linear algebra
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calculus

Relatively advanced topic in Differential equations:

How do I prove that $ \infty$ unstable for the following system:

[img]http://img355.imageshack.us/img355/684/helpxz4.jpg[/img]

Thanks alot to anyone who can help.

I have no idea what you mean by $ ``\infty"$  being unstable.

This is an autonomous system. It has equilibrium points whenever the right hand side vanishes. This system is easy to solve. There are three equilibrium points: $ (0,0,0),\left(\frac1{\sqrt{2}},0,\frac12\right),$ and $ \left(\minus{}\frac1{\sqrt{2}},0,\frac12\right).$

The first step then is a linear stability analysis. The Jacobian matrix of the system is
\[ \begin{bmatrix}0&1&0\\1\minus{}z\minus{}3x^2&\minus{}1&\minus{}x\\ 2x&0&\minus{}1\end{bmatrix}\]
Find the values of this Jacobian at the three equilibrium points. We get these three matrices:
\[ \begin{bmatrix}0&1&0\\1&\minus{}1&0\\0&0&\minus{}1\end{bmatrix}, \begin{bmatrix}0&1&0\\ \minus{}1&\minus{}1&\minus{}\frac1{\sqrt{2}}\\ \sqrt{2}&0&\minus{}1\end{bmatrix},\begin{bmatrix}0&1&0\\ \minus{}1&\minus{}1&\frac1{\sqrt{2}}\\ \minus{}\sqrt{2}&0&\minus{}1\end{bmatrix}\]
Now, what can you say about the eigenvalues of each of these? If any eigenvalue has positive real part, then the equilibrium is unstable. (The equilibrium at $ (0,0,0)$ is unstable because the eigenvalues of the first matrix are $ \minus{}1$ and the roots of $ x^2\plus{}x\minus{}1\equal{}0,$ which are two negative real numbers and one positive real.) If all eigenvalues have negative real parts, the equilibrium is stable. If all eigenvalue have nonpositive real parts but at least one is pure imaginary, then we must go beyond the linear analysis to determine the stability.

for $ x\ne 0$
$ x\equal{}\frac{1}{u}$, $ y\equal{}\frac{v}{u}$, $ z\equal{}\frac{w}{u}$

for $ y\ne 0$
$ y\equal{}\frac{1}{u}$, $ x\equal{}\frac{v}{u}$, $ z\equal{}\frac{w}{u}$

for $ z\ne 0$
$ z\equal{}\frac{1}{u}$, $ x\equal{}\frac{v}{u}$, $ y\equal{}\frac{w}{u}$

The matrices $ \begin{bmatrix}0 & 1 & 0 \\
\minus{} 1 & \minus{} 1 & \minus{} \frac1{\sqrt {2}} \\
\sqrt {2} & 0 & \minus{} 1\end{bmatrix}$ and $ \begin{bmatrix}0 & 1 & 0 \\
\minus{} 1 & \minus{} 1 & \frac1{\sqrt {2}} \\
\minus{} \sqrt {2} & 0 & \minus{} 1\end{bmatrix}$

both have characteristic polynomial $ x^3 \plus{} 2x^2 \plus{} 2x \plus{} 2,$ hence they both have the same eigenvalues.

The roots of this polynomial are approximately $ \minus{} 1.544$ and $ \minus{} 0.338\pm1.115\,i.$ In particular, all three eigenvalues have negative real parts. Hence the two equilibrium points $ \left(\frac1{\sqrt {2}}\,,0,\frac12\right)$ and $ \left( \minus{} \frac1{\sqrt {2}}\,,0,\frac12\right)$ are stable, and solution trajectories that start sufficiently close to these points in phase space will spiral in towards these equilibrium points.

This does not address the question of whether there might be solutions (away from the basins of attraction of the stable equilibrium points) in which one or more of the components might tend to infinity, or even blow up in finite time.

----

V.V., I don't find your post to be self-explanatory. What conclusion does that lead you to? Can you explain your point?

Is this equation system a Lorenz system?

It looks very very similar but some differences like $ x'$ is missing $ \minus{} \delta {x}$ and $ y'$ has as an addition $ \minus{} x^3$.

[url]http://planetmath.org/encyclopedia/AllSolutionofTheLorenzEquationEnterAnEllipsoid2.html[/url]

What do you guys think?

Thanks.

Your system is slightly different but the idea is similar. Try to find the Lyapunov function in the form $ ay^2\plus{}bx^4\plus{}cz^2\plus{}dxy\plus{}ex^2z$ with $ a,b,c,d,e>0$ ;)


Actually, the stuff he's mentioning is probably a bit beyond classroom math.

Just use a calculator...

[quote="shaun ong"]A=(1+0.08/12) with the power of 1200[/quote]
Are you just saying what is $ (1\plus{}\frac{0.08}{12})^{1200}$?

[quote="beans"]Don't quote me on this (it's 3am!) but I vaguely seem to remember that $ \left(1 \plus{} \frac {x}{n}\right)^n \to e^x$ as $ n\to \infty$, so how can you use that to answer your question?[/quote]
I don't mean to be rude, but it's $ (1 \plus{} \frac {1}{n})^n \to e$ as $ n\to \infty$

@Mr Rubix: He's using a varient of that identity you are describing  :roll: .

I dun mean to be rude but what on earth  are u all saying??? I was juz asking how do i do that question with that formula

don't try to understand it now.

come on ppls, no calc in the classroom math forum

as I said, just plug the expression into a calculator and it should give you the answer, which is somewhere around 3000???

1. It's not calculus. It's basic Alg 2.

2. This belongs in HSB.

[quote]I dun mean to be rude but what on earth are u all saying???[/quote]

3. I have no idea what you're saying either.

[hide="Easy Way"]
Just enter it into your calculator, like what alanchou said. You can get a far more precise answer using the $ e$ method but that should be unecessary.
[/hide]


graph theory
combinatorics solved
combinatorics

in each row and column of m*n table we have a star and if we have star in (i,j) { (i,j) means the intersection between i th row and j th column) then ri/cj=(i/j)^a  a is constant real number. and ri is the number of star in i th row and cj is the number of stars in j th column prove m=n

let R be the set of rows, C the set of columns. Construct the bipartite graph connecting row i with column j iff in [i,j] is a star. It splits in some conex components. Let X be one of them.
Our condition we write i^a/r[i]=j^a/r[j]. As X is conex, this quantity is the same for all elements of X. Hence r[i]=c*i^a, c[j]=c^i^a for r[i],c[j] in X. Now let's count the number of edges in X. Summing for rows, it is c*(sum(i^a)|r[i] in X)
Summing for columns, it is c*sum(sum(j^a)|c[j] in X).
Hence this two sums are equal and sum(i^a|r[i] in X)=sum(j^a|c[j] in X)
Summing over all conex components we get 1^a+..+m^a=1^a+..+n^a.
So m=n.

by the way, did you participate at Iran Olympiad, or at the IMO?


function
real analysis
floor function
combinatorics unsolved
combinatorics

Use generating functions to evaluate the following sum:

$ \sum_k \binom{n\plus{}k}{2k}2^{n\minus{}k}$.

I tried writing is as a convolution and also writing $ A(x) \equal{} \sum_{n} a_n x^n$, where $ a_n$ is the desired sum and neither of those ideas worked.

Yes, but you may be disappointed because they don't use generating functions. I write

$ \sum_{k \equal{} 0}^{n} \binom{n \plus{} k}{2k}2^{n \minus{} k} \equal{} \sum_{k \equal{} 0}^{n} \binom{n \plus{} \left(n \minus{} k\right)}{2\left(n \minus{} k\right)}2^{k}$ (here we substituted $ n \minus{} k$ for $ k$)
$ \equal{} \sum_{k \equal{} 0}^{n} \binom{2n \minus{} k}{\left(2n \minus{} k\right) \minus{} k}2^{k} \equal{} \sum_{k \equal{} 0}^{n} \binom{2n \minus{} k}{k}2^{k}$ (because $ 2n \minus{} k\geq 0$ yields $ \binom{2n \minus{} k}{\left(2n \minus{} k\right) \minus{} k} \equal{} \binom{2n \minus{} k}{k}$).

Now, rename $ 2n$ into $ n$ to get a more general question: What is $ \sum_{k \equal{} 0}^{\left\lfloor n/2\right\rfloor}\binom{n \minus{} k}{k}2^k$ ? Denote

$ a_n \equal{} \sum_{k \equal{} 0}^{\left\lfloor n/2\right\rfloor}\binom{n \minus{} k}{k}2^k$,

and use that $ \binom{n \minus{} k}{k} \equal{} \binom{\left(n \minus{} 1\right) \minus{} k}{k} \plus{} \binom{\left(n \minus{} 2\right) \minus{} \left(k \minus{} 1\right)}{k \minus{} 1}$ to show the recurrence $ a_n \equal{} a_{n \minus{} 1} \plus{} 2a_{n \minus{} 2}$. What formula do you obtain for $ a_n$ for this recurrence?

Once you have found the answer, here comes a challenge: find a combinatorial proof by bijection. I have one, but it's one of those proofs that leave me in doubt whether bijections are really the most superior wisdom...

  darij

Find out what $ F(x) \equal{} \sum_{k} {n \plus{} k \choose k} x^k$ is ([url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=184510]hint[/url]) and then consider the value of $ F \left( \frac{1}{2} \right) \plus{} F \left( \minus{} \frac{1}{2} \right)$.  :)


Consider the finite set $ [n]\equal{}\{1,\dots,n\}$. We define a binary relation $ R$ over the set $ [n]$ and we write $ a\sim b$ to indicate $ a$ is related to $ b,\quad a,b\in R$. We have the following properties of $ R$:
(i) $ a\sim b \Rightarrow b\sim a$
(ii)$ a\sim b \quad and \quad b\sim c \Rightarrow a\sim c$

A relation $ R$ together with a set $ [n]$ is a set of unordered pairs denoted by $ ([n],R)$.
Consider the example $ ([n],R_1)\equal{}\{(1,2),(2,3),(4,12),(6,7)\}$. This is [i][b]extendable[/b][/i] to       $ ([n],R_2)\equal{}\{(1,2),(2,3),(1,3),(4,12),(6,7)\}$. In this case we call $ R_1$ to be [i][b]equivalent[/b][/i]  to $ R_2$ (over $ [n]$).(This defines a equivalence relation over the set of relations) Two relations are called [i][b]distinct[/b][/i] if they are [i][b]not[/b] [/i]  equivalent.

Let $ R_1,\dots,R_k$ be [i][b]all[/b][/i] the equivalent relations. A set $ ([n],R_i) \quad \forall 1\leq i\leq k$ is said to be a [i][b]minimal relation[/b][/i] iff the number of unordered pairs in $ ([n],R_i)$ is minimum among the number of unordered pairs of  $ ([n],R_i) \quad \forall 1\leq i\leq k$.

1.Do we have a unique minimum relation $ R_i$ among $ R_1,\dots,R_k$ ?
2.Find the number of distinct relations
   (i) when points are numbered
   (ii) Upto isomorphism.

First off, $ ([n],R)$ is not well-defined. In particular, the example $ ([n],R_1) \equal{} \{(1,2),(2,3),(4,12),(6,7)\}$ contradicts to the given definition of binary relation. Indeed, since $ 1\sim 2$ w.r.t. $ R_1$ and $ 2\sim 3$ w.r.t. $ R_1$, the property (ii) implies that $ 1\sim 3$ and hence,   $ ([n],R_1)\ni (1,3)$, a contradiction.

Usually, a binary relation on a set $ S$ (e.g., $ S \equal{} [n]$) is defined as subsets of $ S\times S$, i.e., a subset of [i]pairs[/i] of elements. Under this definition what you denoted $ ([n],R)$ is simply $ R$ itself. Furthermore, relations satisfying the property (i) are called [i]symmetric[/i] and those satisfying the property (ii) are called [i]transitive[/i].

A binary relation $ R$ on the set $ S$ can be viewed as a graph $ G \equal{} G(S,R)$ with the vertex set $ S$ and the arc set $ R$. Moreover, if $ R$ is symmetric then $ G$ is undirected, and if $ R$ is transitive then so is $ G$.

It seems that what you denoted $ R_1$, $ R_2$, etc. are not transitive relations. And you call two of them equivalent iff their [i]transitive closures[/i] coincide.

