plus{} 16(1 \minus{} a)(1 \minus{} c)(1 \minus{} b \minus{} d \plus{} bd) \equal{} C_1 \plus{} 2A^3 \minus{} 6AB \plus{} 2C_2(1 \minus{} A \plus{} B)\] \[ \equal{} C_1 \plus{} 2C_2 \plus{} 3(A^2 \minus{} 2B)A \minus{} A^3 \minus{} 2C_2A \plus{} C_2A^2 \minus{} C_2(A^2 \minus{} 2B) \equal{} C_3 \plus{} 3(A^2 \minus{} 2B)A \minus{} A^3 \minus{} 2C_2A \plus{} C_2A^2\], where $ C_1 \equal{} 2a^3 \plus{} 2c^3$, $ C_2 \equal{} 8(1 \minus{} a)(1 \minus{} c)$, and $ C_3 \equal{} C_1 \plus{} 2C_2 \minus{} C_2(A^2 \minus{} 2B)$.

Differentiation now gives $ 2f'(A) \equal{} 3(A^2 \minus{} 2B) \minus{} 3A^2 \minus{} 2C_2 \plus{} 2C_2A\le 0$ when $ B\ge 0$(which always holds) and $ A\le 1$.[/hide]


Recently, a number of [b]very[/b] old topics have been reopened.  While this isn't strictly prohibited, it is generally considered bad etiquette to reopen threads that have been idle for many, many months.

If your motivation is to get your post count up, you are a spammer.  I would rather assume that is not anyone's motivation.

If your motivation is to solve interesting problems, we need to discuss the options.  In many cases, the original poster has moved on to higher forums or is no longer active on AoPS.  Please ask yourself why you are reopening the old thread.  Are you posting a unique and interesting solution?  Are you hoping for feedback from the original poster?  Are you trying to run up your post count?  Is it an interesting problem you think a new batch of Middle School participants would like to solve?  Or is there something in the old post you are inquiring about?

Maybe some of these don't need to be posted, at all?  Maybe some of these would be better off with a new thread?

If this forum is short of interesting, fresh problems, maybe we all need to contribute a few more high-quality fresh problems.

Only you know your motivation.  Please reopen old threads responsibly.

Well, if you've answered all of the interesting problems in the subforum...

[quote="hockeyadi23"]But then again, some of the old problems are really good... and just died out for a while....[/quote]

Yeah but most of them have solutions already so there is no need to post anymore.

In the future, if one does have a solution already, I will post it as a new topic with a link to the old. How does that sound?

Ya, you all have a good point... I think the linking idea isn't a bad one tho...

I have an Idea for the moderators... if a topic goes beyond like the fifth of sixth page (or some predetermined date), then it gets locked, and you have to either quote or link to it if you want to revive it.. idk but that seems like a good idea to avoid spamming and to avoid clutter.

It wouldn't be a great idea, because that would mean a ton more work for the moderators. Right now, there are about 10 or so new topics going up each day. We don't want to lock all topics that are inactive. Also, we aren't outlawing reviving old topics. Just posting an answer to an old topic should not be done. If you solve it, keep it to yourself and either check with an older sibling or parent if the answer was not mentioned in the post.


calculus
analytic geometry
graphing lines
slope
algebra
polynomial
Rational Root Theorem

How can I tell if something is unfactorable by grouping?

This problem triggered my asking of this question:

If $f(x) = x^4 - 3x^3 - 9x^2 + 4$, how many real numbers $k$ does $f(k) = 2$?

My approach is to factor $x^4 - 3x^3 - 9x^2 + 2$ and to find the zeros. However, it seems like I cannot factor it.

Easy way: Use calculator since you're dealing with high degree equation.

Hard way: Use Rational Root Theorem and try every choices (which is VERY tedious). I wouldn't recommend it.

[hide="Advaned way.. Calculus"]
Find critical points and inflection points. Also, using those to find concavity and slope change. This will help you to "picture" where it'll reach zero or be factorable. Again, this is not suitable for this forum so that's why I put in hide. [/hide]

Just wondering, any reason why you wonder? (i.e. was this like a contest problem?)

I saw this problem listed in an SAT II Math booklet. I couldn't solve it without a calc, so I was wondering if there was an efficient way to solve high degree polynomials in general.

Yeah, I tested this polynomial with Eisenstein's Criterion, and it is irreducible. (I multiplied it by 2 (since zeros would be same), and found 3 would be the prime that works for that case)

I haven't learned calc yet, so I'm gonna have to use the other calc for now.

The problem only says to find the number of real zeroes, not what they actually are. Often use of Intermediate Value can show a zero exists. (i.e. if $g(x)=x^4-3x^3-9x^2+2$, $g(0)=2$ and $g(1)=-9\Rightarrow$ there exists $r$ between 0 and 1 such that $g(r)=0$. Similarly there is a zero between -1 and 0).

[quote="Silverfalcon"]Easy way: Use calculator since you're dealing with high degree equation.

