If you have a look at subject description,you may think that this is not combinatorial problem,but I assure you that it is.The problem was proposed about months ago,so you have nothing to worry about.
Consider $ n$ houses of a height,$ 1,2\dots,n$.Define $ A_{r,l}$ as follows:
it is the number of arrangements of houses,such that:
you can see exactly $ r$ houses if you look at them from the right and exactly $ l$ of them from the left.
Find an explicit formula of $ A_{r,l}$?
Remark: reccurent relations are OK either,don't know if we can name it explicit.

In other words, how many permutations have $ r$ left-to-right maxima and $ l$ right-to-left maxima?  Then consider the position of $ n$ to get $ A_{r,l}(n) \equal{} \sum_{i \equal{} 1}^n \binom{n}{i}\left(\sum_l A_{r \minus{} 1,l}(i \minus{} 1)\right)\left(\sum_r A_{r,l \minus{} 1}(n \minus{} i)\right)$ or something like that?

[quote="JBL"]In other words, how many permutations have $ r$ left-to-right maxima and $ l$ right-to-left maxima?[/quote]
In other words,"good" permutation is that one with elements placed from left-to-right  $ a_1,a_2,\dots,a_l$ and $ b_1,b_2,\dots,b_r$ so that the first sequence increases and another decreases,where $ a_i,b_j$ are the heights of houses.(well...some houses can be placed among $ a_i$ or $ b_j$,only in case if the next houses in sequence are still visible.

This is from a programming contest, and I think it can be done by dynamic programming (this is not what you inquired about,but well...):

let A(r,l,n) be the number of such sequences of length n. Depending on the position of 1, the lowest horse in the context of the question, we have:
A(r,l,n) = A(r,l-1,n-1)+A(r-1,l,n-1)+(n-2)A(r,l,n-1) (*)
except for boundary cases, like r=0 or l=0 we have A set to 0...

In actual implementation, we can drop n to reduce the memory spaces needed.

Is it possible to deduce the explicit formula from (*)?

Ah, that's much nicer.  \begin{align*}F(x, y, z) & = \sum_{n \geq 1} A(r, l, n)x^r y^l z^n/n! \\
& = xyz + \sum_{n \geq 2}A(r, l, n)x^r y^l z^n/n! \\
& = xyz + \sum_{n \geq 2}A(r, l, n - 1)x^r y^l z^n/(n - 1)! + (A(r - 1, l, n - 1) + A(r, l - 1, n - 1) - 2A(r, l, n - 1))x^ry^l z^n/n! \\
& = xyz + zF(x, y, z) + (x + y - 2) G(x, y, z)\end{align*}
where $ \frac {d}{dz} G(x, y, z) = F(x, y, z)$, $ G(x, y, 0) = 0$.  This gives a differential equation in terms of $ G$ that one can solve ... using Mathematica to skip all the ugly steps, it gives that $ \sum_{r,l} A(r, l, n) x^r y^l = xy \prod_{i = 1}^{n - 1} x + y + i$.  That's not a closed form, but it's pretty darn nice.  I don't know if you can do better.

Hi, that is much nicer indeed. I think you can call that an explicit formula.

For programming purpose, your result can be readily computed in Mathematica using Coefficient[], while the recurrence relation is easier to implemented using arrays in other languages. But i think both will take ard the same amount of time.


trigonometry
function
calculus
integration
real analysis
real analysis unsolved

Suppose that $n_{k}$ is a sequence of positive integers that tends to infinity. Let $A$ be the set of those $x\in [0,2\pi)$ such that $\sin(n_{k}x)$ converges. Prove that $A$ has zero measure, but it can be uncountable.

Thanks, I knew it must have been discussed.

The Eugene's proof in
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=kestelman&t=66841
 is very unclear; its result is as follows: let an, bn be sequences of reals s.t. an->infinity; let A be the set of those real x s.t.
sin(an*x+bn) converges. Then m(A)=0.
 i) Eugene assumes that A is a measurable set (proof?) and that m(A) is finite (no problem because A is a countable union of such sets).
ii)  Eugene assumes that sin(an*x+b) converges to 0, that is not the hypothesis which is: sin(an*x+bn) is pointwise convergent (over A) to f(x) (a measurable function). The Eugene's method leads to:
the mean value of f²(x) is 1/2 , that is not a contradiction.
     In addition can one find an, bn s.t. A is a cantor set?

Indeed, the proof merits some explanations. But anyway, the set of points where a sequence $f_{n}$ of measurable functions has a limit is measurable. Next, he uses a classical result saying that the integral of $\sin(nx)$ over a set of positive measure tends to $0$ for large $n$ (consequence of Riemann-Lebesgue theorem). Now, we can split the set of convergence on three parts: those on which the limit is positive, respectively negative and that on which the limit is zero. Eugene has shown that the third part has measure zero. Using the above classical remark, the other two parts also have measure 0.


calculus
integration
limit
greatest common divisor
number theory

(N-14)/(5N+16) is an irreducible fraction, what is the smallest N for which this is  possible.

Can we use numbers that aren't integers?
I'm assuming this problem wants integral values, but if it doesn't the answer is ${\lim_{k \to 0^{+}}k}$.
Otherwise, we can do it like [hide="this (hint)"]
If either of them are prime, and the other isn't a multiple of that prime, then we're done.
So we want the most negative number for which one is prime, and the other isn't a multiple (or factor) of that prime.
[/hide]

See if that helps.

Ok so we have (N-14)/(5N+16)

reciprical of (5N+16)/(N-14), that is 5+86/(N-14), meaning N's lowest is 100, but I was wondering why N cant be something like 57?

[hide]
$\frac{N-14}{5N+16}$ is irreducible.
That means $\frac{4N-30}{5N+16}$ is irreducible also, vice versa.
So if one of N-14 and 4N-30 is prime, and the other isn't divisible by it, then the fraction is irreducible.  (sorry that's not too in-depth)
I'm pretty sure that's the Euclidean algorithm.  Continue with it to find that
if the gcf of $N-14~ \text{and}~N+12=1$, then the fraction is irreducible.
I think that means you can make N as low as possible, right?
[/hide]

I'm not good with this kind of problem, so my solution is probably all wrong.

i don't get the point of this...n=1 obviously works

for a more general understanding, gcd( n-14,5n+15)=gcd(n-14, 15+5*14)=gcd(n-14,85)...note that 85=5*17, so the fraction reduces only when n-14 is 0,5,17, mod 85, i.e. n={14,19,31} mod 85

[quote="DiscreetFourierTransform"]Which is the most negative value (smallest) for N for which this works?
Or do any of those values that you gave mod 85 work when we're dealing with negative multiples of 85?[/quote]
Unless multiplying/dividing by $-1$ is considered reducing, I think the conditions apply to negative numbers, too.

So then, any negative that is within the bounds specified by Altheman works?
i.e. there is no most negative that this works for, as we can always generate a smaller N for an irreducible fraction?


induction
inequalities
triangle inequality
topology
calculus
calculus computations

Assume that a sequence $ {x_n}$ satisfies $ |x_{n \plus{} 1} \minus{} x_n| \leq \alpha |x_n \minus{} x_{n\minus{}1}|$
for $ n \equal{} 2, 3, . . .$ and some $ 0 < \alpha < 1$. Show that $ {x_n}$ is a convergent sequence.

[hide="Hint"] consider $ |x_{n \plus{} p} \minus{} x_n|$, and Cauchy sequences.[/hide]

This is the middle portion of the proof of the Banach fixed point theorem.

I have no idea what that is ^^

Edit: So I looked it up and found a proof on wiki. Could this not be proved without some sort of induction?

$ |x_{n \plus{} 1} \minus{} x_n| \leq \alpha |x_n \minus{} x_{n \minus{} 1}|$
Repeating this, we get $ 0\leq |x_n \minus{} x_{n \minus{} 1}|\leq {\alpha}^{n \minus{} 1}|x_1 \minus{} x_0|$

Which is only an intermediate step. Arguing that $ |x_n\minus{}x_{n\minus{}1}|$ is small does not yet show that the sequence is Cauchy. We need to write $ |x_{n\plus{}p}\minus{}x_n|$ as a telescoping sum that we can bound and estimate by a geometric series.

all of this gets broken up by the triangle inequality
 $ |x_{n\plus{}p}\minus{}x_{n\plus{}p\minus{}1}|\plus{} [|x_{n\plus{}p\minus{}1}\minus{}...\plus{}...\minus{}x_n|] \geq |x_{n\plus{}p}\minus{}x_{n\plus{}p\minus{}1}\plus{}x_{n\plus{}p\minus{}1}\minus{}...\plus{}...\minus{}x_n| \equal{} |x_{n\plus{}p}\minus{}x_n|$

??

$ |x_{n \plus{} p} \minus{} x_n| \equal{} |(x_{n \plus{} p} \minus{} x_{n \plus{} p \minus{} 1}) \plus{} (x_{n \plus{} p \minus{} 1} \minus{} x_{n \plus{} p \minus{} 2}) \plus{} ... \plus{} (x_{n \plus{} 1} \minus{} x_n)|$

$ \leq |x_{n \plus{} p} \minus{} x_{n \plus{} p \minus{} 1}| \plus{} |x_{n \plus{} p \minus{} 1} \minus{} x_{n \plus{} p \minus{} 2}| \plus{} ... \plus{} |x_{n \plus{} 1} \minus{} x_n|$

$ \leq \alpha^p |x_n \minus{} x_{n \minus{} 1}| \plus{} \alpha^{p \minus{} 1} |x_n \minus{} x_{n \minus{} 1}| \plus{} ... \plus{} \alpha |x_n \minus{} x_{n \minus{} 1}|$

$ \equal{} |x_n \minus{} x_{n \minus{} 1}| \sum_{k\equal{}1}^p \alpha^k$

$ \equal{} |x_n \minus{} x_{n \minus{} 1}| \cdot \frac{\alpha}{1\minus{}\alpha}(1 \minus{} \alpha^p)$

$ < |x_n \minus{} x_{n \minus{} 1}| \cdot \frac{\alpha}{1\minus{}\alpha}$

$ \leq \frac{\alpha^{n\minus{}1} |x_2 \minus{} x_1|}{1\minus{} \alpha}$

$ \equal{} \varepsilon$

Fermat's Little Turtle didn't say what set the $ x_n$ belong to. He was probably tacitly assuming that they were real numbers.

In fact, all of this works more broadly than that, since all we have used are metric space properties (notably the triangle inequality, which is explicitly present in Carcul's post). If we write $ d(x_{n\plus{}1},x_n)$ instead of $ |x_{n\plus{}1}\minus{}x_n|,$ then we have this:

The claim that a sequence which satisfies this hypothesis is a Cauchy sequence holds in any metric space.

The further claim that such a sequence converges requires the additional hypothesis that the metric space is complete.

No I did not say what they were of. Sorry to me it was obvious since it is from a Real Analysis book =)
Sorry


f is continuous on [0,+infinity], for all a>0,
\[ lim_{n \minus{} > \plus{} infinity}f(na) \equal{} 0
\]
Prove that
\[ lim_{x \minus{} > \plus{} infinity}f(x) \equal{} 0
\]

It's been posted before, several times. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70016]here[/url].


analytic geometry
conics
parabola
percent
calculus
geometry
3D geometry

AMC 2006!

I know everyone said wait til today...so I have. Can a mod confirm this, so we can get started discussing the test?

-Jorian Hoover

[color=green]edited:kyuanmathcounts confirmed this. talk away![/color]

[quote="jhredsox"]AMC 2006!

I know everyone said wait til today...so I have. Can a mod confirm this, so we can get started discussing the test?

-Jorian Hoover[/quote]

I hope we can. By the way, who takes it now anyways?

some school hoping someone posts the answers before 11 pm est :rotfl:
-jorian

can a mod still confirm this?

Yes, I believe it is now ok to talk.

Blow yourselves away  :P

[quote="mysmartmouth"]I haven't seen the test yet, but I have heard that it was easier than last years test.[/quote]

I thought it was easier.

[quote="jhredsox"]It was amazingly easy, but due to 2 stupid mistakes (I know how to do them, i just wasn't thinking), I'm 99% sure I got a 23/25.

I got these following ones wrong

$\frac{3}{2}*\frac{4}{3}*\frac{5}{4}...*\frac{2006}{2005}$

I got that wrong because I just wasn't thinking about crossing out!

I got the one wrong where it asked, "What would a graph look like if you posted all the square x and y coordinates on a graph," or something like that, so I forgot it was half of a parabola (darnet!).

-jorian[/quote]

For $\frac{3}{2}*\frac{4}{3}*\frac{5}{4}...*\frac{2006}{2005}$, the answer was $1003$.