Now, what you called minimal relation is called [i]transitive reduction[/i]. In terms of graphs, it is simply a [i]spanning forest[/i] of the graph representing transitive relation; and if the graph is connected, this forest consists of a single [i]spanning tree[/i]. Therefore, the answer to question 1 is NO: generally, the same (transitive) graph may have several spanning forests (or trees). As of the question 2, see
http://www.cs.uwaterloo.ca/journals/JIS/VOL7/Pfeiffer/pfeiffer6.html
http://en.wikipedia.org/wiki/Binary_relation#The_number_of_binary_relations

Hello, I am also having the problem of counting the number of distinct binary relations in a set: 1,2,...,n.

I am trying to find the number for: total / symmetric / reflexive relationships

As I have seen [url=http://en.wikipedia.org/wiki/Reflexive_relation#Number_of_reflexive_relations]here[/url] the total number of relationships are:

$ 2^{n^2}$

And the number of reflexive are:

$ 2^{n^2\minus{}n}$

I'd like to ask for some help (a useful url maybe) with a short proof of the above (plus the number of symmetric relationships in the set 1..n).
I have checked maxal's url but I haven't quite understood a thing. :oops:

Thanks in advance. :)  :)  :)

To build a relation, decide whether each of the $ n^2$ pairs of elements of $ S \equal{} [n]$ is to be included or not.  This gives $ 2^{n^2}$ relations.  If the relation is reflexive, we must include the $ n$ pairs of an element with itself but can choose other pairs arbitrarily, giving $ 2^{n^2 \minus{} n}$.  If the relation is symmetric we may choose for each of the $ \binom{n}{2} \equal{} \frac{n^2 \minus{} n}{2}$ unordered pairs of different elements (since if we include one ordered pair we also include the symmetric pair) and separately for tne $ n$ pairs of an element with itself, giving $ 2^{\frac{n^2 \plus{} n}{2}}$.  Finally, if we want both reflexive and symmetric we actually have no choice about the $ n$ pairs of an element with itself, so it is $ 2^{\binom{n}{2}}$.


[quote="JohnnyCage52"]I haven't spent much time for this, but I think it is not difficult  :D

$ \left\{\begin{array}{l}x \plus{} \frac {x \plus{} 2y}{x^{2} \plus{} y^{2}} \equal{} 2 \\
y \plus{} \frac {2x \minus{} y}{x^{2} \plus{} y^{2}} \equal{} 0\end{array}\right.$
(I think we can multiply y to line 1 and x to line 2  :) )[/quote]

Multiply the upper equation by x and the lower equation by y:

$ \left\{\begin{array}{l}x^2 \plus{} \frac {x^2 \plus{} 2xy}{x^{2} \plus{} y^{2}} \equal{} 2x \\
y^2 \plus{} \frac {2xy \minus{} y^2}{x^{2} \plus{} y^{2}} \equal{} 0\end{array}\right.$

SUBTRACT the lower equation from the upper equation:

$ x^2 \minus{} y^2 \plus{} \frac {x^2 \plus{} y^2}{x^{2} \plus{} y^{2}} \equal{} 2x$

$ x^2 \minus{} y^2 \plus{} 1 \equal{} 2x$

$ x^2 \minus{} 2x \plus{} 1 \equal{} y^2$

$ (x \minus{} 1)^2 \equal{} y^2$ . . . (*)

We need another equation in x and y.

From the original set of equations, multiply the upper equation
by y and the lower equation by x:

$ \left\{\begin{array}{l}xy \plus{} \frac {xy \plus{} 2y^2}{x^{2} \plus{} y^{2}} \equal{} 2y \\
xy \plus{} \frac {2x^2 \minus{} xy}{x^{2} \plus{} y^{2}} \equal{} 0\end{array}\right.$

ADD the upper equation to the lower equation:

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

Divide both sides by 2:

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

$ xy \plus{} 1 \equal{} y$

$ xy \minus{} y \equal{} \minus{} 1$

$ y(x \minus{} 1) \equal{} \minus{} 1$

$ y \equal{} \frac { \minus{} 1}{x \minus{} 1}$ . . . (**)

Substitute the expression for y in (**) into y in (*):

$ (x \minus{} 1)^2 \equal{} (\frac { \minus{} 1}{x \minus{} 1})^2$

$ (x \minus{} 1)^2 \equal{} \frac {1}{(x \minus{} 1)^2}$

$ (x \minus{} 1)^4 \equal{} 1$

Let me work with the real roots for now.

Take the real fourth roots and use (**) to get the y-values:

$ (x \minus{} 1) \equal{} \pm 1$

Split it up:

x - 1 = -1
x = 0,  . . . . y = 1

x - 1 = 1
x = 2, . . . . y = -1

One way of solving for the nonreal solutions is to factor out the real
polynomial factors, after subtracting 1 from $ (x \minus{} 1)^4$, and then set
the remaining polynomial to zero and solve:

$ (x \minus{} 1)^4 \minus{} 1$ =
$ x^4 \minus{} 4x^3 \plus{} 6x^2 \minus{} 4x \plus{} 1 \minus{} 1$ =
$ x^4 \minus{} 4x^3 \plus{} 6x^2 \minus{} 4x$

Factor out x (from the x = 0 solution):

$ x^3 \minus{} 4x^2 \plus{} 6x \minus{} 4$

Factor out (x - 2) [from the x = 2 solution] probably either by long division
or by synthetic division to get:

$ x^2 \minus{} 2x \plus{} 2$

$ x^2 \minus{} 2x \plus{} 2 \equal{} 0$ has the solutions:

x = $ 1 \pm i$

x = 1 + i, y = i
x = 1 - i, y = -i

======================================================

All solutions:

(0, 1)
(2, -1)
(1 + i, i)
(1 - i, -i)

[quote="JohnnyCage52"]Another,................

$ 32x(x^{2} \minus{} 1)(2x^{2} \minus{} 1) \plus{} \frac {1}{x} \equal{} 1$

$ 8x(1 \minus{} 2x^{2})(8x^{4} \minus{} 8x^{2} \plus{} 1) \equal{} 1$

I can't solve it  :([/quote]

JohnnyCage52,

this problem belongs in its own separate thread, because it is separate (it's not a direct variation)
of the above problem that I worked on.

Also, you don't have two variables here, so there are no common points of solution between
them.

And if it is instead the case that you want the separate solutions in x between them,
then you don't type "I can't solve IT [singular]."

hello, your first equation $ 32x(x^2\minus{}1)(2x^2\minus{}1)\plus{}\frac{1}{x}\equal{}1$ have i factorized to
$ (x\minus{}1)(64x^5\plus{}64x^4\minus{}32x^3\minus{}32x^2\minus{}1)\equal{}0$, i think the second factor you can only solve by a numerical way.
Sonnhard.

hello, the equation
$ 8x(1\minus{}2x^2)(8x^4\minus{}8x^2\plus{}1)\equal{}1$ can be written as
$ (2x\plus{}1)(8x^3\minus{}6x\plus{}1)(8x^3\minus{}4x^2\minus{}4x\plus{}1)\equal{}0$
Sonnhard.


Welcome to the forum!!!

[hide="For nr.1 and 2,"] one can deduce a formula to find the sum of the integers, where $n$ is the last number in the sequence: $2n-1$
So nr.1 = $\boxed{63}$ and
nr.2 = $2\cdot{1024}-1 = \boxed{2047}$[/hide]

[hide="For nr.3 and 4,"] one can deduce a formula to find the sum of the fractions, where $\frac{1}{n}$ is the last number in the sequence: $\frac{n-1}{n}$
So nr.3 = $\boxed{\frac{31}{32}}$ and
nr.4 = $\boxed{\frac{1023}{1024}}$[/hide]

[hide="Nr.5"] $2^n-1$[/hide]
[hide="Nr.6"] $\frac{2^n-1}{2^n}$[/hide]

ummmm, can you tell me how you deduced that formula ???i cant understand  that 2n-1 thingy  :blush:  :blush:  :oops:

did you use geometric progressions or something??

If $2^n-1$ is the sum of the first $n$ numbers in the sequence, $2n-1$ is the sum of all the numbers in the sequence where $n$ is the last number in the sequence. Note that $n$ is not the same thing in both formulas. $2^n-1$ is a well known formula and can be proved by induction.

Yeah.  You can also use geometric progressions.  It's just a trivial extension.

Oops :oops:  you're right. My formula was for $1+2+4+...+2^{n-1}$. Didn't read the question properly.

so as you confirmed fourierseries , you can use GP too , right??

thx

Yeah.  To come up with a general formula, just telescope the sum.

Can someone show me the proof? I don't think I did it anywhere near what it should be..

(I started out with adding one to the series $2^0+2^1+2^2+...+2^n$)


$ ( (a\plus{}b) \plus{} (c\plus{}d) \plus{} (e\plus{}f) ) ( \frac{(a\minus{}b)^2}{a\plus{}b}) \plus{} \frac{(c\minus{}d)^2}{c\plus{}d} \plus{} \frac{(e\minus{}f)^2}{e\plus{}f}) \geq ((a\minus{}b)\plus{}(c\minus{}d)\plus{}(e\minus{}f))^2$

This yields $ \frac{(a\minus{}b)^2}{a\plus{}b} \plus{} \frac{(c\minus{}d)^2}{c\plus{}d} \plus{} \frac{(e\minus{}f)^2}{e\plus{}f} \geq \frac{(a\plus{}c\plus{}e\minus{}(b\plus{}d\plus{}f))^2}{a\plus{}c\plus{}e\plus{}b\plus{}d\plus{}f}$

Subtract 3 from both sides and the original inequality is proved.

[quote="Luba1123"]a,b,c,d,e,f>0
prove that:
$ \frac {ab}{a \plus{} b} \plus{} \frac {cd}{c \plus{} d} \plus{} \frac {ef}{e \plus{} f}\le\frac {(a \plus{} c \plus{} e)(b \plus{} d \plus{} f)}{a \plus{} b \plus{} c \plus{} d \plus{} e \plus{} f}$[/quote]

Just use the lemma: $ \frac{ab}{a\plus{}b}\plus{}\frac{cd}{c\plus{}d}\leq \frac{(a\plus{}c)(b\plus{}d)}{a\plus{}b\plus{}c\plus{}d}$


It's really easier to use $\LaTeX$ for something like this. Rephrased, using a standard floor notation.

Suppose $\sum_{i=1}^n\lfloor\sqrt{i}\rfloor=217$. Find $n$.

Let me give it a go.

[hide]As you begin testing values, it looks easier.
It begins 1,1,1,2,2,2,2,2,3,3,3,3,...
Written better: 1(3)+2(5)+3(7)+4(9) ...
The sum to n=48 is 203.  14 more are needed so a limit n of 50 will give the correct sum. [/hide]


To $\text{hoangminh}$: you can post it in the Vietnamese forum ([url]http://www.mathlinks.ro/Forum/index.php?f=256[/url]).

It is the first of mine. I can write the solution for it and it will shorter than you. I have much more inequalities anh them are very very difficult.
You understand?

Here is not as http://www.diendantoanhoc.net or http://www.mathnfriend.net. You post in abroad site, so take care of how it is. I know you quite a lot, and you should change your post in another manner, or other members laugh u.

Now, post your hard inequalites and we will kill it.  :lol:

[quote="Xevarion"][quote="Flove"]Now, post your hard inequalites and we will kill it.  :lol:[/quote]
 :rotfl:

Indeed, please be polite and civil. The purpose of mathematics is not to achieve personal recognition. (At least, it shouldn't be.)

edit: Just for clarification, I mean to address everyone, no one in particular. ;)[/quote]

Yesm of course, and also we will wait his problems.  :P

[quote="Albanian Eagle"][quote]1 secret:
Suranyi and Vasile are one.
Schur( n number) and SuranYi and Vasile are 3 in 1.[/quote]

I [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=89757]agree[/url][/quote]
I don't understand.
What Suranyi and Vasile are one?
What Suranyi and Vasile are 3 in 1?  :huh:

I don't understand a single heck of a Viatnamese word...

[quote="Albanian Eagle"]he means that Vasc's and Suranyi's inequalities are both a special case of a generalisation of schur for n variables. (The one I call Zaimi-Marku) :roll:[/quote]

But I don't think his idea is the same as you, about Zaimi-Marku inequality, Albanian Eagle.  :lol:


[tex]\[
\{ (x,\arctan x),x \in \mathbb{R}\}
\]
[/tex]

Show if the set is bounded,closed,open or compact.
I cant do it formally (Geometrically,if i sketch it,it is possible)

[hide]I would say, not bounded, not open, closed, not compact.[/hide]

I agree with Belenos.  The "not" ones are all pretty easy to show -- just find a counterexample to the definitions of the conditions.