Hard way: Use Rational Root Theorem and try every choices (which is VERY tedious). I wouldn't recommend it.

[/quote]

You can use a calculator on the SAT II.

Luckly, the equation $x^4-3x^3-9x^2+2$ only has 4 possible rational roots, except that none of those work.


Five aluminum cans can be recycled to make a new can. How many
new cans can eventually be made from 125 aluminum cans?
(Remember that the first new cans that are made can then be
recycled into even newer cans!)

[quote="nmani2"]I'm confused. Why don't you add the original 125?[/quote]

[quote="professordad"]It's almost like a pyramid; we can't add $125$ because $5$ of those can make a new one, so $25$.  Then, $5$ of those can make another new one, so add $5$.  Finally, those $5$ can combine to form an ultra pro new one :).  So $25 + 5 + 1 = \boxed{31}$.[/quote]

Actually, you don't add the original 125 since they asked for the number of NEW cans that can be made, not the TOTAL number of cans you make.

It says "How many new cans can [b]eventually[/b] be made from 125 aluminum cans"?  So I think this means that all new cans that were made count (though I might be wrong).


Find all $k\in\mathbb{N^*}$ such that there exists $a,b,c,d\in\mathbb{N^*}$ satisfying the condition:
[b](i)[/b] $(\frac{k-1}{2})^2\leq a<b\leq c<d$
[b](ii)[/b] $d-a=k$
[b](iii)[/b] $d=bc$

Hmm... You can write $a+k\geq d=bc\geq (a+1)^2$. So we obtain estimate $\frac{k-1}{2})^2\leq a\leq \frac{\sqrt{4k-5}-1}{2}$.
Therefore, you have estimate for $k$. Something like $k\leq 6$, I think.

[quote="luuduchoan"]
[b](iii)[/b] $d=bc$[/quote]
Hi Myth,
\[ad=bc\]
Sorry for my hand bad writing! :blush:

[quote="Myth"]Strange.... Take, for example, $a=n^2-m^2$, $b=n(n-m)$, $c=n(n+s)$, $d=n^2$, $k=m^2$. And there are many other examples....[/quote]
Ops... example is wrong, since b<a, c>d  :blush:


[hide]$\left(1+\frac1x\right)\left(1+\frac1y\right)=1+\frac2{xy}$
By AM-GM, $\frac{(x+y)^2}4=\frac14\ge xy\Rightarrow \frac2{xy}\ge 8$
So, $1+\frac2{xy}\ge 1+8=9$[/hide]

[quote="chess64"]Let $x$ and $y$ be positive real numbers such that $x+y=1$. Show that \[ \left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\ge 9.  \][/quote]

[hide]Rearrange to get $1+\frac{x+y}{xy}+\frac{1}{xy}\ge 9$, $\frac{2}{xy}\ge 8$, $xy\le \frac{1}{4}$. By AM-GM $\frac{x+y}{2}\ge \sqrt{xy}$, $xy\le \frac{1}{4}$.[/hide]

ch1n353ch3s54a1l, you wrote the steps backwards.

Um...how does it make a difference? I don't really know the proper way to write out these responses...

Not a big deal, it makes your argument (proof) more elegant and 'sound' (logical sense).

[quote="gotztahbeazn"]another way..

[hide]

$x+y = 1 \implies y = 1-x$

$\Big(1+\frac{1}{x}\Big)\Big(1 + \frac{1}{1-x}\Big) \ge 9$

simplifying eventually gives

$4x^2 - 4x + 1 = (2x-1)^2 \ge 0$ which is obviously true

[/hide][/quote]

Nice solution.  We all forgot that you could just substitute it that way.  And there seems to be no division by 0 as well.

good job.

[quote="mathgeniuse^ln(x)"]So, we were allowed to use AM-GM only because they both were positive real numbers?[/quote]
yes.

[quote="gotztahbeazn"]another way..

$ x \plus{} y \equal{} 1 \implies y \equal{} 1 \minus{} x$

>>> $ \Big(1 \plus{} \frac {1}{x}\Big)\Big(1 \plus{} \frac {1}{1 \minus{} x}\Big) \ge 9$ <<<

simplifying eventually gives

$ 4x^2 \minus{} 4x \plus{} 1 \equal{} (2x \minus{} 1)^2 \ge 0$ which is obviously true

[/quote]

This is what was to be proved.  You could take a false statement, and
manipulate to get a true statement.

If gotztahbeazn had presented the same steps (and missing details)
instead, then I would see it as valid.

It's really not a big deal, since all the steps are reversible (including clearing the denominators, since they are positive).

In terms of [i]presentation[/i], it would've been nicer to write, "which is algebraically equivalent to:" in place of "simplifying eventually gives".