I thought I did everything right, but I got a 23. :censored: :wallbash_red: :furious: :cursing:

[quote="goldendomer"]
For $\frac{3}{2}*\frac{4}{3}*\frac{5}{4}...*\frac{2006}{2005}$, the answer was $1003$.[/quote]

@goldendomer: yes, i know, i found out 10 minutes afterwards (i know the answer to the other question as well :) )
-jorian

[quote="redcomet46"]answers

1. D
2.C
3.A
4.B
5.D
6.C
7.B
8.E
9.C
10.A
11.C
12.D
13.D
14.B
15.C
16.E
17.B
18.D
19.D
20.C
21.A
22.D
23.A
24.A
25.B[/quote]

Thanks, but it's kind of useless without the problems.

Well, a lot of people copy down their answers on a separate sheet of paper.

[quote="kyyuanmathcount"]I don't see how you people can really complain about wording.  Can I have an example?  I thought these problems had no ambiguety (spelling -_-) at all.

However, MATHCOUNTS sometimes has a little weirdness to it.  All and all...AMC's are very well written (though I think AMC8 could be a little harder).[/quote]

aMC 8 should have been harder this year.

HOLY - it's Quevvy... won't post real name for obvious reasons.

it's richard ni from 5th/pi+ :o.
----------------------------------------
My school officially informed that i had a perfect score & gonna have the picture posted on the amc website, etc. U guys?

made a dumb mistake
got a 24.
one guy in my school got a 25

[quote="Fanatic"]HOLY - it's Quevvy... won't post real name for obvious reasons.

it's richard ni from 5th/pi+[/quote]
 :rotfl:  That's so weird!
Sweet, I'm as smart as you!

[quote="jhredsox"]umm.....I THOUGHT I GOT A 23 :mad:  :mad:  :mad:

the worst part is i don't know what i got wrong

-joriang[/quote]

*pat pat* no point in killing yourself.  I thought I got 25 last year and ended up with a 22.

[quote="jhredsox"]umm.....I THOUGHT I GOT A 23 :mad:  :mad:  :mad:

the worst part is i don't know what i got wrong

-joriang[/quote]
Don't they send score reports?

yah, but i probably can't ask my teacher

-jorian

I got an 18
I thought I would get a 10 or a 12    :winner_third:  :winner_third:

I got a 24! I'm so proud of myself! :lol:

ugh, this was 34 days old

please don't do this again

jorian

its ok jhredsox, rafealloa is new to AoPS

to Rafaelloa: its common courtesy on AoPS to not post on month old threads to say something pointless

i know, he did this in another topic :P

-----please do not discuss commentary on the subject about this-----
jorian


abstract algebra
group theory
superior algebra
superior algebra unsolved

Find all groups $G$ such that \[ a^3 = e,\ \forall a \in G.  \]

If $G$ is finite, then by Cauchy's theorem, $|G|=3^{n}$. Now if $G$ is abelian then decomposing it into a product of cyclic $p$-groups shows that $G$ must be isomorphic to $(Z/3Z)^{n}$ (otherwise $G$ would have an element of order $>3$). If $G$ is not abelian, then things go bad since you have an example of a nonabelian group of order $27$ satisfying your hypothesis (can be described by the multiplicative group of upper triangular matrices over $Z/3Z$ with $1's$ on the main diagonal, $0's$ on the second upper diagonal, and elements of $Z/3Z$ on the remaining upper cases). How are you going to describe that easily ? Are you sure you don't miss any hypothesis ?

Well, the question was originally

"Does there exist a non-Abelian group $G$ such that $a^3 = e,\ \forall a \in G$?"

So, indeed, I guess you need to give a concrete example of such a group.

How did you find those matrices, by the way?

I saw it passing by on sci.math one day. Look up the archives, I think the author was Arturo Magidin, he's usually the one answering purely algebraic questions.

[quote="mchis68"]Come on. Those matrices are old stuff. At least 150 years old.[/quote]

Might be, but how should I know, I'm pretty new to Group Theory.


number theory
modular arithmetic
number theory unsolved

Find a prime p, where for a larger prime q, 2q-p and 2q+p are prime. Show that there is only 1 such p.

Well I can see that p=3, some examples are q=5 and 7. My attempt only proved the statement in half the cases.

I said p can't be 2 since that implies 2q+/- p are even. Primes larger of equal to 5 are of the form p= 6k +/- 1. I also let q= 6n+/- 1. Written that way we can see that for the cases when we choose +/+ and -/- for the variable signs in (p,q), it implies divisibility by 2. However I can't prove the last 2 cases. An alternative method? Please help, very urgent!!

In fact, you have already pointed out everything you need for a proof. When $ p\ge 5$, then $ p\equiv\pm 1\bmod 3$, so respectively $ 2q\pm p\equiv\minus{}q\mp 1\bmod 3$. Hence $ 3\mid (2q\minus{}p)$ or $ 3\mid (2q\plus{}p)$. But $ 2q\minus{}p > p\ge 5$, hence $ 2q\minus{}p$ and $ 2q\plus{}p$ can't be both primes.

I'm a little confused sorry. Im bad at this. This test was meant to be able to be completed by 13 year olds who do not know modular arithmetic, so can you express in more elementary terms? So for p> 5, primes alternate between remainders of 2 and 1 when divided by 3. I don't really understand the rest, please explain more simply :(

Since $ q > p\ge$5, then $ p \equal{} 3k\pm 1$ and $ q \equal{} 3h\pm 1$, for some $ h, k\in\mathbb{N}$. Now you have four cases to check: $ i)$ $ p \equal{} 3k\plus{}1$, $ q \equal{} 3h\plus{}1$; $ ii)$ $ p \equal{} 3k\plus{}1$, $ q \equal{} 3h\minus{}1$; $ iii)$ $ p \equal{} 3k\minus{}1$, $ q \equal{} 3h\plus{}1$; $ iv)$ $ p \equal{} 3k\minus{}1$, $ q \equal{} 3h\minus{}1$. Apparently, $ 3$ divides $ 2q\plus{}p$, in the cases $ i)$ and $ iv)$; whereas $ 3$ divides $ 2q\minus{}p$, in the two others. Since $ 2q\minus{}p > p\ge 5$, then it's proved that $ 2q\minus{}p$ and $ 2q\plus{}p$ can't be both primes.

:( So i should have used 3n+/- 1 instead of 6n+/-1. Ok thanks heaps!!!

Well, that's right anyway! In fact, since all (rational) primes $ \ge 5$ are (obviously) odd, then they are necessarily in the form $ 6k\pm 1$, for some $ k\in\mathbb{N}$.


Any help with the following problem (#13 from AIME 1983) would be much appreciated. The problem is:

For each non-empty subset of {1, 2, 3, 4, 5, 6, 7} arrange the members in decreasing order with alternating signs and take the sum. For example, for the subset {5} we get 5. For {6, 3, 1} we get 6 - 3 + 1 = 4. Find the sum of all the resulting numbers.

Either a subset of $\{1,2,3,4,5,6,7\}$ contains 7 or it doesn't. If it doesn't, then it must be a subset of $\{1,2,3,4,5,6\}$. If it does, then the set of elements contains both 7 and a subset of $\{1,2,3,4,5,6\}$.

Use search button.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=368238#p368238


Water leaks out of a tank at a rate proportional to the square root of the depth of water at that time.  The water level starts at 100 inches and drops to 64 inches in 1 hour.
1. set up a differental equation for the depth of water in the barrel at time t
2. solve for the particular solution described.
3. How long will it take for all water for leak out of the barrel

The equation should be
$ \frac{dD}{dt}\equal{}k\sqrt{D}$ where $ D\equal{}D(t)$
where $ k$ is some constant to be determined.
The initial conditions are $ D(0)\equal{}100$ and $ D(1)\equal{}64$ so we can now solve:
The solution to the equation is
$ 2\sqrt{H}\equal{}kt\plus{}C$ (we can square it and divide but I'm holding off until the end).
Then plugging in,
20=C (from the first condition)
$ k\equal{}\minus{}4$ (from the second condition)
Then our solution is
$ H(t)\equal{}4t^{2}\minus{}40t\plus{}100$
Then to find when it runs out,
$ H\equal{}0\Rightarrow t\equal{}5$
So at $ t\equal{}5\ \text{Hours}$ it drains.


Let $ a,\ b,\ c$ be integers such that $ a^2 \minus{} 3b^2 \equal{} c^2$. Prove the following contexts.

(1) At least of $ a,\ b$ are even.

(2) If $ a,\ b$ are even, then at least of $ a,\ b$ are the multiple of $ 4$.

(3) If $ a$ is odd, then  $ b$ is the multiple of $ 4$.

For (1), [hide="hint"] consider $ \pmod4$[/hide].

Sorry, I have to go (I'll expand on this later).

[hide="1"]
Squares are either $ 0$ or $ 1$ mod $ 4$, and an odd square is $ 1$ mod $ 4$. If $ a$ and $ b$ are both odd then $ a^2\minus{}3b^2\equal{}\minus{}2\equal{}2$ mod $ 4$ but $ 2$ is not a square mod $ 4$, contradiction[/hide]
[hide="2"]
If $ a$ and $ b$ are both even then $ c$ must be even. We can let $ a\equal{}2a_1$, $ b\equal{}2b_1$, and $ c\equal{}2c_1$, so we have the original problem with $ a_1$, $ b_1$ and $ c_1$. But then one of $ a_1$, $ b_1$ must be even from part 1, so one of $ a$,$ b$ is a multiple of $ 4$.[/hide]
[hide="3"]
Take the equation mod $ 8$. Squares are $ 0$,$ 1$, or $ 4$ mod $ 8$, and any odd square is $ 1$ mod $ 8$. Suppose $ a$ is odd, then from part $ 1$ $ b$ is even, so it is either $ 0$ or $ 2$ mod $ 4$. If it is $ 2$ mod $ 4$ then $ b^2$ is $ 4$ mod $ 8$ so $ a^2\minus{}3b^2\equal{}\minus{}11\equal{}5 mod 8$, but $ 5$ is not a square mod $ 8$, contradiction. Thus, $ b$ is $ 0$ mod $ 4$, so it is divisible by $ 4$.[/hide]

Your solution for #2 isn't quite valid, the "so we have the original problem with a_1, b_1 and c_1" part.  But I think mod 8 might work.

What is not valid about it? I see nothing wrong with it.


Calculate alpha, where AB = AC

[img]http://img39.imageshack.us/img39/9803/cals.jpg[/img]

a) 2 degrees
b) 5 degrees
c) 10 degrees
d) 12 degrees
e) 13 degrees

Hi Luis. I make it $ 10^{\circ}$.

Angles $ CBE$ and $ CEB$ are both $ 50^{\circ}$, so $ CB\equal{}CE$.

Angles $ CBD$ and $ CDB$ are both $ 80^{\circ}$, so $ CB\equal{}CD$.

Therefore $ CB, CE$ and $ CD$ are radii of a circle centered at $ C$.

Then angle$ DEB\equal{} \alpha$ is an angle at the circumference, standing on the arc $ BD$ and so is equal to half the angle at the centre $ BCD \equal{} 20^{\circ}$, that is

$ \alpha \equal{} 10^{\circ}$.


Hi,

Does it exist a factorisation for : cos a + cos b + cos c - cos (a+b+c) for any a, b and c ?

Thanks.

hello, applying the addition formulas to $ \cos(a + b + c)$ we get:
$ {\cos(a + b + c) = \cos(a)\cos(b)\cos(c) - \cos(c)\sin(a)\sin(b) - \cos(b)\sin(a)\sin(c) - \cos(a)\sin(b)\sin(c)}$.
I think you can't factorize your term in the general case.
Sonnhard.


$ xy\equal{}56, \frac7{x} \plus{} \frac{14}{y} \equal{} 4$

$ x\equal{}\frac{56}{y}$

$ \frac{7}{\frac{56}{y}} \plus{} \frac{14}{y} \equal{} 4 \implies \frac{7y}{56} \plus{} \frac{14}{y} \equal{} 4 \implies \frac{y}{8} \plus{} \frac{14}{y} \equal{} 4 \implies y^2 \plus{} 112 \equal{} 32y$

$ y^2\minus{}32y\plus{}112\equal{}0 \implies (y\minus{}28)(y\minus{}4)\equal{}0 \implies y \equal{} 28 \text{or } 4$

If $ y \equal{} 28$, then $ x\equal{}2$.  If $ y\equal{}4$, then $ x\equal{}14$.  However, since $ y>x$, our answer is $ \boxed{2}$.

Alternatively, utilize the fact that they are positive integers...

We have $ xy = 56$, $ \frac7x + \frac {14}y = 4\implies14x + 7y = 4xy$, $ x,y\in\mathbb Z^ +$.

We factor: $ 4xy - 14x - 7y = 0\implies xy - \frac72x - \frac74y = 0$

$ \implies\left(x - \frac74\right)\left(y - \frac72\right) = \frac {49}8\implies(4x - 7)(2y - 7) = 49$.