By the way, where do these problems come from?


Yes, you want to use areas, where the possible region has area 1.  As for the line - what does a line contribute to area?  (i.e. does it matter if the line is 'dashed' or 'solid'?)


A natural number $n$ is said to have the property $P$, if whenever $n$ divides $a^{n}-1$ for some integer $a$, $n^2$ also necessarily divides $a^{n}-1$. [list=a] [*] Show that every prime number $n$ has the property $P$. [*] Show that there are infinitely many composite numbers $n$ that possess the property $P$. [/list]

[quote="Peter"]A natural number $n$ is said to have the property $P$, if whenever $n$ divides $a^{n}-1$ for some integer $a$, $n^{2}$ also necessarily divides $a^{n}-1$. [list=a] [*] Show that every prime number $n$ has the property $P$. [*] Show that there are infinitely many composite numbers $n$ that possess the property $P$. [/list][/quote]

[b](a)[/b] We have that $n=p$ is a prime number. From Fermat's theorem $p|a^{p-1}-1$ and $p|a^{p}-1$, then $p|a^{\gcd(p;p-1)}-1$ so $p|a-1$. We got that $a=pk+1$ for some $k$ then
\[a^{n}-1=(pk+1)^{n}-1 \equiv C_{n}^{p-1}(pk)=p^{2}k\equiv 0(mod p^{2})\]

[b](b) [/b]Let us prove that the number $n=pq$ possess the property $P$, for every different primes $p$ and $q$. Indeed, we have that if $pq|a^{pq}-1$ then $p|(a^{q})^{p}-1$ thus from (a) follows that $p^{2}|a^{pq}-1$ and similarly we can prove that $q^{2}|a^{pq}-1$. Therefore ,whenever  $pq|a^{pq}-1$ then $(pq)^{2}|a^{pq}-1$.

when i walk in the room, i make the whole place shimmmer


Has anyone qualified for the MathPath camp 2008? I personally haven't, but I am rather confident that I would though.

Isn't it due like May 18th or something?
I'm currently working on the Qualifying test [color=red]<snip>[/color].

[color=red][size=75][Moderator rcv:  Please avoid discussing the qualifying test, including your opinion of whether the test was "easy" or "hard".][/size][/color]

lol, i got [color=red]X[/color] done, and i got bored and havent come back to it for a couple months

[size=75][color=red][Moderator rcv:  See the announcement in this forum.  Do not discuss the qualifying test!][/color][/size]

ehh...where can you find the test and do you have to pay to get the test????i dont think i can go but could i take the test for fun and for free!!!!!!!!

http://www.mathpath.org somewhere on there

I thought you can't discuss?
Anyways, I'll just say that this test is [color=red]<snip>[/color] than last years test.

[color=red][size=75][Moderator rcv:  You shouldn't discuss the qualifying test, including your opinion of whether the test was "easy" or "hard".][/size][/color]

You can discuss questions, but not hints or how to solve them. (I think...)

Anyway, solved [color=red]X[/color] questions, now the logistics problem on how to send everything across half the globe (I live in Singapore) to the US.

Good luck to everyone out there!

[color=red][size=75][Moderator rcv:  [/size][size=125]No you may not discuss the questions![/size][size=75]  Read the announcement in this Forum.  Please avoid discussing the qualifying test, including your opinion of whether the test was "easy" or "hard" or which specific questions you solved.][/size][/color]

Eh, I can't [color=red]do #X[/color] and I didn't start solving [color=red]#Y[/color] yet.

[color=red][size=75][Moderator rcv:  Please avoid discussing the qualifying test, including your opinion of which questions are "easy" or "hard".][/size][/color]

hehe, I don't have to take the test :) I tried it anyways, I like this year's problems better than last year's.

are you allowed to give me this information:

i dont get [color=red]<snip>[/color] for [color=red]#X[/color]. can somebody draw me a pic?

[size=75][color=red][Moderator rcv:  Absolutely not!  The qualifying quiz must be entirely your own efforts.  As stated at the top of the quiz:  [/color][color=black] "You may ask other people to help you understand the statements of the problems, but the actual solutions must be your own. You may not work with others, including friends who may also be applying and online acquaintances." [/color][color=red]  Drawing a picture for you goes beyond "understanding the statement of the problem".  If there are specific terms whose definition you don't understand, you can ask for help like that.  Much beyond that, including understanding the mathematical significance of a question or sketching a picture, which makes the answer more obvious is not allowed.  The work *really* does need to be your own.][/color][/size]

I dunno

this was a problem from last year's mathpath challenge, which was given to some of the teachers at the camp, in which there were 7 problems, all written by students in the first problem writing seminar.

We can't draw you a pic because that would make it sorta obvious...

[size=125][color=red][Moderator rcv:  This was way too much help.  Please do not solve, partially solve, or rephrase questions for other applicants!][/color][/size]

As an aside, where does everyone live? I mentioned that I live in Singapore (which is 12 hours ahead of Vermont after Daylight Savings XD).

The one about [color=red]<snip>[/color] -- I haven't worked on it much, but it looks [color=red]<snip>[/color]....

Just for the fun of it (I went to MathPath last year) I solved [color=red]<snip>[/color] and it just required [color=red]<snip>[/color].

[size=75][color=red][Moderator rcv:  Please don't tell people what problem-solving strategy you used to solve a problem.  If the question-writers wanted to provide a hint as to the appropriate problem-solving strategy, they would have done so.][/color][/size]

Can't we just [color=red]<snip>[/color]?
Which is what I'm doing...

[size=75][color=red][Moderator rcv:  Please don't tell people what problem-solving strategy you used to solve a problem.  If the question-writers wanted to provide a hint as to the appropriate problem-solving strategy, they would have done so.][/color][/size]

Discussion of the Qualifying Test is completely unacceptable.  Most posts to this thread have been edited in some minor ways to eliminate the unfair discussion.  One thread contained a fairly complete outline on how to solve a problem.  This must stop!

Thread locked.


topology
parameterization
analytic geometry
calculus
derivative
advanced fields
advanced fields unsolved

1. Suppose $ M$ is a compact smooth n-dimensional manifold without boundary. Prove that $ M$is orientable iff there is a non-vanishing n-form $ w$ on M.

Is this okay?
if: Around each point of the manifold we pick a parametrization so that f*w(e_1,...,e_n) > 0
only if: Suppose we have orientation-compatible parametrizations covering the manifold. Assume that there are finitely many of them by compactness. Then we can pullback the volume form on M for each of these coordinate neighborhoods. Use partition of unity to add them up.

If this is fine, is there any place where I implicitly used the condition "manifold without boundary"? I think the whole thing is true for manifold with boundary too, am I right?

2. Is there any geometric interpretation of exterior derivative? I only have a vague idea:
change of flow in each direction (k-dim parallelepiped) describing a k-form w versus rate of change in one extra dimension ((k+1) dim parallelepiped) describing the (k+1)-form dw (the signs account for the orientation thing)

Is this the correct way to see it?

What is your definition of orientable? My book defines it to be precisely what you are trying to prove.

One of the standard definitions of "orientability" is the existence of an atlas such that the determinant of the differential of the coordinate change from every map to every other map is positive. Probably, that is the one that is to be used for this problem.

fedja is right. That is the definition in my book.

HilbertThm90, is this the way your book defined orientation on manifolds with boundary?

Hmm...it only defines it for a manifold $ M$ (I'm using Boothby and it is definition 7.5), which I seem to remember that all manifolds are assumed to be without boundary unless otherwise noted. Maybe that was a different text, though, I can't seem to find that explicitly written anywhere.

Ah, but you're definition is the very next theorem the way this book develops.


algebra
polynomial
calculus
calculus computations

find all polymial $ P$ satisfying $ P(0)\equal{}0$ and $ 0 \leq P'(x) \leq P(x)$ for all $ x \in (0,1)$

$ g(x) \equal{} e ^{\minus{}x}  P(x)$ has $ g'(x) \leq 0$ so g is decreasing, $ 0 \equal{} g(0) \geq g(x)$, $ 0 \geq P(x)$
On the other hand, $ P'(x) \geq 0$ so P is increasing, $ P(x) \geq P(0) \equal{} 0$ on [0,1]. Because of continuity, points 0 and 1 are included.
P has an infinity of roots, all interval [0,1], P = 0


Romeo wanted to throw an apple to Juliet, who was on a balcony 40 feet above him, so he tossed it upward with an initial speed of 56 ft/s. Juliet missed it on the way up, but then caught it on the way down. How long was the apple in the air?

I got this as a question on my summer assignment for my math class but I can't figure out how to do it with the given info. Is there like an equation? Can someone please help me.

It seems that this is about a vertical throw.

$t_{1}=\frac{v_{0}}{g}, H=\frac{v_{0}^{2}}{2g}, t_{2}=\sqrt{\frac{2(H-h)}{g}}, t=t_{1}+t_{2}$

$t_{1}$ and $H$ are respectively the time and the height of the ascent with the given initial speed, and $t_{2}$ is the time needed for the apple to travel $H-h$ during the free fall with no initial speed, where $h=40\,\text{ft}$ is a given.

With $g=32.174\,\text{ft/s}^{2}$ we get $t=2.478\,\text{s}$

What's the equation for the height given the initial speed? Can you state it in both m/s and ft/s?

The formula is given above. For unit conversion use $1\,\text{ft}=0.3048\,\text{m}$

I'm assuming $g$ is the force of gravity.

I just want to confirm that my equations are approximately right based on your formulas.

with height $h$ and time in seconds $t$,

and $v$ ft/s

$h\approx vt-16t^{2}$

and $v$ m/s

$h\approx vt-4.9t^{2}$

Is all that right?

Yes, that's the height in any given moment, i.e. $h(t)=v_{0}t-\frac{gt^{2}}{2}$. For the maximum height you must calculate the time of ascent, which is basically the moment when the apple stops. The velocity in any given moment is $v(t)=v_{0}-gt$, hence the time of ascent is $t_{0}={v_{0}\over g}$ and then just plug that into the formula for the height to obtain the maximum: $h_{max}=\frac{v_{0}^{2}}{2g}$

If you are using $x = ut-(1/2) g t^{2}$
(I am taking positive direction as up, you know $x$ (height of Balcony),  $g$    and  $u$  (initial velocity) , just solve for $t$

[hide]Yes  its a quadratic equation, so would get two  roots for $t$, larger value is what you need if you know that going down of the apple happens later. [/hide]


Let $\{a_n\}$ be the sequence which is defined by:
  $a_1=2,a_{n+1}=[\frac{3}{2}a_n]\forall n\geq 3$
Note that  $[x]$ denotes the greatest integer less than or equal to  $x$.
Prove that $\exists \alpha \in R $ such that:
   $a_n=[\alpha (\frac{3}{2})^n] + 1$
My method: I used Canto's theorem (but my solution was complicated).
Can anybody try this one?Sorry if it was posted before!

By the difinition of $[x]$, $\frac{3}{2}a_n-1<\left[\frac{3}{2}a_n\right]\leqq\frac{3}{2}a_n$,thus we have ${\frac{3}{2}a_n-1<a_{n+1}\leqq}\frac{3}{2}a_n\ (n=3,4,\cdots)$

kunny

Have a look here :

http://www.mathlinks.ro/Forum/viewtopic.php?t=16470

Pierre.


ratio
geometry
similar triangles
geometry proposed

a) This is obvious. Drop perpendiculars from the vertices and do the same thing with similar triangles as you did for the triangle case. The ratios all (when multiplied out) result in 1.
b) I'm not sure. Let me think about it.

[quote="ephraim"]1)Let $ABCD$ be the quadrilateral and four given points $M,N,P,Q$ such that: $M\in\ AB$, $N\in\ BC$, $P\in\ CD$, $Q\in\ DA$ and $M,N,P,Q$ are collinear. Prove that: $\frac{MA}{MB}$.$\frac{NB}{NC}$.$\frac{PC}{PD}$.$\frac{QD}{QA}$=1.

2)Is the problem true with a polygon (n sides) and n points?[/quote]

As for the question $(2),$ the Menelaus's theorem is generalized for a n-polygon, but it is not true (in all cases), inversely.

Kostas Vittas.