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.CMO1971Problem2

Vo Duc Dien

[quote="chess64"]Let $x$ and $y$ be positive real numbers such that $x+y=1$. Show that \[ \left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\ge 9. \][/quote]

$(1 + \frac{1}{x})(1 + \frac{1}{y}) \ge 9$
$1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{xy} \ge 9$
$\frac{xy + x + y + 1}{xy} \ge 9$
$\frac{xy + 2}{xy} \ge 9$
$xy + 2 \ge 9xy$
$2 \ge 8xy$
$\frac{1}{4} \ge xy$
$\frac{1}{4} \ge x(1-x)$
$\frac{1}{4} \ge -x^2 + x$
The maximum value for any quadratic is when x is -b/2a. In this case -b/2a is 1/2, so the maximum value of the expression $-x^2 + x$ occurs when $x = \frac{1}{2}$ which means the maximum value is $\frac{1}{4}$.


The volume is $\frac{4000\pi}{3}$
the surface area is $400\pi$

[quote="SplashD"]Extension.

A mouse found the cheese ball and started eating it. It can eat $\pi$ cube inches in a 17 seconds. How many minutes will it take him to eat the entire cheese ball?[/quote]

$\frac{\pi}{17}= \frac{4000\pi}{3x}$

$x \approx 22667$ seconds $\approx \boxed{378}$ minutes

[quote="Arvind_sn"]Billy Bob has a big cheese ball. It is a perfect sphere. The radius of the sphere is 10 inches.

1. What is the volume of the sphere?

2. What is the surface area of the sphere?[/quote]

4/3pi r^3
4/3 pi 1000
4000/3 pi

4 pi r^2
4 pi 100
400 pi


group theory
abstract algebra
Euler
geometry
3D geometry
number theory unsolved
number theory

Find integers $ (x,y)$,
$ y^2 \equal{} x^3\plus{}1$

The way of solving is important.

It is an elliptic curve of rank 0 and torsion subgroup isomorphic to multiplicative abelian group isomorphic to C6. I think this helps to find the solutions $ (x,y)\in \{(0,\pm 1), (2,\pm 3)\}$.

Euler gave a proof by decsent to find all rational solutions of this equation.
You can find it along with a more modern treatment in terms of elliptic curves in [url]http://www.math.leidenuniv.nl/~jdaems/scriptie/Catalan.pdf[/url] section 3.3

There might be a simpler proof for integers.

well, there do exist elementary proofs of such equations ('cause i've found them  :wink: )

for now, i am just posting it for $ \pm 1$.

The proof given is incomplete. It claims that $ g(3g\minus{}2) \equal{} h^2$ has only solution g = 1.
In fact it has infinitely many solutions. The next one is $ g\equal{}9$.
At some point you would have to use the fact that $ n$ is a square to rule out the other solutions.

By the way, Euler's proof for rationals is elementary.
I think you have to generalise to rationals so that you can use infinite descent to complete the missing step.

i am aware of that fact...and also that i ve missed writing it..  :roll:  !
well i ve also mentioned that "for satisfaction of the given condition" only 1 satisfies..
well.. why not take as an exercise to prove the same?  :)

infact..do you know that the only numbers that exactly fits in between a square and a cube are 0 and 26? try proving it..i ve proved it using quite different method than what i posted...

Does this gives all points? If rank is 0 then every integer point is a torsion point. The cyclic group C6 means that there are five integer points plus a point at infinity. One can find by checking that $ (x,y)\in \{(0,\pm 1), (2,\pm 3), (\minus{}1,0)\}$.

[quote="Stephen"]Well, if $ x, y$ are intergers, the problem is too easy.

How about trying $ x, y$ are rational numbers?[/quote]

in real numbers there is infinitely many pairs for  $ y \equal{} k$  and  $ x \equal{} \root 3 \of {k^2 \minus{} 1}$, for real  $ k$.
I think... :blush:

[quote="theSA"][quote="Stephen"]Well, if $ x, y$ are intergers, the problem is too easy.

How about trying $ x, y$ are rational numbers?[/quote]

in rational numbers there is infinitely many pairs for  $ y \equal{} k$  and  $ x \equal{} \root 3 \of {k^2 \minus{} 1}$, for real  $ k$.
I think... :blush:[/quote]

$ \root 3 \of {k^2 \minus{} 1}$ is real, not rational. :wink:

[quote="theSA"]Sorry [b]Stephen[/b].
I edited it.
Now it is real... :lol:

Any solution for rational???[/quote]

Well, the only rational solutions for $ x\ge0$ is only $ (x, y) \equal{} (0, 1), (0, \minus{} 1), (2, 3), (2, \minus{} 3)$.

I have just solved it, but the solution is too long... :(

Any solution for $ x\le0$?


if I remember, you got it backwards: (x,y) is the same as (abcissa, ordinate)

so you have
$y=3+x$

because the ordinate (y) is always 3 bigger than the abcissa (x)

[quote="Interval"]What is an equation of the locus of points whose ordinates exceed their abscissas by 3?

Does ordinate = the x in the point (x,y)?

Does abscissa = the y in the point (x,y)?[/quote]

Yep that is incorrect as k8reindeer pointed out.

This was unfair in a MathCounts Question 1999 National Target.

Thanks for your help and correction.

interval