$ x,y\geq1$, so we have $ 4x - 7\geq - 3$ and $ 2y - 7\geq - 5$.

Thus, it is impossible for either factor to be negative, so they must both be positive.

$ 4x - 7 = 49,2y - 7 = 1\implies(x,y) = (14,4)$, which is invalid.

$ 4x - 7 = 7,2y - 7 = 7$ has no integer solutions.

$ 4x - 7 = 1,2y - 7 = 49\implies(x,y) = (\boxed2,28)$, which is our answer.

[quote="GameBot"]Positive integers $ x$ and $ y$ have a product of 56 and $ x < y$. Seven times the reciprocal of the smaller integer plus 14 times the reciprocal of the larger integer equals 4. What is the value of $ x$?[/quote]
There are only so many factors of 56.  Guessing and checking will also get you there quickly.


Isn't this just simply [hide]the pigeonhole principle[/hide]?

OK, [hide]since the problem doesn't state that each person has different number of hair, you can't determine anything. The answer is 0.[/hide]

yes u can
think about it carefully
if all have the same, then there are a lot witht eh same,
but since hte question asks for the MOST that MUST be the same, then think about what u must do

That's what I thought too, but then it says "MUST"...I mean everyone could have no hair, so you can't really be sure about the answer.
Again, the key word here (that's not here) is "different" The answer is what you said if all people had different number of hair, but since it doesn't say that you can't really be sure.

You have to assume the worst case scenario.

[hide]

There can be at most 300,001 people with all different numbers of hair. You keep repeating this cycle. so it's floor (8,008,278/300,001) = 26

[/hide]

[hide]Oh, right, its 301 since a person can have no hair.[/hide]

Though it could, with different numbers, potentially have.

Tare, "MUST" is exactly what the pigeonhole principle tells you -- given n+1 pigeons and n holes, you MUST have 2 pigeons in at least 1 hole.

Couldn't there be cases in which everyone had no hair or all the pigeons went into one hole?

yes.  And then there would be MANY pigeons in one place.  But there won't HAVE to be that many pigeons in one place.

"If no one person has more htan 300,000 hairs, and NY has a population of 8,008,278, what is the greatest value of n such that there must be n people in NY with the same number of hairs?"

"the greatest value of n such that there must be n people in NY with the same number of hairs"

There MUST be n people with the same number of hairs -- but there don't have to be any more.

Is this helping?  Read the problem twice more, and work out exactly what the wording means.


What is your goal on the upcoming AIME?  Answer the poll (anonymous) or post (not anonymous).

There is another thread on this...What poll??


MATHCOUNTS
Support
trigonometry
analytic geometry
geometry
probability
similar triangles

Tricky MC Topics?
Poll on what you think

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

I was wondering what people considered the hardest part of mathcounts, so I am creating a poll to see what you think. Also, post here to announce what you voted, and, if you can, show a MC problem of that subject that supports your reasoning.

EDIT: Note that I did not include trig becuase no MC problem REQUIRES the use of trigonometry

Coordinate Geometry problems can almost always be solved using similar triangles, so there is no reason to include that as a separate topic.

[quote="batteredbutnotdefeated"]I voted for NT only because it's MY weak point...

How many 0's are in the expansion of 120! in base 8?[/quote]
I don't know if this is correct but is it $ \boxed{15}$?

[quote="pytheagle"]It's 16.

120/15 + 15/8 = 16[/quote]
Why did you divide by 15?
I thought you had to divide by 8...

[quote="pytheagle"]It's 16.

120/15 + 15/8 = 16[/quote]
Wouldn't it be 38? There are 116 factors of 2 and 38*3+2=116.

[quote="batteredbutnotdefeated"]
How many 0's are in the expansion of 120! in base 8?[/quote]

I got 38

How did you guy get 16?

why 120/15? I thought we should use 120/4 to get 30 to start with.

I don't know the answer, but unless your talking about nats(I don't have any copies of tests, but I have seen problems on this forum), Number Theory problems don't get that hard.  Except maybe on the team round with those killer problems.  Of course, except in chapter, you really shouldn't care as much about the team round as much as the sprint and target rounds because those are the only ones that count toward your individual score.  All the team score does is help give you a big shiny triangle(if you do well).  In fact, I don't think I'm going to practice for the team round at all(except during the practice sessions at my school for it) because
1. For my skill level, it's mostly too hard at the state level.(but I can handle sprints and targets)
2. It just doesn't matter unless you can't get in the top ten and you need to be in one of the top three teams.
3. You have three partners to ban on if you need help.
If you have to be hopeless in one written round, the team round is probably the best one.

My opinions:

Algebra: Can usually be solved easily if you can set up equations properly.
Combinatrics: Sometimes can get hard to choose what kind of strategy to use (complementary, casework, permutations, ...).
Probability: Part of combinatrics, which is my second weakest point.
Coordinate Geometry: Basically the mix of Algebra and Geometry. Focus on the Algebra or Geometry instead of this.
Number Theory: Know the tricks and when to use them. My weakest point.
Sequence and Series: I don't think these come up very often. Just think about them and be logical.
Regular Geometry: Know the theorems, formulas, and when/how to use them. As long as you [i]understand[/i] all those, you should be good.

I think the hardest are Combinatrics and NT, because they aren't taught in school very well.

[quote="PowerOfPi"]My opinions:

Algebra: Can usually be solved easily if you can set up equations properly. [b]At nats and below, yes. Otherwise, its just silly casework.[/b]
Combinatrics: Sometimes can get hard to choose what kind of strategy to use (complementary, casework, permutations, ...). [b]No... seriously, once you have done 10 mc combo problems in each topic it will be obvious to you which one to use. [/b]
Probability: Part of combinatrics, which is my second weakest point.
Coordinate Geometry: Basically the mix of Algebra and Geometry. Focus on the Algebra or Geometry instead of this.
Number Theory: Know the tricks and when to use them. My weakest point. [b]While this mindset works for nats, get out of this mindset. It will not be good. At all. Later on. most of MC nt is either ad hoc or trivial.Learn how to use modular arithmetic and you should be fine... just remember nt transitions to other stuff easily.[/b]
Sequence and Series: I don't think these come up very often. Just think about them and be logical.
Regular Geometry: Know the theorems, formulas, and when/how to use them. As long as you [i]understand[/i] all those, you should be good. [b]This. Is. True. For. Everything. Else. Especially for counting and number theory. [/b]

I think the hardest are Combinatrics and NT, because they aren't taught in school very well.[/quote]

My comments in BOLD.

Also: I hate it when people call counting/enumerative combinatorics "combinatorics". Its a subclass of combinatorics. It's not actual combinatorics. This subclass has pretty ugly problems compared to higher combinatorics.

Like, really, really ugly problems.

MC should really have bettetr problem.s

Hardest part of mc:

(A part the above poster did not mention:)

Arithmetic: (Apparently I can't do 125-64 in 3 seconds.)

You think that's bad?  Perhaps you should think a little harder about that.

On topic: I personally feel that it varies from test to test too much.  So I can't really pick a subject that would be the hardest in MC...

If you make a good sense in number theory, probably the result in probability comes out nicely although it is tricky and requires high standard of understanding rules as well as regulations in each probability problems.

[quote="pythag011"]
Hardest part of mc:
Arithmetic: (Apparently I can't do 125-64 in 3 seconds.)[/quote]

Arithmetic skills are an important part of MC tests.
However, the Arithmetic you mentioned here is not hard. It is just speed and accuracy.

I am sure you can do 125-64 in 3 seconds if it is not on the national stage and the championship is on the line.

[quote="pythag011"][quote="PowerOfPi"]My opinions:

Algebra: Can usually be solved easily if you can set up equations properly. [b]At nats and below, yes. Otherwise, its just silly casework.[/b]
[quote]

What do you mean "at nats and below"?  That covers everything!

No, it only covers MATHCOUNTS. He is referring to the AMCs - there is more to math than MATHCOUNTS. :)

When I meant "everything", I meant "all of Mathcounts".  And for me, anything involving tons of brute force is yucky.

The hardest part is not making stupid mistakes. If no one made stupid mistakes, like 50 bazillion people would get perfect scores.

I agree with Mewto.

Since we are talking about what subjects are hard, how should we get better? I posted my opinions in my older post.

@PowerOfPi: I'm practicing with the school handbooks, old tests, etc. I'm also reviewing the AoPS Intro. books on my weak spots. (Mine varies with the problems. :) )

I believe Algebra is the easiest.

you forgot to include reading comprehension.  :D

Number Theory. I never liked the different bases.
Algebra is the easiest for me too.

I agree with Mewtwo55555, MC is pretty straightforward as long as you don't make stupid mistakes....

Probably Number Theory/Combinatrics.  Of course, for the last page of the sprint round, all the subjects I included on the top page would be represented as "very tough problems" either because they are:
a) hard to get your head around
b) involve tons of brute force
c) easy to oversimplify or make a conceptual error about
d) are just plain hard

And "sequences and series" questions definitely do show up on the test, sometimes even more than 3 times per sprint round(directly or indirectly).


It is said that the 22nd of December is the shortest day of the year. According to a pocket calendar, the earliest sunset is on the 11th day of December (at 15:53 in Budapest) and the latest sunrise is on the 2nd day of January (at 7 h 32 in Budapest). Explain the phenomenon.

December 22 $\pm$ 1 day, depending on the position of the year with respect to the leap years, is the winter solstice in northern hemisphere, i.e., the shortest day. On January 3 $\pm$ 1 day, depending on the position of the year with respect to the leap years, the Earth is in perihelion, i.e., in the closest approach to the Sun on its elliptical orbit, the Sun being at one ellipse focus. The average distance to the Sun throughout the year is $\overline r \doteq 149.6 \times 10^6\ \text{km}$ and the excentricity of the Earth's orbit is $e \doteq 0.017$.

The angular momentum $\vec L = m \vec v \times \vec r$ of the Earth, where m is the mass of Earth and $\vec v$ its orbital velocity around the Sun, is conserved (one of Kepler's laws). Hence,

$v = \frac{L}{mr},\ \ \frac{\text dv}{\text dr} = -\frac{L}{mr^2},\ \ \frac{\text dv}{v} = -\frac{\text dr}{r}$

The average year has

$t = 1\ \text{yr} \cdot 365.2422\ \text{days/yr} \cdot 24\ \text{hr/day} \cdot 3600\ \text{sec/hr} \doteq 31. 557 \times 10^6\ \text{sec}$

and the average orbital and angular velocity of the Earth are

$\overline v = \frac{2 \pi \overline r}{t} \doteq \frac{2 \pi \cdot 149.6 \times 10^6}{31.557 \times 10^6} \doteq 29.78\ \text{km/sec}$

$\overline \omega = \frac{\overline v}{\overline r} \cdot 0.199 \times 10^{-6} \text{/sec}$

$\frac{\text d\omega}{\text dr} = \frac{r \frac{\text dv}{\text dr} - v}{r^2} = \frac{-r \frac v r - v}{r^2} = -\frac{2v}{r^2} = -\frac{2 \omega}{r},\ \ \frac{\text d\omega}{\omega} = -\frac{2\text dr}{r}$

If we now put $\frac{\triangle r}{\overline r} = -e \doteq -0.017$, we get the increase of Earth's angular velocity at its perihelion, compared to the average angular velocity $\overline \omega$ of its orbital motion.

$\frac{\triangle \omega}{\overline \omega} = 2e \doteq 0.034$

The angular velocity $\Omega$ of Earth's surface with respect to a fictious observer on the Sun, which is the same as the angular velocity of Sun's apparent orbital motion with respect to a real observer on the Earth, is given by the constant angular velocity $\Omega_s$ of Earth's own spin around its axis, take away the slightly variable angular velocity $\omega$ of Earth's orbital motion around the Sun.

$\Omega = \Omega_s - \omega$

The period $T_s = \frac{2\pi}{\Omega_s}$ associated with the angular velocity $\Omega_s$ of Earth's own spin is called the sidereal day (a day with respect to the stars), while the period $T = \frac{2 \pi}{\overline \Omega}$ associated with the angular velocity of Sun's apparent orbital motion with respect to an observer on the Earth is called the solar day (a day with respect to the Sun). Of course, the mean period $\overline T = \frac{2 \pi}{\overline \Omega}$ is called the mean solar day. By the original definition of time units, $\overline T = 24\ \text{hr}$ and it is almost exactly true with the current definition of time units. Hence,

$T_s = \frac{2 \pi}{\Omega_s} = \frac{2 \pi}{\overline \Omega + \omega} = \frac{}{\frac{1}{\overline T} + \frac 1 t} = \frac{t \overline T}{t + \overline T} = \overline T \cdot \frac{1}{1 + \frac{\overline T}{t}}$

Substituting $\overline T = 24\ \text{hr},\ t = 1\ \text{yr},\ \frac{\overline T}{t} = \frac{1}{365.2422}$,[color=white].[/color] the sidereal day comes out about 4 min shorter than the mean solar day.