[quote="ephraim"]2)Is the problem true with a polygon (n sides) and n points?[/quote]

Let $A_{1}...A_{n}$ be the n-gone and let $B_{1}, B_{2}...B_{n}$ be the set of collinear points such that $B_{1}\in A_{1}A_{2}, B_{2}\in A_{2}A_{3}... B_{n}\in A_{n}A_{1}$. Consider the n stretches with centers $B_{1}, B_{2}...B_{n}$ which map $A_{1}$ onto $A_{2}$, $A_{2}$ onto $A_{3}$... $A_{n}$ onto $A_{1}$ respectively. The chain of these n stretches is a stretch which maps $A_{1}$ onto $A_{1}$ and the line $B_{1}B_{2}...B_{n}$ onto itself. So this stretch is the identity and the stretch factor is 1, qed.

Benjamin


Given a graph G with n vertices and independece number at most 2 (there does not exist three pairwise disjoint vertices) what is the minimum number of edges of G?

if we consider the complement $\overline{G}$ of such a graph $G$ then it is triangle free;hence $G$ has minimum numer of edges iff $\overline{G}$ has maximum number of edges,and the maximum such number is $\Big[\frac{n^2}{4}\Big]$,where $|V(G)|=n$.so min number of edges in $G$ is $\frac{n(n-1)}{2}-\Big[\frac{n^2}{4}\Big]$.


inequalities
function
inequalities proposed

Given positive reals a, b, c. Prove that:
$ \frac{a}{1\plus{}a(1\plus{}b)}\plus{}\frac{b}{1\plus{}b(1\plus{}c)}\plus{}\frac{c}{1\plus{}c(1\plus{}a)} \leq 1$

Function $ f: (0,\infty)\rightarrow\mathbb{R}$
\[ f(x) \equal{} \frac {x}{1 \plus{} x(1 \plus{} b)} \plus{} \frac {b}{1 \plus{} b(1 \plus{} c)} \plus{} \frac {c}{1 \plus{} c(1 \plus{} x)}\]
is increasing on $ (0,\frac {1}{bc}]$ and decreasing on $ [\frac {1}{bc},\infty)$ , so \[ f(a)\le f(\frac{1}{bc})\equal{}1\]


Prove that from 52 positive integers, we can select two such that their sum or difference is divisible by 100. Is the assertion also valid for 51 positive integers?

[hide]In $ \pmod{100}$, we can define our numerical values to range from $ -49$ through $ 50$.

If a number is $ \equiv 50$, then it's paired number must be $ \equiv 50$, and if a number is $ \equiv 0$, then it's paired number must be $ \equiv 0$. For all other $ n$, it's paired number can be either $ n$ or $ -n$, since $ n-n \equiv 0$ and $ n+(-n) \equiv 0$.

Thus, we can construct categories $ \equiv 0$, $ \equiv \pm 1$, ... $ \equiv \pm 49$, $ \equiv 50$. This is a total of 51 categories.

If there are at least 52 numbers, then by pigeonhole, there are two numbers in one category, and the assertion is valid. If there are 51 or fewer numbers, we cannot make this assertion since there are only 51 categories. [/hide]


topology
advanced fields
advanced fields unsolved

I have had problems with this exercise:
A compact surface M is nonorientable if and only if it contains a subset homeomorphic to the Möbius band.
[/quote]

Do you know the classification of surfaces theorem? Cause if you do you know how nonorientable surfaces are constructed by gluing down mobius strips along their boundary circles.


group theory
abstract algebra
function
superior algebra
superior algebra unsolved

This is a problem i've been trying the last days and i didn't come up with a solution. Your help would be more than precious.

A subgroup H<=G is said to be characteristic if phi(h) belongs to H for all h belonging to H and phi belongs to Aut(G). If this is the case, we write H char G.
I)Show that if H char G then H is a normal subgroup of G.
II)Find an example of a group G with subgroups H and K, such that H is normal subgroup of K which is normal subgroup of G, but H is not a normal sugroup of G.
III)Show that if H char K and K is normal subgroup of G then H is a normal subgroup of G.

Ok, but i don't see how i should use this fact..
Thanks

[quote="georgech"]
II) Find an example of a group G with subgroups H and K, such that H is normal subgroup of K which is normal subgroup of G, but H is not a normal sugroup of G.[/quote]

Take a look at the dihedral group $ D_{8}$... This group has a subgroup of order 2 that satisfies the condition that you require.

[quote="zeb"]Hint: The function $ h \rightarrow g^{ \minus{} 1}hg$ is an automorphism of $ G$, for any fixed $ g \in G$.[/quote]

The hint solves problem [b]I[/b] at once, [b][color=red]Georgie boy[/color][/b]... Why?

1- Fix $ g \in G$ and suppose that H [b]char[/b] G. You wish to prove that H is a [b]normal subgroup[/b] of G, right?

In order to do that, you [b]only[/b] have to show that $ ghg^{ \minus{} 1} \in H$ for every $ h \in H.$

Consider the homomorphism $ \phi : G \to G$ given by $ \phi(x) \equal{} gxg^{ \minus{} 1}$. It is straightforward to show that $ \phi \in \mathrm{Aut(G)}.$

Since H [b]char[/b] G (by hypothesis) it follows that $ ghg^{ \minus{} 1} \equal{} \phi(h) \in H$ for every $ h \in H$ and the proof terminates.

Hope things are clearer now...  :wink:

[quote="georgech"]

III)Show that if H char K and K is normal subgroup of G then H is a normal subgroup of G.

[/quote]

Last but not least...

Fix $ g \in G.$

Since K is a normal subgroup of G, we have that the homomorphism $ f: K \to K$ given by $ f(k)\equal{}gkg^{\minus{}1}$ is in fact an automorphism of K.

The fact that H is a characteristic subgroup of K implies now that $ H \equal{} f(H) \equal{} gHg^{\minus{}1}$ and the normality of H follows.

Regards.


algebra
polynomial
integration
inequalities
Irrational numbers

For each integer $n \ge 1$, prove that there is a polynomial $P_{n}(x)$ with rational coefficients such that $x^{4n}(1-x)^{4n}=(1+x)^{2}P_{n}(x)+(-1)^{n}4^{n}$. Define the rational number $a_{n}$ by \[a_{n}= \frac{(-1)^{n-1}}{4^{n-1}}\int_{0}^{1}P_{n}(x) \; dx,\; n=1,2, \cdots.\] Prove that $a_{n}$ satisfies the inequality \[\left\vert \pi-a_{n}\right\vert < \frac{1}{4^{5n-1}}, \; n=1,2, \cdots.\]

[quote="Peter"]Is there a nice solution for this, and why is it in chapter irrational numbers??  :maybe:[/quote]

It is because this is an example of 'rational' approximation of the irratioanl number $ \pi$.
I'm not so sure.. but the solution is not that hard..


Find all functions $f$ satifies ${{f(x\sqrt{y}})+f(y\sqrt{x}})=f(x+y)$ for all $x,y$ are non-negative reals

y=0 give $f(x)=2f(0)$. x=0 give f(0)=0. Therefore $f(x)=0$ for all x.


geometry
perimeter
calculus
integration
quadratics
analytic geometry
conics

It's a mock NYML, so naturally most of the questions are easy to the point of stupidity (and may not be suited for Intermediate Topics), but the later ones are more difficult than your average NYML, which is why I put them here.

$ 1.$  $ a \plus{} b \equal{} 3$
$ a \plus{} c \equal{} 4$
$ b \plus{} c \equal{} 5$
$ a \plus{} b \plus{} c \equal{} ?$

$ 2.$  Equally spaced tick marks are drawn on a loop of rope and three of these marks are pinned down, forming a right angle.  Assuming the rope does not slack, what is the minimum number of tick marks necessary for this to be possible?

$ 3.$  A secant intersects a unit circle at points $ A$ and $ B$ and is perpendicular to the radius it bisects.  If the perpendicular to the secant at $ A$ and the tangent to the circle at $ B$ meet at $ C$, compute the area of triabgle $ ABC$.

$ 4.$  Compute the number of real solutions to
$ p^2 \equal{} 25 \minus{} q^2$
$ p^2 \plus{} 2p \equal{} q^4 \plus{} 8q^2 \plus{} 15$

$ 5.$  Compute the largest positive integer value $ k$ such that for all primes $ p > 5, p^4 \equiv 1 (\text{mod} k)$

$ 6.$  Anna and Bob play a game involving 2008 chips.  They take turns removing a number of chips equal to a power of 2 (not including 1).  Assuming Anna plays perfectly, what is the largest number of moves necessary for her to win in?

I have possible answers for 1-5.

[hide]1.  Add them together to get 2a+2b+2c=12.  So a+b+c= 6
2.  It's basically asking for the smallest perimeter of a right triangle with integral sides, so 12.
3.  The secant is the perpendicular bisector of the radius.  This implies that the intersection point is the center of a square with A, B, and C as three of its vertices.  Since each side of the square is 1, the area of ABC is 1/2.
4.  We can rewrite the second equation to (P+1)^2=(q^2+4)^2.  If we sqrt this and substitute 25-p^2 for q^2, I think we have four real values for p and q.
5.  Using Euler's Theorem, p^(k-1)= 1 mod (k).  This implies that k is 5.  But there may be a larger solution if it says all primes p instead of all integers p...
6.  The strategy to win is to always leave Bob with 2^n +1 chips as long as n is odd.  That's as far as I got.[/hide]

You got 1 and 2 (of course), but the rest are wrong:[hide="Problem 3"]There's no squares involved here.[/hide][hide="Problem 4"]Using that method is nasty, and you didn't actually finish the problem anyway.[/hide][hide="Problem 5"]It does say all primes p.  You misread the problem.[/hide][hide="Problem 6"]It is impossible to get to an odd number of chips.[/hide]

[hide="Problem 3"]
Let $ O$ be the center of the circle and $ M$ be the intersection of the secant and the radius.
$ AB \equal{} \sqrt 3$, and because $ m\angle OBC \equal{} 90^\circ$ and $ \triangle OMB$ is a 30-60-90, it follows that $ m\angle ABC \equal{} 60^\circ$ and $ \triangle ABC$ is a 30-60-90 triangle. Thus $ AC \equal{} 3$ and the area is $ \boxed {\frac {3\sqrt 3}2}$.[/hide]

[hide="4"]
$ p^2 \plus{} 2p \equal{} q^4 \plus{} 8q^2 \plus{} 15 \implies (p \plus{} 1)^2 \equal{} (q^2 \plus{} 4)^2$

$ \therefore \pm (p\plus{}1) \equal{} p^2 \minus{} 21 \because q^2 \equal{} p^2 \minus{} 25$

Solve the quadratic equations and discard solutions where $ p^2 < 25$.
[/hide]

[hide="Alternate Solution to 4"]The first equation when graphed on a coordinate plane yields a circle of radius $ 5$ centered at the origin.  As shown above, the second equation simplifies to $ p\plus{}1 \equal{} \pm (q^2\plus{}4)$, which is the graph of two parabolas, one concave up with a vertex of $ (3,0)$ and the other concave down with a vertex of $ (\minus{}5,0)$.  Graphed on the same plane as the first equation, it is clear that the upper parabola intersects the circle twice, while the lower parabola is tangent to the circle at the point $ (\minus{}5,0)$, bringing the total number of solutions to $ \boxed{3}$.[/hide]

[hide="5"]
$ p^4 \equiv 1 \pmod {k} \implies (p \plus{} 1)(p \minus{} 1) \equiv 0 \pmod {k}$

It is well-known that all primes $ p > 3$ can be written in the form $ 6n \pm 1$ for some positive integer $ n$, and it can be shown that 6 is in fact the largest such number.
[/hide]

[hide="Problem 5"]$ p^4\minus{}1 \equal{} (p\plus{}1)(p\minus{}1)(p^2\plus{}1)$.  All three of those are even, and one of $ p\plus{}1$ or $ p\minus{}1$ is a multiple of $ 4$, giving us $ 16$.  Also, one of them is a multiple of $ 3$ as well.  Finally, take the whole thing mod $ 5$ and you get that $ 5$ is always a factor too.  Thus the answer is $ \boxed{240}$.  I'm not sure how you got $ 6$.[/hide]

[hide="Solution (#6)."]
I assume that the player who takes the last chip wins the game.  Thus, that player must avoid being left with, say, $ 6$ chips, as any move would allow his opponent to win on his next turn.  You can create a list of such numbers to be avoided and call those "bad" numbers.  All other numbers are "good" numbers, as they allow you to win by taking enough chips to land your opponent with a bad number.  Consequently, adding a power of two to a bad number results in a good number.  The first bad number is $ 0$, so $ 2$ and $ 4$ are good numbers.  However, $ 6$ is a bad number.  By adding a power of $ 2$ to a multiple of $ 6$, you can never obtain another multiple of $ 6$.  Thus, [i]all[/i] multiples of $ 6$ will be bad numbers.  As a result, taking, say, $ 4$ chips at the beginning would be a winning move.  I'm not sure, but I believe that only $ \boxed{10}$ moves are needed if $ 1024$ chips are taken on the first move, resulting in $ 984$ chips being left.[/hide]

