trigonometry
algebra
polynomial
Vieta
analytic geometry

[hide="Solution"]
Notice that the $ 6$ solutions to $ x^6 \plus{} x^5 \plus{} x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1 \equal{} 0$ (i.e. all seventh roots of unity other than $ 1$) are

$ \cos\left(\frac {2\pi}7\right) \plus{} i\sin\left(\frac {2\pi}7\right)$,

$ \cos\left(\frac {4\pi}7\right) \plus{} i\sin\left(\frac {4\pi}7\right)$,

$ \cos\left(\frac {6\pi}7\right) \plus{} i\sin\left(\frac {6\pi}7\right)$,

$ \cos\left(\frac {8\pi}7\right) \plus{} i\sin\left(\frac {8\pi}7\right) \equal{} \cos\left(\frac {2\pi}7\right) \minus{} i\sin\left(\frac {2\pi}7\right)$,

$ \cos\left(\frac {10\pi}7\right) \plus{} i\sin\left(\frac {10\pi}7\right) \equal{} \cos\left(\frac {4\pi}7\right) \minus{} i\sin\left(\frac {4\pi}7\right)$, and

$ \cos\left(\frac {12\pi}7\right) \plus{} i\sin\left(\frac {12\pi}7\right) \equal{} \cos\left(\frac {6\pi}7\right) \minus{} i\sin\left(\frac {6\pi}7\right)$, so that by Vieta, the sum of the roots of the equation is equal to

$ \minus{} 1 \equal{} \cos\left(\frac {2\pi}7\right) \plus{} i\sin\left(\frac {2\pi}7\right) \plus{} \cos\left(\frac {4\pi}7\right) \plus{} i\sin\left(\frac {4\pi}7\right) \plus{} \cos\left(\frac {6\pi}7\right) \plus{} i\sin\left(\frac {6\pi}7\right) \plus{} \cos\left(\frac {2\pi}7\right) \minus{} i\sin\left(\frac {2\pi}7\right) \plus{} \cos\left(\frac {4\pi}7\right) \minus{} i\sin\left(\frac {4\pi}7\right) \plus{} \cos\left(\frac {6\pi}7\right) \minus{} i\sin\left(\frac {6\pi}7\right)$

$ \equal{} 2\left(\cos\left(\frac {2\pi}7\right) \plus{} \cos\left(\frac {4\pi}7\right) \plus{} \cos\left(\frac {6\pi}7\right)\right)$

$ \implies\cos\left(\frac {2\pi}7\right) \plus{} \cos\left(\frac {4\pi}7\right) \plus{} \cos\left(\frac {6\pi}7\right) \equal{} \minus{} \frac12\implies\boxed B$.
[/hide]

So is there no like....um trig ways other then knowing that its part of the 6 solutions to that polynomial?

There's a really lame guessing way to do it:
I'd probably do it this way if this were a contest. Draw out the unit circle

You know that the first angle is 1/7th (= 0.14 ish) of the unit circle, the second is 2/7ths (= 0.28 ish) of the unit circle, and the third is 3/7ths = (0.42 ish) of the unit circle.

So you could plot roughly these angles on a unit circle like the following:
 [geogebra]251a895e05eca9f735d278d67fb6a970973ef3cd[/geogebra]

Then, because you know the cosine of each of those points is the x-coordinate, you know you're going to just add up the x-coordinates. That alone (the fact that two of them are negative) should tell you that you're going to be looking at a negative quantity, and then from there it's sort of a guess to tell that their sum is further than 0.25 to the left... so that leaves -1/2.

As I said, it's a cheap, stupid way to do it, but it's how I'd do it if I didn't know the really clever roots of unity solution.

[hide]$ \sum_{j \equal{} 0}^6 \cos \frac{2j \pi}{7} \equal{} 0$

$ \sum_{j \equal{} 1}^6 \cos \frac{2j \pi}{7} \equal{} \minus{} 1$

But then note that $ j \equal{} 3, 4$ both give the same value, $ j \equal{} 2,5$ give the same value, $ j \equal{} 1,6$ give the same value, since $ \cos x \equal{} \cos \minus{} x \equal{} \cos 2 \pi \minus{} x$

So, $ 2 \sum_{j \equal{} 1}^{3} \cos \frac{2j \pi}{7} \equal{} \minus{} 1$

$ \sum_{j \equal{} 1}^{3} \cos \frac{2j \pi}{7} \equal{} \minus{} 1/2$

[/hide]

idea fizzled out.

By process of elimination C and D are not the answer, but A and B are not easy to choose between by estimating points on the unit circle.

[quote="kunny"]$ \theta \equal{} \frac {2\pi}{7}\Longrightarrow \cos \theta \equal{} \cos 4\theta\ \Longrightarrow \cos \frac {2\pi}{7}\in{8x^4 \minus{} 4x^3 \minus{} 8x^2 \plus{} 3x \plus{} 1 \equal{} 0}$.

$ \therefore \cos \frac {2\pi}{7}\neq 1\Longrightarrow \cos \frac {2\pi}{7}\in{8x^3 \plus{} 4x^2 \minus{} 4x \minus{} 1 \equal{} 0}$, which follows Vieta's formula.[/quote]
$ \theta \equal{} \frac {2\pi}{7}\Longrightarrow \cos \theta \equal{} \cos 4\theta\implies \cos (\frac{2\pi}{7})\equal{}\cos(\frac{8\pi}{7})\equal{}\cos(\pi\plus{}\frac{\pi}{7})\equal{}\minus{}\cos(\frac{\pi}{7})$

Is it right? :lol:

[quote="kunny"]$ \theta \equal{} \frac {2\pi}{7}\Longrightarrow \cos \theta \equal{} \cos 4\theta\ \Longrightarrow \cos \frac {2\pi}{7}\in{8x^4 \minus{} 4x^3 \minus{} 8x^2 \plus{} 3x \plus{} 1 \equal{} 0}$.

$ \therefore \cos \frac {2\pi}{7}\neq 1\Longrightarrow \cos \frac {2\pi}{7}\in{8x^3 \plus{} 4x^2 \minus{} 4x \minus{} 1 \equal{} 0}$, which follows Vieta's formula.[/quote]

Are you sure?
$ \boxed{\theta \equal{} \frac {2\pi}{7}\Longrightarrow \cos \theta \equal{} \cos 4\theta\ }$

Sorry, the right thing is $ \boxed{\cos 3\theta \equal{} \cos 4\theta }$. :oops:

Since the problem offers us 4 variants of the answer, we don't really have to calculate the exact number. It would be enough to show the range within which one of the proposed variants (A, B, C or D) would fit.

My solution doesn't require any sophisticated formula, except for the values for sines and cosines of $ \pi/6$, $ \pi/4$, $ \pi/2$, $ \pi/2\plus{}\pi/6$ and the formula for the cosine of a difference of two numbers.

[hide]Here's what I did:
$ \cos(2\pi/7)\plus{}\cos(4\pi/7)\plus{}\cos(6\pi/7) \equal{}$
$ \cos(2\pi/7)\plus{}\cos(6\pi/7\minus{}2\pi/7)\plus{}\cos(6\pi/7) \equal{}$
$ \cos(2\pi/7)\plus{}\cos(6\pi/7)\cos(2\pi/7)\plus{}\sin(6\pi/7)\sin(2\pi/7)\plus{}\cos(6\pi/7)$

$ \sqrt(2)/2 < \cos(2\pi/7) < \sqrt(3)/2$
$ \minus{}1 < \cos(6\pi/7) < \minus{}\sqrt(3)/2$
$ \minus{}\sqrt(2)/2 < \cos(6\pi/7)\cos(2\pi/7) < \minus{}3/4$

Expressing remaining sines through cosines we get:

$ 1/2 < \sqrt(1\minus{}\cos(2\pi/7)^2) < \sqrt(2)/2$
and
$ 0 <  \sqrt(1\minus{}\cos(6\pi/7)^2) < 1/2$
which gives us:
$ 0 < \sin(6\pi/7)\sin(2\pi/7) < \sqrt(2)/4$

Finally we summarize the calculated ranges and see that:
$ \minus{}1 < \cos(2\pi/7)\plus{}\cos(4\pi/7)\plus{}\cos(6\pi/7) < \minus{}(6\minus{}\sqrt(2))/8$ which in turn is $ < 1/4$

This proves that the answer is B.[/hide]

The answer is $ \minus{}(cos(\frac{\pi}{7})\plus{}cos(\frac{3 \pi}{7})\plus{}cos(\frac{5 \pi}{7}))$. Now read page 62 of [url]http://books.google.com/books?id=sFsP0kfrxVcC&pg=PA62&dq=olympiad+properties+of+a+heptagon&lr=#PPA62,M1[/url], there are three good proofs, the second of which I think is the best.


Each point of a circle is painted with one of three colours. Prove that there exist three points on the circle which have the same colour and are vertices of an isosceles triangle.

For the 3 color problem, you probably need Van der Waerden's theorem. By that theorem, there exists a V s.t. given V consecutive integers colored with 3 colors, there will be and A.P. of length 3. Then consider a regular V-gon in the circle. We then have 3 in A.P. that are all the same color, implying the result. This problem is readily generalized to n colors.

[quote="moldovan"]I have a solution for a similar problem:

[i][b]Each point of a circle is painted with one of two colours. Prove that there exist three points on the circle which have the same colour and are vertices of an isosceles triangle.[/b][/i]

[u][b]Solution[/b][/u]

Consider the regular pentagon $ ABCDE$ like in the figure. It results that at least three of the points $ A,B,C,D,E$ have the same color. Since every three vertices of $ ABCDE$ are the vertices of an isosceles triangle, the conclusion is that  there exist three points on the circle which have the same colour and are vertices of an isosceles triangle.[/quote]

yes, but i think that modifiyng it to 3 colors alters the problem greatly.Is it really that easy to find a polygon that will solve it as well?I´m not so sure, i didn´t spend enpugh time on this...

It doesn't work for $ 3$ colors. My solution is only for $ 2$ colors.


Suppose $p(x)\in\mathbb Z[x]$ is a polynomial which if $t\in\mathbb Z$ then $\sqrt[n]{p(t)}$ is an integrer. Prove that there exist $g(x)\in \mathbb Z[x]$ that $p(x)=g(x)^n$

[hide="Hint"]It is related to [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=19873#p19873]this topic[/url] [/hide]

see [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=326560#p326560]this post[/url] and post #10.


real analysis
topology
calculus
calculus computations

Can we have disjoint closed intervals I1, I2,... all in [0,1]
such that the sum of their lengths equals 1?

any help is appreciated.

assume there are such $ A\equal{} \cup_{k\equal{}1}^\infty I_k$

closed interval: $ I_k \equal{} [a_k ,b_k]$

disjoint: $ I_j \cap I_k \equal{} \emptyset  \to a_j,b_j \notin [a_k,b_k] \forall j \ne k$

$ \mu (I_k) \equal{} b_k \minus{} a_k$

$ \mu (A) \equal{} \sum_{k\equal{}1}^\infty \mu (I_k) \equal{}\sum_{k\equal{}1}^\infty  b_k \minus{}a_k \equal{} 1$

$ 1\equal{}\sum_{k\equal{}1}^\infty \mu (I_k) \le \sup b_k \minus{} \inf a_k \le 1 \minus{} 0 \equal{}1$

$ \to \sup b_k \equal{} 1, \inf a_k \equal{} 0$

having problems proving theres an open set in $ [0,1]\setminus A$

It seems really easy to construct such a sequence informally. Just choose one interval of length 1/2, then one or more of total length 1/4, then one or more of total length 1/8 etc... At each step there are only finitely many intervals and enough space so I don't see what the issue is


calculus
integration
topology
complex analysis
function
complex analysis unsolved

Not sure this is an appropriate question for this forum but here goes:

When is:

$\left(\int_{C_{1}}f(z)dz\right)\left(\int_{C_{1}}g(w)dw\right) \overset{?}{=}\iint_{C_{1}XC_{1}}f(z)g(z)dzdw$

and what would the double contour integral even mean?  It does pertain to a genuine problem I'm working on.

[quote="owk"]A contour integral $\int_\gamma f(z)\,dz$ can be interpreted as the integral over the circle of the pull-back $\gamma^{*}(f(z)\,dz)$ of the 1-form $f(z)\,dz$. In the case of two circles you get a torus and an exterior product of two 1-forms. Fubini may be applied locally. owk[/quote]

My goodness . . .  now where is that forum for beginners?  Really I don't mind being a beginner.  I accept that.   I just though this was not something usually encountered for beginners so I didn't post it there.  You wouldn't mind giving me an example  would you owk?.  In a way someone just out of a first semester Complex Variables class could understand.   Please don't tell me I'm suppose to know what you just said right out of class.  It was a very disappointing experience for me to begin with.  Let me see . . . I don't know  what "pull-back" means, "1-form", "exterior product of two 1-forms, or even "applied locally".  Sorry.  Thanks for your reply though.  I wouldn't mind if you attempted to explain these things either but as you wish I'm ok. :)

Ok, sorry. If you forget about contour integrals and are content with simple line integrals (i.e., non-closed curves), then Fubini tells you that
$\int_{a}^{b}f(\gamma(t))\gamma'(t)\,dt\int_{c}^{d}g(\delta(u))\delta'(u)\,du=\int_{[a,b]\times[c,d]}f(\gamma(t))g(\delta(u))\gamma'(t)\delta'(u)\,dt\,du$
for paths $\gamma\colon[a,b]\to\mathbb C$ and $\delta\colon[c,d]\to\mathbb C$. Does that answer your question? owk

The language owk used comes from the theory of manifolds- that is normally taught independently from complex analysis, and it's hard to find at all at an undergraduate level.