$T_s \doteq 24\ \text{hr} \cdot \frac{1}{1 + 0.00273} \doteq 24\ \text{hr} \cdot (1 - 0.00273) = 24\ \text{hr} - 3\ \text{min}\ 55.9\ \text{sec}$

For example, recall that the angular velocity of Moon's surface with respect a real observer on the Earth is given by the constant angular velocity $\Omega_s$ of Moon's own spin around its axis, take away the slightly variable angular velocity $\omega$ of Moon's orbital motion around the Earth. Since these are equal (on the average), the angular velocity of Moon's surface with respect the observer on the Earth is zero and the Moon faces the Earth with the same side. But due to the existence of Moon's perigee and apogee, Moon's face appears to wobble slightly and we can see more than 50% of Moon's surface from the Earth (almost 60%).

At Earth's perihelion, the angular velocity of Earth's orbital motion is greater than its average, $\omega > \overline \omega$, hence, the angular velocity of Earth's surface with respect to a fictious observer on the Sun is smaller than its average, $\Omega < \overline \Omega$. At the winter solstice, the length of the day goes through its minimum (the 1st derivative with respect to time is zero), hence, it changes very slowly and the greater angular velocity of Earth's orbital motion at its perihelion and consequently, the lower angular velocity of Earth's surface with respect an observer on the Sun become apparent; both sunrise and sunset lag behind the mean solar time.


calculus
derivative
function
integration
calculus computations

How can I find the first partial derivatives to the following function involving an integral? Do I simply invoke Fundamental Theorem of Calculus Part 1?

$f(x,y) = \displaystyle\int^x_y cos(t^2)\,dt$

I got the answers:

$\frac{\partial f}{\partial x} = cos (x^2)$ and $\frac{\partial f}{\partial y} = -cos(y^2)$, but I not sure if they're right. Could someone verify? Thanks.


1. Let a and b be two distinct positive integers such that ab(a+b) is divisible by a^2+ab+b^2. Prove that |a-b| > (ab)^(1/3).

2. Let a, b, c be positive real numbers such that abc = 1. Prove that
(a – 1 + 1/b) (b – 1 + 1/c) (c – 1 + 1/a) ≤ 1.

Number 2 :
Set
$ a\equal{}\frac{x}{y}$
$ b\equal{}\frac{y}{z}$
$ c\equal{}\frac{z}{x}$

Our inequality becomes $ xyz \ge (x\plus{}y\minus{}z)(y\plus{}z\minus{}x)(z\plus{}x\minus{}y)$

Which is exaclty Schur Inequality..


algorithm
abstract algebra
function
email
MIT
college

Ok, so, to add to my hectic schedule I want to learn how to program.  Is there a nice easy book to start out with anyone can recommend?  Interested mainly in C/C++. I want to learn the basics at least.  Thanks!

I almost hate to admit this, but I used "[url=http://www.amazon.com/exec/obidos/tg/detail/-/0672310708/104-2516765-5808735?v=glance]Teach Yourself C++ in 21 Days[/url]".  If you can get past the the rather lofty title, it's a well-written book with lots of examples.  I never got to the end, but I imagine it might take longer than 21 days.

Mmmmm.... Functional programming.  :P  Granted I learned LISP instead of Scheme, but you can come up with some really interesting (and elegant) solutions, expecially recursive, when you program functionally.

Also don't underestimate the power of interpreted languages. I would check out  http://www.python.org/ before choosing to start with C or C++. Granted you dont learn as much about the internals, and you take a speed hit, but thats is something you realistically won't be affected by. Python is really elegant, has an wonderful community, great resources and modules (libraries), and is a fun language.

[quote="Elyot"]
I should warn you that everything for first-years at Waterloo is in Java (although Java is syntactically very similar to C)... just cuz I know you're gonna end up coming here!!!  :P
[/quote]

Hahaha, its so true. Thanks for the advice guys!

Alot of people try and dig in with C or C++ and get fed up, while it is possible to learn them first, many don't recomend it.  The problem with learning them, is they are really only usefull if you already know/ want to know more of how you program low level. True, most modern program are still written in C/C++, but that is becuase of the power you get of writing on such a low level, not the ease.

What I reccommend: Lean python first. Python is a great language, look it up on http://www.python.org, or pick up a book like "Learning Python" by Orilley. Thats what I did after I got fed up with C.

The real selling point? Python is fun right away! It is fully functional, with a great standard library, and tons of built in extras ( called modules in python ).

Once you feel you can write whatever you want in python, you're ready to tackle C / C++, thats what I did, and I found it much easier the second time around. and plus: once you learn C / C++, your python proggraming is much better! This is a testment to the rigor needed to write in C.

btw, if you're going to go straight to C++, and skip C, then take the time to learn about OOP ( object oriented programing ) in Python first, the jump can be big for some people.

ALot of people like C++ better than C, but I reccomend C first, you get a better respect for your computer the more time you spend learning how true linear programming works. also: while C and C++ are very simmilar, don't fool yourself that you can write in one just becuase you know the other. Someone should correct me if I'm wrong, but most of the stndlib is named differently, and some of it functions differently, there are a few minor changes in syntax I think too. Not fun little things to slip up.

Though if you run Linux, having the man pages to help you with C is handy. But I would go with Python first, or Lisp if you feel adventurous. I used C++ over my summer internship and it was dull coding in it. If a had a quarter for every time I did "for(int n =0; n < Limit; n++)" ...

After coming back from IMO 2005, I decided to learned C++ from scratch. At first, you'll find coding EXTREMELY frustrating, because none of your codes will seem to work (or even compile). But once you get the hang of it, programming can be quite addicting, especially with [url=http://www.usaco.org]USACO[/url] training, which features very interesting problems. I remember spending my whole days on USACO during those two weeks because it was soooo addicting.

USACO problems don't really require you to know a lot of technical coding, rather they look for ideas and methods (or what they call algorithm). It's really similar to math in some ways.

Again to emphasize my point: during the first few days or weeks of programming, you'll gonna want to pull your hair out, but it will get a lot more fun as you go on.

Though, I have to envy you for having the time to learn programming (or even the thought of it). I wish I have that much time right now ... *sigh*

[quote="billzhao"]
Though, I have to envy you for having the time to learn programming (or even the thought of it). I wish I have that much time right now ... *sigh*[/quote]

Right now its the thought of it.  But I figure I can fit in time between the running around on the weekends.  I hope. lol  If worse comes to worse, I'll learn it while I'm in the car driving from one place to another, heh.

[quote="billzhao"]After coming back from IMO 2005, I decided to learned C++ from scratch.[/quote]

Where from?

If one of you programmers wanted to make good money, a beginners' book with respectable title would sell really well.. :P

My first language was actually Pascal, which is as low-level as C.  I really learned a lot, and learned quite fast.

The problem with learning something like Python first is that you lose a bit of the technical knowledge that comes with learning a much lower language.  I know some people who learned to program on Visual Basic... yes, they could create applications very quickly that could do very elegant things (thanks to the hordes of libraries and pre-written code), but it took them much longer to grasp the fundamental programming principles that allow things to happen.  It really is best to start in Pascal or C and work your way up.  Learning the proper form and rigour required to write good C or Pascal code gives you the foundation that will make coding in other languages less sloppy.

And yea... you will write "for(int n =0; n < limit; n++)" quite a bit.

The most important thing to do when you start is to just get something to compile.  Even a program as simple as reading in a number from a file, adding 1 to it, and writing it to an output file can be a pain in the ass if you don't know any syntax.  You can PM or email me if you want some help getting started.  Don't try USACO until you've mastered loops, arrays, string manipulation, and recursion though.  They don't teach you any of that.

I should warn you... it DOES get addictive.  Make sure you stay on top of your schoolwork because you will waste hours and hours on the computer getting stuff to run.

[quote="Elyot"]If you just want to learn to code, any book like that will do fine.  My girlfriend learned to code from a book called "C FOR DUMMIES", and after checking it out, it actually seemed pretty good for a first programmer.  You can cover the basics of datatypes, arrays, loops, conditional statements, recursion, string manipulation, etc. pretty fast.  Don't worry about the really technical details.  If you want to learn object-oriented programming (ie the ++ part of C++) then you can learn topics like inheritance, polymorphism, etc (but that's less important than the core programming topics).

A lot of those books contain sections on language specific features, GUI development, API stuff (ie coding real applications for windows, etc) that you probably don't need to learn unless you're really interested.

Also, don't worry about C if you're gonna learn C++.  C is more-or-less a subset of C++.

I should warn you that everything for first-years at Waterloo is in Java (although Java is syntactically very similar to C)... just cuz I know you're gonna end up coming here!!!  :P
Don't take the CS135/functional programming course.  It's in scheme (a weird, complicated, old, and never-used language).  Take the other first-year courses instead.  (CS133 or CS134 if you get good enough to skip 133.)

If you want to learn more computer science type concepts (ie if you want to learn algorithms and data structures) then you'll probably need more than what these books have to offer.  Let me know and I can recommend lots of very very good resources for that.[/quote]

Hmm....my first year at UW was that we learned to code in C++ first term than we did data structures and algorithims in C in second term.  Actually they didn't teach us C they just told us to look it up on the internet and you will learn it!   :P   His philosophy was that no one could come out of university knowing C++ without knowing C?  He said it is like getting a degree in English with out taking a course in english!

Anyways my suggestion is that you try to learn C++ or C.  Biggest difference is that in C++ you use classes and in C you use structs.  Learning how to code is pretty easy, in fact you can learn it online or you can read a book made by the  [url=http://www.prenhall.com/deitel/]Deitels[/url].  Once you learned C++ or C you can easily learn other programs like Java, C#, python, perl, PHP and other modern programming languages.  In fact the syntax of all these programing languages are pretty much similar.  Learning how to code is pretty easy, once you learn it, you find that coding is somewhat boring (In my opinion).  The fun part is where you decide what data structures to use (link-lists, stacks, queues, trees, hash tables, ... etc) or what algorithims.  Anyone can learn how to code, but it is knowledge of data structures and algorithim that will give you an advantage in programming.

I recommend you learn C++, learn about pointers, classes, ...etc.  Then take time to learn Java (it is much more "cleaner", I am also learning it too). You can go on the Sun Microsystem website for documents on it. Once you get the hang of coding, spend more of your time on learning about data structures and algorithims.  They will show you the real power of programming!  There are lots of good books on algorithims, there is one big one made by this guy from MIT (I forgot the name).

If you go to University in an techanical program (cs, math, physics, eng, etc.) you will walk away with a knowledge of a couple programming languages.  Also if you come to UW there are LOTs of co-op jobs from software industries, so try to learn as much as you can now. Also learn how to do software testing and debuging it will help you in finding a job.

Have Fun


inequalities
search
logarithms
inequalities theorems

How many ways to prove this inequality ?
$ \sum_{i\equal{}1}^n a_i   \geq  n \sqrt[n]{a_1a_2...a_n}$
all $ a_i \geq 0$ and  $ i \equal{} 1 , 2 , ..., n$
 Can you post a way ?

You will search it in "arithmetic-geometric mean" or "AM-GM".

Can you prove this ?
I know this inequality is "AM-GM" but why don't we post prof this inequality on this topic for everyone to know together

Simple proof
$ \log \left(\frac{\sum x_i}{n}\right) \ge \frac{\sum (\log x_i)}{n} \equal{} \frac{\log (\prod x_i)}{n} \equal{} \log \sqrt[n]{\prod x_i}$
by Jensen's inequality
So, $ \frac{\sum x_i}{n} \ge \sqrt[n]{\prod x_i}$

Another proof
[url=http://en.wikipedia.org/wiki/AM-GM]Inequality of arithmetic and geometric means[/url]

I read that there are more than 50 elementary ways to prove that. I think the shortest is that one:

First take:
$ A\equal{}\frac{\sum_{i\equal{}1}^{n}a_{i}}{n}$
Then:
$ (\prod_{i\equal{}1}^{n}a_{i})^{\frac{1}{n}}\equal{}A(\prod_{i\equal{}1}^{n}(1\plus{}\frac{a_{i}\minus{}A}{A}))^{\frac{1}{n}}
\leq A(\prod_{i\equal{}1}^{n}e^{\frac{a_{i}\minus{}A}{A}})^{\frac{1}{n}}\equal{}A*e^{\frac{\sum_{i\equal{}1}^{n}(\frac{a_{i}}{A}\minus{}1)}{n}}\equal{}A$
Proof by Lev Kourliantchik :wink:

Thanks so much
so intersting . Other ways ? Can you ?
Oh , do you know the way to prove this inequality by inductive method of Cauchy ?
If you know , can you post this way ? thanks a lot

[quote="BanishedTraitor"]What about:

$ (1,0,0,.......,0) \succ (\frac {1}{n}, \frac {1}{n}, \frac {1}{n}, ........, \frac {1}{n})$

 :rotfl:[/quote]
I don't understand what you mean ? What do you mean ?

Muirhead Inequality:

http://en.wikipedia.org/wiki/Muirhead_inequality#Deriving_the_arithmetic-geometric_mean_inequality

Jacobstahl's inequality gives direct solution to the proof of A.M.-G.M.

I remember that someone said Rado's inequality. :lol:

A example: 1978 Kyoto University entrance exam/Humanities and Science

Let $ a,\ b,\ c$ be positive numbers. Prove that : $ 2\left(\frac{a\plus{}b}{2}\minus{}\sqrt{ab}\right)\leq 3\left(\frac{a\plus{}b\plus{}c}{3}\minus{}\sqrt[3]{abc}\right)$.

When does the equality hold?

Who can prove this inequality by  inductive method of Cauchy  ? I really need to know this way . Can you hepl me ?


Christina and Toni took 24 minutes to walk 3/4 of the way home from school. At a constant rate, how many minutes would the whole trip take from school to home?


An ammonia and water mixture fills a five-gallon container. Eighty percent of the mixture is ammonia but some of the mixture will be drained and replaced with pure water. If a five-gallon mixture of fifty percent ammonia is desired, how many quarts of the mixture need to be drained before the water is added? (A quart is one-fourth of a gallon.) Express your answer as a decimal to the nearest tenth.

If 80% of our mixture is currently ammonia, there are 4 gallons of it. We want a mixture which contains 50%, or 2.5 gallons, ammonia. We need to remove 1.5 gallons, or 6 quarts, of ammonia to have the desired quantity. However, since the mixture is dilluted with water, we will remove 1.25 quarts of the mixture for every quart of ammonia. In the end, we drain $ 6 \times 1.25 \equal{} \boxed{7.5}$ quarts.


linear algebra
matrix
linear algebra unsolved

Given an $ n\times n$ matrix $ B$, does there always exist an $ A$ with $ \text{adj}A=B$? If no, what conditions on $ B$ should be set?(I can answer the first part with the aid of a theorem of Jacobi about non-singular matrices.)

"adjoint" is an overused word; do you mean the classical adjoint ($ \det(A)\cdot A^{-1}$ if $ A$ is invertible) or another definition?

If this is what you mean, the answer depends on $ n$ and the field; we can't get anything with negative determinant if the field is a subfield of $ \mathbb{R}$ and $ n$ is odd, for example. Anything with rank strictly between 1 and $ n$ is never possible; that answers the first question for $ n\ge 3$. For $ n=2$, the answer is trivially yes, and adj is undefined for $ n=1$.

Yeah I mean the classical adjoint. As you said, the cases $ n=1,2$ are trivial. So let me put it this way; Given an algebraically closed field and an $ n\times n, \ n>2$ matrix $ B$ with $ \text{rank }B\leq1 \text{ or rank}B=n$, what conditions (if any) are required to have an $ A$ with $ \text{adj}A=B$?


you know that the chapter command automatically numbers the chapter (talking about koma-script here, but same with the book class).

However,  is it possible to label it with alphabet?? (not appendix, of course).  For example, Apostol's Calculus volume 1 starts off with a chapter "I Introduction", then 1, 2, 3, ....  Any possible ways to do the same thing in LaTeX??

well I guess I can do \addchap and type up the I myself.  But the trouble is that when I number the theorems/problems in this chapter it gives me 0.a.b, which is quite annoying.  Of course, I can ignore it and just go with 0.

Also, any possible ways to skip section numbers? like jump from section 1.3 to 1.7.  it sounds weird but just wondering because it sorta helps for my project thingy.

Sorry for all the weird questions and thanks for any help.  And any place where I can read up more LaTeX?

\renewcommand{\thechapter}{\Alph{chapter}} will label chapters as Chapter A, Chapter B etc and \renewcommand{\thechapter}{\arabic{chapter}} will go back to Chapter 1, Chapter 2, Chapter 3 etc.

You can change the numbers you want the chapters or sections to continue with by using
\setcounter{chapter}{0}
starts the next chapter as Chapter 1
and
\setcounter{section}{4}
starts the section as something.5

[quote="pkerichang"] And any place where I can read up more LaTeX?[/quote][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=117360]LaTeX How-Tos[/url]

Dang I KNEW $ \text{\LaTeX}$ is the most powerful word-processing document EVER.  It beats the sissy Word which can't even handle numbering your list in a simple way...

How do you tell TeXnicCenter to print the value of a counter (e.g., printthe section counter in the header)?

\thesection prints the current value of the section counter. Similarly for other counters.
If you find that it has been redefined so that it shows both the chapter and section counter then use \arabic{section} instead.


function
geometry
superior algebra
superior algebra unsolved

It is well-known that a quintic equation is not solvable by "radical".
Isn't there a quintic equation which is solvable by sin, cos, or elementary functions... etc. ?
Why considered only by radical?
It's clear that elementary functions are computable.

Good question, I hope someone will still read this and give an answer , the famous theorem only states that there is not always a formula for quintics, using the normal field operations plus taking roots

The quintic can be solved using trancendental functions, i.e generalized non-radical functions such as hyperegometric series and theta functions. Mathworld has a great page on this;
http://mathworld.wolfram.com/QuinticEquation.html
http://library.wolfram.com/examples/quintic/main.html

There is facinating geometry behind this theory and I would suggest you take a look at Shurmans "Geometry opf the quintic".


Let  $a,b,c>0$ and satify $\left(a+\frac{1}{2}\right)\left(b+\frac{2}{3}\right)\left(c+\frac{1}{4}\right)=5$. Prove that $a^{2}b^{3}c^{4}\leq 1$

Thank you for show me that. What a pity of wrong problem!

What about it?
Let $a,b,c>0$ such that: $\left(a+\frac{1}{2}\right)\left(b+\frac{2}{3}\right)\left(b+\frac{1}{4}\right)=5$. Prove that $a^{2}b^{3}c^{4}\leq 8$

[quote="leecom"]I don't know where is c?[/quote]

I think he meant $\left(a+\frac{1}{2}\right)\left(b+\frac{2}{3}\right)\left(c+\frac{1}{4}\right)=5$.

By AM-GM, we have:
$a+\frac{1}{2}\geq  3 (\frac{a^{2}}{8})^{\frac{1}{3}}= \frac{3}{2}a^{\frac{2}{3}}$
$b+\frac{2}{3}\geq  5 (\frac{b^{3}}{3^{5}})^{\frac{1}{5}}= \frac{5}{3}b^{\frac{3}{5}}$
$c+\frac{1}{4}\geq  5 (\frac{c^{4}}{4^{5}})^{\frac{1}{5}}= \frac{5}{4}c^{\frac{4}{5}}$.

Hmm... for some reason I don't think this is going to work out.

That is Ok. The problem has solved completely. Thank you everybody

But when i have checked the condition of problem then the condition is also have changed some invents.


probability
inequalities
geometry
3D geometry
algorithm
algebra
polynomial

about fermat last theorem,what do you think?
was fermat really dumb and forgot to write the proof,or did
he not have any proof at all and he bragged allot about it
to make us all curious.
if in the first case(he had a proof) then wasn't the proof
of Wiles too big and much too complicated ?
What is the probability that a problem in an elementary form
can be solved through elementary methods?
How does Fermat last theorem (in wich there was allot of time
involved by much researchers)influence mathematics nowadays.
Do we see practical results,is it applied anywhere?
What is the link between number theory and crystals from chemestry?

I think he thought to have a proof, of the type like with unique factorisation in $\mathbb{Z}[\zeta_p]$. But we know today that this is only true for small $p$.
It's also possible that he didn't have anything more than lets say exponents $3,4$ and just claimed that the other cases are not harder... (well, there should more problems he claimed to have proven, but also some he really proved).

About the probability: I would claim that a randomly chosen equation (however one should define "randomly" exactly :D) has probability $0$ to be solved by elementary methods (but some of these proofs could be elementarized).

Influence: Wiles didn't only prove FLT, but a lot more, e.g. also the same for $x^n+y^n=pz^n$ for some $p$, all consequences of the partial proof of the Taniyama-Shimura conjecture (now totally proven).

Applied: outside of mathematics, there should be none :lol:
Inside Mathematics: it kills olympiad problems ;)

For the last one: I know nothing about crystals and would suspect that there is no close relation to FLT; but some sum of square stuff etc. could be important there.

I think that he have a proof for some interesting numbers (it is known that he had a proof for $n=4$), and maybe he thoguht it wouldn't be really difficult to find the proof........he was wrong.....
Remember that he post a result without posting the proofs and the others have to post it, so maybe he had an "interesting" part of a proof, but he couldn't finish it

i'm seeing that allot of work has been done in the ideea that one
day this theorem would eventually be prooven.is it luck or certainty
about the time that it took for it to be prooven ?
also i think that the ammount of study that has been put into the
branches of mathematics are not really equal.why is this ?
is it because there is someone somewhere wich needs some bits
of mathematics and pays for them ((i'm sure that you may take this as a
fabulation but take it as a posibility only)and this really is not good because
there is a natural way of discovering things and this thing really leads
to an inequality in the advance of every branch, i think they should be
all equal)?
i mean...allot of time has been invested in this theorem and after it
was prooven no application to the real world ? there should be some
connection to the real world.for example what geometrical model would
you associate to $x^n$ for $n \geq 4$ ?
about an elementary problem beeing solved with elementary methods,
i think that the reason that we cannot solve them tghrough elementary
methods is that elementary methods haven't been studied enough,
and probably people rushed to getting to methods for generalizing
things,or tried to bring maths to other directions,wich it was not
supposed to go to yet,if this sounds stupid then look at this
practical example :if you tangle a knot and you remember all
the steps,is it not you that can rearange the knot in the original order?,
if you make a lego is it not you,the creator that by the same methods are
able to deconstruct the lego (this example actually relates to the fact
that if man invented numbers and he tghought up what they really are,
then he should manage to tackle all problems about them with ease)?
there is obviously something not right.
i don't have your experience ZetaX,i dont have your knowledge so maybe
what i'm saying is just gibberish,sorry if i'm mistaking,altough i would
like to see your oppinion

Your are right with the Knot-part, just an example: Find all integers $x,y,z$ such that the following is prime:
$-176+54x^3+224y^3+107z^3-18x^4+96y^4-385z^4-812y-140z-69x^3y+45x^3z$
$+124y^3x-484y^3z+59z^3x+358z^3y-23x^2y^2+299x^2z^2+415y^2z^2$
$-216x+116x^2+152xy+488xz-382y^2+1382yz+1728z^2+345x^2y-379x^2z$
$+585xy^2-1222xz^2-543y^2z-28yz^2-1435xyz+78x^2yz-416xy^2z+353xz^2y$
[hide="I got this ugly thing by..."]expanding $(x+y-z-2)(3x-4y+5z-11)(2x+3y-11z+4)(3x+8y+7z+2)$[/hide]
Well, for this one, it's still not too complicated to refind the solution, but I think it shows the idea one more time.
But I don't think we can prove most stuff by elementary methods by one simple reason: when you study more and more elementary stuff, this new stuff will then also be called nonelementary since you need also some work (and prerequisite) to understand it. When you do the "nonelementary" things, you start seeing them more elementary and natural after some time.

It's likely that the "time was there" to solve FLT; when there wouldn't been Wiles, I think it would have needed some more years, but it would (have) happen(ed).
The only real world application I can (except of very unnatural problems like packing cubes into containers :D ) imagine is some stuff in quantum physics, but I know nothing about it. But I wouldn't see FLT as anything geometrical, even when seen as some box of rocks (in cubic form), since also then it's more number theory than really (euclidean) geometry (but that depends on a definition of "geometry", for that none is known to me). And to $x^4$, I would either associate hypercubes or multiplication like in the complex plane ;)
Seriously: I don't think one should associate geometrical objects since it's only a number theoretic property (and e.g. Catalan would have a very funny interpretation then). Thus also no "canonical" interpretation for the real world.

To the payment: yes, there probably is money from companies e.g. for informatical mathematics (algorithms, cryptology, ...). But this is happening in all types of science and should be not that bad in mathematics (it's a lot worse in physics, chemistry, biology etc. I think). It should be changed, but is there really any way to do so¿ Companies will not spend money for something that doesn't give them any use. And I dislike any "use" of mathematics ;)

@the tackle of all problems: we can never proof all theorems, nor even show the coherency of ZF(C); there are even polynomials, that have only solutions iff ZF(C) is coherent, thus proofing that it has none would spoil everything, but disproving the existence of solutions would do so, too :D
Concluding, I think we will get further and further, but never reach the end (and it's good so).

There are some good books out there about Fermat's Last Theorem - Wiles says he was inspired to work on it by reading one of those books.

My personal opinion is that Fermat did not have what mathematicians today would call a 'rigorous' proof, but that he may have had either an incomplete or a haphazard proof.

spx2 - I see your point, but keep in mind that mathematics came about as a way for ancient Mesopotamians to keep track of property and resources.  Mathematics has never truly been a stand-alone study until approximately the last two to three hundred years, maybe a little more. (excluding the ancient Greeks)  Number theory is critical to modern cryptography (modern encryption relies very much on Fermat's [i]Little[/i] Theorem).  n-dimensional/non-Euclidean geometry is coming to the forefront of quantum mechanics and our understanding of the fabric of the universe.  Mathematics has been, and will be, first and foremost a tool.  Maybe an incredibly complex tool that we don't completely understand, but you don't have to understand metallic bonding to use a hammer, do you?

Also, companies do not really have that much effect on mathematics.  They are ephemeral, and do not normally last for very long.  Also, the primary source of research mathematicians' sustenance and encouragement comes from collegiate mathematics departments and specialized institutes. (e.g., the Clay Math Institute with its Millennium Problems)  There will always be someone willing to spend there life looking at the music of the spheres and trying to understand how mathematics works.

[quote="spx2"]
about an elementary problem beeing solved with elementary methods,
i think that the reason that we cannot solve them tghrough elementary
methods is that elementary methods haven't been studied enough,
...
(this example actually relates to the fact
that if man invented numbers and he tghought up what they really are,[/quote]

1.  You have to define what an elementary method is, though.  It's generally agreed (at least from my limited experience) that "elementary" means "without the tools of analysis."  Many problems that can be easily formulated elementarily honestly do not have elementary solutions.

2.  I think this is a little arrogant on the part of Man.  The integers are a group that obey a certain set of rules.  Just because the rules are known does not mean that everything interesting to be known about the integers can automatically be discovered.  In fact, Godel's Incompleteness Theorem makes this impossible.

[quote="t0rajir0u"]It's generally agreed (at least from my limited experience) that "elementary" means "without the tools of analysis."[/quote]
Hmm....  I would say not only analytic ones.
What about algebraic number theory¿ I wouldn't call it elementary.

here they say lies an analytic proof of fermat-wiles theorem
http://arxiv.org/pdf/math.GM/0510081

i opened it and it is an elementary proof
does this mean that the proof of wiles...that big big proof,
can we throw that to the bin ?


Are there any distinct natural numbers $m,n,p,q$ which satisfy $m+n=p+q$ and $\sqrt{m}+\sqrt[3]{n}=\sqrt{p}+\sqrt[3]{q}\ge2004$. :P

Any quadruplet where $m=p$ and $n=q$. So I assume the point is in the numbers being pairwise distinct.

[quote="cefer"]You are right Farenhajt.I have seen this problem and there is $m,n,p,q$ are distinct and $\sqrt{m}+\sqrt[3]{n}=\sqrt{p}+\sqrt[3]{q}\ge2004$[/quote]
Posted here:[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=17846#p17846]www.artofproblemsolving.com/Forum/viewtopic.php?p=17846#p17846[/url]


Find $ n\in Z$ smallest, so that $ \tau (n)\equal{}14$.

[hide="Solution"]$ n \equal{} p_1^{a_1}...p_n^{a_n}$ has number of divisors $ (a_1\plus{}1)(a_2\plus{}1)...(a_n\plus{}1)$.

Two cases:

Case 1: $ n$ is the power of a prime. $ 2^{13} \equal{} 8192$.

Case 2: $ n$ is the product of the power of two primes. $ 2^6 \times 3 \equal{} 192$.

Our answer is $ \boxed{192}$.[/hide]


limit
topology
vector
real analysis
real analysis unsolved

1)Let $ x_i \equal{} (x_i^1, x_i^2, x_i^3, .... ) \in \mathbb{I^\infty}$ for i = 1,2,... Prove that $ \lim_{ i \to \infty} x_i \equal{} x_0$ iff $ \lim_{i \to \infty} x_i^k \equal{} x_0^k$ for k = 1,2, ...

2)Show that 1) isn't true if we replace $ \mathbb{I^\infty}$ with $ \mathbb{R^\infty}$

$ \mathbb{I}^{\infty}$ isn't well-known standard notation. Please define it.

As for $ \mathbb{R}^{\infty}$, what topology do you assume? In the product topology, this is true.

$ \mathbb{R^\infty}$ - set of real sequences $ (x_n)_{n\equal{}1}^\infty$ such that $ \sum_{n\equal{}0}^\infty (x_n)^2 < \infty$

$ \mathbb{I^\infty}$ - similar but $ 0 \leq x_n \leq 1$ for all $ n$

If $ x \equal{} (x^1 , x^2 , ...)$ and $ y \equal{} (y^1, y^2 , ... )$ , $ x,y \in  \mathbb{R^\infty}$ we define
$ d(x,y) \equal{} \sqrt{\sum_{i\equal{}0}^\infty(x^i \minus{} y^i)^2}$

What you write as $ \mathbb{R}^{\infty}$ is usually written as $ \ell^2$. I would always take $ S^{\infty}$ to be the set of all sequences with elements in $ S$- this conflicts with your notation in both cases.

The same counterexample works in both cases: $ x_i\equal{}e_i$, the vector with $ 1$ in the $ i$th spot and zeros elsewhere. This sequence converges pointwise to zero, but is bounded away from zero in any standard norm.


geometry
geometric transformation
homothety
parallelogram
geometry proposed

[color=darkred]Given $\triangle ABC$. Let $Q$ and $P$ points on sides $\overline{BC}$ and $\overline{AB}$ respectively. Let $F$ a point which lies on $\overline{AC}$ such that $\overline{FP}\parallel\overline{BC}$ and let $G$ a point on $\overline{BC}$ such that $\overline{GP}\parallel\overline{AQ}$. If $M$ and $N$ are the mid points of $\overline{AQ}$ and $\overline{BC}$ respectively. Prove that[/color]

\[\frac{{\overline{FP}}}{{\overline{BC}}}+\frac{{\overline{GP}}}{{\overline{AQ}}}=1\]
also $M,\ N$ and mid point of $\overline{FG}$ are collinear.

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=145355]this[/url] problem too (it has nothing to do with this one).

Let $AQ\cap FP=O$. by homothety, $\frac{FP}{BC}=\frac{AO}{AQ}$
and since $OPGQ$ is a parallelogram, $\frac{GP}{AQ}=\frac{OQ}{AQ}$
adding the above results we get the answer.

Just left to prove the collinearity and we're done.

well i'm not so good at geometry. so you should better count on somebody else for that.  :blush: but i'll try my best  :D

[quote="コリネアル"][color=darkred]Given $\triangle ABC$. Let $Q$ and $P$ points on sides $\overline{BC}$ and $\overline{AB}$ respectively. Let $F$ a point which lies on $\overline{AC}$ such that $\overline{FP}\parallel\overline{BC}$ and let $G$ a point on $\overline{BC}$ such that $\overline{GP}\parallel\overline{AQ}$. If $M$ and $N$ are the mid points of $\overline{AQ}$ and $\overline{BC}$ respectively. Prove that $M,\ N$ and mid point of $\overline{FG}$ are collinear.[/color]
[/quote]
[color=indigo][b]Proof.[/b] Denote $R$ be the midpoint of $FG$ and $S=AN\cap FP,$ thus $S$ is the midpoint of $FP$

We have: $\frac{SR}{AM}=\frac{\frac{1}{2}PG}{\frac{1}{2}AQ}=\frac{PG}{AQ}=\frac{BP}{BA}=\frac{NS}{NA}$

Hence: $M,\,N,\,R$ are collinear.[/color]


$\text{Solve for }x\text{: }x=\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$

$x=\sqrt{x+x}$
$x=\sqrt{2x}$
$x^{2}=2x$

$x=0$ or $x=2$ are the solutions.

[quote]Can't X be all real numbers?[/quote]

How?
//////

Oh, nevermind. I screwed up. The answer is already stated.

[hide]from the given information, we have $x=x^{2}-x$, so x=2,0[/hide]

wasn't this question on the log1 contest last week?

It was a variation of a problem I saw in a book.  I think it's called [i]More Games for the Superintelligent[/i].  Here's the solution I got:
[hide]Square both sides to get:
$x^{2}=x+\sqrt{x+\sqrt{x+\ldots}}$
Substitute:
$x^{2}=2x$
$x^{2}-2x=0$
$x=0$ or $x=2$[/hide]

Is zero really a solution?
Consider:
$1=\sqrt{x+\sqrt{x+\sqrt{x+...}}}$
What are the roots?
Also consider:
$\sqrt{0+\sqrt{0+\sqrt{0+...}}}$

Substitution gives some funny stuff eh?

$1=\sqrt{x+\sqrt{x+\sqrt{x+...}}}$
$1^{2}=x+1$
$x=\sqrt{2}$

$\sqrt{0+\sqrt{0+\sqrt{0...}}}=0$, surely?

[quote]i think besides that one (an exception), it works for all.[/quote]

Which equation are you referring to?

[quote="tim1234133"]$1=\sqrt{x+\sqrt{x+\sqrt{x+...}}}$
$1^{2}=x+1$
$x=\sqrt{2}$
[/quote]

Your last step is wrong. I don't know how you got to $x=\sqrt{2}$, but actually it seems that x = 0...

[quote="k8reindeer"]$\sqrt{0+\sqrt{0+\sqrt{0+...}}}=x$

$0+\sqrt{0+\sqrt{0+...}}=x^{2}$

$\sqrt{0+\sqrt{0+...}}=x^{2}$

$x=x^{2}$

$0=x^{2}-x$

$0=x(x-1)$

$x=0, 1$

this doesn't make sense, that expression can have 2 values:?:[/quote]

ok so x=2
x^2=4
x=2,-2

the reason is that you cannot reverse your steps (as i cannot in my example)

:oops: yeah, I added rather than subtracted

[quote="Max300"][quote="k8reindeer"]

this doesn't make sense, that expression can have 2 values:?:[/quote]

squaring often gives exterraneous roots. check to see that 1 isn't right.

and i think tim1234133 merely added 1 instead of subtracting[/quote]

But from $1 = x+1$, adding still gives $x = 2$...

So is there just no x which satisfies $1=\sqrt{x+\sqrt{x+\sqrt{x+...}}}$?

ah yes for some reason i thought it said $1^{2}=x^{2}+1$
my bad

it's not as bad as me- I thought that it said
$1^{2}=x^{2}-1$, and I was the one who wrote it.  :rotfl:
ooops.

$1=\sqrt{x+\sqrt{x+\sqrt{x+...}}}$

$x+\sqrt{x+\sqrt{x+...}}$ obviously has to equal 1 for the equation to be true.

Let's break it into two sides

$x$ and $\sqrt{x+\sqrt{x+...}}$

If the left side is 1, the right side will need to be 0 which won't be the case.

If the left side is any less than 1, the right side will be bigger than $1-x$ simply because $\sqrt{y}>y$ where $0<y<1$ so $1=\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ does not have any solutions.


hey guyz- i've never been on this forum (but im an indian)

i currently live in america and i wuz jw wat u guyz talk about on this forum

evrything probably have a look at the  metathread........it wud help u a lot if u know tamil  :P   :P  :rotfl:

im gujarti  :P :D

but thx

ill take a look  :ninja:

yes...literally everything    :P   :rotfl:

A little of maths and loads of fun!  :rotfl:  :rotfl:

yep... but nowadays its quietened down here...

and beleive me by a huge huge huge huge huge margin... i remeber of days(before jee :P ) when i wud sit here for almost 12-13 hrs a day :rotfl:  :rotfl:  :D

Hey, who are you really? What's your name?

[quote="Da Vinci"]Tell me also [u]man[/u], who are you really? :P  :D  :rotfl:[/quote]

Man or what????? :rotfl:  :rotfl:  :rotfl:

[quote="Da Vinci"]Tell me also man, who are you really? :P  :D  :rotfl:[/quote]
pengu eater coming to B'lore  :P

[quote="pardesi"]
pengu eater coming to B'lore  :P[/quote]

nice try!!  :harhar:   :D

[quote="Da Vinci"][quote="rituraj007"]and the [b] FSM[/b] as well!!!!!![/quote]

what is that???  :maybe:  :maybe:[/quote]
"FLYING  SPAGHETTI MONSTER" :rotfl:
(read 1st page of intel. humor & one of the pages of METATHREAD(can't tell which 1 :) ))

rt ,who originated in antarctica (wrt ur prev post?????)

think da vinci
@da vinci_plz take it as a joke!
bye!


function
algebra
floor function
combinatorics unsolved
combinatorics

Hi

Can anyone help me with this problem?

 Show that in an n*n chessboard ,one can put all the n 2

squares under attack, by using k queens.(k=ceiling (2n-1)/3).Show that k

is the least possible number of queens needed.

[quote="tripleJ"]Hmm, I'm not much of a chess player but for n=3, you can put all 9 squares under attack with just one queen. But ceiling ((2n-1)/3)=2. Perhaps you meant the floor function instead.[/quote]
Yes but for n=6 you can't put all 36 squares under attack using only [2*6-1]/3=3. You need at least 4=((2*6-1)/3) queens. That's also true for 12 and 13.

[quote="S_Alex"][quote="Leonardo"]Then it should be ceiling of (2n-1)/3 for n>3. Am I right?[/quote]
Not quite. For n=5,n=8 we only need [2n-1]/3. I've checked this using C++.[/quote]
Sorry. I wanted to say [(2n-1)/3]

Very bad mistake !!!!!!!!!!!!!!
Wow !!!! Hossein or Amir !!!!
This problem was in step 2 of Iran Computer Olympiad, 2000, problem 3

In the problem [b]the queens must be in diagonal[/b] .
And we must prove the $ k \geq [ \frac {2n-1}{3} ]$

I know the solution.

There are many unsolved problems, so please post complete solutions/answers if you have one....

Pierre.

If the queens must be in diagonal then :

If it's not an empty square in diagonal , we have n queens and $n \geq \frac {2n-1}3$
Else number the squeres of diagonal from 1 to n from top,
and let i , the first empty squere of diagonal and j , the last empty squere in diagonal.
then let $A = \{a_{i+1,i}, ..., a_{j,i}, ..., a_{j,j-1} \}$
because all of squeres of A must attacked and non of queens before i and queens after j cannot attack them and each queen after i and before j can olny attack at most 3 of them then we must have at least $(i-1) + (n-j) + \frac {2(j-i) + 1}3$
and it is at least $\frac {2n-1}3$

If $i=2$ and $j=n$ then there is another arithmetics: $(2-1)+0+\frac{2(n-2)-1}{3}=\frac{2n-2}{3}$.
Right? :huh:


How can I show that in an open connected set $\Omega \subset \mathbb{R}^{2}$, every two points can be joined by a a path made up solely of a (finite number of) vertical and horizontal line segments lying in $\Omega$?

I want to say something like that we know that $\Omega$ is path-connected, so we have a continuous curve connecting the two points. The range of this curve is compact, so it can be covered by a finite number of open balls in $\Omega$. These should overlap, so it seems intuitively clear that you can construct the required path. However, I don't know how to motivate this.

For each point $a\in\Omega$ let $U_{a}$ be the set of all points of $\Omega$ that can be connected to $a$ in the way that you want. Note that (1) $U_{a}$ is open; (2) two sets $U_{a}$ and $U_{b}$ are either equal or disjoint. This makes $\Omega$ into a disjoint union of nonempty open sets -- but $\Omega$ is connected...


logarithms
symmetry
algebra unsolved
algebra

If $x^y+y^x=1+xy$,then
$\boxed{x=1}$ or
$\boxed {y=1}$

It is very easy $\frac{\partial }{\partial y}(x^y+y^x-1-xy)=x(x^{y-1}\ln x +y^{x-1}-1)$. It give no solution if (x>1,y>1) or (x<1,y>1). By symmetry no solution if (x>1,y<1). If x<1,y<1 this expression $x^y+y^x-1-xy>0$.

Oh i see..but it doesn`t use elementary thinks..cause I don`t understant what $\partial$ or $\ In$ means.Still thanks for posting a solution.


Find, without a calculator, the value of cot80  + cosec40.
The angles are measured in degrees


Prove there are no integer solutions $x,y \geq 2$ to the equation

$\frac{x^{2000}-1}{x-1}= y^{2}$

For which positive integers n does $\frac{x^{n}-1}{x-1}= y^{2}$ have integer solutions $x,y \geq 2$?


function
real analysis
real analysis unsolved

$f: \mathbb{R}^2 \to \mathbb{R}$ a function:
1/ suppose that $x \to f(x,y_0)$ and $y \to f(x_0,y)$ are continuous for all fixed $(x_0, y_0)$
2/ moreover, $f$ vanishes on a dense subset of $\mathbb{R}^2$

Is $f$ necessary the null function ?

The set of points where $f$ is continuous is dense on every vertical line and every horizontal line (I'll leave this as a very nice exercise :)), so the set of points where $f$ vanishes has the same property (it's dense on every vertical and horizontal line), which implies that $f$ must indeed be the identically null function.

I really think the part you left as an exercice is the most interesting part  :)

Ok then, let's solve the "exercise" :).

It suffices to prove that the set of continuity points is non-empty on the vertical line $x=0$ (since we could then repeat the reasoning for a small interval of the line, proving that it's non-empty in every interval of the chosen vertical line). Also assume WLOG that $f$ vanishes on this vertical line.

Let $A_{m,n}=\{y\in\mathbb R\ |\ |f(x,y)|\le\frac 1m,\ \forall |x|\le\frac 1n\}$. If we fix $m_0,\ \mathbb R=\bigcup_nA_{m_0,n}$. Now, all the sets $A_{m,n}$ are closed, so, by Baire's Theorem, there is some $A_{m_0,n_0}$ with non-void interior $V$. If $I$ is some compact subinterval of $V$, then for all $(x,y)\in K_0=\left[-\frac 1{n_0},\frac 1{n_0}\right]\times I$ we have $|f(x,y)|\le\frac 1{m_0}$. We apply this procedure again, restricting our attention to $K_0$ and fixing some $m_1>m_0$ to find a compact $K_1\subset K_0$ (a compact $K_1$ of that special form: the product of two compact intervals) in which $|f(x,y)|\le\frac 1{m_1}$, and so on. The intersection of all the $K_t$'s will contain a continuity point of $f$ situated on the line defined by $x=0$, as desired.

I hope it's Ok.


Prove or disprove:

For every odd $ k\in \mathbb{N}$ exist prime $ p\equiv 3\; ({\rm mod }\;4)$ such that
 $ p|5^k\plus{}1$ and $ p^2\not{|} 5^k\plus{}1$

I have showed by myself that if a counterexample $ k \in \mathbb{N}$ exists then $ 3 \cdot 7 \cdot 43 \cdot 127 \cdot 2287 \cdot 7603 \cdot 9547 \cdot 7866608083$ divides $ k$.
I have a idea how to continue, but before I want to ask you if this one is really taken from a olimpic exercise or is it only a your conjecture.. :huh:

If k odd, then $ 5^k\plus{}1\equal{}6\mod 8$, therefore exist prime $ p\equal{}3\mod 4$, suth that $ ord_p(5^k\plus{}1)$ is odd. If $ 3\not |k$, then $ ord_3(5^k\plus{}1)\equal{}1$. If  $ 3|k$ we can find prime factor $ \Phi_k(5)$.


This problem was partially discussed here earlier.

Positive numbers $\alpha$, $\beta$ and $\gamma$ are sides of triangle and satisfy $\alpha+\beta+\gamma<\pi$.
Prove that there exists triangle with sides $\sin\alpha$, $\sin\beta$, $\sin\gamma$ and its area is at most
\[
\frac{1}{8}(\sin 2\alpha+\sin 2\beta + \sin 2\gamma).
\]

Here's the earlier discussion (the easy part, though): http://www.mathlinks.ro/viewtopic.php?t=5606

for the second part, after doing work it comes down to proving: (if my calculations arent wrong  :) )
2(x^2*y^2 + x^2*z^2 + y^2*z^2)
<= (x^2 + y^2 + z^2) + 2*(xy*sqrt(1-x^2)(1-y^2) + yz*sqrt(1-y^2)(1-z^2) + xz*sqrt(1-x^2)*(1-z^2).
where x=sina, y=sinb, z= sinc ; (so 0<= (x,y,z) <= 1)
any ideas?


Russia 1995: Solve

$ \cos (\cos (\cos (\cos x))) \equal{} \sin (\sin (\sin (\sin x)))$

So here's my solution:
[hide]
This equality can be written as $ \sin (\sin (\sin (\sin x))) \minus{} \cos (\cos (\cos (\cos x))) \equal{}0.$ The left hand side can be rewritten as $ 2\sin (\frac{\sin (\sin (\sin x)) \plus{} \cos (\cos (\cos x))}{2} \minus{} \frac{\pi}{4}) \cos (\frac{\sin (\sin (\sin x)) \minus{} \cos (\cos (\cos x))}{2} \plus{} \frac{\pi}{4})$ And, we know that this is equal to 0. So, we can say that $ \frac{\sin (\sin (\sin x)) \plus{} \cos (\cos (\cos x))}{2} \minus{} \frac{\pi}{4}$ should be $ 0$ for this to hold. Rewriting, we have $ \sin (\sin (\sin x)) \plus{} \cos (\cos (\cos x)) \equal{} \frac{\pi}{2}$. But, we could rewrite the left hand side into the form $ \sqrt(2) \sin (x\plus{}A)$ where $ A$ is the angle brought about by this change. Now, this shows that the maximum value of the left hand side is $ \sqrt(2)$, but this is less than $ \frac{\pi}{2}$. There's a contradiction, and thus, no solutions.

[\hide]

please check. thanks!!!

Hello Jakey !
[quote="Jakey"]Russia 1995: Solve

$ \cos (\cos (\cos (\cos x))) \equal{} \sin (\sin (\sin (\sin x)))$

... The left hand side can be rewritten as $ 2\sin (\frac {\sin (\sin (\sin x)) \plus{} \cos (\cos (\cos x))}{2} \minus{} \frac {\pi}{4})$ $ \cos (\frac {\sin (\sin (\sin x)) \minus{} \cos (\cos (\cos x))}{2} \plus{} \frac {\pi}{4})$

And, we know that this is equal to 0. So, we can say that $ \frac {\sin (\sin (\sin x)) \plus{} \cos (\cos (\cos x))}{2} \minus{} \frac {\pi}{4}$ should be $ 0$ for this to hold.[/quote]

1) you must explain why you dont consider the case $ \cos (\frac {\sin (\sin (\sin x)) \minus{} \cos (\cos (\cos x))}{2} \plus{} \frac {\pi}{4})\equal{}0$

[quote="Jakey"]Rewriting, we have $ \sin (\sin (\sin x)) \plus{} \cos (\cos (\cos x)) \equal{} \frac {\pi}{2}$. But, we could rewrite the left hand side into the form $ \sqrt(2) \sin (x \plus{} A)$ where $ A$ is the angle brought about by this change. ...[/quote]

2) How could you write $ \sin u \plus{} \cos v\equal{} \sqrt 2 \sin z$??? (with $ u\equal{}\sin(\sin x)\neq v\equal{}\cos(\cos x)$ ??

[quote="pco"]Hello Jakey !
[quote="Jakey"]Russia 1995: Solve

$ \cos (\cos (\cos (\cos x))) \equal{} \sin (\sin (\sin (\sin x)))$

... The left hand side can be rewritten as $ 2\sin (\frac {\sin (\sin (\sin x)) \plus{} \cos (\cos (\cos x))}{2} \minus{} \frac {\pi}{4})$ $ \cos (\frac {\sin (\sin (\sin x)) \minus{} \cos (\cos (\cos x))}{2} \plus{} \frac {\pi}{4})$

And, we know that this is equal to 0. So, we can say that $ \frac {\sin (\sin (\sin x)) \plus{} \cos (\cos (\cos x))}{2} \minus{} \frac {\pi}{4}$ should be $ 0$ for this to hold.[/quote]

1) you must explain why you dont consider the case $ \cos (\frac {\sin (\sin (\sin x)) \minus{} \cos (\cos (\cos x))}{2} \plus{} \frac {\pi}{4}) \equal{} 0$

[quote="Jakey"]Rewriting, we have $ \sin (\sin (\sin x)) \plus{} \cos (\cos (\cos x)) \equal{} \frac {\pi}{2}$. But, we could rewrite the left hand side into the form $ \sqrt(2) \sin (x \plus{} A)$ where $ A$ is the angle brought about by this change. ...[/quote]

2) How could you write $ \sin u \plus{} \cos v \equal{} \sqrt 2 \sin z$??? (with $ u \equal{} \sin(\sin x)\neq v \equal{} \cos(\cos x)$ ??[/quote]

ok, i'll check 1.

for 2) oh right! the angles aren't equal....hm...any hints on how to procees then?


Heres the problem "A circle is incribed inside a square that has an area of 49 sq. cm. What is the area of the circle?"

[hide]The square has side length 7, from which you can determine (draw a picture) that the circle's diameter is 7.  Then the circle's area is 3.5^2*pi, or 12.25 pi[/hide]
note that the circle is inside the square, not the other way around.

[hide]Each side of the square is 7 cm, so the radius of the circle is 3.5 cm.

Area of circle=pi(3.5)^2. Just simplify as usual, on your own, it's easy enough. The hint is good enough here.[/hide]

[hide="My Solution"]
If the area of the square is 49, then the side length is $ \sqrt49$, which is 7.
The radius of the circle would be half of that, so it would be 3.5.
And since the area of a circle is $ \pir^2$, the area of the circle would be $ \pi3.5^2$
Which is $ 12.25\pi$
[/hide]
Seriously, stop spamming the MC fourms with questions that could easily be on the homework of an 8th grader.

Yeah, it's not really spamming, unless you define spamming to include everything too easy for you.

then that would mean so many people are spamming you alan

haha

[quote="tpulak"]This is definitley not spamming. And many people might find this a bit hard(like me) , so by the way you are saying it, these people will feel bad. I posted this to see if there was a shorter way to do it. I know the answer, I just wanted the solution. I don't have the intention to put junk in this forum.[/quote]
Ok Ok fine...
I apologize.
 :lol:
Happy?

[quote="hunter34"]Now remember [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98085]this[/url][/quote]
Ok Hunter, I got it.
See?
More than 5 words, more than 1 sentence, and more words faces.
 :lol:

[quote="Identity"]Fractional answers are much better  :wink:[/quote]
Orly?
How so?
 :o

cuz for instance in multiple choice competitions the possible answers you see for geometry/area questions 9 out of 10 times will be expressed as fractions. I guess it's more [i]to the point[/i] (ahahahaha).

plus  it's annoying to write out the decimal (other than the easy fractions(1/2))

here is a good problem to know and it helps you ftw
mathcounts
what is the length of a cube with volume of 343
7*7*7 so it is 7

the last post in this thread was in 2007, and your post had nothing to do with it. THINK before posting.


calculus
integration
logarithms
real analysis
real analysis unsolved

Compute

$\int_{0}^{+\infty}t^{x-1}e^{-(\ln t)^2}dt$ x real number

$\sqrt{\pi}e^{x^2/4}$.

Substitute $s=\ln t$.
The integral becomes
$\int_{-\infty}^\infty (e^s)^{x-1}e^{-s^2}\cdot e^s\, ds=\int_{-\infty}^\infty e^{sx-s^2}\, ds=\int_{-\infty}^\infty e^{x^2/4}e^{-(s-\frac x2)^2}\, ds.$ Using the well-known fact that $\int_{-\infty}^\infty e^{-u^2}\, du=\sqrt{\pi}$, we have the answer.


quadratics
superior algebra
superior algebra unsolved

[quote="Abel"]Let c,d be positive integers such that cd is not a square. Show without using advanced results that there exist infinitely many primes q such that (c/q)=1 and (d/q)=-1.[/quote]

I think you also need the condition $\gcd(c,p)=\gcd(d,q)=1$. No?

Well, I forgot to add that d is not a square either.
 :oops:

No answers? I thought the problem was non-trivial. Was I wrong?

Aftersome stupid factoring, we may assume $\gcd(c,d)=1$.

Case $d \equiv 1 \mod 4$:
There is an $x_{0}$ such that $\left( \frac{x_{0}^{2}-c}{d}\right)=-1$ (otherwise all residues $r \mod d$ would fulfill $\left( \frac{r}{d}\right)=0,1$, wrong). Additionally we may assume $\gcd(x_{0},c)=1$. Consider prime divisors $p$ of $(cx+x_{0})^{2}-c$ for some integer $x$. For at least one of these $p$ we have $\left( \frac{p}{d}\right)=-1$ as otherwise their product $(cx+x_{0})^{2}-c$ would have jacobi symbol $1$ over $d$.
Fix this $p$ and see that $\left( \frac{c}{p}\right) = \left( \frac{(cx+x_{0})^{2}}{p}\right) = 1$. By quadratic reciprocity we get $\left( \frac{d}{p}\right)=\left( \frac{p}{d}\right)=-1$, too.
Now if there would be only finitely many primes of the type we want, let $P$ be their product (leaving out those primes that divide already $x_{0}^{2}-c$) and construct $p$ as above with $x=P$, giving a contradiction as $p \nmid P$.

Other cases of $d$ are similar.


Hi folks, I would like to know how many people in here had attempted the computer science subject GRE? I have some queries regarding the computer science subject GRE, they are as follows:

[b] Query 1 [/b]

I have heard some of the top schools (in the USA) that offer an M.S. in computer science like to scrutinize a person's subject GRE score (provided he/she had attempted it), they place more "weightage" on the subject GRE score. Is this true?

[b] Query 2 [/b]

I have also heard that an extraordinary score (say a very good score) on the CS subject GRE can offset a rather ordinary GPA, and that one can gain admission at places like Georgia Tech, UT Austin provided one has a great subject GRE score (in CS), is this true?

I would also like to know the the difficulty level of the computer science subject GRE, and how many months of preparation is necessary to get a an outstanding/very good score?

[b] P.S: [/b] I ask people in here to answer these queries of mine (provided they have time), I need elaborate replies;It would be very nice if people who had attempted the computer science subject GRE found some time to answer these queries of mine.

Thanks!

Aren't there anyone here who will respond to my queries? Didn't anyone attempt the computer science subject GRE here? (If it's so, I really can't believe it!)

You speak American! You must live in one of our colonies! Like Scotland! Or Australia! Or Puerto Rico!

[color=white]/sarcasm[/color]

It would be highly beneficial for me (and for people who have similar queries too) if someone responded to these queries of mine in an earnest manner rather than posting meaningless and irrelevant posts which serve no fruitful purpose at all.


inequalities
induction
number theory
relatively prime
number theory proposed

[quote="enndb0x"]Let $ p_n$ be $ n$'th prime number.Prove the inequalities:

1.$ p_{n \plus{} 1} < p_1\cdot p_2 \cdot ...\cdot p_n$
2.$ p_n < 2^{2^{n \minus{} 1}}, n > 2.$[/quote]
For number 2: Consider the sequence $ a_1  \equal{} 2$, $ a_{n\plus{}1} \equal{} a_1a_2...a_n \plus{} 1$...
For number 1: Use bertrands postulat.

Bertrand's Postulate is much stronger than needed (and could you prove it for someone on the spot? :p)
Just use the fact that there's a number ($ >1$) less that $ p_1 \cdots p_n$ which is relatively prime to $ p_1 \cdots p_n$:
$ \phi(p_1 \cdots p_n) \equal{} \prod_{i \equal{} 1}^{n}\phi(p_i) \equal{} \prod_{i \equal{} 1}^{n}p_i \minus{} 1 > 1$ for $ n > 1$, so there is a number that is divisible by a prime other than $ p_1,\cdots,p_n$ and is smaller than their product (it can't equal their product because its not prime to them), thus $ p_{n \plus{} 1}$ satisfies the inequality. (we needed strict inequality in $ \phi(p_1 \cdots p_n) > 1$ because $ 1$ is always prime to anything)
(I tried to modify Euclid's proof of the infinitude of primes, and couldn't - there's the case when $ \prod_{i \equal{} 1}^{n} p_i \plus{} 1$ is prime, though bigger than $ p_{n \plus{} 1}$, but I couldn't show that it is indeed can't be $ p_{n \plus{} 1}$)

2 is induction (and uses the identity $ \sum_{i \equal{} 0}^{n \minus{} 1} 2^i \equal{} 2^n \minus{} 1$)

[quote="bambaman"]Bertrand's Postulate is much stronger than needed (and could you prove it for someone on the spot? :p)[/quote]
I was aware of that. (No.)

hmm ...Thanks Guys for replying me , but I was wondering if anyone of you can give us more elementary solytion.

lol.(thnx bambaman )

*Guys XD

I think my solution is elementary enough - it is basically counting the numbers that are relatively prime to a product of distinct primes, and less than it. It follows from exclusion-inclusion (and many other arguments). I recommend reading about Euler's totient function.

Sorry to bring this up again- just wanted to say that I overcomplicated my proof-
Take $ m \equal{} (\prod_{i \equal{} 1}^{n} p_i) \minus{} 1$. It is clearly not divisble by $ p_1, \cdots, p_n$. Thus, if it's not 1, it has a prime divisor other than $ p_1,\cdots,p_n$ and it is smaller than $ \prod_{i \equal{} 1}^{n} p_i$ which implies the inequality. But it is 1 iff $ n \equal{} 1$, and indeed - the inequality doesn't hold for $ n \equal{} 1$...

@mods: can you move this to NT?
[mod edit: yes, we can.]


Let $a,$ $b$ and $c$ are positive numbers such that $abc=1$ and $0\leq k\leq3.$ Prove that:
$\sqrt{\frac{a}{b+k}}+\sqrt{\frac{b}{c+k}}+\sqrt{\frac{c}{a+k}}\geq\frac{3}{\sqrt{1+k}}.$

I think it follows the same idea as here :   http://www.mathlinks.ro/Forum/viewtopic.php?t=142431 in arqady's post

[hide]
Substitute $a=\frac{x}{z},\ b=\frac{y}{x},\ c=\frac{z}{y}$ so that
\[\sum_\text{cyc}\sqrt{\frac{a}{b+k}}=\sum_\text{cyc}\frac{x}{\sqrt{z(y+kx)}}\]
But by Holder's Inequality,
\[\left( \sum_\text{cyc}xz(y+kx) \right) \left( \sum_\text{cyc}\frac{x}{\sqrt{z(y+kx)}}\right)^{2}\geq (x+y+z)^{3}\]
Therefore, it remains to prove
\[(1+k)(x+y+z)^{3}\geq 9(3xyz+k(x^{2}z+y^{2}x+z^{2}y))\]
After multiplying by 3, it's equivalent to
\[k[4(x+y+z)^{3}-27(xyz+x^{2}z+y^{2}x+z^{2}y)]+(3-k)[(x+y+z)^{3}-27xyz]\geq 0\]
Of course $(x+y+z)^{3}\geq 27xyz$ by AM-GM, but we have to deal with
\[4(x+y+z)^{3}\geq 27(xyz+x^{2}z+y^{2}x+z^{2}y)\]
So multiply by 2, expand everything, and use $\sum_\text{cyc}xy^{2}-x^{2}y=(x-y)(y-z)(z-x)$, $\sum_\text{cyc}2x^{3}-x^{2}y-xy^{2}=\sum_\text{cyc}(x+y)(x-y)^{2}$, and $\sum_\text{cyc}x^{2}y+xy^{2}-2xyz=\sum_\text{cyc}z(x-y)^{2}$. It reduces to
\[\sum_\text{cyc}(4x+4y+z)(x-y)^{2}\geq 27(x-y)(y-z)(z-x)\]
Since the inequality is cyclic, we can WLOG let $z=\min \{ x,y,z\}$, then let $p=y-z$ and $q=x-z$, so that it becomes
\[(4p+4q+9z)(p-q)^{2}+(p+4q+9z)p^{2}+(4p+q+9z)q^{2}\geq 27pq(p-q)\]
Well, we can just get rid of  everything with $z$ in it now; the remaining inequality is the result of replacing $x$ with $q$, $y$ with $p$, and $z$ with $0$, so it's equivalent to
\[4(p+q)^{3}\geq 27p^{2}q\]
But now we can just use AM-GM on $\frac{p}{2}$ and $q$, so it's true.

Whew.
[/hide]

Very nice, scorpius119!
What about $k\geq8$ $?$ :wink:

For $k\geq 8$, there's the stronger inequality
\[\sum_\text{cyc}\sqrt{\frac{a}{a+k}}\geq \frac{3}{\sqrt{1+k}}\]

We know it's stronger because for any permutation $\{ A,B,C \}$ of $\{ a,b,c\}$, we have
\[\sum_\text{cyc}\sqrt{\frac{a}{A+k}}\geq \sum_\text{cyc}\sqrt{\frac{a}{a+k}}\]
which is just Rearrangement.

[hide="proof of stronger inequality"]
Let $a=\frac{x^{2}}{yz},\ b=\frac{y^{2}}{xz},\ c=\frac{z^{2}}{xy}$ so that \[\sum_\text{cyc}\sqrt{ \frac{a}{a+k}}=\sum_\text{cyc}\frac{x}{\sqrt{x^{2}+kyz}}\]
Now use Holder's Inequality, so that
\[\left( \sum_\text{cyc}x(x^{2}+kyz) \right) \left( \sum_\text{cyc}\frac{x}{\sqrt{x^{2}+kyz}}\right)^{2}\geq (x+y+z)^{3}\]
This time the inequality that remains is
\[(1+k)(x+y+z)^{3}\geq 9(x^{3}+y^{3}+z^{3}+3kxyz)\]
which follows from
\[9\left[ \sum_\text{cyc}z(x-y)^{2}\right]+(k-8)[(x+y+z)^{3}-27xyz]\geq 0\]
[/hide]