[hide="6?"]
We have $ 2^n \equiv 1,2 \pmod{3}$. If Anna takes $ 1024$ chips on the first move, there will be $ 984 \equiv 0 \pmod{3}$ chips left; and Bob cannot make the chips another multiple of $ 3$, while Anna can reduce whatever number of chips Bob ends up with to a multiple of $ 3$. Since $ 0 \equiv 0 \pmod{3}$, Anna has the winning strategy. Now Bob's goal is to try to keep the pile size as large as possible and that Anna's goal is to reduce the pile size as quickly as possible, eg $ (1)\ 984 \minus{} 2 \minus{} 256 \equal{} 726\ (2); \minus{} 2 \minus{} 256 \equal{} 468\ (3); \minus{}4 \minus{} 128 \equal{} 336\ (4);$ $ \minus{}4 \minus{} 128 \equal{}  204\ (5); \minus{}2 \minus{} 64 \equal{} 138\ (6); \minus{} 2 \minus{} 64 \equal{} 72\ (7);$ $ \minus{}4 \minus{} 32 \equal{} 36\ (8);$ $ \minus{}2 \minus{} 16 \equal{} 18\ (9);$ $ \minus{}4 \minus{} 8 \equal{} 6\ (10);$ $ \minus{}2 \minus{} 4 \equal{} 0\ (\boxed{11})$.
[/hide]

[quote="azjps"][hide="6?"]
We have $ 2^n \equiv 1,2 \pmod{3}$. If Anna takes $ 1024$ chips on the first move, there will be $ 984 \equiv 0 \pmod{3}$ chips left; and Bob cannot make the chips another multiple of $ 3$, while Anna can reduce whatever number of chips Bob ends up with to a multiple of $ 3$. Since $ 0 \equiv 0 \pmod{3}$, Anna has the winning strategy. Now Bob's goal is to try to keep the pile size as large as possible and that Anna's goal is to reduce the pile size as quickly as possible, eg $ (1)\ 984 \minus{} 2 \minus{} 256 \equal{} 726\ (2); \minus{} 2 \minus{} 256 \equal{} 468\ (3); \minus{} 4 \minus{} 128 \equal{} 336\ (4);$ $ \minus{} 4 \minus{} 128 \equal{} 204\ (5); \minus{} 2 \minus{} 64 \equal{} 138\ (6); \minus{} 2 \minus{} 64 \equal{} 72\ (7);$ $ \minus{} 4 \minus{} 32 \equal{} 36\ (8);$ $ \minus{} 2 \minus{} 16 \equal{} 18\ (9);$ $ \minus{} 4 \minus{} 8 \equal{} 6\ (10);$ $ \minus{} 2 \minus{} 4 \equal{} 0\ (\boxed{11})$.
[/hide][/quote]I believe that to be correct, but I was counting each move by a player to be a move.  [hide="Thus the answer is"]$ \boxed{23}$[/hide]


:P yes but i look for a generalization of that .okthes is is one of n case ok amigo :stretcher: :pilot:


function
real analysis
real analysis unsolved

does there exist continuous function f from [0,1] to [0,1] such that for any y in [0,1] there exist uncountably many x such that f(x)=y.
  p.s. i think peano curve may work but am not sure.

It does work, as I mentioned [url=http://www.artofproblemsolving.com/Forum/index.php?f=67]recently[/url]. The Cantor-staircase construction from that thread only gives countably infinite preimages, and I do not know if it can be improved to uncountable.  :maybe:


Consider $ a,b,c>0$ such that $ a\plus{}b\plus{}c\equal{}1$. Prove that: $ \sum \frac{a}{abc\plus{}a^2\plus{}1}\le \frac{27}{31}$

Here is an adapted solution by [color=orange][b]S.O.S.[/b] [/color] (sum of squares) inspired from [color=orange][b]can_hang2007[/b][/color]'s solution:

For $ a,b,c >0$ and $ a\plus{}b\plus{}c\equal{}1$ it is true the following inequality $ \boxed {9abc\plus{}1 \ge 4(ab\plus{}bc\plus{}ca)}$ which can be deduced easily.

Using this inequality, we get:

$ \sum \frac{9a}{4(ab\plus{}bc\plus{}ca)\plus{}8\plus{}9a^2} \le \frac{27}{31}$

Denote by $ s\equal{}\frac{8\plus{}4(ab\plus{}bc\plus{}ca)}{9}$

$ \sum \left (\frac{1}{31}\minus{}\frac{a}{a^2\plus{}s} \right )\ge 0$

$ \sum \frac{(c\minus{}a)(7a\plus{}8c\plus{}10b)\minus{}(a\minus{}b)(7a\plus{}8b\plus{}10c)}{31(a^2\plus{}s)} \ge 0$

$ \sum (a\minus{}b)^2 \frac{8a^2\plus{}8b^2\plus{}15ab\plus{}10c(a\plus{}b)\plus{}s}{(a^2\plus{}s)(b^2\plus{}s)} \ge 0$

which is true.


geometry
3D geometry
function
calculus
integration

[quote="TheShadowWithin"]Hello well these are some of the tougher math problems in this contest.

Anyways these are some problems my teacher gave me in class for fun. I solved the first few, but these last couple are puzzling me. If you guys want to just answer 1 or 2 its fine. I just want to learn so that I can be better at prob solving (one of my weak points) as usually my math teacher only goes over 1 or 2 of them.

7) If x^2+5x+6 is a factor of x^4+ax^2+b then a+b equals:

8) If x is a solution of the equation x^2 = 8x +13, then x^3 = ax+b where a and b are integers. The sum of a and b equals:

9) The volume of a right circular cone is given by V = 1/3(pi)(r^2)(h) where h is the height and r is the radius of the base. The top is removed by slicing the cone with a plane parallel to its base at a distance 1/2h from it. The volume of the remaining piece is given by:

10) How many positive integers less than 100 have exactly 4 factors (including the number itself and 1 as factors)?[/quote]

For number 8,
first, since it is chopped off at 1/2h, we have the volume of the slice to be $ \frac{\pi r^2}{3}\times{\frac{h}{2}}$, subtracting this from the original volume we have: $ \frac{\pi r^2 h}{6}$ I am still working on the number 7.

[hide="Less computational alternative for 7"]$ x^4 \plus{} ax^2 \plus{} b$ is divisible by $ x^2 \plus{} 5x \plus{} 6 \equal{} (x\plus{}2)(x\plus{}3)$, so it has $ \minus{}2,\minus{}3$ as roots. But it is also an even function, so it must have $ 2,3$ as roots. Then $ x^4 \plus{} ax^2 \plus{} b \equal{} (x\minus{}2)(x\minus{}3)(x\plus{}2)(x\plus{}3)$ and putting $ x \equal{} 1$ in gives $ 1 \plus{} a \plus{} b \equal{} (1\minus{}2)(1\minus{}3)(1\plus{}2)(1\plus{}3)$. Now just isolate $ a\plus{}b$.[/hide]

For #8 (I think Poincare did #9 instead),

$ x^2 \equal{} 8x \plus{} 13$
$ \to x^3 \equal{} 8x^2 \plus{} 13x$
$ \to x^3 \equal{} 8(8x \plus{} 13) \plus{} 13x$
$ \to x^3 \equal{} 77x \plus{} 104$.

Thus, $ (a, b) \equal{} (77, 104)$, so $ a \plus{} b \equal{} 181.$

For #10, note that a number has exactly 4 factors if and only if it is the product of exactly two distinct primes. We wish to count the number of ordered pairs $ (p, q)$ where $ p < q$ and $ pq < 100$.

If $ q > 50$, then $ p < 2$ (otherwise $ pq > 100$.) Since there's no integral prime less than 2, $ q \leq 50$.

We see that the solutions are (2, 47), (2, 43), (2, 41), (2, 39), (2, 37), (2, 31), (3, 31), (2, 29), (3, 29), (2, 23), (3, 23), (2, 19), (3, 19), (5, 19), (2, 17), (3, 17), (5, 17), (2, 13), (3, 13), (5, 13), (7, 13), (2, 11), (3, 11), (5, 11), (7, 11), (2, 7), (3, 7), (5, 7), (2, 5), (3, 5), and (2, 3).

This yields 31 solutions. Sorry if I counted wrong/forgot a solution. A better solution for #10 is welcome.

Zhero I believe #10 the correct answer is 32, but thanks for the explanation I get it now. Though a little tedious. Its the exact same method my teacher used, but there is probably a much quicker way.

That solution is incorrect; you forgot cubes of primes.

Woops. =/ My bad.

By the way, I don't think there's a faster way. Due to the odd distribution of primes, I think the only way is to list them. Correct me if I'm wrong, though.

[quote]For number 8,
first, since it is chopped off at 1/2h, we have the volume of the slice to be , subtracting this from the original volume we have:  I am still working on the number 7.
[/quote]

I believe your answer is incorrect from the answer key, which states it as 7/24(Pi)(r^2)(h), but since I can't solve this myself I am not sure if the key is wrong to your answer.

hmmm... I am pretty sure that key is wrong, but then again, the answer came too easily. Any suggestions?

For number 9, I think the answer is $ \frac{7}{8}*\frac{1}{3}\pi r^2*h$.

Notice that the smaller cone and the larger cone are similar.  Thus, every length is cut in half in the new cone.  Volume is essentially something cubed, so the new volume is $ (\frac{1}{2})^3\equal{}\frac{1}{8}$ the original volume.  Subtraction yields the desired answer.


Let f be a function defined in the set of integers such that f(1)=3 and f(x)f(y) is equal to f(x+y)+f(x-y). Find f(7).


geometry
geometric transformation
rotation

carcul, i have a basic doubt in stereochemistry.why is chirality a necessity for optical activity? what makes chiral compounds special that they can polarize light? why can't an achiral compound(for eg CHCl3)exhibit optical activity?

Instead of explaining you that right away, I will try to walk you through the reasoning. Start by answering the following questions:

1) What is chirality?

2) What is optical activity?

Not all compound with chiral centres exhibit optical activity. Compounds that are meso for instance have chiral centres but are no optical activity.

First, Physici olympiaci have not referred to "chiral centres", but to chirality. Second, let's wait until Physici olympiaci thinks/investigate for himself, shall we?

sometimes chiral molecules can be optically inactive(or a at least their rotation has not been measured till now bcoz the diff bbetween the groups is negligible) for eg. ethyl-npropyl-nbutyl-n-hexylmethane


Hi. I'm Matteo, I live in Italy (I am sorry for my horrible english. Correct me if you want please.)  and unforunately I do not like how things are going in my country. Obviously we have got beautyful places to visit and other touristic things of this kind. But I do not care of them. I would like going studying in a foreign country because I feel that here I have no oportunities. I would like to become a physiscian but in Italy it is really hard to find a work in this field.
I think that the problem comes from Italian politicians that I think are seen, in the rest of the world, but also here in Italy, like a funny joke and probably they are.
In my opinion another huge problem we have is the church. In fact it is always concerned about "non-spirituals" matters and it inflouences the public opinion.
I would be really pleased to hear from you all how Italy is seen in the rest of the world.

(I want to apologize another time for my horrible english  :blush: , I hope that writing in this forum will improove my skills)
Bye  :)

Both Italian politicians and Catholic church lie squarely outside of my area of concern. The achievements of  Italian mathematicians in geometric measure theory and partial differential equations are of considerable interest to me.
But I don't know how you can use such input in your career decisions.  :maybe:

I'm aware of the huge influence the church wields in Italian politics, but it doesn't seem to cause as much of a problem as it would in the U.S.--or at least, that's my impression.  I've also heard that there's a good deal of corruption in Italian politics, but I don't know much about this.  In general, I think Italy is seen favorably, at least in the West.  Certainly, you shouldn't encounter any problems or obstacles in your life because of popular perceptions of Italy.