The original question you asked doesn't seem very interesting- a product of two independent integrals is just a product of integrals, after all. If that's $\int_{C_{1}}\int_{C_{1}}f(z,w)\,dw\,dz$, being able to apply Fubini's theorem is more interesting, since interchanging the order actually does something nontrivial.

If that $f(z,w)$ is holomorphic in each variable separately (on the torus), we have more than enough to apply Fubini, and the order can be reversed freely. One note: the torus (product of two circles) doesn't actually enclose a region- it's associated to the unit polydisk $D(0,1)^{2}$, but it's only the "corners" rather than the whole boundary.
The argument for applying Fubini is simple: just parametrize with a product of real intervals, and everything is continuous.

I'm so out of my league in here . . . she calls though. :blush:

This is my speed:

Let:
$f(z)=z$  and $g(z)=z$ (I'm a beginner remember) and wish to evaluate these analytic functions along the parabolic contour defined in the complex plane by $y=x^{2}$ from x=0 to x=2.

So on $C_{1}: z(t)=t+it^{2}$ and on $C_{2}: z(t)=t+it^{2}$

and form the product:

$\left(\int_{0}^{2}(t+t^{2}i)(1+2ti)dt\right)\left(\int_{0}^{2}(t+t^{2}i)(1+2ti)dt\right)=-28-96i$

but:

$\int_{0}^{2}\int_{0}^{2}(t+t^{2}i)(1+2ti)(u+u^{2}i)(1+2ui)dudt=-28-96i$

Edit:  Suppose that's really obvious now that I look at it.

 :oops:  :blush:

Now . . . what about the zeta function?

$\zeta(s)=\frac{1}{2\pi i}\int_{C}z^{-s}e^{-z}dz\int_{\multimap}\frac{w^{s-1}}{e^{-w}-1}dw;\quad Re(s)>0$

Seems to me I could do the same with it.  That's the original problem I was working on.  Thanks guys. :)

[quote="jmerry"]There's nothing wrong with writing that product as a double integral- it just doesn't [i]do[/i] anything, since you evaluate it as an iterated integral anyway, and that process naturally breaks up the integral as a product again.[/quote]

Yea.  You're right . . . started with my realization recently that the Gamma function is really a specific case of the more general contour integral function:

$\Omega(C,s)=\int_{C}z^{s-1}e^{-z}dz$

and I thought maybe we could combine both the integrals in the zeta expression (for $Re(s)>0$ anyway) into a single integral.

That whole idea: PNT,  Riemann's 1859 paper on primes, zeta, vonMangoldt's formula, Perrons's formula, to me is the most beautiful concept I have ever worked on in mathematics. :)


Estimate how far would the Earth move away from the Sun because of its radiation in a $100$ years. The power of the radiation of the Sun is $P=3.8\times 10^{26}W$.

Why does radiation make something move? I don't understand.

Is it something like this?

First find the energy of the radiation in $100$ years, then use $E=\frac{1}{2}M_{E}v^{2}$, where $M_{E}$ is the mass of Earth. Then use $S=(\frac{v}{2})t$, where $t=100$ years.


geometry
parallelogram
vector
geometry proposed

[quote="who"]let A1A2A3 given tri. T1T2T3 its extouch tri.with T1 on A2A3 etc.further H1
orthocenter of A1T2T3,similarly define H2,H3.show HiTi concurr i=1,2,3. :cool:[/quote]

Don't understand a word. Could you please rephrase it more clearly

Daniel

The only thing I did not understand was 'extouch triangle'. Searched in google but nothing came up. That's got to be one of his custom-made definitions....

Maybe it means the triangle that is formed by the intersection of a side and the ex-circle on it.

Let $X$ be such a point that its projections on the sides of $\triangle{ABC}$ are $T_1,T_2,T_3$ respectively.
Then the segments $XH_i$ pass through the midpoints of $T_{i+1}T_{i+2}$, thus $H_iH_jT_iT_j$ ($i\neq{j}$;$i,j\in\{1,2,3\}$) are parallelograms and the conclusion follows.

my  :| inelegent sol. used vectors
 but nice sailor- :first:


Prove that if two homogeneous systems of linear equations in two unknowns have the same solutions, then they are equivalent.

I'd like to discuss the problem in two cases:
the first, the systems have only zero solution;
the second, the systems have also non-zero solutions.
Am I right? I've no further ideas.

Recently discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=145026]here[/url]. Why is this popping up everywhere on this forum all at once?


When I write 36 in $\LaTeX$ I get the same thing I got when writing n, i.e. $36$, which normally is achieved by writing $[ABCDEF] = 6 \frac{a^2 \sqrt 3}{4}$.

Does this happen when writing other things, too?

Now that's peculiar! $36$ :| It's starting to look like a bug.

Okay this is a bug. perfect_radio in which post did this happen the first time?

Fixed temporary, but still uncertain as to how it happened  :?

Can someone show an example of such a duality? Seems interesting... It's extremely unlikely that 2 "normal" math expressions would have the same md5 hash.

They don't. None of them, but rather the image is generated wrong. As it was also perfect_radio that had the first error, in the same forum, I tend to belive there is a strange bug. We'll just have to observe further.

[quote="Peter VDD"]Then what is wrong? I see a perfect 36? What's the link with md5?[/quote]I fixed the image. If I knew what was wrong I would have fixed it :D


floor function
inequalities
combinatorics unsolved
combinatorics

[b]Problem[/b]:A graph G with n vertices is called  [i]great[/i] if we can label each vertex with a difference positive integer$\leq\frac{n^2}{4}$and there exists set D of positive integers so that there is an edge between 2 vertices iff the difference of their number label is in D.Show that if n is sufficiently large we can find a graph with n vertices which is not [i]great.[/i]
 I have no idea,can anyone help me?

Well, my idea is to show that the possible choices for numbering the vertices and for the set D are too few to cover all the possible graphs with n vertices, if n is sufficiently large.

Can you post your idea more clearly?Thanks.

'course.

In order to simplify the notation, define $X = \left\lfloor \frac{n^2}{4} \right\rfloor$.

It is clear from the definition that, if a set D and a numbering are given, then there is only one way to construct a great graph with that particular numbering and subset D: just draw an edge iff its difference is in D.

I will call this graph the great graph [i]associated[/i] to D and the numbering. It is obvious that a great graph is associated to at least one choiche for numbering + set D, by definition.

The possibilities for numbering the vertices are simply $\frac{X!}{(X-n)!}$ (here and in what follows I consider [i]different[/i] two graphs, each obtained by permutation of the other's vertices).

Obviously, it makes sense to consider only subsets D with elements < X. Infact, the maximum value for the difference is clearly X - 1, so if we add integers greater than X, we are not changing the graph associated. So, the possible choices for the numbering and D are no greater than

$A = \frac{X! \cdot 2^X}{(X-n)!}$

and so there cannot be more than A great graphs. This is a very rough estimate: in this way we are counting the same great graph a lot of times. Nevertheless, it is sufficient to prove the claim.

The total number of possible graphs are $B = 2^{\binom{n}{2}}$. If we prove that A < B, we have finished. Now it is a simple matter of algebra:

$\frac A B \leqslant \frac{X! \cdot 2^{\frac{n^2}{4}}}{(X-n)! \cdot 2^{\frac{n(n-1)}{2}}} \leqslant \frac{ \left( \frac{n^2}{4} \right)^n}{\left( 2^{\frac{n-2}{4}} \right)^n}$.

So, whenever $\frac{n^2}{4} < 2^{\frac{n-2}{4}}$, the claim holds. But this inequality can be written in the form

$n < C k^n$,

for positive constants C and k>1, so it is true for every n sufficiently large. []

M.

Nice solution it's use probabilistic method,right?
Where can i find an article about it

...er... well, no.

I simply say: the number of great graphs is no more than the possible different choices for (set D + numbering of vertices). Then I count how many different sets D I can have ($2^X$) and how many ways are there to number the vertices ($\frac{X!}{(X-n)!}$). In this way I obtain an upper bound for the number of great graphs. Then I compare this estimate with the total of possible graphs ($2^{\binom{n}{2}}$). I find that the possible graphs are too much (more than the possible great graphs), so there must be some graph not great. It is just counting.


A random variable X takes the values 0, 1, 2,.......... and   $ p_n$ =  Pr(X=n).

Let P(s) be the generating function    $ \sum_{0}^{\infty}p_ns^n$

If P(0) =0   and   $ P(\frac{1}{3}$) = $ \frac{1}{3}$  ,   find {$ p_n$}


superior algebra
superior algebra solved

Should be in College Playground -> Superior Algebra -> uUnsolved problem

Are you saying that di are ALL the divisors of 0 in the ring?

I think i saw something similar in a RMT (from 2001 i think) It belonged to Dorel Mihet.

I think it was like this : prove that if a finite ring has a finite number of divisors $d_1, d_2, ... , d_n$ of 0 and D is the set of these ,  $\sum d_i  \in D  \cup \{0,1\}$.

Proof goes like this : let $S =  \sum d_i$ . If S is invertible, then SD = D, so taking the sum of elements we get that $S^2 = S$, therefore S=1. Otherwise since $\sum d_i$ is not invertible, it is either 0 or a divisor of 0.

Proof of the fact that if b is in A, A finite and b is not invertible and non-zero then it is a divisor of 0. Consider f:A-->A , f(x) = bx. Suppose b isn't  a divisor of 0, then f is one to one on A. Therefore there is c such that bc=1. Consider the set M={x in A|bx=1}. M is non-void because c is in M. Also since M is included in A, it's finite. Consider g:M->A , g(x) = xb + c - 1; bg(x) = b(xb+c-1) = (bx)b + (bc) - b = b + 1 - b = 1.  g(x)=g(y) implies xb=yb ==> xbc = ybc , so x=y. Thus g is one to one on M, so there is a in M such that g(a)=c ==> ab + c - 1 = c ==> ab = 1. But abc = a(bc) = (ab)c = a = c ==> bc=cb=1, so b is invertible, contradiction.

[quote="valica"]In a finite ring we have d1,d2,..,dn divisors of 0.Prove that d1d2...dn is a divisor too.[/quote]

Lets solve valicas problem. I see he deals with $D=d_1d_2\cdot \ldots \cdot d_n$ which I think it is a product.

As iandrei showed, if D is not a divisor of 0, then D is invertible (I add 0 to the divisors of 0, but let it out of the $d_i$s). He also proved that if D is left or right invertible, it is left and right invertible.
So assume D is not a divisor of 0. Then there exists c in the ring such as cD=Dc=1.
So $(c\cdot d_1\cdot d_2\cdot \ldots\cdot d_{n-1})\cdot d_n=1$, so $d_n$ is invertible (a contradiction).


inequalities
number theory proposed
number theory

Solve for $ p,q,n$ where $ p(p \plus{} 1) \plus{} q(q \plus{} 1) \equal{} n(n \plus{} 1)$ where $ p,q$ are primes and $ n$ is a natural number.
If this is a well known problem, please give the source for this problem.

we get $ p(p \plus{} 1) \equal{} (n \minus{} q)(n \plus{} q \plus{} 1)$.obviously we can't have $ p|n \minus{} q$ because in this case we'll have $ p \plus{} 1 \equal{} \frac {n \minus{} q}{p}.(n \plus{} q \plus{} 1) \ge n \plus{} q \plus{} 1 \implies p \ge n \plus{} q$ which is a clear contradiction with $ p | n \minus{} q$.
in fact we can prove it in the same way $ p \ge n \minus{} q \implies p \plus{} q \ge n$.so we must have $ n \minus{} q | p \plus{} 1,p |n \plus{} q \plus{} 1$.
similarly we musthave $ q | n \plus{} p \plus{} 1 \implies q|n \plus{} p \plus{} q \plus{} 1,p |n \plus{} p \plus{} q \plus{} 1$.the case $ p \equal{} q$ is trivial.
so we must have $ pq | n \plus{} q \plus{} p \plus{} 1 \implies pq \le n \plus{} q \plus{} p \plus{} 1$.the rest is trivial by using inequalities and the fact that $ p \plus{} q \ge n$.

Superb. Had the same solution.
Infact I assumed $ n\plus{}q\plus{}1\equal{} kp$ and $ p\plus{}1\equal{} k(n\minus{}q)$.
Also, we can conclude that $ p\plus{}q \ge n$ by simply saying that  $ (p\plus{}q)(p\plus{}q\plus{}1) \ge n(n\plus{}1)$.
 Using the earlier divisibiliy bounds and by without loss of generality $ q\le p$ bound along with  $ p\plus{}q \ge n$ bound, we can say that $ 2\le k \le 3.$
CHecking both cases and solving the diophantines, we get the solutions.


function
limit
real analysis
real analysis unsolved

$ f$ is a function on $ \mathbb{R}$.
\[ E_n\equal{}\left\{x\in \mathbb{R}\left|\omega_f (x)\geq \frac{1}{n}\right.\right\}\]

I got. only to prove $ E_n^c$ is open as the usual approach.