Well I'll try to express what I and people in general think. :maybe:

I had the opportunity to visit Italy with my school in 2002. So most of what I say is from a tourist's perspective.
One of the first things I noticed was that so many people were trying to rip us off. :huh:

Italian men have a reputation of being charmers/machos.  I don't know if this is true for "grown" men, but when we were in Napoli, there were tons of young kids driving around near the beach making noise and making proposals to the girls in our group.
Mafia is also one thing people tend to think of, a few years later a 100 people were killed in one year in Napoli. Had I known about that, then I wouldn't have been walking by myself in Napoli at night as a 17-year old foreigner. :ninja:

Society is way more catholic there but that comes as no surprise, since in Europe southern countries like Spain, Portugal, Malta....are like that. And of course Rome is the heart of catholicism.
In one city our teacher told us there were more than a dozen parishes there, and each year they would (fiercely!)  compete with each other in championships.
In Malta I also had the chance to witness a "festa" when everyone parties and people carry a statue of their saint around.  And I had the impression I had in Italy : religion is really part of life.  Where I live, people don't care about going to church, and if they do, then it's never to actually have a good time. :!:

On a political level, Italy didn't really make a good impression. In the eighties there was prime minister Craxi who actually fled the country to settle in Tunisia to evade imprisonment.  Berlusconi is sort of the same problem, it was hard to take him serious.
Actually, when it comes to politics, Italy doesn't seem to very stable.  I am referring (for instance) to the fact that Italian communist obtained 34.4% in the parliamentary elections in 1976, which is amazingly high for a western country (and this is well known here since it is often referred to when it comes to blocking "radical" parties from power)

But don't get me wrong.  :wink:  I liked Italy. It is obviously the center of the ancient Roman Empire and that has left its traces, and also the land where the renaissance etc..was born.

To tell you the truth, I know universities here have good relations with Italian universities, mathematicians like Segre are well known here and it happens from time to time that I stumble upon books in Italian in the library.

Okay you have got the experts opinion from fred. It's the best available on this forum. :roll:

Firstly, to clear a point, you write perfect english for a non-british european. It is simple and lucid, and the grammar is almost flawless, which is often more than can be said about other italians, germans etc on this forum. The reason simply is that you are not taught english as a first or second language. So do not worry about your english, it is way above readable. :) (As for corrections, you have made some spelling mistakes (beautyful, physiscian, inflouence etc.) but they are very probably typing errors! :P )

Coming to Italy, she has by far the richest history among the european countries, comparble to the asian civilizations, owing mainly to the Roman empire and the Renaissance period. Assuredly, the era of the popes has left an indelible black mark on her history since the dark ages, and Mussolini onwards, her history is a bit clouded in the recent years, but a country which can produce an enrico fermi can produce, frankly, anything. So cheer up! I won't ask you to love your country, for I am an anti-nationalist, but do be proud of its culture.

God (if He/She/It exists) bless you!

[quote="bubka"]Okay you have got the experts opinion from fred. It's the best available on this forum. :roll: [/quote]
I'm not an expert, the only times I understand Italian is when I fake it using French/Latin. Therefore I cannot be an expert. :(

I didn't want to offend anyone, maybe I focussed a bit too much on the weird/bad things. :D

Italy is a typical school trip for people here in the last months before they finish high school.  It's considered THE country to go if you want a cultural overdose. :P

Italians are obviously cultural people when it comes to food, I ate delicious pasta all the time. :wink:

The stereotype we have here is that Italians are typically "Romance" people, meaning more "laid back" than for instance Germans.

[quote]Coming to Italy, she has by far the richest history among the european countries, comparble to the asian civilizations, owing mainly to the Roman empire and the Renaissance period. [/quote]
Well actually the Romans were pretty good at copying stuff from other people like the Greeks too. :)

Aside to fred: [quote="fredbel"] I'm not an expert, the only times I understand Italian is when I fake it using French/Latin. Therefore I cannot be an expert. :( [/quote] I have a friend who navigated us for a week or a little more through Italy, using only a combination of the Spanish and Latin he had learned in school.  It was amazing.

I would like to second Bubka in saying that your English skills are quite good.  I've added corrections for the errors I noticed in your original post below the fold, if you want to compare.
[hide][quote="Matteo_89"]Hi. I'm Matteo, I live in Italy (I am sorry for my horrible english. Correct me if you want please.)  and unfor[i]t[/i]unately I do not like how things are going in my country. Obviously we have got beaut[i]i[/i]ful places to visit and other touristic things of this kind. But I do not care [i]about[/i] them. I would like [i]to go study[/i] in a foreign country because I feel that here I have no op[i]p[/i]ortunities. I would like to become a physician but in Italy it is really hard to find work in this field.
I think that the problem comes from Italian politicians that I think are seen, in the rest of the world, but also here in Italy, like a funny joke and probably they are.
In my opinion another huge problem we have is the church. In fact it is always concerned about "non-spirituals" matters and it influences the public opinion.
I would be really pleased to hear from you all how Italy is seen in the rest of the world.

(I want to apologize another time for my horrible english  :blush: , I hope that writing in this forum will improve my skills)
Bye  :)[/quote][/hide]

As for the actual content of your post: I think Italian politicians are viewed as a joke in much of the rest of the world -- Italy is seen as something of a mafia-run state.  Unfortunately, it's not clear to me what sorts of things one can do to counter-act this fact (either the image or the reality).

Actually, a funny story: I studied in Hungary for a semester and took a history class there.  One day, we were discussing what the distinction in between "Eastern Europe" and "Western Europe," since from the point of view of the Hungarians (and, probably, the Czechs, Slovaks, Slovenes, etc.) they belong to "Central Europe" (which they certainly do geographically, if you take Europe to extend all the way to the Urals).*  Someone suggested that bad governance (corruption, mafia influence, etc.) were the explanation, and our teacher responded, "Well, certainly, many of the Eastern European states are poorly governed.  But look at Italy!  No one is disputing that Italy belongs in Western Europe, and we're no more corrupt than they are."

Now, back to the discussion at hand.  The other major view of Italy, at least in America, is what fred and bubka mentioned: Italy is viewed as the cradle of a very large amount of European culture, and responsible for what it means to be "in the West."  (Maybe this view is less-common among the average American -- we are hampered in our appreciation of culture by having been a British colony, after all. :wink: )  Also, as the place where pizza and pasta were invented.

So, I'm going to stop these ramblings for now, but perhaps I will stumble across something intelligent to say later.

* As a totally unrelated point, I happen to believe that the distinction is based almost entirely on whether one spent longer being ruled by Fascists or by communists -- it's the only way to explain how Greece is viewed as western.

I'm a student of Italian, and have visited twice.  I loved the country, but for all the things that are of no consequence to you- the food, the atmosphere, and a chance to speak italian.  The only negative aspect was the reality of pick pockets in Rome.  As for politics we never talked about them much in my italian classes, but when we did we never took it seriously.  The political situation is seen as something of a joke.  A me, Italia e` una bella  paese, ma non devo vivere li.

As for being a doctor, I don't believe it is difficult to find work.  Doctors are always in demand.  If you want to go study in a foreign country, I doubt you would have any problems because you are Italian- especially because your english is so good.

Oh I forgot to mention another fascinating gentleman of yours, to wit, Al Capone!
@JBL: [hide="To be fastidious"]:[quote="Matteo"]Hi. I'm Matteo, I live in Italy (I am sorry for my horrible english. Correct me if you want [i]to[/i] please.) and unfortunately I do not like how things are going in my country. Obviously we have got beautiful places to visit and other touristic things of [i]that[/i] kind. But I do not care about them. I would like to go study in a foreign country because I feel that here I have no opportunities. I would like to become a physician but in Italy it is really hard to find work in this field.
I think that the problem comes from Italian politicians that I think are seen, in the rest of the world, but also here in Italy, [i]as[/i] a funny joke and probably they are.
In my opinion another huge problem we have is the church. In fact it is always concerned [i]with[/i] "non-spirituals" matters and it influences the public opinion.
I would be really pleased to hear from you all how Italy is seen in the rest of the world.

(I want to apologize another time for my horrible english  :blush: , I hope that writing in this forum will improve my skills)
Bye  :) [/quote]
And that's about it. And if with only ten minor corrections (three of them possibly typos) your english stands thus, it can't be horrible can it?[/hide]

By the way I just noticed this:
[quote="Matteo"]I would like to become a physiscian but in Italy it is really hard to find a work in this field. [/quote]
Won't that suggest something rather along positive lines about general Italian health? :rotfl:

[quote="JBL"]Aside to fred: [quote="fredbel"] I'm not an expert, the only times I understand Italian is when I fake it using French/Latin. Therefore I cannot be an expert. :( [/quote] I have a friend who navigated us for a week or a little more through Italy, using only a combination of the Spanish and Latin he had learned in school.  It was amazing.[/quote]
However, Italians -well at least those I needed to talk to- do tend to speak English quite well (and French too, but I didn't really try that) so there was no problem. Come to think of it, there weren't many occasions when we talked with local people. :|

[quote]As for the actual content of your post: I think Italian politicians are viewed as a joke in much of the rest of the world -- Italy is seen as something of a mafia-run state. [/quote]
Well Berlusconi was very extraordinary. I heard he recorded his own songs, he was rude to women, he enforced regulations concerning bringing spiders into the country just because he didn't like them....
Romano Prodi seems a rather serious man (compared to Berlusconi)

Here are some quotes from Berlusconi:

[quote]To German MEP Martin Schulz, at start of Italy's EU presidency in July 2003:
"I know that in Italy there is a man producing a film on Nazi concentration camps - I shall put you forward for the role of Kapo (guard chosen from among the prisoners) - you would be perfect." [/quote]

[quote]"Mussolini never killed anyone. Mussolini used to send people on vacation in internal exile."[/quote]

[quote]"The West will continue to conquer peoples, even if it means a confrontation with another civilisation, Islam, firmly entrenched where it was 1,400 years ago." [/quote]
(from   [url]http://news.bbc.co.uk/1/hi/world/europe/3041288.stm[/url])

Of course, I only hear about Italy from time to time, when I read BBC articles or even Dutch articles about my country, I usually find one mistake after the other. When a leader messes up, they often don't show his subordinates disagreeing with him.

[quote] Unfortunately, it's not clear to me what sorts of things one can do to counter-act this fact (either the image or the reality).  [/quote]
Well some time ago I read about traders in Napoli jointly refusing to keep paying "protection mafia" to the mafia. Maybe small things like that can improve things..

[quote]Actually, a funny story: I studied in Hungary for a semester and took a history class there.  One day, we were discussing what the distinction in between "Eastern Europe" and "Western Europe," since from the point of view of the Hungarians (and, probably, the Czechs, Slovaks, Slovenes, etc.) they belong to "Central Europe" (which they certainly do geographically, if you take Europe to extend all the way to the Urals).*  [/quote]
Here we tend to describe that used to be on the other side of the Iron Curtain as "Eastern Europe".

[quote]Someone suggested that bad governance (corruption, mafia influence, etc.) were the explanation, and our teacher responded, "Well, certainly, many of the Eastern European states are poorly governed.  But look at Italy!  No one is disputing that Italy belongs in Western Europe, and we're no more corrupt than they are."[/quote]
Well again I'm no expert, but I think there are some regional differences.  The south is  poorer, I wouldn't dare to say things about the entire country in general. We should keep in mind that this country is only about 150 years old.

[quote]The only negative aspect was the reality of pick pockets in Rome[/quote]
Yes, Rome can seem unsafe at times.  The traffic is crazy.  There are cameras in lots of places watching the people.  There were actually some people in our group who walked from one hotel room to another without locking their doors, and some people (nonitalians) had gotten into the hotel and stole several things. :| When walking at night, you might want to check on what kind of neighbourhood you are in from time to time.

I just want to add something, since I'm a violinist:

Italy, in my opinion is where classical music is at its fullest. Along with Vienna, Italy creates the focal point of many aspects of European classical music.

Italy also makes many beautiful sounding violins. :)

I've always wanted to visit either Italy or Vienna, have fun with great musicians...

Ah, music. The reason the world is full of people who don't read or speak Italian, but do know the meaning of phrases like [i]poco piu mosso[/i] or [i]senza sordino[/i] or [i]col legno[/i].

Thank you very much for all your answers. I have been really glad to see that a lot of people posted on my topic. I appraciate all your answers. Fred, you have probably highlighted all the most important things and I share all your ideas.
I have probably been too much pessimistic, but you all make me appraciate more all the good aspects of my country which you have just said: history, music and other cultural things. What still let me down is the fact that a large amount of people have voted Berlusconi and a large amount of people will vote Berlusconi in the future. The "politician" who has been [b]perfectly[/b] described by Fred  :lol: (I smile even if I should cry  :roll: ) .
Thank you also for all the corrections. I think they will be really useful to me.

(Who wants to add something else is welcome)

There's another thing I've learnt from life, and since you seem to be a year younger to me, I'll pass it on: do not make hasty opinions about politicians; and if possible, do not make opinions at all. India as you know is reportedly full of some of the world's most corrupt politicians, but beyond saying that they are all dashed stupid, I seldom go out of my way to abuse them. I don't know about Berlusconi, but I believe that either a guy should take up politics actively, or stay away fully; building your own career while abusing politicians is low.

And as for your history and cultural things, we've only just begun mentioning them, (which is clear from the fact that Galileo hasn't found a mention here yet :P ), and I'd advise you to study history: of your country and the world in general. Not only does that make you broader, it helps you provide answers like freds'. :roll:

Another point: at least here in the USA, our image of Italy has been shaped by Italians who immigrated here, and those immigrants are disproportionately from the poorer south. Where in Italy are you from, Matteo?

And since I like soccer:
As the current #1 soccer country!  :wink:
Forza Italia  :first:

I come from Arezzo in Tuscany, it is near to Florence.

My friend used to live in Italy. He sayed it was Ok. He hated the weather. Luckly for him, he found somekind of english school in his city (rome). I don't know anything about Italy except about it's history. As you said you want to be a physician (ahy cunt spel wery godd), there is no way that there are no good medical schools in Italy. Not, to be too critical, buy I think you should accept Italy as it is and to appreciate it. It's not everybody who gets to be raised in the land of one of the greatest empires!

The Pope and many famous cathedrals makes Italy a religious  tourist attraction. Many people see Italy only as simply the head of the church. Others see it as a place filled with beautiful buildings and art. For many of these and other people Italy is just one big tourist attraction with no other point but being the capital of the church.

actually, many people see it just as a place with good food.  :D


geometry
parallelogram
number theory
least common multiple
circumcircle
geometric transformation
reflection

Let $K$, $L$, $M$, $N$ be the midpoints of the sides $AB$, $BC$, $CD$, $DA$, respectively, of a cyclic quadrilateral $ABCD$.
Show that the orthocenters of triangles $NAK$, $KBL$, $LCM$ and $MDN$ are the vertices of a parallelogram.

Label the orthocenters of the triangles $\triangle NAK, \triangle KBL, \triangle LCM, \triangle MDN$, as $E, F, G, H$ and the quadrilateral circumcenter as $O$.

The altitudes $KE, MH$ are parallel (both $\perp$ to $DA$), the altitudes $KF, MG$ are also parallel (both $\perp$ to $BC$), hence, $\measuredangle EKF = \measuredangle GMH$.
The altitudes $LF, NE$ are parallel (both $\perp$ to $AB$ ), the altitudes $LG, NH$ are also parallel (both $\perp$ to $CD$), hence, $\measuredangle FLG = \measuredangle ENH$.

The segments $KE = ON$ are equal, because $KE \parallel ON$ (both $\perp$ to $DA$), $NE \parallel OK$ (both $\perp$ to $AB$), and the quadrilateral $NEKO$ is a parallelogram.
The segments $MH = ON$ are equal, because $MH \parallel ON$ (both $\perp$ to $DA$),  $NH \parallel OM$ (both $\perp$ to $CD$), and the quadrilateral $MHNO$ is a parallelogram.
Therefore, $KE = MH$.

The segments $NE = OK$ are equal, because $KL \parallel OL$ (both $\perp$ to $BC$), $LF \parallel OK$ (both $\perp$ to $AB$), and the quadrilateral $KFLO$ is a parallelogram.
The segments $MG = OL$ are equal, because $MG \parallel OL$ (both $\perp$ to $BC$),  $LG \parallel OM$ (both $\perp$ to $CD$), and the quadrilateral $LGMO$ is a parallelogram.
Therefore, $KF = MG$.

The quadrilaterals $NEKO, KFLO$ are both parallelograms (see above), therefore $NE = OK = LF$.
The quadrilaterals $MHNO,  LGMO$ are both parallelograms (see above), therefore $LG = OM = NH$.

Since $\measuredangle EKF = \measuredangle GMH$, $KE = MH, KF = MG$, the triangles $\triangle EKF, \triangle GMH$ are congruent by SAS and $EF = GH$.
Since $\measuredangle FLG = \measuredangle ENH$, $LG = NH, LF = NE$, the triangles $\triangle FLG, \triangle ENH$ are congruent by SAS and $FG = HE$.

Consequently, $EFGH$ is a parallelogram.

Another solution:

Let O be the circumcenter of the cyclic quadrilateral ABCD, and let X, Y, Z, W be the reflections of this circumcenter O in the sides AB, BC, CD, DA, respectively. Now I will prove that the midpoint of the segment WX is the orthocenter of the triangle NAK. When this will be proven, it will follow by analogy that the midpoints of the segments XY, YZ, ZW are the orthocenters of the triangles KBL, LCM, MDN, respectively, and since the midpoints of the segments WX, XY, YZ, ZW form a parallelogram (actually, the midpoints of the sides of an arbitrary quadrilateral form a parallelogram), it will follow that the orthocenters of triangles NAK, KBL, LCM, MDN form a parallelogram, so that the problem will be solved.

So it remains to prove that midpoint of the segment WX is the orthocenter of the triangle NAK. For the proof, we denote by S the midpoint of the segment WX.

Being the circumcenter of the quadrilateral ABCD, the point O lies on the perpendicular bisector of its side AB. Thus, the midpoint K of this side AB is the orthogonal projection of the point O on the side AB. On the other hand, the point X is the reflection of the point O in the side AB. Hence, the point K is the midpoint of the segment OX. Combining this with the fact that the point S is the midpoint of the segment WX, we see that the line KS is a midparallel in the triangle WOX and thus parallel to its side OW. But since the point W is the reflection of the point O in the line DA, we have $OW\perp DA$, and thus, KS || OW becomes $KS\perp DA$. In other words, $KS\perp AN$. Thus, the point S lies on the altitude of triangle NAK issuing from its vertex K. Similarly, the point S lies on the altitude of triangle NAK issuing from its vertex N. Hence, the point S is the point of intersection of two altitudes of triangle NAK, and thus it is the orthocenter of the triangle NAK. This completes the proof that midpoint of the segment WX is the orthocenter of the triangle NAK.

  Darij

Orthocenters of the triangles $\triangle ABC, \triangle BCD, \triangle CDA, \triangle DAB$ make a quadrilateral congruent to the cyclic quadrilateral ABCD.

[quote="darij grinberg"][quote="Armo"]I am not sure, but isn't it sufficient to prove that the orthocentres of the triangles $ABD, ACD, BCD, ABC$ form parallelogram? ;)[/quote]

Actually, they don't. ;)

  Darij[/quote]
[quote="yetti"]The orthocenters of the triangles ABC, BCD, CDA, DAB make a quadrilateral congruent to cyclic quadrilateral ABCD.

...[/quote]
Some year in balkan math. olympiad was a problem same as [b]yetti[/b] said, but I remembered it wrong.I remembered that the problem was to prove that the orthocentres form parallelogram and I was wrong.     :(


How can I make my header show the section name in the upper right corner and the page number in the upper left corner? (in a different font size than what appears in the body of the text)  Can someone also show me how to create that large line directly below the header?

Thanks!

See the [url=http://www.tug.org/tex-archive/help/Catalogue/entries/fancyhdr.html]fancyhdr[/url]package documentation for more details about handling headers and footers in LaTeX.
For your question:
Put this code in the preamble[code] \usepackage{fancyhdr}
\pagestyle{fancy}
\rhead{\rightmark}
\lhead{\thepage}
\renewcommand{\headrulewidth}{0.5pt}
[/code]
\headrulewidth to handle the thikness of the line.
\rightmark to display the section number and name.

Thanks!

I can't seem to download the fancyhdr package though.  Does anyone know another download site?  I tried to google it but I only find links back to ctan.

Um, what's wrong with downloading from CTAN?
First though, if you are using MiKTeX run the MiKTeX Package Manager, right-click on fancydhr, choose Install and it will all be done for you.

Even better, the latest version of MiKTeX (version 2.5) will even offer to install fancyhdr once you compile a tex file with \usepackage{fancyhdr} in it.

Failing that fancyhdr.sty is downloadable from CTAN [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/fancyhdr/]here[/url]

I tried to compile a TeX file with the fancyhdr package, and got the "failed fancyhdr.stv" error.  When I go to CTAN, I get a "failed to establish connection" error when trying to download the fancyhdr.stv file.  I tried other mirrors, too.

The CTAN link work. For example: Right click [url=ftp://ftp.sun.ac.za/CTAN/macros/latex/contrib/fancyhdr/fancyhdr.sty]here[/url] and "saves as..."

That link doesn't work either  :(  :(  :( I tried turning my firewall off and disabling norton internet security, and it still doesn't work.  A friend of mine sent me the fancyhdr package over AIM, but when I go to the Miktex package manager to install it from a directory (the package is saved on my desktop) I get an error saying that the directory is not a local repository.

Just let the package manager install from its own repository [i]not[/i] from the desktop where you put it. When you installed MiKTeX you put packages onto your computer which MiKTeX doesn't install but will do so on command. So all you need do is [b]right-click on fancyhdr in the list in the package manager and choose install[/b] and MiKTeX does the rest - no questions asked.
(It may use a cab file on your computer to install from or may download it from MiKTeX's servers).

If you really can't get that to work you can put fancyhdr.sty in the same directory as your tex file and that will work.


analytic geometry
probability
modular arithmetic

$S$ is the set of all points with coordinates $(x,y)$ such that $x$ and $y$ are positive integers such that $x<n$ and $y<n$ where $n$ is a positive integer greater than 2. Two distinct points from $S$ are chosen at random. What is the probability that the midpoint of the segment of the two points is also an element of $S$? Express your answer in terms of $n$.

[hide="Idea"] Find the probability, with the 2 points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$, that $x_{1}\equiv x_{2}\pmod 2$ and $y_{1}\equiv y_{2}\pmod 2$. [/hide]

I think you also have to consider two different cases. When $n$ is even and when it's odd.

Umm...  :maybe: If this is too hard, can someone move it to the appropriate forum? :ninja:


How to change the fontsize in latex(WinEdt).

I have tried \documentclass[a4paper, 12pt]{report}
But it is not working.
Kindly help

What's not working? Does \documentclass[a4paper, 12pt]{report} give a larger font than \documentclass[a4paper, 10pt]{report}?


AMC
AIME
trigonometry
number theory
relatively prime
geometry
angle bisector

14.  In triangle $ABC$, $AB=13$, $BC=15$, and $CA=14$. Point $D$ is on $\overline{BC}$ with $CD=6$. Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD$. Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q$.

[hide="tiger"]Let $DE=x$, so $BE=9-x$. Let $AX$ be a triangle angle bisector of $\angle CAB$. Apply Angle bisector theorem we obtain that $CX=\frac{70}{9}$, hence
$DX=\frac{16}{9}$, and
$EX = x-\frac{16}{9}$.
Note that $AX$ also bisects $\angle DAE$, so

$\frac{AD}{AE}=\frac{DX}{EX}=\frac{16}{9x-16}$.

$[CAD]=\frac{AC(AD)\sin CAD}{2}$

$[BAE]=\frac{AB(AE)\sin BAE}{2}$

$\sin CAD = \sin BAE$, so

$\frac{[CAD]}{[BAE]} = \frac{AC}{AB}(\frac{AD}{AE})$

$=\frac{14}{13}(\frac{16}{9x-16})$.

Now $\triangle CAD$ and $\triangle BAE$ share the same altitude, hence

$\frac{[CAD]}{[BAE]}=\frac{CD}{EB} = \frac{6}{9-x}$

So we have

$\frac{14}{13}(\frac{16}{9x-16}) = \frac{6}{9-x}$

Now we solve for $x$ we get

$x=\frac{8(204)}{463}$,
BE = $9-x = \frac{9(463)-8(204)}{463}$

hence $q=463$.[/hide]


Let x, y, z be three positive real numbers such that

$\frac{x^{2}}{y}+\frac{y^{2}}{z}+\frac{z^{2}}{x}=\frac{x^{2}}{z}+\frac{y^{2}}{x}+\frac{z^{2}}{y}$.

Prove that at least two of the numbers x, y, z are equal.

  Darij

Easy! The expression is equivalent to $(x - y)(x - z)(y - z)(x + y + z) = 0$.