In other words, the function $ \omega_f(x): \equal{} \lim_{\epsilon\to 0}(\sup_{[x \minus{} \epsilon,x \plus{} \epsilon]}f \minus{} \inf_{[x \minus{} \epsilon,x \plus{} \epsilon]}f)$ is [url=http://en.wikipedia.org/wiki/Semi-continuity]upper semi-continuous[/url]. Next step: prove that for any upper semicontinuous function $ g\colon\mathbb R\to [0,\infty)$ there exists $ f\colon \mathbb R\to\mathbb R$ such that $ \omega_f \equal{} g$. Or disprove.  :)

Incidentally, I edited the definition of $ \omega$ twice because I was not sure what the "canonical" definition is. Two other possibilities are $ \limsup_{y\to x}|f(y) \minus{} f(x)|$ and $ \limsup_{y\to x}f(y) \minus{} \liminf_{y\to x}f(y)$, which are both different from the above definition..


find all positive integers $n$ such that the following is a perfect squre:

$2^8+2^{11}+2^n$

There are an infinite number of solutions for this. Do you mean find all integral $n$? :huh:

You can easly find all $n\leq 11$. If $n\geq 12$, we have $2^8+2^{11}+2^n=2^8(1+8+2^k)$, so number $9+2^k$ must be perfect square. So $2^k=m^2-9=(m-3)(m+3)$, and both $m-3$ and $m+3$ must be of the form $2^l$, so $m=5$, and $n=12$.

[quote="me@home"]i don't see how n<11 is easy to find...[/quote]

Substitute and check.  ;)

But you are not proving that that is the only solution, sen, you are just finding one solution and that's it.

since $2^8+2^{11}$ is a square....

using the  [hide="generating formula "] $m^2-n^2, 2mn, m^2+n^2$  [/hide]for pahygorean triples might be useful

[quote="JBL"]That does not generate all Pythagorean triples -- in particular, 9-12-15 cannot be expressed in that form.

Goblin's solution is most likely the nicest anyone is going to find, however.[/quote]

If you really insist, we could always use [hide="the slightly expanded Pythagorean triple generating formulae"]$2cmn, cm^2 - cn^2, cm^2 + cn^2$, where $gcd(c, m, n) = 1$ and $c$ cannot be expressed as a sum of two squares[/hide], but I doubt it would lead to a simple solution.   :P

yes, the [i]primitive[/i] pythagorean triple formula is much nicer, where 9,12,15 isn't primitive because it is similar to the primitive 3-4-5.

[quote="JBL"]That does not generate all Pythagorean triples -- in particular, 9-12-15 cannot be expressed in that form.

Goblin's solution is most likely the nicest anyone is going to find, however.[/quote]
Yes it's a nice solution ..

But ..  why even worry about n>11 etc..  n>8 would do.
Even more, simpler, perhaps would be:

To  notice that to have $9 \cdot 2^8 + 2^n = m^2$
 $2^n = (m+3\cdot 2^4)(m-3\cdot 2^4)$
 each has  factor has to be a power of 2
so ..
If $2^a$ is first power and $2^b$ is second factor ..(a >b)  and $2^a-2^b = 6 \cdot 2^4$
and we $2^b(2^{a-b} -1)= 3 \cdot 2^5$ making    (a-b)= 2,  , b= 5 so n = a+b = 12 ..etc..


read the question  ;)

I just really like listening to music. I don't like playing music, just the idea of doing something physical doesn't fit with me. I like writing and studying it too.

[quote="AGSpammer"]You should add another choice:

"All of the above."

I'm like that.[/quote]

Listening...

I like to play music a lot more than listening, and I have tried and failed at writing music.

I attempt writing music to play on my violin or my piano, but they never sound good.

I write music, but I should do more of the others...

Writing really comes after years of playing and listening.

I like playing a bit more than listening. ♪♫

I prefer listening: when playing, you need to focus on the technicality of what you're playing, and only on the part of the piece you're playing, but when listening, you can enjoy the song as a whole.

[quote="José"]Let me add another option: Listen to music, singing and dancing... That's really nice too! You can sing it in a loud volume (at a party) and no one can stop you! Yeah, fantastic option lol![/quote]

u can dance?

i cant, maybe i dont know how to do it. :D

I like doing all of the poll options, but listening is something that is most easily done. When you're really tired you don't feel like practicing, but listening to music for a while is relaxing.

I cannot write music at all Ive tried  :blush:
Though, i can play music but also listening to music gives you more inspiration to play.

Playing and listening, both...but I said playing because it was the first on the list

Noting that I know nothing but how to play, I would say playing music.  (You wonder how one fails to listen.)  :ninja:

I enjoy doing all of them. I am not a good music writer and my pieces sound awkward, but i enjoy it. I play a lot and it tends to relax me. I listen to something most of the time and I REALLY enjoy learning about and studying music. I am clicking on the studying music link since it had been neglected.

I hate playing music(and i don't know why i applied for band in high school this year).
But I just love listening to nice music(rock, classical,country,etc) :lol:

I've tried writing a few violin songs but none of them are that great.  :(
I like listening the best though by far.  :D


geometry
circumcircle
incenter
inequalities
vector
triangle inequality

Let ABC be a triangle with AB=2a, AC=2b, BC=a+b. Let D be a point on BC such that AD bisects angle BAC. Prove that OI is perpendicular to AD where O, I are the circumcenter and incenter of ABC, respectively.

[quote="probability1.01"]I'm pretty sure that's not true...[/quote]

why?(of course a does not equal to b, and the three sides satisfy triangle inequality)

[quote="probability1.01"]Based on a fairly accurate sketch, it's way off... consider a right triangle for example.[/quote]

I don't see why a right triangle won't work... Perhaps you are thinking the hypotenuse can be a+b? But the hypotenuse can't be a+b because a+b can't be the longest side.

[quote="probability1.01"]Consider a right triangle, where angle A is right.  The incenter lies on AD.  The extension of AD hits the circumcircle at E, and since OE is perpendicular to BC, you can see pretty easily that the point on AD that makes a perpendicular when connected to O is not inside the triangle.  There are other counterexamples.[/quote]

Angle A can't be right angle, you are forgetting that  AB=2a, AC=2b, BC=a+b, and BC can't be the longest side(WLOG assume a>b, 2a>a+b). Remember this is a specific triangle(not just any triangle)!

[hide]
The problem can be restated as follows: OI is perpendicular to AD if b+c=2a where a,b,c are the side lengths of the triangle.

This is a typical vector problem, let the circumcenter O be the origin, then D=(bB+cC)/(b+c) and I=(aA+bB+cC)/(a+b+c). AD is parallel to bB+cC-(b+c)A and OI is parellel to aA+bB+cC. Then just calculate the dot product $(bB+cC-(b+c)A)\cdot(aA+bB+cC)$. Proving this to be 0, you will need $A\cdot A=R^2$ and $2B\cdot C=2R^2-a^2$, etc.
[/hide]

[quote="probability1.01"]Hmmm... I convinced myself that some right triangle would satisfy the conditions though.  And any right triangle does not work... maybe I lost brain cells after the USAMO.[/quote]

I hope you still remember a triangle has three angles :-)

The condition of the problem implies that A is not the largest angle (unless it's a equilateral).

[quote="probability1.01"] maybe I lost brain cells after the USAMO.[/quote]

Hehe that's me! I have trouble adding recently lol.


algebra
polynomial
complex numbers
algebra proposed

I rate this quickie at [i]3 minutes[/i]: find, for each $k$, a linear recurrence in which every integer appears exactly $k$ times.

Many will find this too boring. Here is a slightly more interesting (but still trivial) question that I rate at [i]30 minutes[/i]: prove that a linear recurrence of complex numbers [b]cannot[/b] contain all rational numbers.

Nevertheless, for those who [i]really[/i] want to challenge themselves, try this (I rate it at [i]3 days[/i]  ;) ): [i]prove that there does not exist a linear recurrence in which each integer occurs infinitely often. [/i]

--Vesselin

[quote="vess"]I rate this quickie at [i]3 minutes[/i]: find, for each $k$, a linear recurrence in which every integer appears exactly $k$ times.[/quote]

Well,  if $a_n = a_{n-k} + 1$ the sequence also satisfies a homogeneous linear recurrence...   (30 seconds)

[b] Added: [/b]  I see from Pascual's solution that you meant all positive and negative integers.  Well, one can look at a second sequence with $b_k = b_{n-k} - 1$ for the negatives, and there is certainly a sequence with $c_{2n} = a_n$ and $c_{2n+1} = b_n$ (for any pair $a,b$ of linear recursive sequences).

[quote]Many will find this too boring. Here is a slightly more interesting (but still trivial) question that I rate at [i]30 minutes[/i]: prove that a linear recurrence of complex numbers [b]cannot[/b] contain all rational numbers.[/quote]

We may assume the recurrence has rational coefficients (by "specialization", in case of transcendental coefficients, and "restriction of scalars", in case some of the algebraic coefficients are nonrational).   Multiplying $a_n$ by appropriate rational $c^n$ we can reduce to the case where characteristic polynomial is monic with integer coefficients.   Then  $a_n c^n $ is some integer linear combination of the first several $a_i$'s, so only a finite number of primes can appear in the denominator of $a_n$.
So it cannot be all rational numbers.   (2 minutes)

[quote]Nevertheless, for those who [i]really[/i] want to challenge themselves, try this (I rate it at [i]3 days[/i]  ;) ): [i]prove that there does not exist a linear recurrence in which each integer occurs infinitely often. [/i] [/quote]

May take me 33 days...

Dear Vess, do you have a solution for the problem with the recurrent sequence that contains each integer infinitely many times and that does not use the Skolem-Mahler-Lech theorem? I would be highly interested. I managed to prove using that theorem that for a recurrent sequence there are just a finite number of values that are takeen infinitely many times. This implies your problem, but the problem is that I used the atomic bomb ( hope Pierre is not angry if I use his words) to prove that.

:lol:  I didn't manage to solve it without the Skolem-Mahler-Lech theorem (or at least without the ideas & techniques used to prove it)   :(. This is why I said this problem is extremely difficult (on its own, otherwise it's an immediate corollary of the theorem  :) ). I, too, would be very curious to see an alternative solution (at least one that does not plunge into $p$-adic analysis, though I doubt that such exists ... ). Anyway, an elementary solution to this is an open problem for me. Fresh ideas are (always) welcome!   :D


[quote="bigknick"]Prove that if the medians m[size=59]1[/size],m[size=59]2[/size], and m[size=59]3[/size] of a triangle satisfy the following equation m[size=59]1[/size] squared + m[size=59]2[/size] squared equals 5m[size=59]3[/size] squared the this triangle is a right triangle.[/quote]

$ m_{1}^{2}\plus{}m_{2}^{2}\equal{}5m_{3}^{2}$  <->  $ \frac{1}{4}(2(a^{2}\plus{}b^{2})\minus{}c^{2})\plus{}\frac{1}{4}(2(b^{2}\plus{}c^{2})\minus{}a^{2})\equal{}5\frac{1}{4}(2(a^{2}\plus{}c^{2})\minus{}b^{2})$ ---> $ b^{2}\equal{}a^{2}\plus{}c^{2}$  <->    right triangle


AMC
USA(J)MO
USAMO
AIME
ARML
AMC 10
probability

I am just wondering: How is the state qualifier chosen? Is is AMC 12+10*AIME or both AMC 10 or AMC 12 + 10*AIME?

[quote="solafidefarms"][quote="Iversonfan2005"][quote="Phelpedo"]Wyoming's highest AIME score was a 2, second place as a 1. If you really wanted to take the USAMO, I guess you could move there :P[/quote]

You have got to be kidding me! :o , 3 AIME qualifiers and the scores were a 1 and a 2!!!!

Did only like 2 schools take the AMC or something????[/quote]

You are correct. Actually, only 2 schools [i]took[/i] the AMC in Wyoming, and both had at least 1 AIME qualifier. :)[/quote]

Wow, that IS just sad...... must be because of the NCLBA  :roll:  ;)

[quote="h_s_potter2002"]It's not really that suprising, considering that Wyoming also has the smallest population out of all the US states.[/quote]

But Alaska, which has barely a larger pop. had about 20x as many qualifiers.......  :roll:

[quote="h_s_potter2002"]Alaska actually has a population about 25% higher than Wyoming's, and has the same number of qualifiers (but that's only because of the every state has to have a qualifier thing.  Otherwise, Alasks would probably have 1 qualifier, and Wyoming would have none).[/quote]

I kind of meant aime qualifiers, but whatever... :lol:

[quote="Xantos C. Guin"]Both.

I probably had the highest non-qualifing index from GA.  :)[/quote]

Close, but not quite. Guy from Macon got a 115/10.

When and if (with a 213 index, that oughta be "when") you get on the ARML team you'll be nigh-omniscient.  :ninja:

Just move almost anywhere that isn't Asia, Russia, or Eastern Europe and I imagine you'll have a better chance of making IMO than from the US.

And yes, Xantos, I am stalking you and replying to every post you make.  :ninja: I coulda sworn you'd posted a lower score than that somewhere... Ah well.

[quote="Xantos C. Guin"]Actually, I had a 146.5/7 = 216.5 index. But that is an AMC10 index, not as important. My point with that post was that if the 8 qualifiers from GA weren't there (:lol:, like that would happen) I would be the state qualifier.

Somehow when I saw that you replied to this topic I had a feeling that you would quote my post.[/quote]

Why don't you move to Wyoming then? I've heard it's lots of fun there. My father doesn't wish for us to move out there quite yet though. :( :D

[quote="SamE"][quote="Fermatprime"]Just move almost anywhere that isn't Asia, Russia, or Eastern Europe and I imagine you'll have a better chance of making IMO than from the US.[/quote] Isn't Russia in Asia? Or if it isn't, it's in Eastern Europe? Or both?[/quote]

Most of it is in asia, but it's capital is in Europe, so it's officially in Europe (eastern).

Either way, it's a subset of $\{\text{Asia}\}\cup\{\text{Eastern Europe}\}$. :D

[quote="MysticTerminator"]yeah and you know what if you want to make IMO just move to like Ireland or something actually I dunno how good the Irish math team is but judging by their olympiads...

then again I probably shouldn't judge by their olympiads; I think the Canadian one is pretty easy but they're really good (at least from what I understand of these penguins and tsimerman and all)[/quote]

Fiachra posts in the olympiad section and is (was) on the Irish IMO team.  He's obviously very bright.

Well, I'm sure if you move somewhere other than the Americas, Asia, Europe, Africa, Australia, and Antarctica, you'll easily qualify for the IMO team.

[quote="Xantos C. Guin"]Those are the only places to qualify for the IMO.

Antarctica should have an IMO team, but the UN doesn't consider us as a country.

Oh well, our IMO team wouldn't be good anyway since most penguins are unable to hold pens.[/quote]

Knowing Probability, I think he was kidding around :rotfl:  :|


If $a,b,c \ge0$ prove or disprove that
$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}}\ge\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{2}$
and If $a,b,c\ge0$ and $a+b+c=1$ prove or disprove that
$ab^{2} +bc^{2}+ ca^{2}\ge ab+bc+ca -\frac{2}{9}$

[quote="chidat"]If $a,b,c \ge0$ prove or disprove that
$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}}\ge\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{2}$
[/quote]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=56071]http://www.mathlinks.ro/Forum/viewtopic.php?t=56071[/url]

For the second, just homogenize and it should be straightforward from there.

It becomes

$2(a+b+c)^3+9(ab^2+bc^2+ca^2) \ge 9(ab+bc+ca)(a+b+c)$
$2(a^3+b^3+c^3)+6(ab^2+bc^2+ca^2) \ge 3(a^2b+b^2c+c^2a)+15abc$

which can be finished off with AM-GM and Rearrangement:

$\frac{3}{2}a^3+\frac{3}{2}ab^2 \ge 3a^2b$ and so forth.


A nickel is placed on a table. The number of nickels which can be placed around it, each tangent to it and to two others is:

$ \textbf{(A)}\ 4 \qquad\textbf{(B)}\ 5 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12$

[hide="Solution"]$ 6$ other nickels can be placed around it in a hexagonal formation, so the answer is $ \mathrm{(C)}$.[/hide]


induction
inequalities
limit
real analysis
real analysis unsolved

Let $ a_1 > 0, a_2 > 0$ and $ a_{n \plus{} 1} \equal{} \frac {2}{a_n \plus{} a_{n \minus{} 1}}$ for $ n\geq2$.Show that the sequence $ \{a_n\}$ is convergent.

Hello,I am sorry for my ugly proof,I hope someone can post a proof nicer :)

First I may prove some primary results:
1.If $ a_1 \equal{} a,a_2 \equal{} b$,then:$ \min\{a,b,\frac {2}{a \plus{} b}\} \le a_n \le \max \{a,b,\frac {2}{a \plus{} b}\}$
Proof by induction,If $ \min\{a,b,\frac {2}{a \plus{} b}\} \le a_{n \minus{} 1},a_n \le \max \{a,b,\frac {2}{a \plus{} b}\}$
Then,$ \min\{a,b,\frac {2}{a \plus{} b}\} \le a_{n \plus{} 1} \le \max \{a,b,\frac {2}{a \plus{} b}\}$(distinguish the case for which of $ a,b,\frac {2}{a \plus{} b}$ will be the maximum or minimum,it is not hard.)
We may assume:$ m \equal{} \min\{a,b,\frac {2}{a \plus{} b}\},M \equal{} \max \{a,b,\frac {2}{a \plus{} b}\}$
2.for every positive integer $ n$,$ a_n \minus{} 1,a_{n \plus{} 1} \minus{} 1$ have different signs,or $ a_{n} \minus{} 1,a_{n \minus{} 1} \minus{} 1$ have different signs.(include one of them is $ 0$)
The proof of $ 2$ is obvious.
Return the problem,
$ a_{n \plus{} 1} \minus{} 1 \equal{} \frac {1 \minus{} a_n \plus{} 1 \minus{} a_{n \minus{} 1}}{a_n \plus{} a_{n \minus{} 1}}$
$ a_{n \plus{} 1} \plus{} 1 \equal{} \frac {1 \plus{} a_n \plus{} 1 \plus{} a_{n \minus{} 1}}{a_n \plus{} a_{n \minus{} 1}}$
Thus:
$ |\frac {a_{n \plus{} 1} \minus{} 1}{a_{n \plus{} 1} \plus{} 1}| \equal{} |\frac {1 \minus{} a_n \plus{} 1 \minus{} a_{n \minus{} 1}}{1 \plus{} a_n \plus{} 1 \plus{} a_{n \minus{} 1}}| \le \max\{|\frac {a_{n} \minus{} 1}{a_{n} \plus{} 1}|,|\frac {a_{n \minus{} 1} \minus{} 1}{a_{n \minus{} 1} \plus{} 1}|\}$

Then let  $ b_n \equal{} \max\{|\frac {a_{n} \minus{} 1}{a_{n} \plus{} 1}|,|\frac {a_{n \minus{} 1} \minus{} 1}{a_{n \minus{} 1} \plus{} 1}|\}$
If,$ a_n \minus{} 1,a_{n \minus{} 1} \minus{} 1$ has the different signs,then the above inequality can be stronger by:
$ |\frac {a_{n \plus{} 1} \minus{} 1}{a_{n \plus{} 1} \plus{} 1}| \equal{} |\frac {1 \minus{} a_n \plus{} 1 \minus{} a_{n \minus{} 1}}{1 \plus{} a_n \plus{} 1 \plus{} a_{n \minus{} 1}}| \le \frac {1 \plus{} M}{2 \plus{} m \plus{} M}\max\{|\frac {a_{n} \minus{} 1}{a_{n} \plus{} 1}|,|\frac {a_{n \minus{} 1} \minus{} 1}{a_{n \minus{} 1} \plus{} 1}|\}$
(This is because
$ |\frac {1 \minus{} a_n \plus{} 1 \minus{} a_{n \minus{} 1}}{1 \plus{} a_n \plus{} 1 \plus{} a_{n \minus{} 1}}| \le |\frac {1 \minus{} a_n}{1 \plus{} a_n \plus{} 1 \plus{} a_{n \minus{} 1}}| \le \frac {1 \plus{} M}{2 \plus{} m \plus{} M} |\frac {1 \minus{} a_n}{1 \plus{} a_n}|$ or
$ |\frac {1 \minus{} a_n \plus{} 1 \minus{} a_{n \minus{} 1}}{1 \plus{} a_n \plus{} 1 \plus{} a_{n \minus{} 1}}| \le |\frac {1 \minus{} a_{n \minus{} 1}}{1 \plus{} a_n \plus{} 1 \plus{} a_{n \minus{} 1}}| \le \frac {1 \plus{} M}{2 \plus{} m \plus{} M} |\frac {1 \minus{} a_{n \minus{} 1}}{1 \plus{} a_{n \minus{} 1}}|$)
Thus:if $ a_n \minus{} 1,a_{n \minus{} 1} \minus{} 1$ have different signs,then,$ |\frac {a_{n \plus{} 1} \minus{} 1}{a_{n \plus{} 1} \plus{} 1}| \le cb_n$,where $ c \equal{} \frac {1 \plus{} M}{2 \plus{} m \plus{} M} < 1$
Then if  $ a_n \minus{} 1,a_{n \plus{} 1} \minus{} 1$ also have different signs,we can still prove:$ |\frac {a_{n \plus{} 2} \minus{} 1}{a_{n \plus{} 2} \plus{} 1}| \le cb_n$
Otherwise suppose $ a_n \minus{} 1,a_{n \plus{} 1} \minus{} 1$ has negative sign(You can do similar when they all have positive sign)
$ \frac {a_{n \plus{} 2} \minus{} 1}{a_{n \plus{} 2} \plus{} 1} \equal{} \frac {1 \minus{} a_n \plus{} 1 \minus{} a_{n \plus{} 1}}{1 \plus{} a_n \plus{} 1 \plus{} a_{n \plus{} 1}}\le \frac {b_n(1 \plus{} a_n) \plus{} cb_n(1 \plus{} a_{n \plus{} 1})}{1 \plus{} a_n \plus{} 1 \plus{} a_{n \plus{} 1}}$
$ \le b_n \frac {1 \plus{} M \plus{} c(1 \plus{} m)}{1 \plus{} M \plus{} 1 \plus{} m} \equal{} db_n$,where $ d \equal{} \frac {1 \plus{} M \plus{} c(1 \plus{} m)}{1 \plus{} M \plus{} 1 \plus{} m}$
Now,combined to 2.,we have actually prove that $ b_{n \plus{} 3} \le db_{n}$ and $ b_n$ decreasing,then it is not to see $ \lim_{n \to \plus{} \infty}b_n \equal{} 0$,
Thus: $ \lim_{n \to \plus{} \infty}a_n \equal{} 1$,

Here is basically the same idea but realized in a slightly different way.

1) If $ a_n,a_{n\plus{}1}\in[x,1/x]$, then all $ a_k$ ($ k\ge n$) lie in this interval. (simple induction)

2) If $ a_n\in[x,1/x]$ and $ a_{n\plus{}1}\in(x,1/x)$, then $ a_{n\plus{}2}\in (x,1/x)$.

3) If $ a_n,a_{n\plus{}1}\in[x,1/x]$ and $ a_{n\plus{}1}\ne a_n$, then $ a_{n\plus{}2}\in (x,1/x)$.

4) Let now $ X$ be the supremum of such $ x$ that $ a_n\in[x,1/x]$ for all sufficiently large $ n$.  (1)-(2) imply that all $ a_n$ should be either approximately $ X$ or approximately $ 1/X$ (if any $ a_n$ with large $ n$ lies deep inside $ [X,1/X]$, then $ a_{n\plus{}1}$ also does and $ X$ is not a supremum). $ (3)$ implies that all $ a_n$ with large indices should be approximately the same. This is already enough to conclude that the limit exists (and is either $ X$ or $ 1/X$ but it is nice to note that if $ a_n$ and $ a_{n\plus{}1}$ are about $ X$, then $ a_{n\plus{}1}$ is about $ 1/X$ and vice versa, so the only possibility is $ X\equal{}1$, which immediately tells that the limit is $ 1$.


calculus
integration
algebra
function
domain
abstract algebra
Ring Theory

If $\phi : A\to B$ is an surjective homomorphism of integral domains, is it possible to define a homomorphim $\hat{\phi}: K(A)\to K(B)$?

you need to assume $\phi$ injective for the natural choice of $\hat{\phi}$.

Are $K(A)$ and $K(B)$ the quotient fields?

As an example, what about letting $\phi$ be the quotient map from $\mathbb{Z}$ to $\mathbb{Z}/p\mathbb{Z}$? If $K$ are quotient fields, there's definitely nothing there.

hi there,
I presume the ones u were talking about are quotient fields,so is'nt it like localising at the prime ideal zero,so wo'nt it be true that u could extend the map iff kernel is zero  :maybe:

Yes, you are right.. so we have that isomorphic groups give isomorphic quotient fields.. not very exciting really. Thanks anyway..


How many integers from 1 to 1000, inclusive, are divisible by neither 2 nor 5?

[hide] There are $ 500$ positive even integers $ \le1000$, and $ 200$ positive multiples of $ 5$ which are $ \le1000$. However, there are also $ 100$ positive multiple of $ 10$ which are less than $ \le1000$. Thus, by PIE, $ 1000\minus{}500\minus{}200\plus{}100\equal{}\boxed{400}$.[/hide]

[quote="gaussintraining"]  Thus, by PIE, $ 1000 \minus{} 500 \minus{} 200 \plus{} 100 \equal{} \boxed{400}$.[/quote]
What's PIE?


Find the minimum value of
$ 2a^2\plus{}2b^2\plus{}10c^2\minus{}6a\minus{}4c\minus{}2ab\minus{}6bc\plus{}20$
for $ a,b,c$ are real.


what's "non-Euclidean geometry"?
i read its definition from the wikipedia, but i couldn't understand it clearly

please give me a short and easier explanation with some examples...

thanks in advance!  :)

Roughly speaking, in non-Euclidean geometry, the parallel postulate doesn't hold.  That is, within a two-dimensional plane, for any given line l and a point A, which is not on l, there is exactly one line through A that does not intersect l.  An nice example of a non-Euclidean surface is a sphere.  You should know that if you draw a triangle, it will always contain $ 180^{\circ}$.  Now try drawing that same triangle on the surface of a sphere; you'll find it impossible to construct a triangle with $ 180^{\circ}$.  Rather, it will always have more.  If you construct a triangle out of 3 perpendicular lines (yes, it's possible), it will have $ 270^{\circ}$.  Hope that clears it up a little.  In AoPS Volume 2, there is a nice little piece on non-Euclidean geometry.

Although, one problem with spherical geometry is that it violates axiom 2 (lines may be extended indefinitely). An example which follows the first four but not the fifth is hyperbolic geometry. In here, the angles of a triangle always add up to less than $ 180^\circ$. In general, the first four axioms can prove that the angles of a triangle add up to at most 180, and the fifth completes it to exactly 180.


[hide="Solution"]Let the numbers on the cards be $ a, b, c$ and $ d$.  We have
\[ a \plus{} b \plus{} c \equal{} \minus{} 8.5 \\
b \plus{} c \plus{} d \equal{} \minus{} 6.5 \\
c \plus{} d \plus{} a \equal{} \minus{} 5.5 \\
d \plus{} a \plus{} b \equal{} \minus{} 3.5\]
Adding all four equations together yields us
\[ 3(a \plus{} b \plus{} c \plus{} d) \equal{} \minus{} 24\implies a \plus{} b \plus{} c \plus{} d \equal{} \minus{} 8\]
Now subtracting each of the first four equations from our final one (one at a time) yields us
\[ d \equal{} 0.5 \\
a \equal{} \minus{} 1.5 \\
b \equal{} \minus{} 2.5 \\
c \equal{} \minus{} 4.5\]
Obviously the largest number is $ \boxed{0.5}\implies \fbox{\textbf{c)}}$.[/hide]

[quote="joshisunny16"]Tom has 4 cards. Each card has a different number written on it.
The totals of any 3 cards are -8.5, -6.5, -5.5, and -3.5
what is the number on the card with largest value?
a)-4.5
b)-1
c)0.5
d)5[/quote]

Hint: Just set $ a\plus{}b\plus{}c\equal{}\minus{}8.5, a\plus{}b\plus{}d\equal{}\minus{}6.5, b\plus{}c\plus{}d\equal{}\minus{}5.5, a\plus{}c\plus{}d\equal{}\minus{}3.5$ and you have a symmetric system.

well lets right them down.

a+b+c=-8.5
a+b+d=-6.5
a+c+d=-5.5
b+c+d=-3.5

well what we do is we add the equations all up and we get

3a+3b+3c+3d=-24

now we take the equations and what we do is substitute it into other equations

c=-8.5-a-b
d=-6.5-a-b
a=-5.5-c-d
b=-3.5-c-d

so now we choose the the equations we want to substitute them in

d=-6.5-(-5.5-c-d)-(-3.5-c-d)

d=2.5+2c+2d

and you continue to do it to 1 or two more than you use elimination to find the letters.


$ ABC$ is an acute triangle and $ AB$is not equal to $ AC$.the circle that $ BC$ is it 's diameter intersect $ AB$,$ AC$ at $ M$,$ N$ respectively.$ O$ is the midpoint of $ BC$.the bisectors os $ <MON$ and $ <BAC$ meet each other at $ R$.prove that circumcircles of $ BMR$ and $ CNR$ meet each other at a point on $ BC$ Side.

it's imo 2004 #1: http://www.mathlinks.ro/viewtopic.php?t=14020


Fie $a,b,c$ numere pozitive astfel incat $abc=1$. Demonstrati ca:

$\sum{\frac{a}{a^{2}+2b^{2}+3}}\leq \frac{1}{2}$

                                                                            [i]Mircea Lascu,Zalau[/i]


Let $\omega (n)$ denote the number of prime divisors of the integer $n>1$.
Find the least integer $k$ such that the inequality $2^{\omega (n) } \leq k \cdot  n^{\frac 1 4}$ holds for all $n > 1.$

Pierre.

k=5.

Consider n = 30030, the product of all primes up to 13.  If n has any more primes, RHS increases by a factor of > 2 while LHS increases by a factor of 2.  Taking factors out of n will also only make the inequality stronger.

30030 > 13^4.
5 * 13 > 2 ^ 6.


Sry i do not know how to use latex.

so Determine the aveerage rate of change of the function between the given values of the variable

 f(x) =x^2+x; x=h-1, x=3

If u surround the math with dollar signs, u will get the LaTeX to appear, and for simple expressions like these, no extra work is needed:

$x^2+x; x=h-1, x=3$

But what is the problem asking?

calculus does not belong in this forum, go to calculus computations and tutorials

[url]http://www.artofproblemsolving.com/Forum/index.php?f=296[/url]

i am not that strong in calculus, but i think that the answer is just

$\frac{3^2+3-[(h-1)^2+(h-1)]}{x-(h-1)}$ i.e.

$\frac{f(x)-f(h-1)}{x-(h-1)}$


abstract algebra
vector
superior algebra
superior algebra theorems

hello

in this proof of the hilbert nullstellensatz, in which i am interested now,
they speak of a field K which is a field extension of the field k , and finitely generated as a k-algebra
the first theorem states that it automatically is algebraic over the subfield k
what do they mean with , finitely generated as a kalgebra
what is a kalgebra? :?
do they mean that it is generated by a finite set of elements
(i'd doubt it because Q( e) is also generated by one element and is certainly not algebraic)
do they mean that it is a finite dimensional vectorspace over k( in which case the theorem seems very logic to me)
can anyone explain me?

I think an algebra over a ring $A$ is an $A$-module which is also a ring and satisfies the extra axiom $a(xy)=(ax)y=x(ay)$. I think that by finitely generated they mean finitely generated as a module (or, in your case, vector space, since you have a field instead of simply a ring).

Let $F$ be a field. An $F$-algebra (also called "an algebra over $F$") is a nonempty set $\mathcal{A}$, together with three operations, called addition(+), multiplication and scalar multiplication (both denoted by juxtaposition), for which the following propertities hold
(1) $\mathcal{A}$ is a linear space under addition and scalar multiplication
(2) $\mathcal{A}$ is a ring under addition and multiplication
(3) if $r\in F$ and $a,b\in\mathcal{A}$ then $r(ab)=(ra)b =a(rb)$

A finitely generated R-algebra (R ring) means that it is naturally isomorphic to $R[X]/I$ where $X$ is a finite set of indeterminates.


inequalities
integration
inequalities proposed

Let $k(n)$ be the smallest constant such that for any positive numbers $ a_1,...?a_n$ whose sum of inverses is 1 we  have the inequality $\frac{1}{a_1+a_2}+\frac{1}{a_2+a_3}+...+\frac{1}{a_{n-1}+a_n}\leq k_n$. Prove that $ k_n-\frac{1}{2}=O(\frac{1}{n})$. Open question: find this constant!

Ok! First part of your question is obvious.

I just adore how you make people feel they are the most idiot creatures in the world, Myth. But do you know why I like this? Because I am really an exponent of that category. Ok, to increase just a bit the level ( altough I am sure you will also find this more than trivial), I propose the following: $k_n$ is at most ${ \frac{1}{2}cos\frac{pi}{n+1}}$.  Hope this time it will take more than 2 minutes...

[quote="harazi"]I just adore how you make people feel they are the most idiot creatures in the world, Myth.[/quote]
If you think like this, solve your problem yourself.

Myth: I think harazi is right - because what is obvious and what is not depends on a certain person and if you say that something is obvious you make other people feel stupid.

BTW: I don't remember who has said this but it's funny.
He asked his students: "Do you know what's the definition of a mathematician". Then he wrote on a blackboard: $\int_{-\infty}^{\infty}e^{-x^2} \ dx=\sqrt{\pi}$ and added: "A mathematician is a guy for whom it's obvious"  :)

Sorry, Myth, I am not going to continue this useless discusion. Anyway, I do not remember to have told you to solve this problem. If this does not interest you, ok.


let f is a operator linear of V. PROVE exist a operator linear g  of V satify: fgf=f

You'd better consult a book describing the [i]pseudoinverse[/i] or [i]Moore-Penrose inverse[/i] subject, I suppose.


Determinati functiile $f: \mathbb{R}\to\mathbb{R}$ care admit o primitiva $F: \mathbb{R}\to\mathbb{R}$ astfel incat \[ x-F(x)=(1+x^{2})(f(x)-1). \] ;)

$g=f-1\Longrightarrow |f(x)-1|=k(1+x^2)-e^{\arctan x},\ k>0$ (brut) $\Longrightarrow$ finisare (slefuire)....


Alright I wanna see how many AoPS members do marching band. I played in my school's marching band last year as lead trombone, and this year as tuba.

Next year I will be trying out for a Drum Corp too (The Jersey Surf).

Unfortunately, it seems as if the majority of AoPSers seem to play stringed instruments, including me...
You may not get as many results as: What Orchestra/Symphony do you play in?

;]

Maybe they should make a marching orchestra?  :D

I'm in my school's marching band... I play oboe normally, but for marching band I either play alto or tenor sax.  Well, actually, not any more, since I'm drum major!

[quote="Quickster94"]hmm, that would be so interesting, marching orchestra.  I pity the timpani player...[/quote]

It's not that bad. The timpanis have wheels right? Now what about the bass player? :D

Being in a marching band sounds like fun, except my school does not have one.  :( I used to play percussion in band though.

true, they do have wheels, r u referring to bass drum or bass (strings)?

Alto Sax Pride ftw :)  (And some years Baritone if I feel like it...);  Moi = Woodwind Captain for both my Concert & Marching Band

Yeah, I play the violin too, but marching band looks pretty cool.   :P
I pwned everyone in my school orchestra so bad it is not even funny. :winner_first:

Instead of starting a new thread I'm bumping this.

So I got into the Jersey Surf hornline on Bari 2, and should be marching in the Corp this summer. I'm curious has anyone on AoPS ever heard of Drum Corp before, if so post your thoughts, and if you march in a corp say your corp.

Anyways I can't wait for the summer and Surf's debut in World Class. There are gonna be some great shows this year. The Cadets are playing a show with all Leonard Bernstein music which should be cool. Can't wait until some of the other top corps announce their shows, this is gonna be a great year for DCI.

Other things to discuss: What was your favorite show last year? What is your favorite show of all time? What is your favorite corp?

Last year: I'd say Karneval (Glassmen) was my favorite, followed closely by Samurai (Cavaliers) and Spartacus (Phantom Regiment).
Favorite show: Machine (Cavaliers 2006) marching wise, and Angelus (Crown 2005) playing wise
Favorite corp: Jersey Surf!!!

I play trumpet, but I'm not in marching band.  I thought about joining, but I don't have the time nor the physical ability.  (I can already see myself tripping and ruining my trumpet.)


induction
number theory unsolved
number theory

The sequence $\{{a_n}\}$ is defined by $a_0 = a_1 =1$ and $(2n+1)a_n+3na_{n-1} = (n+1)a_{n+1}$. Prove that the numbers in this sequence are integers.

$a_n$ is the number of sequences of length $n$ which only use the symbols $0,1,-1$ and such that the sum of the $n$ numbers is $0.$
Thus, I believe it is an integer... :D

Pierre.

Pierre, how could you find this solution? I mean, what made you think at this approach? Really incredible solution. And when I see my super-ugly solution... :(

Could you give me an idea what your super-ugly solution is like?

It is the ugliest solution I've ever seen. Introduce $ f(x)=a_0+a_1x+a_2 x^2+...$, find a monstruous differential equation, solve it and you find the generating function. Don't ask me to write all here, since it wil never finish.

Harazi : I cheat. :P

Do you remember about the math journal Quadrature I told you about?
One of the problems from 'my' problem corner was to prove that the number of sequences I defined does satisfy the given induction relation...   :D

Pierre.

I don't know why, but I suspected you were cheating. Of course, I did not know it's from Quadrature. And so that the chat should be complete, please post the proof that the number of these sequences is given by our initial sequence. ;)

Anyway, I think it's impossible to find that solution. It's just outrageous.


I know I posted a similar topic :oops: , but this one's a bit more specific...

What is a good way to study for the last 5 questions on the AMC10 test? If I could get just 3 of those right...

I need help! :help:

Do the first few problems (like #1-5) on the AIME. The first few problems are at the level of #20-25 (and probably somewhat higher). Do especially the earlier AIMEs (1980s) because they are on the easy side, but still at the same level of AMC 10.

The specific strategy doesn't change much from the general one. Do problems that are just on the edge of your skills and you will improve.

Doing early AIME problems actually did help a lot for me. I used to get 1 or 2 rarely right on the last 5 for AMC10, but after doing 80 or so early AIME problems and understood their solutions, the end AMC problems seemed a lot easier for me.

For me, when I don't know about a problem. I find out what kind of problem it is, and then I try to learn more about that particular subject.  :P


aqui ,  un  problema
   A una convencion interplanetaria asistieron 2000 embajadores . Durante los 6 dias que duro la convencion,  cada uno de ellos se entrevisto con cada uno de los otros 1999 y las entrevistas son mutuas(es decir , si "a" se netrevista con "b" . se considera que "b" ya se entrevisto con "a"). Demuestre que existen 3 embajadores tales que las tres entrevistas entre dos de ellos ocurreiron el mismo dia .

fabiola ;)
peru  :w00tb:  :juggle:

Este problema es un caso particular del " teorema de Ramsey " para 6 colores y grupos de 3 .

En este momento no se cual es el numero exacto de ramsey para n = 6 y k = 3 pero debe ser menor o igual que 1958 .

El nuevo problema a resolver seria asi :

Si se tiene 1958 puntos en un plano y unimos los puntos con sgementos y pintamos los segmentos con 6 colores entonces habra al menos un triangulo con sus tres lados del mismo color .
Nota : el cambio hecho fue : ( puntos = embajadores ; colores = dias ; segmmentos coloreados = entrevistas ) .

Solucion :

Eligimos arbitrariamente uno de los 1958 puntos y trazamos los 1957 segmentos restantes  y los pintamos de los 6 colores disponibles . por el principio de casillas existiran
ceil(1957/6)=327 segmentos del mismo color ( sea C1 dicho color ) . Ahora estos 327 puntos ( extremos de los 327 segmentos )  los unimos por segmentos y los coloreamos y si algunos de estos segmentos es del color C1 entonces ya habria un triangulo con sus tres lados del color C1 y no habria nada que probar . caso contrario todos los segmentos que resultan al unirse estos 327 puntos se pintaran solo con 5 colores distintos .

Se tiene ahora 327 puntos : se procede de igual manera . Se elige un punto de ellos y se trazan 326 segmentos los cuales se pintaran con 5 colores .  por el principio de casillas existiran
ceil(326/5)= 66 segmentos del mismo color ( sea C2 dicho color ) . Ahora estos 66 puntos ( extremos de los 66 segmentos )  los unimos por segmentos y los coloreamos y si algunos de estos segmentos es del color C2 entonces ya habria un triangulo con sus tres lados del color C2 y no habria nada que probar . caso contrario todos los segmentos que resultan al unirse estos 66 puntos se pintaran solo con 4 colores distintos .

Se tiene ahora 66 puntos cuyos segmentos se colorearan con 4 colores distintos . procediendo de manera similar llevaria a probar que 17 puntos a ser coloreados con 3 colores y estos a 6 puntos a ser coloreados con 2 colores y finalmente a 3 puntos a ser coloreados con un solo color y en este ultimo siempre habra un triangulo con sus tres lados del mismo color .

Bueno como 2000 > 1958 entonces con 2000 puntos se cumplira lo mismo .

NOTA : Si alguien tiene buena informacion sobre el teorema de Ramsey y los numeros de Ramsey . por favor cuelguense algo referente para bajarlo yo y todos los interesados del asunto .


I was watching a lecture on titration and the lecturer was explaining what happens to the phenolphthalein molecule in the presence of a base. He showed a series of electronic rearrangements which resulted in the $ sp^{3}$ hybridized carbon to become a $ sp^{2}$ hybridized one. He said that originally the carbon prevents the $ \pi$ system from 'communicating' yet after the rearragement the three bezene rings are able to 'communicate' with each other. Why is that so?

Also try and sythesize Phenolphthalein :D .

Phenolphthalein = Phenol + Pthalic anhydride
And the structure of it explains its colour in basic medium
There are  2 theories to explain   quininoid theory and ionisation theory

Probably I will try and explain to you the quininoid theory because if it is damn interesting.
Coloured substances like dyes are coloured due to extended conjugation. The Quinone and other
substitued quinones have extended cojugation like phenophthalein has phenol entity which reacts
with the base to form the phenoxide which will get drawn into resonance to form quininoid
intermediate which attributed to its colour.

One more very interesting dye is Flouracene=Pthalic anhydride + Resorcinol
It gives you a nice flouracene colour in basic medium and again due to the quininoid

An expression can also be got from kinetics for ionisation theory

pKIn = pKa + log[In-]/[HIn] if it is an acid.
pKIn = pKb + log[In+]/[InOH] if it is a base.

Yes and there is also something like phenolphthalein works in the range of 6 - 8
so the log value can be between -1 and 1

That is just the Buffer capacity value as the indicator can be thought of as a buffer(weak acid
and salt(phenophthalein) or weak base and salt(mehyl orange)).

Phenolphthalein shows four colour ranges, and the coloured ones are due to conjugation between the three aromatic rings - this is only possible if the carbon connected to the three rings is sp2.


Let's assume that $n>0$ and $s=x+y+z$. Then the system becomes
$\left\{\begin{array}{c}{(s-x)^{n}=x}\\{(s-y)^{n}=y}\\{(s-z)^{n}=z}\end{array}\right.$
We can WLOG assume that $x=\max(x,y,z)$. But then $(s-x)^{n}= \max\left((s-y)^{n},(s-y)^{n}\right)$, and since the function $f(t)=(s-t)^{n}$ is decreasing, we have $(s-x)^{n}= \min\left((s-y)^{n},(s-y)^{n}\right)$, so $(s-x)^{n}=(s-y)^{n}=(s-z)^{n}$, which gives $x=y=z$. Thus $x,y,z$ are equal to one of roots of equation $(s-t)^{n}=t$. But there's no efficinet way to find such roots, try for example find a root for $s=1$ and $n=5$, i.e. solve $-t^{5}+5t^{4}-10t^{3}+10t^{2}-6t+1=0$.

[quote="srulikbd"]can't you find for $n=3$?[/quote]
It can be solved by Cardano's formulas, but the solutions look very ugly.

[quote="mathnerd314"][quote="DanK"]What's WLOG?[/quote]

"Without loss of generality". For example, if you have the equation $a+b+c=3$, and you assume that $a>b>c$, it's not going to change the result. Basically you just get rid of all the cases that are identical to the one you're dealing with.[/quote]
actually it should have a condition $a\not= b\not= c$ or the wlog should be $a\geq b\geq c$


Find all positive integers $n$ such that $n^{2}+78n-2006$ is divisible by the product of digits of $n$.

The only solutions are $1,\ 11,\ 111,\ \dots$ and $2,\ 12,\ 112,\ \dots.$

First notice that $n^{2}+78 n-2006$ is never divisible by 3, 5, or 7 because $n^{2}+78 n-2006 = (n+39)^{2}-3527$ and $3527$ is NOT a square modulo 3, 5, or 7.

It is easy to see that $n^{2}+78 n-2006$ may be divisible by 2 but never by 4. Moreover, if it is divisible by 2, n must be even.

Therefore, the only allowed digits in $n$ are $1$ and $2$, and if $2$ is present, it is the last digit. That leads to the solutions shown above.


quadratics
algebra
arithmetic sequence
system of equations

5. The table below represents the relationship between $ x$ and $ y$ in a quadratic equation of the form $ y\equal{}ax^2\plus{}bx\plus{}c$, where $ a$,$ b$, and $ c$ are integers. What is the value of $ a$?

I see that $ y$ increases by 13, 13+6n... or does this even help?

[hide="Answer"]$ 3$[/hide]

[hide]
We would expect y to increase by an arithmetic sequence, as you discovered, since y is a quadratic.  The easiest way, in my opinion, is to use this to generate values of y for x = 0 and x = -1.
You should find that at $ x \equal{} 0$, $ y \equal{} \minus{}15$ and at $ x \equal{} \minus{}1$, $ y \equal{} \minus{}16$.

The point at $ x \equal{} 0$ ggive us $ y(0) \equal{} c$.
Similarly $ y(1) \equal{} a \plus{} b \plus{} c$ and $ y(\minus{}1) \equal{} a \minus{} b \plus{} c$.

Now that you know the values of y at $ x \in \{\minus{}1, 0, 1\}$, finding the coefficients should be straight up algebraic manipulation to solve the system of linear equations.

Note that you can technically do this using any three points of the quadratic, but I find that using $ x \in \{\minus{}1, 0, 1\}$ makes for easier computations.
[/hide]

Using the method that haoye described, we have $ c\equal{}\minus{}15$, because that's the value of $ y(0)$.
Furthermore, at $ x\equal{}1$ and $ x\equal{}\minus{}1$, we get the following system:
\[ a\plus{}b\minus{}15\equal{}\minus{}8
\\
a\minus{}b\minus{}15\equal{}\minus{}16\]Solving, we get $ a\equal{}3$ and $ b\equal{}4$.
Thus, the equation we are seeking is $ 3x^2\plus{}4x\minus{}15$, and $ \boxed{a\equal{}3}$.

[hide="another way"]
Well, it's not that different.

For $ x\equal{}1,2,3$, we have

$ a\plus{}b\plus{}c\equal{}\minus{}8$
$ 4a\plus{}2b\plus{}c\equal{}5$
$ 9a\plus{}3b\plus{}c\equal{}24$

Subtracting the second equation from the third, and subtracting the first equation from the second, we get the equations
$ 3a\plus{}b\equal{}13$
$ 5a\plus{}b\equal{}19$
From which $ \boxed{a\equal{}3}$ can quickly be determined
[/hide]

and if this was on the target round you could plug that system of equations into your calculator and easily find $ a$ using matricies or polysimult app.

You could also plug three of the points into a list, and use QuadReg.


integration
function
trigonometry
calculus
calculus computations

i really confused bout this problem...$ \int_0^{\infty} e^{\minus{}x^2}dx\equal{}....$

Let $ I=\int_0^\infty e^{-x^2} dx$. So, we put $ I=\int_0^\infty e^{-y^2} dy$ too.
Hence, $ I^2=\int_0^\infty e^{-x^2} dx\int_0^\infty e^{-y^2} dy=\int_0^\infty\int_0^\infty e^{-(x^2+y^2)}dxdy$
$ \therefore I^2 = \iint_D e^{-(x^2+y^2)} dxdy \ \ (D: x\ge0, y\ge0)$

Therefore,
\begin{eqnarray*}
I^2
& = & \iint_D r e^{-r^2} dr d\theta \\
& = & \int_0^{\frac{\pi}{2}}\int_0^\infty r e^{-r^2} dr d\theta \\
& = & \int_0^{\frac{\pi}{2}}d\theta \int_0^\infty r e^{-r^2} dr  \\
& = & \left[\theta\right]_0^{\frac{\pi}{2}} \left[-\frac{1}{2}e^{-r^2}\right]_0^\infty \\
& = & \frac{\pi}{4}
\end{eqnarray*}

\[ \therefore I=\boxed{\int_0^\infty e^{-x^2} dx=\frac{\sqrt{\pi}}{2}}\]

hello
$ I \equal{} \int_{0}^\infty e^{ \minus{} x^{2}}dx$
putting $ x^2 \equal{} t$
$ I \equal{} \frac{1}{2} \int_{0}^\infty t^{\frac {1}{2} \minus{} 1}{e^{ \minus{} t}}dt$
we know
$ \int_{0}^\infty e^{ \minus{} x} x^{n \minus{} 1} dx \equal{} Gamma function of n$
$ Gamma function(\frac {1}{2}) \equal{} \sqrt {\pi}$
$ I \equal{} \frac {\sqrt {\pi}}{2}$
thank u

[quote="kabi"]hello
$ I \equal{} \int_{0}^\infty e^{ \minus{} x^{2}}dx$
putting $ x^2 \equal{} t$
$ I \equal{} \frac {1}{2} \int_{0}^\infty t^{\frac {1}{2} \minus{} 1}{e^{ \minus{} t}}dt$
we know
$ \int_{0}^\infty e^{ \minus{} x} x^{n \minus{} 1} dx \equal{} Gamma function of n$
$ Gamma function(\frac {1}{2}) \equal{} \sqrt {\pi}$
$ I \equal{} \frac {\sqrt {\pi}}{2}$
thank u[/quote]

I think that we must prove $ \Gamma\left(\frac{1}{2}\right) \equal{} \sqrt{\pi}$.
So we must prove 2 proof.
$ \forall m>0, \ \Gamma (m) \equal{} 2\int_0^\infty x^{2m\minus{}1}e^{\minus{}x^2} dx$,
$ \Gamma (z) \Gamma (1\minus{}z) \equal{} \frac{\pi}{\sin \pi z}$.


Let $S$ be the set of all reals greater than one.Find all the functions $f:S->R$ such that
$f(x)-f(y)=(y-x)f(xy)$ for all $x,y>1$


Hi.
In this Topic lets talk about the definition of Terrorism in order to know who is terrorist.

Lets have a friendly disscusion.

start.

"Terrorist" means different things when used by different people. It seems to me that many people (in the USA at least) now use "terrorist" to mean "a person who uses lethal violence against the USA or one of its allies, but is not a member of a national organized military force."

Unfortunately this is a very broad definition, and I think the purpose of this definition is to force people to associate the very worst deeds ascribed to "terrorists" with anyone who is given the name, even those who don't deserve it.

Perhaps a better definition would be to the effect of "someone who commits acts internationally recognized as war crimes but who is not a member of a national organized military force." Then at least someone who is resisting foreign occupation by fighting against only the occupying military won't be unfairly associated with someone who blows up himself and a bunch of civilians.

It's interesting to note that as terrorists seem to be defined by the national media in the USA (and our government), if you just drop the part about how they have to be fighting against us, the people whom we consider WWII heroes for resisting against the Nazi occupation would have to be called terrorists as well!

It's all based on opinions. Using Xevarion's better definition, you could say the colonists were terrorists in that they commited war acts against England.

Ah, but you missed the nuance in Xevarion's definition. He didn't say "acts of war;" he said "war crimes."

Let's look at this. Was John Brown a terrorist when he attacked Harpers Ferry just before the U.S. Civil War? He was certainly the leader of an irregular force not sanctioned by any government, and he was committing acts of violence. But his target in Harpers Ferry was a military installation - an armory. They did kill several civilians, but it seems that those civilians confronted them or opposed them in some way.

By Xevarion's definition, I'd be inclined not to use the label "terrorist" for Brown and his men: "guerrilla insurgent," perhaps. They committed acts of war and answered to no government, but I don't see in this instance that they committed war crimes. Note that when Brown and those of his men who were captured were hanged, the specific charge was treason against the Commonwealth of Virginia.

On the other hand, Brown came from the Kansas conflict, and there probably were acts of terrorism committed in Kansas, on both sides.

Of course he was a terrorist. So were all outlawed militant revolutionaries in history.
The problem lies in the fact that people are inclined to think that terrorists are always bad.

Perhaps a way to identiify bad terrorists would be to see if they are targetting civilians.
By this, the attack on WTC would be terrorism, the attack on Pentagon would not (that would be war).
Again, the Afghan and Iraqi mass murders by the US army would be terrorism, but the attack on the Taliban army or Saddam's forces would not.

But such definitions lead to problems, as I'll later throw light on when the discussion has proceeded further.

That definition excludes most terrorists. Why shouldn't people with the same intentions, but e.g. placing bombs away from their body not be terrorists¿
Your definition probably excludes some so-called top-terrorists like Bin Laden, too.

All right, let me rephrase that: any one who does that, or who convinces/forces other people to do that (kill themselves to kill innocent people).

[quote="Valentin Vornicu"]For me a terrorist has a simple definition: a person that is willing to kill himself only to hurt a bunch of other (innocent usually) people.[/quote]
That's not a good definition, who is to define "only to"? :huh: Can you give examples?

nice avatar val
but i don't see what terrorism has got to do with suicide bombing
people who simply bomb market places (as they're now doing in the middle west frequently) are also terrorists, right?

Actually, the definition of terrorism differs from country to country.
In Romania, the definition of terrorism is :
" Terrorism represents all the actions with violent character with the following cummulative properties:
1. They are premeditated.
2. They imply distructive or lethal methods.
3. The targets of terrorism are representative and wide known( individuals,symbols,etc)
4. They have a political character (explicit or implicit)
5. They are commited outside the state of war or the zone where the laws of war apply "

Also, an act of terrorism in Romania may not be considered the same way in the USA or other country. For example, Itzhak Rabin and Yasser Arafat former high ranked members of Hagana which is known to be the most violent terrorist organization that has ever existed (along with Fatah) have received the Noble prize for peace in 1992. Also it is known that Arafat was behind the Berlin attempt in the 70's.
So the main thing which I would like to stress out is that everything is relative. It is only a matter of perspective.

You guys are getting pretty bureaucratic with your definitions of "terrorism", and this is coming from someone who just spent approximately 50 hours discussing technicalities at the Princeton Model United Nations Conference.

Terrorism - it's pretty much a moral belief that varies throughout distinct demographics, but it's probably a general consensus that terrorists are bad. Most people are also willing to admit that, you know, terrorism is something they're morally opposed to.

What I'm implicitly trying to get at is you can't strictly define terrorism without totally alienating certain beliefs that happen to be morally and horizontally challenged with respect to your point of view.

A terrorist is a person will to cause mass chaos, kill innocent people or assassinate a specific person, usually because of their belief. Most terrorist kill because they feel that their belief is being threatened by a certain group of people that oppose them. For instance, radical islams hate American's because they feel that the U.S. is trying to convert everyone to Christianity and we are trying to rid the world of people that oppose us. A lot of people consider people of Mid-Eastern descent are terrorist, however less than 1/1000 people Mid-Eastern is acctually a terrorist. Usually terrorist feel that they aren't the terrorist, but that the people that oppose them are terrorist. Usually these people are severely misled and under educated. Another cause for someone to become e terrorist is because they have been the victim of violence and are just plain angry. So what I'm trying to say is a terrorist is not just a person that's bored and is just killing other's but they usually have a reason, no matter how misled, to kill. Happy leap day everyone.

I don't think its possible to discuss terrorism without the quote

"One man's terrorist is another man's freedom fighter"

I think that the behaviour of almost every country in the world has been bad enough to warrant the title of "terrorism" but there was little resistance to it at the time (Great Britain and the US I'm looking at you). I think that indiscriminate acts of violence towards large numbers of people is terrorism, and it doesn't matter who authorises it.


geometry
incenter
angle bisector
geometry proposed

Dear Mathlinkers,
let ABC be a rectangular triangle at A,
A' the foot of the A-altitude on BC,
K the incenter of the triangles AA'C,
Fe the Feuerbach's point of ABC
and B" the midpoint of AC.
Prove : Fe, K and B" are collinear.
Sincerely
Jean-Louis

i'll change the name of some points $ B'' \rightarrow B',A' \rightarrow P$.$ C'$ is the midpoint of $ AB$.
Then let $ A'$ be the midpoint of $ BC$,$ J$ the incenter of $ ABP$ and $ Q$ the midpoint of $ JK$.
it's known that these points lie on the circle $ (Q,QJ)$:$ J,K,P,F_{e}$.
All that we must prove is that $ \angle KB'A' \equal{} \angle A'C'J$.(the explanation is given)
but see that $ \Delta APB \sim \Delta APC$.so we easily get $ \angle JC'A \equal{} \angle KB'C$(because of similarity).
but $ \angle A'C'A \equal{} \angle A'B'C \equal{} 90^\circ$.so we deduce $ \angle KB'A' \equal{} \angle A'C'J$.as we wanted.
Let $ S$ be the intersection of $ C'J,B'K$.
we deduce that $ S$ (which is in fact $ F_{e}$) lies on the nine-points circle and in the other hand the angle $ C'SB' \equal{} \angle C'A'B' \equal{} 90^\circ$.or in other words $ \angle JSK \equal{} 90^\circ$.it means that $ S$ lies also on the circle $ (Q,QJ)$ which means that $ S$ is either $ P$ or $ F_{e}$(because the intersections of the nine-points circle and the circle $ (Q,QJ)$ are $ P,F_{e}$).But assume that $ S \equal{} P$ it means that the angle bisector of the right angle $ APC$ passes through the midpoint of the opposite side which means that $ AP \equal{} PC$.We deduce that in this case(if $ S \equal{} P$ then)$ AB \equal{} AC$.but if the triangle is isocele then $ P \equal{} F_{e}$,obviously.
so in both cases,$ S \equal{} F_{e}$. DONE!!

Dear [b]Jayme[/b],
Let $ L$ be the incenter of triangle $ AA'B, N$ be the midpoint of $ BC$, $ M$ be the midpoint of $ AC$. From your topic we have $ Fe$ lies on $ (LK)$ therefore $ \angle A'FeK \equal{} \angle ALK \equal{} \angle B \equal{} \angle MNC \equal{} \angle MFeA'$
We are done.

Dear Livetolove and Mathlinkers,
yes for your short and nice proof.
It is possible to avoid the angles by using http://www.mathlinks.ro/viewtopic.php?t=299708
and a convese of the Reim's theorem.
Sincerely
Jean-Louis

Dear Mathlinkers,
a synthetic proof of this result can be see
on: http://perso.orange.fr/jl.ayme    vol. 5  Le théorème de Feuerbach-Ayme p. 21
Sincerely
Jean-Louis


AMC
USA(J)MO
USAMO
AIME
trigonometry
AMC 10
complex numbers

Would you guys recommend online courses? ive never taken one, how is it and how far does it go toward prepping for the AMc's...ps my level is about AMC 10 : 130 any reccommedations about online courses to take?
my goal is to Make usamo next year.. should I take the Aime online as well as the Amc 10?

If your goal is USAMO, I would recommend focusing on the AIME rather than the AMC 10.  I found the AIME Problem Series class to be pretty helpful.

thanks. I just want to make sure I do well on the 10 so it takes the pressure of the AIME.

If you have time, taking AMC-12 course also will help. I'm saying this because since you're taking AMC-10, you probably are not that acknowledged with trig, complex numbers, and other AMC-12 stuffs, which do come up on AIME.

So, it might help to understand AIME series. Keep in mind that not everyone who takes AIME series are AMC-10 or AMC-12 qualifiers. There've been students who took the class and scored 5 or 6 and yet unfortunately missed the cutoff. So, yeah, it's an intense course so it might be a big challenge to take this and AMC-12 problem series but if you do, you'll definitely improve!


According to the sheet, 1 team got 0, yet all 57 teams got #1 right. Could someone explain this? Thanks.

on the rest of the competition they did so bad they got -1 points

lol. You're right. I hadn't noticed that before. I have no idea as to what it means.

Someone must have a resonable explaination.
No offense, but I don't think kpavery's theory is correct.

There was a secret 58th team that nobody knew about that scored 0. Well, maybe the MATHCOUNTS people knew, but other than that...  :?  :D

No offense taken, I come to the same conclusion: it is incorrect.

My new theory involves an old saying: No one is perfect, and I beleive that there is a mistake in the score sheet.

That makes sense.
Does anyone know how to contact MATHCOUNTS?

info@mathcounts.org

[url=http://www.mathcounts.org/webarticles/anmviewer.asp?a=184&z=8]Contact Mathcounts Page[/url]

Soo...maybe I was really 1st but graded incorrectly ;)  :D

They probably put two extra 5's after the 1 on my rank.... :P

[quote="Xantos C. Guin"]I got a reply.
Apparently the system they use accidentally enters in a score of 0 for an imaginary team by mistake.[/quote]

Aha! I was right!  :lol:

[quote="09husbri"]Meaning...?[/quote]
Their system "accidentally" put a 58th team in.

Win.

We happened to be the only team to get #10 right :P .


A circle of radius $ 2$ is inscribed in a semicircle, as shown. The area inside the semicircle but outside the circle is shaded. What fraction of the semicircle's area is shaded?
[asy]unitsize(6mm);
defaultpen(linewidth(.8pt)+fontsize(8pt));
dotfactor=4;

filldraw(Arc((0,0),4,0,180)--cycle,gray,black);
filldraw(Circle((0,2),2),white,black);
dot((0,2));
draw((0,2)--((0,2)+2*dir(60)));
label("$2$",midpoint((0,2)--((0,2)+2*dir(60))),SE);[/asy]$ \textbf{(A)}\ \frac{1}{2}\qquad
\textbf{(B)}\ \frac{\pi}{6}\qquad
\textbf{(C)}\ \frac{2}{\pi}\qquad
\textbf{(D)}\ \frac{2}{3}\qquad
\textbf{(E)}\ \frac{3}{\pi}$

[hide]
The circle has area $ 2^2\pi\equal{}4\pi$.

The semicircle has radius $ 4$ and thus area $ \frac{4^2\pi}{2}\equal{}8\pi$.

Thus $ 8\pi\minus{}4\pi\equal{}4\pi$ of the area is shaded and the fraction is $ \frac{4\pi}{8\pi}\equal{}\frac{1}{2}\equal{}A$.
[/hide]

the grey semicircle has a radius that is $2$ times the white circles radius

therfore the ratio of the areas is $1-\frac{1}{\frac{2^2}{2}}=\boxed{\frac12}$

[quote=brendan_cape]nice 12 year bump[/quote]

thank you :)

The semicircle has a radius which is the diameter of the white circle which is $4$. So the area of the semicircle is $\frac{4^2\pi}2=8\pi$. The shaded region has area $8\pi-2^2\pi=4\pi$ by complementary counting. So the answer is $\frac{4\pi}{8\pi}=\frac12$.


Ok. Suppose one equals five, two equals twenty-five, 3 equals one hundred twenty-five and 4 equals six hundred twenty-five. What is five equal to?

Since you're redefining the relation $=$, it isn't necessarily symmetric anymore. So it could be possible that $1=5$ but $5\neq1$.

Not enough information is given in my opinion. We don't know whether or not equality is symmetrical in this new definition of "equals" and assuming that it follows a pattern, well, there are infinite polynomials that yield the first few numbers, so there are infinite answers in that case.

lol you guys...

all i stated was
1=5
2=25
3=125
4=625

this isnt a function or anything. assuming that those 4 equations are the only ones true, then 5=1 according to the first equation.
pi can equal dog turds here, its not stated

i dunno... stop ruining the question

It all depends on your definition of "="

And besides, if you use the regular definition of "=", the antecedent is already false. So you can't suppose anything. As a camp counselor once said, all five-legged dogs can fly.

The problem statement makes it clear that you aren't using the regular definition of $=$, and you haven't stated anywhere that $a = b \implies b = a$.

but it still is a good brainteaser because it tricks you into thinking powers of five, when really it already told you what five was. technicalities aside, it's good!

Like I already said, nowhere in the problem does it tell you that you should think $1 = 5 \implies 5 = 1$.  Clearly the $=$ sign here is supposed to be a bijection between two sets.

[quote="K81o7"]If you can define equality, then I can define addition.  $5=1+4=630=2+3=150$, etc...
You need to give more information about the new number system.[/quote]

Well if you can define addition, then I can define multiplication.

5 = 1*5
but since 1=5,
5 = 5*5
Thus, 5=0 or 5=1

woah it actually works o_O

And so on.  WHen you say $1=5$, you're already creating a new number system with the same names.  In fact, equality in this number system could mean $a=b$ in that system $\implies f(a)=b$ in our regular number system for a fixed function f.  In which case there are an infiniten umber of solutions.

omg whatever lets close this thread lol
you guys are getting too serious about this >_>

The point is that the answer was wrong, and we were correcting it. That's not being 'too serious'.


Find all positive integers $n$ such that $n!$ (in decimal expansion) ends with exactly 2006 zeros.

Well, we have $\sum_{n=1}^{\infty}\left\lfloor{\frac{a}{5^{n}}}\right\rfloor$ for a!.  So let's write:

$\left\lfloor{\frac{x}{5}}\right\rfloor+\left\lfloor{\frac{x}{25}}\right\rfloor+\left\lfloor{\frac{x}{125}}\right\rfloor+\left\lfloor{\frac{x}{625}}\right\rfloor+\left\lfloor{\frac{x}{3125}}\right\rfloor$

$\left\lfloor{\frac{625x}{3125}}\right\rfloor+\left\lfloor{\frac{125x}{3125}}\right\rfloor+\left\lfloor{\frac{25x}{3125}}\right\rfloor+\left\lfloor{\frac{5x}{3125}}\right\rfloor+\left\lfloor{\frac{x}{3125}}\right\rfloor$

Basically $\frac{781x}{3125}\approx 2006$, so $x \approx 8026$.  Then we need to compensate for any additional 5s, so dividing it out we find that 8026 gives 2004 zeros, meaning our range is from 8,035 to 8,039.

$2006 = \sum_{n=1}^{\infty}\left\lfloor{\frac{a}{5^{n}}}\right\rfloor \approx \sum_{n=1}^{\infty}{\frac{a}{5^{n}}}\implies \frac{a/5}{1-\frac{1}{5}}=\frac{a}{4}\approx 2006 \implies a \approx 8024$

We can quickly evaluate the actual number of zeros at the end of $8024!$ and adjust accordingly.


Sorry for the stupid question,

Does exist a universal formula for solving expressions like:

x^(2n) + ax^n + b = 0

Where
x /in R
a, b /in N \ {0}
n /in N \ {0,1}

?

Thanks.

Substitute \[ y = x^n  \]

Thus \[ y^2 + ay + b = 0  \]

This has solutions \[ y = \frac{{ - a \pm \sqrt{a^2 - 4b} }}{2}  \]

Thus\[ x^n = \frac{{ - a \pm \sqrt{a^2 - 4b} }}{2}  \]

giving \[ x = \sqrt[n]{{\frac{{ - a \pm \sqrt{a^2 - 4b} }}{2}}}  \]

Though some roots are complex you can ignore them. :D

[quote="bugmenot"]
Does exist a universal formula for solving expressions like:

$x^{2n} + ax^n + b = 0$

Where
$x \in \mathbb{R}$
$a, b \in \mathbb{N} / \{0\}$
$n \in \mathbb{N} / \{0,1\}$[/quote]

TeX'd.  And in fact none of those conditions are necessary; only $n \neq 0$ :D


Prove that for any positive integer [i]n[/i] there exists a positive integer, which is divisible by 2^n and whose decimal representation consists of only digits 1 and 2.

We will show that for any positive integer n there exists an n-digit positive integer $x_n$, which is divisible by $2^n$ and whose decimal representation consists of only digits 1 and 2.

For $n=1$ it is true; there is the 1-digit number $2$, which is a multiple of $2^1$.

We assume that it is true for $n$.
That is, there exists an n-digit number $x_n$ which is divisible by $2^n$ and all its digits are 1 and 2

We will show that it is true also for $n+1$

$2^n \mid x^n \Rightarrow x_n=y_n \cdot 2^n$

The number $10^n$ has $n+1$ digits
we define the numbers
$a = 10^n + x_n$
$b = 2\cdot 10^n + x_n$

The numbers $a,b$ have $n+1$ digits, and also all their digits are all $1$ and $2$
It's easy too see that $2^{n+1}$ divides exactly one of them. Indeed:

$a = 2^n5^n + 2^ny_n = 2^n(5^n+y_n)$
$b = 2 \cdot 2^n5^n + 2^ny_n = 2^n(2\cdot 5^n+y_n)$

The numbers $5^n+y_n, 2\cdot 5^n+y_n$ are different in modulo $2$, so some of them is even. So we have:

if $y_n \equiv 1 \mod{2}\Rightarrow 2^{n+1} \mid a$
if $y_n \equiv 0 \mod{2}\Rightarrow 2^{n+1} \mid b$

We choose the number $x_{n+1}$ using the following algorithm:

$x_{n+1}=\{  \begin{array}{ll} 10^n + x_n, & \mbox{ if }y_n \equiv 1 \mod{2} \\ 2\cdot 10^n + x_n, & \mbox{ if }y_n \equiv 0 \mod{2} \end{array} $

at any time, $x_{n+1}$ has $n+1$ digits, it consists of $1$ and $2$, and it is dividible by $2^{n+1}$

So we get the sequence $2,\ 12,\ 112,\ 2112,\ 22112,\ 122112,\ 2122112,\ 12122112,\ 212122112,\ 1212122112,\ 11212122112,...$

All these numbers have the desired properties.

These are not the only solutions for the original problem.

Notice, for example, that $8 \mid 112$ but $8 \mid 1000,$ so $8$ divides any number $a_1a_2\cdots a_k112 \ \ (a_i=1$ or $2)$

In general, $2^n \mid 10^n \Rightarrow 2^n$ divides any number which has only the digits $1$ and $2$ and ends at $x_n$


M is a point inside a square ABCD, such that AM=1, BM=2 and CM=3. Find DM.
Can the problem be solved if AM=a, BM=b and CM=c?
[hide]Post solutions not answers[/hide]

Hint: prove that $AM^2 + CM^2 = BM^2 + DM^2$.

Arne could you show me how to prove that, please?

Try using Pythagoras theorem (firstly draw a line through $M$ perpendicular to one pair of the sides)


If one penny is 0.06785 inches thick, what is the approximate value of a stack of pennies one mile high?

The answer is
[hide]
9,338.25 dollars
[/hide]

Answer & Explain, plz

Thanks a bunch,
Jimmy V

A mile is 5,280 feet, which is 63360 inches. Since a penny is 0.06785 inches thick, there are about $ 63360 \div 0.06785 \approx 933825$ or 9,338.25 dollars.


Find the number of real solutions of the system:
$ x \plus{} y \plus{} z \equal{} 3xy$
$ x^{2} \plus{} y^{2} \plus{} z^{2} \equal{} 3xz$
$ x^{3} \plus{} y^{3} \plus{} z^{3} \equal{} 3yz$

multiply them we get that:

$ (x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2)(x^3 \plus{} y^3 \plus{} z^3) \equal{} 27x^2y^2z^2 (I)$

now by AM-GM we know that:

$ x \plus{} y \plus{} z\geq3\sqrt [3]{xyz}$

$ x^2 \plus{} y^2 \plus{} z^2\geq3\sqrt [3]{x^2y^2z^2}$

$ x^3 \plus{} y^3 \plus{} z^3\geq3xyz$

$ \Rightarrow (x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2)(x^3 \plus{} y^3 \plus{} z^3)\geq27x^2y^2z^2 (II)$

now from $ (I),(II)$ we get that in our three AM-GMs equality occurs,which means $ x \equal{} y \equal{} z$,now plug these in we get that:

$ 3x \equal{} 3x^2$

$ \Rightarrow x \equal{} y \equal{} z\in\{0,1\}$

Excellent solution. It took me almost the same time as yours. I though this problem is far more difficult

Similar approach may be used.
1. Multiply first and third equation.
2. From  the second follows $ y^{2} \leq xz$
3. Then it follows that $ (x^{2}\plus{}y^{2}\plus{}z^{2})^{2} \leq (x\plus{}y\plus{}z)(x^{3}\plus{}y^{3}\plus{}z^{3})$
But it is possible only if $ x\equal{}y\equal{}z$ What do you think about this?

[quote="delegat"]But how can you apply $ AM \minus{} GM$ inequality if it is given that $ x$, $ y$ and $ z$ are just reals :?:

$ AM \geq GM$ can be applied just to nonnegative reals.[/quote]
actually AM-GM can be applied when ever $ a \plus{} b \plus{} c\geq0$ because we have:

$ \frac {a^3 \plus{} b^3 \plus{} c^3}3\geq abc\iff (a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)(a \plus{} b \plus{} c)\geq 0$

Ya You are right [b]BaBak Ghalebi[/b]

But also you have not proved that $ \sqrt [3] x \plus{} \sqrt [3] y \plus{} \sqrt [3] z \geq 0$ for first $ AM \geq GM$ step and also $ x \plus{} y \plus{} z\geq0$ for your third $ AM \geq GM$ step.


modular arithmetic
number theory proposed
number theory

Let $ n$ be an odd positive integer. Prove that $((n-1)^n+1)^2$ divides $ n(n-1)^{(n-1)^n+1}+n$.

Nice!
As usual, let $ v_p(n)$ be the unique integer satisfying $ p^{v_p(n)}\parallel{}n$ ($ p$ - prime), and $ ord_p(a,b) \equal{} \min \{ x \ge 1: p|a^x \minus{} b^x\}$ (the set $ \min \{ x \ge 1: p|a^x \minus{} b^x\}$ consists of all the multiplies of $ ord_p(a,b)$) ($ a,b$ are coprime non-zero integers).
A well-known theorem states: for odd prime $ p$, if $ p|a^n \minus{} b^n$: $ v_p(a^n \minus{} b^n) \equal{} v_p(a^{ord_p(a,b)} \minus{} b^{ord_p(a,b)}) \plus{} v_p(\frac {n}{ord_p(a,b)})$.
We'll use this theorem with $ b \equal{} \minus{} 1$. Notice that $ ((n \minus{} 1)^n \plus{} 1)^2$ is odd, hence divisible only by odd primes.

Let $ x \equal{} (n \minus{} 1)^n \plus{} 1$. We need to show $ x^2|n((n \minus{} 1)^x \plus{} 1)$. Cleary, $ x \equal{} (n \minus{} 1)^n \plus{} 1|(n \minus{} 1)^x \plus{} 1$, since $ n|x$ ($ n$ is odd).
It is enough to show that $ v_p(x) \equal{} t\ge 1 \implies v_p(n((n \minus{} 1)^x \plus{} 1)) \ge 2t$. Let $ ord_p(n \minus{} 1, \minus{} 1) \equal{} k$.
$ v_p(x) \equal{} t\ge 1 \implies t \equal{} v_p(\frac {n}{k}) \plus{} v_p((n \minus{} 1)^{k} \minus{} ( \minus{} 1)^{k})$ $ (*)$.
$ v_p(n((n \minus{} 1)^x \plus{} 1)) \equal{} v_p(n) \plus{} v_p(\frac {x}{k}) \plus{} v_p((n \minus{} 1)^{k} \minus{} ( \minus{} 1)^{k})$

Hence, we need to show that
$ v_p(n) \plus{} v_p(\frac {x}{k}) \plus{} v_p((n \minus{} 1)^{k} \minus{} ( \minus{} 1)^{k}) \ge 2(v_p(\frac {n}{k}) \plus{} v_p((n \minus{} 1)^{k} \minus{} ( \minus{} 1)^{k}))$
Or (using $ v_p(ab) \equal{} v_p(a) \plus{} v_p(b)$, etc.):
$ v_p(x) \ge v_p(\frac {n}{k}) \plus{} v_p((n \minus{} 1)^{k} \minus{} ( \minus{} 1)^{k})$
And from $ (*)$, we even have equality! Which means that $ (\frac {RHS}{LHS}, LHS) \equal{} 1$.

If $ n$ is even, must of our arguments fail. Already for $ n \equal{} 4$ we get $ 41^2 | 9^{41} \plus{} 1$, which doesn't hold since by Fermat's Little Thereom: $ 9^{41} \plus{} 1 \equal{} 9 \plus{} 1 \equal{} 10 \pmod {41}$.

[b]bambaman[/b] could you give me a link where I can find a proof of this well-known theorem?

It's lifting the exponent lemma. Google it :wink:

Just to make it a bit more clear - let $ f_{a,b}(n) \equal{} a^n \minus{} b^n$. Then this problem is:
\[ f_{a,b}(x) | \frac {xf_{a,b}(f_{a,b}(x))}{f_{a,b}(x)}\]
with $ a \equal{} n \minus{} 1, b \equal{} \minus{} 1, x \equal{} f_{a,b}(1)$
Claim: $ f_{a,b}(x) | \frac {xf_{a,b}(f_{a,b}(x))}{f_{a,b}(x)}$ holds for $ x$ such that $ x|f_{a,b}(x)$ and $ a \neq b \pmod 2$.
Proof: for $ p|f_{a,b}(x)$ we have: $ v_p(\frac {xf_{a,b}(f_{a,b}(x))}{f_{a,b}(x)}) \equal{} v_p(x) \plus{} v_p(\frac {f_{a,b}(x)}{x}) \equal{} v_p(f_{a,b}(x))$, and that's enough.
In our case, $ x \equal{} f_{a,b}(1) | f_{a,b}(x)$ and $ n \minus{} 1 \neq \minus{} 1 \pmod 2$ so we can apply the claim.

We don't need the strong version of 'Lifting the exponent': it is enough to use $ v_p(\frac{a^n\minus{}b^n}{a\minus{}b})\ge v_p(n)$ (for odd $ p$ and $ a,b$ s.t. $ p \nmid a,b$ and $ p|a\minus{}b$) which follows directly from $ p|a\minus{}b \implies p|\frac{a^p\minus{}b^p}{a\minus{}b}$.

Added problem:
 Find all even n such that the statement is true for it.
[hide="click text"] the only solution is 2[/hide]

Prove that for any odd prime $ p$ that divides $ (n \minus{} 1)^n \plus{} 1$ we should have $ ord_{p}(n \minus{} 1) \equal{} 4$. Then deduce that $ 2\parallel{}n$ and also conclude that the set of primes dividing $ (n \minus{} 1)^2 \plus{} 1$ and $ (n \minus{} 1)^n \plus{} 1$ must be the same.then because of this well-known problem (see  http://www.mathlinks.ro/viewtopic.php?t=150413) we will have $ n \equal{} 2^k$ for some integer k. but we had $ 2\parallel{}n$. so $ n \equal{} 2$. DONE!

Notice $(n-1)^n+1\equiv (-1)^n+1\equiv 0\pmod{n},$ so we can let $Sn=(n-1)^n+1,$ where $S\in\mathbb{Z}.$ We note $S$ is odd as $(n-1)^n+1$ is odd. We also see \begin{align*}n(n-1)^{(n-1)^n+1}+n&=n\left((n-1)^{(n-1)^n+1}+1\right)\\&=n((n-1)^{Sn}+1)\\&=n((n-1)^n+1)\left(\sum_{k=0}^{S-1}(-1)^k((n-1)^n)^k\right)\\&=((n-1)^n+1)n\left(\sum_{k=1}^{S-1}(-1)^k(Sn-1)^k\right)\end{align*} by sum of powers for odd exponents. Since $$\sum_{k=1}^{S-1}(-1)^k(Sn-1)^k\equiv\sum_{k=1}^{S-1}(-1)^{2k}\equiv 0\pmod{S},$$ we see $(n-1)^n+1=nS\mid n\sum_{k=1}^{S-1}(-1)^k(Sn-1)^k.$ Multiplying both sides of the division by $(n-1)^n+1\mid (n-1)^n+1$ finishes. $\square$