This was exactly my solution, and I don't think that there exists any substantially different way. A problem to be solved by exactly the same method (but one has to make one more step at the end):

Let x, y, z be [i]arbitrary real[/i] numbers (not necessarily positive anymore) such that

$\frac{x^{3}}{y}+\frac{y^{3}}{z}+\frac{z^{3}}{x}=\frac{x^{3}}{z}+\frac{y^{3}}{x}+\frac{z^{3}}{y}$.

Prove that at least two of the numbers x, y, z are equal.

  Darij

The given expression factors to

$(x - y)(y - z)(z - x)(x^2 + y^2 + z^2 + xy + yz + zx) = 0$.

So either two of the numbers are equal, or

$(x + y)^2 + (y + z)^2 + (z + x)^2 = 0$.

Then $x = -y,~y = -z,~z = -x \Rightarrow x = y = z = 0$.

QED.

I'm curious... how'd you know that would factor?

I wonder if the problem can be generalized for x^n/y+y^n/z+z^n/x=x^n/z+y^n/x+z^n/y with positive x,y,z.

It's clear that (x-y)(y-z)(z-x) stays a factor because the equality holds for x=y and so on. It is also clear that a generalization with real x,y,z is not true because the other factor in the factorization can be a polynomial of odd degree. But the statement above seems quite likely to me although I haven't prove it yet.

Hi Misha!

[quote="Misha123"]I wonder if the problem can be generalized for x^n/y+y^n/z+z^n/x=x^n/z+y^n/x+z^n/y with positive x,y,z.[/quote]

Indeed it can (see below). The key is that for every integer $m\geq 0$, we have the identity

$x^{m+2}\left( y-z\right) +y^{m+2}\left( z-x\right) +z^{m+2}\left( x-y\right)$
$=\left( x^{2}\left( y-z\right) +y^{2}\left( z-x\right) +z^{2}\left( x-y\right) \right) \cdot C_{m}\left( x;\;y;\;z\right) $,

where $C_{m}\left( x;\;y;\;z\right) $ is the "complete homogeneous polynomial" of m-th degree of the three variables x, y, z, defined as:

$C_{m}\left( x;\;y;\;z\right) =\sum x^{A}y^{B}z^{C}$ over all nonnegative integers A, B, C satisfying A + B + C = m.

Or, asymmetrically but more explicitely, we can define $C_{m}\left( x;\;y;\;z\right) $ through

$C_{m}\left( x;\;y;\;z\right) =\sum_{A=0}^{m}\left( \sum_{B=0}^{m-A}x^{A}y^{B}z^{m-A-B}\right) $.

We have

$C_{0}\left( x;\;y;\;z\right) =1$;
$C_{1}\left( x;\;y;\;z\right) =x+y+z$;
$C_{2}\left( x;\;y;\;z\right) =x^{2}+y^{2}+z^{2}+xy+xz+yz$;
$C_{3}\left( x;\;y;\;z\right) =x^{3}+y^{3}+z^{3}+x^{2}y+x^{2}z+y^{2}x+y^{2}z+z^{2}x+z^{2}y+xyz$;
....

The nice thing is that the formula

$x^{m+2}\left( y-z\right) +y^{m+2}\left( z-x\right) +z^{m+2}\left( x-y\right)$
$=\left( x^{2}\left( y-z\right) +y^{2}\left( z-x\right) +z^{2}\left( x-y\right) \right) \cdot C_{m}\left( x;\;y;\;z\right) $,

gives an explicite representation of the polynomial $C_{m}\left( x;\;y;\;z\right) $ in all cases except of the case when two of the numbers x, y, z are equal. [I was so glad when I found this formula, and now I see that it's not new... Maple knows it!]

Another interesting question is how many coefficients the polynomial $C_{m}\left( x;\;y;\;z\right) $ has. Since all of its coefficients are 1, this question comes down to finding the sum of its coefficients, i. e. to finding the value $C_{m}\left( 1;\;1;\;1\right) $. In fact, it turns out that

$C_{m}\left( 1;\;1;\;1\right) =\frac{\left( m+1\right) \left( m+2\right) }{2}$

(easy to prove).

Incidentally, the identity

$x^{m+2}\left( y-z\right) +y^{m+2}\left( z-x\right) +z^{m+2}\left( x-y\right)$
$=\left( x^{2}\left( y-z\right) +y^{2}\left( z-x\right) +z^{2}\left( x-y\right) \right) \cdot C_{m}\left( x;\;y;\;z\right) $

can be proven just by expanding the right side, but I find this proof somehow ugly. Does anybody know a better proof? Peter?!

And now to your problem: Given positive numbers x, y, z such that $\frac{x^n}{y}+\frac{y^n}{z}+\frac{z^n}{x}=\frac{x^n}{z}+\frac{y^n}{x}+\frac{z^n}{y}$ for some natural $n\geq 1$. Prove that at least two of these numbers x, y, z are equal.

In fact, multiplying the equation $\frac{x^n}{y}+\frac{y^n}{z}+\frac{z^n}{x}=\frac{x^n}{z}+\frac{y^n}{x}+\frac{z^n}{y}$ with xyz, we get $x^{n+1}z+y^{n+1}x+z^{n+1}y=x^{n+1}y+y^{n+1}z+z^{n+1}x$. And this is equivalent to

$x^{n+1}\left( y-z\right) +y^{n+1}\left( z-x\right) +z^{n+1}\left( x-y\right) =0$,

i. e. to

$\left( x^{2}\left( y-z\right) +y^{2}\left( z-x\right) +z^{2}\left( x-y\right) \right) \cdot C_{n-1}\left( x;\;y;\;z\right) =0$.

But since the numbers x, y, z are positive, the polynomial $C_{n-1}\left( x;\;y;\;z\right) =\sum x^{A}y^{B}z^{C}$ is positive, too, and thus it can't be = 0. Hence, we must have

$x^{2}\left( y-z\right) +y^{2}\left( z-x\right) +z^{2}\left( x-y\right) =0$.

But $x^{2}\left( y-z\right) +y^{2}\left( z-x\right) +z^{2}\left( x-y\right) =\left( y-z\right) \left( x-z\right) \left( x-y\right) $, and hence we have (y - z) (x - z) (x - y) = 0, what implies that two of the numbers x, y, z are equal. Proof complete.

What an interesting problem!

  Darij

[quote="TripleM"]Well, other than the fact that you could expect it to factor being a problem like that..
If you want to prove two of x,y,z are equal, then you know one of (x-y), (y-z), or (z-x) must be 0. Its quite likely that this means (x-y)(y-z)(z-x) must be a factor. Not necessarily, but its pretty likely.[/quote]

It's necessary if the polynomial you are factoring is cyclic :)

Yeah, generally plugging in x=y, y=z, or z=x works nicely in factoring multi-variable polynomials, especially cyclic ones.


this was posted because there was no nick on tteh last poll


Are the teams mostly comprehensive? Or are a large amount of people going and entering in the Geometry and Alg II division

Unfortunatly, Wheeler's Robotics team is going to Auburn that weekend. So, I'll be in the same state, but not close to Vestavia.

Most teams from GA are probably comprehensive, although maybe a few of the JV folks from Lassiter will want to come; who knows?

OMG! i wanna go but it seems too far though my mother wouldn't think it was safe....would anyone be interested in carpooling with someone who is in the Forsyth/Fulton area?Or perhaps I could get dropped off in Chamblee...

[quote="Boy Soprano II"]It's on December 9.  [url=http://math.gosuapm.com/tourn]Website (not updated very recently).[/url]  You don't have to be on any team except your school's team, and even then, you can pretty much go as an individual, I think, but I don't really know.  It's in Alabama, by the way.  Anyhow, I've heard it's really well-attended, but I can't say for sure, never having been there before.[/quote]

Quite well-attended :); I'd say more than Cobb, but I'm not sure.

This year is the biggest year we've ever had, by some margin too.

Hmm, well, I've never been to Cobb either... :rotfl:   I should stop having scheduling conflicts.

Congrats. on the attendance.  I think we're doing well in that realm, also.  Saturday should be fun.

what about sleep.  i think it might take precedents.  that or english

can't wait till vhhs. I want to own some vestavia booties.

Now, I await the Vestavia Attack on forums. Elemm nop and jhcreinhard and Cicatriz shall destroy me!

AHHHHHH!  :lol:

its not like you'll beat us, so we dont care.

i wished that could sound as arrogant as i meant it to, but we're not even in the same division

yea a state of about 8 million people versus a high school with roughly 1700 students.  the sad thing is, we would crush you even if you brought the arml team to vestavia

[quote="Karth"]DUDE

NOBODY BEATS GEORGIA A

WE PWNED VESTAVIA LIKE ELEPHANTS AT DUKE

WOOT

The only reason georgia a lost to east chapel is because they have jeremy, arnav, etc. grr[/quote]

LIKE ELEPHANTS?

That is one of the most random expressions that I have ever heard.

BTW: IS CAPS LOCK STUCK ON YOUR KEYBOARD, OR DO YOU JUST LIKE YELLING?

Just wondering.

Good point.

[quote="Boy Soprano II"]Who from Georgia is coming?  I know that Lassiter is sending a team, and that I am on it. :D   (But come to Lassiter first!)[/quote]

<tone="sarcastic">
Sorry, I was planning to go to the Vestavia tournament before I go to the Lassiter tournament even though the Lassiter tournament occurs before the Vestavia tournament.
</tone>

[quote="Xantos C. Guin"]
<tone="sarcastic">
Sorry, I was planning to go to the Vestavia tournament before I go to the Lassiter tournament even though the Lassiter tournament occurs before the Vestavia tournament.
</tone>[/quote]

The proper way to do it is like this: [quote="Xantos C. Guin"]
<sarcasm>
Sorry, I was planning to go to the Vestavia tournament before I go to the Lassiter tournament even though the Lassiter tournament occurs before the Vestavia tournament.
</sarcasm>[/quote]
or even:
[quote="Xantos C. Guin"]
Sorry, I was planning to go to the Vestavia tournament before I go to the Lassiter tournament even though the Lassiter tournament occurs before the Vestavia tournament.
</sarcasm>[/quote]

And in fact I have gone to Vestavia before Lassiter even though the Lassiter meet occurs before Vestavia :P.

[quote="Elemennop"][quote="Karth"]DUDE

NOBODY BEATS GEORGIA A

WE PWNED VESTAVIA LIKE ELEPHANTS AT DUKE

WOOT

The only reason georgia a lost to east chapel is because they have jeremy, arnav, etc. grr[/quote]

glad it takes your whole state to beat one school...[/quote]

Yea that's true... you do have a point.

And @ santosh:

now that you mention it my brother typed the above quote, not me. He's Arvind_sn... i usually don't type in all caps unless it's necessary.

@ jhcreinhardt:

And I highly doubt that if we brought the ARML team to Vestavia, you could even touch us... after all, you can't beat what you can't touch  :P

But your school is pretty awesome, I have to admit.

maybe not at arml, but at vestavia maybe

[quote="Xantos C. Guin"][quote="solafidefarms"]
The proper way to do it is like this: [quote="Xantos C. Guin"]
<sarcasm>
Sorry, I was planning to go to the Vestavia tournament before I go to the Lassiter tournament even though the Lassiter tournament occurs before the Vestavia tournament.
</sarcasm>[/quote][/quote]

Oh. It was one of those 2. I'm bad at HTML.[/quote]

It's XML, actually, my friend.  :wink:

Bad news—we apparently haven't been able to get a sufficient team together, so it looks like I won't be there after all. :(  :(

(And I knew they were on different dates :  I just wanted to make sure you realized what was important/couldn't resist an opportunity to advertise.)

I [i]hope[/i] it is the teachers, since the students participate in the competition :P

[quote="Arvind_sn"]Who makes the vestavia test? The students or the teachers? I know Lassiter was made by some students... :|[/quote]

Tis made by student alumni

haha, I make up my english rules... after someone made up every language.

 :lol:

jk

[quote="ragnarok23"][quote="Go Around the Tree"]and geometry test tis very formula heavy  :lol:[/quote]
'tis=it is so you could only say "is" there not tis[/quote]
But "and the geometry test: 'tis inordinately formula heavy" would be better. I'd have gone with "The geometry test? 'Tis incredibly formula-heavy."

Well, in French, a mere comma would be acceptable—or perhaps even no comma.  But I think there should be an apostrophe.

there can be a new rule for anything.

look at our "war" and foreign policies  :lol: