geometry
geometric transformation
modular arithmetic
graph theory
combinatorics unsolved
combinatorics

Hi!, Ive been trying to solve this problem since a long time and I cant. It is  a special case of a unsolved problem; of course this  case has a known solution, and I just asked if somebody could solve it. Any help is thanked  :)

A complete graph G is given. We need to color both, its vertices and edges in such a way that:
 1) Two edges coinciding in a vertex cannot be the same color
 2) Two vertices connected by a edge cannot be the same color
 3) One edge coinciding at a vertex  cannot have the same color as the vertex

Its obvios that at least we need grade(G) +1 colors, but in some graphs we need grade(G)+2. Find a necesary and sufficient condition for knowing that it just needs grade(G)+1  colors.

Thx...

Leo, Its nice to have you here, Saludes a los chamos!

:lol: pascual, Valentin has set rules for these forums. English please.
[quote]Saludes a los chamos[/quote]
Approximate translation of quote: Welcome to the 'chamos', where 'chamos' means 'kiddos'?!?!? I'm really not that sure how to translate that, it's quite a Hispanic slang.

Leo: it's not called 'grade(G)' in English texts although in Spanish it's 'grado'. It's called the [i]degree[/i] of graph G.

Sorry, it is my fault.......Y would not do it again :D

By the way "chamos" means people from Venezuela.

Thanx for telling me that rule![/quote]

Let $n=|G|$ be odd number. Consider folowing coloring of $G$ into $n$ colors. First of all, color all vertices into colors 1, 2, ..., $n$.
Vertices $i$ and $j$ are connected with edge of color
1) $i+j$ iff $i+j\leq n$,
2) $i+j-n$ iff $i+j>n$, $i,j\neq n$,
3) $2i\pmod{n}$ iff $j=n$.

I think, for even $n$ we can't find appropriate coloring. Maybe Pierre knows about that.

Oh my God! I am really blind! :blush:
Suppose $n$ is even. It is clear that if we color edges and vertices into $n$ colors then amounts of edges of different colors are equal! But there are $\frac{n(n+1)}{2}$ edges and this number is not divisible by $n$ for even $n$!

So problem is solved. Pierre can move it to Solved Section ;)

Cool.
I was keeping that problem to think about it later, but now I'm free of it. :D
I should always let Myth answer before thinking at some problem (if he ever let me time for it). :P

Pierre.


What is your Rank and how can I get up my rank?

Thanks

Rank has to do with post count, see the topic called "stars" in Fun and Games since this is talked about a lot in that thread.

Where can i find that?  Sorry but I am new here :(

[quote="Hexahedron"]It is in the forum list, just scroll down untill you see it.[/quote]

Okay thank you :D


Prove that there doesn't exists a smooth($ C^{\infty}$) surjective map $ f : \mathbb{R}\to \mathbb{R}^{k}$.

Whereas,it is well known that there does exists such a continuous map, namely Peano curve.

Where $ k\ge 2$, of course.
The restriction of $ f$ to $ [-N,N]$ is Lipschitz, which implies that $ f([-N,N])$ has $ k$-dimensional measure zero. [Lipschitz maps can only increase measure up to a multiplicative constant.] Taking the union over $ N$, we conclude that $ f(\mathbb R)$ also has $ k$-dimensional measure zero.


function quadratics search conics parabola

I've been given some random data.
X	Y

0	440
6	450
12	480
18	570
24	680
30	800
36	980
42	1090
48	1520
54	1920
60	1670
66	1440
72	1380
78	1300
84	1150
90	1060
96	970
102	900
108	850
114	800
120	780
126	740
132	710
138	680
144	660

My task : To create a function which can model this data.  Now my problem is, this function does not follow any one pattern.  What I did instead, was break it up into 5 sections, and make a pseudo 'piece wise function.'

How would i go about creating one continuous function?

Here is a picture of the data :
http://img118.imageshack.us/my.php?image=68941387hc2.jpg

Thanks for the help.[/img]

Offhand, it looks like a parabola.  I'd go with a [url=http://www.google.com/search?hl=en&q=quadratic+regression&btnG=Google+Search]quadratic regression[/url].

Wouldhave gone with a bell curve myself  :lol:

how does it look anything like a quadratic...

Sorry, I only looked at the data and not the graph; it's rather too pointy to be a parabola  :oops:

Your use of the word "continuous" is perhaps a bad choice.  Any continuous function which approximates this set of data is going to be rounded at the maximum, so it won't look as sharp as it does.  I'd separate it into two functions at the maximum.  There isn't a particularly good choice of continuous function that would mimic the spike (unless the spike is an approximation of a singularity!).


Determine all pairs of integers $(x,y)$ which satisfy the equation

\[ 6x^2-3xy-13x+5y = -11 \]

[hide="My solution"]

This is a very nice problem. My first idea was to see if this problem had no solution and prove it with mod but that didn't work out very nicely. So, I solved for $y$ in terms of $x$.

After moving things around, I got $y = \frac{6x^2-13x+11}{3x-5}$. Using Long Division, this becomes $2x-1 + \frac{6}{3x-5}$. For $y$ to be an integer, we need to have $3x-5$ as an integer divisior of 6. So, $3x-5 = \pm 1, \pm 2,\pm 3,\pm 6$. Solving for $x$, we only get 2 integer $x$ values and plugging them gives $y$ values as well. Those are $(2,9), (1,-2) \blacksquare$. [/hide]


linear algebra
matrix
function
integration
calculus
calculus computations

Solve the differential equations

$ \frac {dy(t)}{dt} \equal{} x(t)$
$ \frac {dx(t)}{dt} \equal{} \minus{} y(t)$

Please show work.

(By inspection, $ y(t) \equal{} sin(t)$ and $ x(t) \equal{} cos(t)$ works, but how do you solve it without guess and check?  And what is the general solution for the parametric differential equations (simultaneous)?)

EDIT: Yes, I have heard of using matrix to solve this type of differential equations.  How do I reduce it to a single ODE?  Thanks!

Generally speaking, what you have here can be called a system of differential equations. There are more than few ways to do it. One is reducing it to just one differential equation, with degree equal to the number of equations in your system (in your case, reducing your system to $ Ax''\plus{}Bx'\plus{}Cx\plus{}D\equal{}Ef(t)$ or something like that, solving that equation and finding $ y$ by simply plugging in the $ x$. That's generally the best way if you're dealing with a system of 2 equations (3 at most). If more equations are involved, then use matrices (brief descripiton given on wikipedia: [url]http://en.wikipedia.org/wiki/Ordinary_differential_equation[/url]).

Differentiate second equation,
\[ \frac {d^2x(t)}{dx} \equal{} \minus{} \frac {dy(t)}{dt} \equal{} \minus{} x(t),
\]
now you have one single differential equation in one single variable, with general solution $ x(t) \equal{} Ae^{jt} \plus{} Be^{ \minus{} jt}$, substitute in this equation and get $ y(t) \equal{} \minus{} \frac {dx(t)}{dt} \equal{} \minus{} jAe^{jt} \plus{} jBe^{ \minus{} jt}$, substitute both in the first equation to show that it works, done!  Note that in general $ A,B$ are complex constants, and if you're interested only in real solutions, $ A^* \equal{} B$, where the asterisk denotes complex conjugation.  Then, calling $ A \equal{} re^{j\theta}$, ($ r\geq0$ and $ \theta$ real constants) you find the most general REAL solution
\[ x(t) \equal{} 2r\cos(t \plus{} \theta),
\]

\[ y(t) \equal{} 2r\sin(t \plus{} \theta),
\]
of which your solution found by inspection is a particular case with $ r \equal{} \frac {1}{2}$ and $ \theta \equal{} 0$.

One of the usual ways to deal with systems of differential equations is to try to reduce them to one differential equation in one of the functions, solve it, substitute in the rest to find the remaining functions, as outlined in this case.

Other methods are possible according to the particular form of the system of differential equations, this being "a world" in itself.  One typical way is to reduce the system to a system of regular equations by use of Laplace or Fourier transformation, since the transform of $ \frac {dx(t)}{dt} \equal{} j\omega X(\omega)$ in Fourier, $ \frac {dx(t)}{dt} \equal{} sX(s)$ in Laplace, where
\[ X(s) \equal{} \int_{0}^{\infty}x(t)e^{ \minus{} st}dt
\]
for functions such that $ x(t) \equal{} 0$ for $ t < 0$, or
\[ X(\omega) \equal{} \frac {1}{2\pi}\int_{ \minus{} \infty}^{ \plus{} \infty}x(t)e^{ \minus{} j\omega t}dt,
\]
valid at least when $ x(t)$ is an absolutely integrable function with a finite number of maxima, minima, and discontinuities, all discontinuities being finite (Dirichlet conditions), althoug it may work for other types of solutions as well.  Finally, periodic solutions with fundamental period $ T_0 \equal{} \frac {2\pi}{\omega_0}$ may be searched for by means of Fourier series, defined such that
\[ x(t) \equal{} \sum_{n \equal{} \minus{} \infty}^{\infty}c_ne^{jn\omega_0t}.
\]
You may write $ x(t)$ in this form and solve for the $ c_n$.  This "family" of methods is particularly useful when the system is not homogeneous, since you may apply the corresponding transformation both to the target functions and to the seed functions ("independent" terms).

You may also write always the functions as McLaurin series with unknown coefficients and solve for them after substitution of the general form of the functions in the differential equations, but this is in general a more complex method except in particular cases, since it usually involves the solution of an infinite system of coupled recursive equations on the coefficients of the different functions.


linear algebra
matrix
function
real analysis
linear algebra unsolved

[quote="shack"]Hello!

I have a minor problem, that wont get solved...

Its been posted in an old competition at our university, but I didnt manage to solve it in time. Maybe someone in here could help me?

[url=http://img504.imageshack.us/my.php?image=optimeringts0.jpg]Assaintment[/url]

Where D is a symmetric n x n matrix, and x is an unknown n x 1 vector.

I can only figure out how to test for a maximum later on, but is there something im missing with the charataristic roots?

Regards[/quote]This is a standard result. You are maximizing the smooth function $ f(x) \equal{} x^TDx$  over the compact (closed and bounded) set $ \{x | \|x\| \equal{}1\}$, the unit sphere. Hence the maximum is attained. Write down the Lagrange multiplier equations. The maximizing $ x$ is an eigenvector of $ D$.


geometry probability trigonometry \/closed

I'm new to this website, and I was wondering if I could still take the Introduction to Geometry class that started already. If I can't when do I enroll for classes I want to take?

Enrollment for the Intro Geometry class has closed.  (This is already Week 4 of the class.)  If you're interested, I would recommend our [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=9][i]Introduction to Geometry[/i][/url] textbook.

Currently only our [url=http://www.artofproblemsolving.com/Classes/AoPS_C_PSeries.php]Problem Series[/url] classes are open for enrollment.  We should have more subject classes in early 2007.  Please join our [url=http://www.artofproblemsolving.com/StayInformed/AoPS_Stay_Informed.php]mailing list[/url] to be informed when new classes are announced.

[quote="rrusczyk"][quote="funcia"]is there to be one in the spring?  :)[/quote]

We have Intro Counting & Probability in the spring; the next time Geometry will be offered is fall, 2007.[/quote]
Approximately when's the next time intermediate trig/complex numbers will be offered?
Thanks,
bpms

Here is one general question I have. Which would you suggest more, taking a problem class online, or getting the book?

[quote="13375P34K43V312"]I'd prefer the book, it's cheaper and you can go at your own pace.[/quote]
Yes, but I was wondering about quality of the class compared to book, how good are the problem sets, which way do people learn more from, etc.

[quote="rrusczyk"]This really depends on how you learn.  If you learn fine from a good book and are self-motivated enough to work through it on your own, then that's the way to go (and try to negotiate with your in-school teachers to let you have time during the school day to do this, rather than having to listen in classes you already know).[/quote]

including US History teacher?? :rotfl:
generally I just finish my homework in class for the first 30-40 minutes (if not faster), then work on my own math stuff (currently 103 trig problems).

And yes I do kind of prefer reading book over classes, as you can buy multiple books on different subject with the same prize of one online class.  But for something you just couldn't get no matter how much you read (Umm, combinatorics), I'll take an online class for it since you can be annoying and ask people like crazy.  For now I'm doing both since I'm a senior and don't care about grades (in fact I have 2 free periods, or math periods :D )


In a given arithmetic sequence the first term is $2$, the last term is $29$, and the sum of all the terms is $155$. The common difference is:

$\text{(A)} \ 3 \qquad \text{(B)} \ 2 \qquad \text{(C)} \ \frac{27}{19} \qquad \text{(D)} \ \frac{13}9 \qquad \text{(E)} \ \frac{23}{38}$

Here's my Kalva version of a solution. :P (very brief)

[hide]Use arithmetic series sum formula (both versions). $\frac{n}{2}(29+2)=155\implies n=10$. Now, $\frac{10}{2}(2+29)=\frac{10}{2}(4+9d)\implies d=3$[/hide]

We know the Sum Formula is $S=\frac{n(a+l)}{2}$.

The we just plug in values:

$a=2$
$l=29$
$S=155$

Let's start with $155=\frac{n(2+29)}{2}$, so simplifying, we get $n=10$.

Now let us use Soni's Theorem, namely:

$a+dn-d=l$, where $a$ is the first term, $d$ is the common difference, $n$ is the number of terms, and $l$ is the last term.

Now we plug in values once again:

$2+10d-d=29$
Simplifying gives $d=\boxed{3}$.

Don't know the sum formula, usually just use common sense.

[hide]
$2 + (n-1)d = 29$
$d = \frac{27}{n - 1}$
$\text{Solving gives: } n = 10, d = 3$
[/hide]

$\frac{n}{2}(2+29)=155$
$\implies n=10$
It's common sense from there.
$d = 3$
$\boxed{A}$


In parallelogram $ PQRS$, the measure of angle $ P$ is five times the measure of angle $ Q$. What is the measure of angle $ R$, in degrees?

Let the measure of angle $ Q$ be $ x$.  Adjacent angles are supplementary in a parallelogram, so $ 5x\plus{}x\equal{}180$.  Therefore, $ x\equal{}30$.  That means the measure of angle $ P$ is $ 5\times 30\equal{}150$.  In a parallelogram, opposite angles are congruent, so the measure of angle $ R$ is $ \boxed{150}$

Due to the properties of a parallelogram, $ \angle P\cong\angle R$, so if we find $ m\angle P$, that same numerical value will also be $ m\angle R$. Another property of parallelograms is that adjacent angles are supplementary, so $ m\angle P\plus{}m\angle Q\equal{}180$. It is given in the problem that $ m\angle P\equal{}5m\angle Q$. Substituting the second equation into the first, we get $ 5m\angle Q\plus{}m\angle Q\equal{}180$, so $ 6m\angle Q\equal{}180$, and $ m\angle Q\equal{}30$, making $ m\angle P\equal{}5m\angle Q\equal{}5(30)\equal{}\fbox{150}$, and since this is $ m\angle P$, this is also $ m\angle R$.


Let $ (O)$ be the circumscribed circle of the triangle $ ABC$ , $ D,E$ are the midpoints of the sides $ AB,AC$ , draw the circles  $ \gamma_1$ and $ \gamma_2$ internally tangent to $ (O)$ and tangent to $ AB,AC$ at $ D,E$ . Show that one of the external common tangents to $ \gamma_1$ and $ \gamma_2$  is parallel to $ BC$

Sincerely
T.H


calculus
integration
inequalities
number theory unsolved
number theory

[quote="Myth"]Find the number of the ordered integral number groups $(x, y, z, u)$ such that $1\leq x,y,z,u\leq 10$ and
\[\frac{x-y}{x+y}+\frac{y-z}{y+z}+\frac{z-u}{z+u}+\frac{u-x}{u+x}>0.\][/quote]

This equation is very simple, the solutions satisfy:

1.  $x>z$  and  $(y-u)(zx-yu)>0$, or
2.  $x<z$  and  $(y-u)(zx-yu)<0$.

then is too simple to get this number which is a combinatorial problem.

Best regards.
RX the Crazy Math.

[quote]
RobertuX wrote
This equation is very simple, the solutions satisfy:

1.  and , or
2.  and .

then is too simple to get this number which is a combinatorial problem.
[/quote]

Could you explain how to calculate result? I had no idea to find solutions of above two inequlities simplely :blush:

He is right. I have just checked it with MathCad. :)

[quote="zhaoli"]
Could you explain how to calculate result? I had no idea to find solutions of above two inequlities simplely :blush:[/quote]
You only need to add the firt and second factors, and third and fourth and you get the cases easily.

Best Regards.
RX the Crazy Math


function
inequalities
inequalities proposed

Let $ a,\ b$ be real numbers with $ 0<a<b$ and $ x,\ y,\ z$ satisfy : $ a\leq x\leq b,\ a\leq y\leq b,\ a\leq z\leq b$.
If $ x\plus{}y\plus{}z\equal{}a\plus{}2b$, then prove that $ xyz\geq ab^2$.

[quote="kunny"]Let $ a,\ b$ be real numbers with $ 0 < a < b$ and $ x,\ y,\ z$ satisfy : $ a\leq x\leq b,\ a\leq y\leq b,\ a\leq z\leq b$.
If $ x \plus{} y \plus{} z \equal{} a \plus{} 2b$, then prove that $ xyz\geq ab^2$.[/quote]
See my solution at here: http://batdangthuc.net/forum/showthread.php?p=2351&posted=1#post2351

[quote="kunny"]Let $ a,\ b$ be real numbers with $ 0 < a < b$ and $ x,\ y,\ z$ satisfy : $ a\leq x\leq b,\ a\leq y\leq b,\ a\leq z\leq b$.
If $ x \plus{} y \plus{} z \equal{} a \plus{} 2b$, then prove that $ xyz\geq ab^2$.[/quote]
$ f(x,y,z)\equal{}(a\plus{}2b)^3xyz\minus{}ab^2(x\plus{}y\plus{}z)^3$ is a concave function of $ x,$ of $ y$ and of $ z.$ :wink:

Here is an elementary solution
Consider the following expansion

$ (x\minus{}a)(y\minus{}b)(z\minus{}b) \ge 0$

$ \Leftrightarrow xyz \plus{}xb^2\plus{}ab(y\plus{}z) \ge ab^2\plus{} ayz \plus{} b(xy\plus{}xz)$

Similarly, we obtain the same evaluations

$ xyz \plus{}yb^2\plus{}ab(x\plus{}z) \ge ab^2\plus{} axz \plus{} b(xy\plus{}yz)$

$ xyz \plus{}zb^2\plus{}ab(y\plus{}x) \ge ab^2\plus{} ayx \plus{} b(zy\plus{}xz)$

Summing up these inequalities, we get

$ 3xyz\plus{} b^2(x\plus{}y\plus{}z) \plus{}2ab(x\plus{}y\plus{}z) \ge 3ab^2 \plus{}(2b\plus{}a)(xy\plus{}yz\plus{}zx)$ (*)

Futhermore, we have

$ (x\minus{}a)(x\minus{}b) \le 0$

$ \Leftrightarrow x^2 \plus{}ab \le x(a\plus{}b)$

It follows that

$ x^2 \plus{}y^2\plus{}z^2 \le a^2\plus{}2b^2$(**)

Meanwhile, we got $ (x\plus{}y\plus{}z)^2\equal{} x^2\plus{}y^2\plus{}z^2 \plus{}2(xy\plus{}yz\plus{}zx) \equal{} a^2\plus{}4b^2\plus{} 4ab$(***)

From (**) and (***), it yields $ xy\plus{}yz\plus{}zx \ge b^2 \plus{} 2ab$
Hence, from (*) it is derived

$ 3xyz \ge 3ab^2 \plus{}(2b\plus{}a)(xy\plus{}yz\plus{}zx \minus{} b^2\minus{}2ab) \ge 3ab^2$

Thus, $ xyz \ge ab^2$ as desired

We complete our proof!


calculus
integration
Putnam
college contests

Problem:  Evaluate \[\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^{2}n+n^{2}m+2mn}\] This is a follow up to [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=563862#563862[/url] since this solution is also not in the book or on Kalva, AND it involves an integral from 0 to 1.

Solution:

\begin{eqnarray*}\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^{2}n+n^{2}m+2mn}&=& \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{mn}\cdot\frac{1}{m+n+2}\\ \ &=& \int_{0}^{1}x\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{x^{m+n}}{mn}dx\\ \ &=&\int_{0}^{1}x\left(\sum_{m=1}^{\infty}\frac{x^{m}}{m}\right)^{2}dx\\ \ &=&\int_{0}^{1}x\bigl(\ln(1-x)\bigr)^{2}dx\\ \ &=&\int_{0}^{1}(1-x)\bigl(\ln(x)\bigr)^{2}dx\\ \ &=&\frac{7}{4}\end{eqnarray*}

Could you please explain your second step (the transformation from summation to integral)? I admit I am not good at these things. Thanks in advance.

$\int_{0}^{1}x^{m+n+1}dx = \frac{1}{m+n+2}$.

I was confused a bit at first until I noticed the $x$ that Negative_3 moved out of the sum.

It's a very nice solution, I think.

The official solution is to rewrite \[\frac{1}{m^{2}n+n^{2}m+2mn}=\frac{1}{m(m+2)}\left(\frac{1}{n}-\frac{1}{m+n+2}\right)=\frac{1}{2}\left(\frac{1}{m}-\frac{1}{m+2}\right)\left(\frac{1}{n}-\frac{1}{m+n+2}\right)\] Then use 3([b]THREE[/b]) telescoping areguements to get the answer.  Not very pretty.


modular arithmetic
number theory
relatively prime

(a) Consider the positive integers $ p,q,r,a$ such that where $ r$ is a prime and $ gcd(p,q)\equal{}1$.Prove that $ p$ or $ q$ is a perfect square of a positive integer.

(b) Find all primes $ p$ for which the number $ p(2^{p\plus{}1}\minus{}1)$ is a perfect square of a positive integer.

a) What, we have 1 prime, 2 relatively prime integers, so what?
b) If it's a square, then $ 2^{p\plus{}1}\equiv1\pmod{p}$, but by FLT $ 2^p\equiv2\pmod{p}$, so $ 4\equiv1\pmod{p}$, so $ p\equal{}3$. But even then the number isn't  a square. So, no solutions.


is there any method to find a big number(of say 10-12 digits) is prime or composite?

to check if $ n$ n is prime or not u have to check every prime number up to the integer part of $ \sqrt{n}$ I don't think a faster way exists, certainly not a simple one, so basically without a fast computer and without knowing some programing u can't :(


How many positive factors of 4000 are multiples of 100?

we can list them out

100, 200, 400, 500, 800, 1000, 2000, 4000

we see that there are 8 positive factors of 4000 that are multiples of 100

We can also divide everything by $ 100$: How many positive factors of $ 40$ are multiples of $ 1$.  Since $ 40$ has $ 8$ factors, the answer os $ \boxed{8}$.


Three women A,B,C at once cleaned the plates.It's known that three ones are equally [b]skilful[/b].In a month,there were 4 broken plates.Find the probability of :
a.The woman A made three plates broken and the woman B made only one plate broken.?
b.One of three women made these 4 plates broken?
Thanks in advanced :blush:

This is Middle school, MOEMS level problem, not High school, Pre-olympiad...


calculus
integration
function
algebra
domain
LaTeX
real analysis

Theorem : if f is monotonic on [a,b], and if a (a is the function not the domain a) is continuous on [a,b], then f is integrable with respect to a. (we still assume that a is monotonic)

pf : let s >0 be given. For any positive integer n, choose a partition s.t
                             |a i+1  - a i | = (a(b)-a(a))/n, i =1,2,....n-1
This is possible because a is continuous. (WHY? I DON'T UNDERSTAND WHY THIS IS POSSIBLE, CAN YOU EXPLAIN PLS?)

we suppose that f is monotonically increasing, then, Mi = f(xi), mi =f(f xi-1)
s.t
                U(P,f,a) -L(P,f,a) = ( a(b)-a(a) )/n    (f(b) -f(a)) <s for every s >0 if n is big enough.

I'm confused about why you can always find n s.t you get even space between a i+1 and a i,
also, in order to prove the last line is true, you have to prove that you have always find a n that is large enough for a particular s, so, you kinda have to prove that there are INFINITELY many n that can make the space between a i+1 and ai a constant number........Can any1 pls explainn how that works?  :)  thxx! [/i]

This is utterly awful notation. Using the same name for both a function and a relevant point is horribly confusing, and $ \text{\LaTeX}$ for subscripts would make a huge difference.

I won't say much unless it gets cleaned up, but here's one idea: integration by parts. We're working with [url=http://en.wikipedia.org/wiki/Stieltjes_Integral]Stieltjes integrals[/url], and if $ f$ is continuous and $ g$ isn't, $ \int f\,dg$ is easier to work with than $ \int g\,df$.


Please post your AOL messenger ID but NOT YOUR REAL NAME!!!
This is for practices I'm thinking to do for Mathcounts State Competition upcoming.. I'm planning to do this week before Friday so more people, better it be..
I'll explain more after seeing how many people can attend it..

The only thing is, many states have already had their state competitions. But sure, i'll try to help. My aol screen name is caocoprez

One thing to say is that we won't talk about this year's state competition becasue it's not allowed to do it and is not fair for other students.. So, we're only talk about past years' plus this year's chapter only.. And thank you for accepting it... I hope we could get Tare or Mr.Orcax(not sure how to spell it) or Syntax Error or other good mathletes..

intellectual121, i might not be able to do it though

Handsome B Good <---aim screenname

although I've already taken the state comp I'd be glad to teach concepts and stuffs.

Goodie~~
So far:
me: (I have to make another one, now.. I'll tell one when before Tuesday)
yif man 12: caocoprez
eryaman: intellectual121
Syntax Error: Handsome B Good

For Syntax Error, can you tell Tare or other known matheletes join?
We'll appreciate!!

I forgot to add Rep123max's ID which is: Rep345max
Are you guys all on EST time period 'cause I am...
I'm thinking to do first practice on Tuesday, 3/16/04 at 7:00 P.M. in EST time period..

I won't accept anymore people because that's gonna take a lot time to organize and etc...

So, let's organize our memebers:
me: jwec05
yif man 12: caocoprez
eryaman: intellectual121
Syntax Error: Handsome B Good
Rep123max: Rep345max

Please post any adjustments you want before Tuesday and here's some rules going to be on the practice:

Nobody is moderator in here but I'll begin with few instructions given at the beginning.. I'll say this beginning at the beginning of the practice, too but here you can look at first..

Thanks to all people came to join the practice and I must inform you that everyone must have their pencils, papers, and calculator(not for all uses, be honest) and their past year's state questions.  Since I'm the one who organized this, I'll begin with a question. For the rule for doing this, it will be done similar to MathJam practice which if you got the question right, you'll get point.. I'm not sure how to determine the first correct answer but I have to look on that now.. Question is first started off by me, but then next questoin will be passed to other person who get my question or question from prior person who got the question correct.  All questions must come from state but it doesn't matter whether it's sprint, target, or countdown. You could post the team round question but I rather hope that you don't post any question for team.  Like in my case, I'm only going for the individual..
Here are scores:

Sprint: 1 point for maximum of 2 minutes given to answer (no calculator, please.)

Target: 2 point for maximum of 3 minutes given to answr (calculator may used)

Team: 2 point for maximu of 3 minutes given to answer (calculator may used)

Countdown: 1 point for maximum of 45 seconds to answer the question (no calculators permitted)

If you think something won't work, please post the advice fast.. I'll really appreciate.. Next practice after this isn't decided yet.. I'll set up after seeing this one first..

For this practice, I guess no one except Syntax Error is available..

So, anyone who wants to do must sign up for next one..

..

Read spoiler please:

[hide] It is your choice of practicing by AOL messenger but what I'm thinking is practicing all rounds not just countdown rounds.. It is better to practice with other good mathletes than alone.. So, that's point I want to point out..[/hide]

hmmm, I'm not in mathcounts so I can't do this right?

Only thing is that if you have some questions with you, you can..
But after Tuesday because that's already set up...
So, other than and you have problems to share, yes, you can participate..

hmm, I don't know if I can find problems.  I will try to get them though.  After Tuesday got it.

Do you participate in any math competitions?
Because if you are in really high competition, don't pick those hard questions from them.. They won't appear for us the dear Mathcounts state-going people.. You can post them if any of us made to national, though.. And by the way, I didn't take my test yet! It's this weekend.. Wish me luck..

I am in a program called tmsca and it's a lot easier than mathcounts.

I think I heard it before... Post one of your question here and let's look at the difficulty level... If you really want to practice, you can do now to... Just post your AOL messenger ID, i'll contact with you..
Got it?

Do anyone wants join more?
So far we have:

me: jwec05
yif man 12: caocoprez
eryaman: intellectual121
Syntax Error: Handsome B Good
Rep123max: Rep345max
BHorse Math: BHorse Math

Please if you see this, contact me if you can.. I wasn't able to talk to with Rep345max at all...

Not quite sure... I don't think I can't do tomorrow because my school is taking a huge tests through out the week...
How about this Friday? I'm fine with that time you know...

me: jwec05
yif man 12: caocoprez
eryaman: intellectual121
Syntax Error: Handsome B Good
Rep123max: Rep345max
BHorse Math: BHorse Math

Not only them, I can get MathFiend and Chinaboy because I know their sn.. So, please post your thoughts...But since many people took their state tests and not succeed, we'll get some questions from state competition and look over some terms... I have no ideas so far but I'll post more thoughts throughout the week...

[i]Remember math is not just competition stuff, it is the thing you have to do throughout your life...[/i]

hmm... I can try but can't guarantee a hundred percent...
Do you have any good problemsj?

i've got some problems to share. dunno if i can make it or not though

my AIM is jiggityjustin08
so when you've got a final list compiled, just post it, and we can all add it to our buddy lists, so when we see a large group of us on, we can just hold a practice then maybe?
(best days for me are monday, wednesday, and friday)

Final posters:

justinian: jiggityjustin08
me: jwec05
yif man 12: caocoprez
eryaman: intellectual121
Syntax Error: Handsome B Good
Rep123max: Rep345max
BHorse Math: BHorse Math

Plus
MathFiend21 (her AIM had been posted in another topic before so I think it's okay to say; but not sure whether she's gonna do or not

and I know chinaboy's AIM but I'm not sure whether I have right to post his AiM without asking since this site requires high quality of security..

Since all these competitions are nearly over...
Do you want to look at some problems from mathcounts competition if we have practice or have lessons like in mathjams?
I'm planning to do on this weekend, so what do you thinK?


What percent, rounded to the nearest half percent, of the numbers from 1 to 53 can be written as the sum of two or more members of the list 1, 1, 3, 3, 9, 9, 27?

Note that any number taken base $ 3$ can be expressed with at most $ 2$ ones, $ 2$ threes, $ 2$ nines, etc. Clearly, then, all numbers $ 2\minus{}53$ can be expressed by the given conditions, so the percent is $ \frac{53}{54}\approx\boxed{98.0\%}$.

I'm not sure on how to do this, but math154, your answer is wrong - the real one is $ 92.5$....

All the numbers [i]except[/i] the powers of $ 3$ can be made with two or members of that set. Hence, this is $ \frac{53\minus{}4}{53}\approx\boxed{92.5\%}$.


Prove that if the midpoints of the sides of a quadrilateral are vertices of a rectangle, then this quadrilateral is orthodiagonal.

Orthogonal means it's a rectangle, right? :?

edit: oh, orthodiagonal... :blush:

Orthodiagonal means that the diagonals are perpendicular to each other.

AOPS Volume 2?

use the fact that the segment joined by the midpoints is parallel to the base.

One way is trivial, the other...just find parallel lines.

Happy solving!

[quote="MCrawford"]This problem is clearly beyond Getting Started in difficulty.  Please stick to the forum posting guidelines.[/quote]
Well, then what is Getting Started difficulty? This is the High School section, high school students should be able to solve this kind of problem in a few seconds.

Well its got me stumped, and I'd like to think of myself as a problem solver - former USAMO qualifier, heh. Haven't thought too deeply yet, maybe I should treat it like a brainteaser and think outside the box?  :)

[color=blue]In quadrilatera $ABCD$, $E, F, G, H$ are mid-points of $AB$, $BC$, $CD$, $DA$. If $EFGH$ is a rectangle,then prove: $AC \perp BD$.[/color]

Proof: Let J=EF $\cap$ BD.
          Since E, H are mid-points of AB and AD, so EH // BD.
          Similarly we have EF // AC.
          Hence: $\angle{DOC}=\angle{OJF}=\angle{HEJ}=90^{o}$.


inequalities
number theory unsolved
number theory

Is there any elementary way to prove that for any integer $n$  the equation $ 2^x-3^y=n$ has only finitely many solutions in natural numbers? I've seen various proofs using Bker's inequalities, but they were more than far from elementary.

What is this inequality? :? I never ask before googling, but I couldn't find anything. Maybe Bker is not spelled correctly? :?

His name is Alan Baker and he received a Fields medal for his work on transcendental number theory.


vector
algebra
polynomial
linear algebra
linear algebra unsolved

let A be nxn matrix. Want to show that the matrices I, A, A^2, A^3,..... are linearly dependent.

any hints?
cheers!

The set of all $ n\times n$ matrices is a vector space of dimension $ n^2$.

And the dimension of the span of the powers of $ A$ is the degree of the minimal polynomial of $ A.$ The argument implied by mlok's comment is an overly generous upper bound for this dimension, but it is an elementary argument (as opposed to the decidedly non-elementary Cayley-Hamilton theorem.)

ouch...so simple :)

big thanks to both of you!


modular arithmetic
induction
number theory unsolved
number theory

Let $M=\{x^2+x \mid x\in \mathbb N^{\star} \}$. Prove that for every integer $k\geq 2$ there exist elements $a_{1}, a_{2}, \ldots, a_{k},b_{k}$ from $M$, such that $a_{1}+a_{2}+\cdots+a_{k}=b_{k}$.

This is obviously true. We denote $n_x = x^2 + x$. Note that $n_x \equiv 0 \pmod{2} \forall x \in \mathbb{Z}^+$, thus $\frac{n_x}{2} \in \mathbb{Z}^+ \forall x \in \mathbb{Z}^+$. We now make use of the identity $n_{a+b} = n_a + n_b + 2ab$. If $n_m = a_1 + a_2 + \ldots + a_k$ for some $k \in \mathbb{Z}^+, a_i \in M$ for $i = 1, 2, \ldots , k$, then from our identity, we have \[ n_{(\frac{n_m}{2} + 1)} = n_{\frac{n_m}{2}} + n_1 + n_m = n_{\frac{n_m}{2}} + n_1 + a_1 + a_2 + \ldots + a_k \] This means that if $n_p$ can be expressed as a sum of $k$ $n_i$'s, then we can find another $n_q$ that can be expressed as a sum of $(k+2)$ $n_i$'s. Finally, we note that $n_3 = n_2 + n_2$ and that $n_2 = n_1 + n_1 + n_1$ and we are done.

I wonder if the problem really wants distinct elements? Otherwise the case $ k\geq 3$ can just be solved by the fact that $ n_{k\plus{}1} \equal{} n_k \plus{} (k\plus{}1)n_1$

[quote="prowler"]Let $ M \equal{} \{x^2 \plus{} x \mid x\in \mathbb N^{\star} \}$. Prove that for every integer $ k\geq 2$ there exist elements $ a_{1}, a_{2}, \ldots, a_{k},b_{k}$ from $ M$, such that $ a_{1} \plus{} a_{2} \plus{} \cdots \plus{} a_{k} \equal{} b_{k}$.[/quote]
Let $ x_i \equal{} \sqrt {4a_i \plus{} 1}$. Then $ x_i \in \mathbb{R}$ and $ x_i$ can be any odd natural number $ \ge 3$. Likewise let $ y \equal{} \sqrt {4b_k \plus{} 1}$.
So it suffices to prove that there exists odd natural numbers $ x_1,x_2,...,x_k,y \ge 3$ such that:
$ x_1^2 \plus{} x_2^2 \plus{} \cdots \plus{} x_k^2 \equal{} y^2 \plus{} (k \minus{} 1)$
Or equivalently:
$ y^2 \minus{} x_k^2 \equal{} x_1^2 \plus{} x_2^2 \plus{} \cdots \plus{} x_{k \minus{} 1}^2 \minus{} (k \minus{} 1)$
Let $ y \equal{} x_k \plus{} 2$. Then $ y^2 \minus{} x_k^2 \equal{} 4x_k \plus{} 4$
So we have:
$ x_k \equal{} \frac {x_1^2 \plus{} x_2^2 \plus{} \cdots \plus{} x_{k \minus{} 1}^2 \minus{} (k \minus{} 1)}{4} \plus{} 1$
It is apparent that $ x_i^2 \equiv 1 \bmod 8$ and therefore $ \frac {x_1^2 \plus{} x_2^2 \plus{} \cdots \plus{} x_{k \minus{} 1}^2 \minus{} (k \minus{} 1)}{4}$ is even and hence $ x_k$ is odd. If we just make $ x_1,x_2,...,x_{k \minus{} 1}$ "large" and define $ x_k$ as above and $ y \equal{} x_k \plus{} 2$ we find a solution where all the numbers are different.

It is obvious that $ 2 \in M$.

For $ k \equal{} 2$, we have that $ (2^2 \plus{} 2) \plus{} (2^2 \plus{} 2) \equal{} (3^2 \plus{} 3)$.

For each $ k \geq 3$, we have that $ 2 \plus{} 2 \plus{} .. \plus{} 2 \plus{} (k \minus{} 2)^2 \plus{} (k \minus{} 2) \equal{} (k \minus{} 1)^2 \plus{} (k \minus{} 1)$
(there are $ k \minus{} 1$ 2's)

Assuming the elements must be distinct, we can prove it by induction on $ k$.
The base case is obvious as $ 12 \plus{} 30 \equal{} 42$
Now assume the problem is true for up to $ k \equal{} n$. Then,
$ a_1 \plus{} a_2 \plus{} ...a_n \equal{} y^2 \plus{} y$ for some $ y$. To prove the original statement for $ k \equal{} n \plus{} 1$, take $ a_{n \plus{} 1} \equal{} (\frac {y^2 \plus{} y}{2} \minus{} 1)^2 \plus{} (\frac {y^2 \plus{} y}{2} \minus{} 1)$ and leave all the other $ a_n$'s the same. Then,
$ a_1 \plus{} a_2 \plus{} ...a_n \plus{} a_{n \plus{} 1} \equal{} y^2 \plus{} y \plus{} (\frac {y^2 \plus{} y}{2} \minus{} 1)^2 \plus{} (\frac {y^2 \plus{} y}{2} \minus{} 1) \equal{} \frac {(y^2 \plus{} y)^2}{4} \plus{} \frac {y^2 \plus{} y}{2}$ which is of the desired form.


Help me solve this problem.

A sequence A is called an integer sequence of length N if all its elements A1 A2 .. AN are non-negative integers less than 2000 000 000. Consider two integer sequences of length N, A and X. The result of their multiplication (A*X) is an integer number R=A1*X1 + A2*X2 + .. + AN*XN. Your task is to solve the equation A*X=B (mod P), given the integer sequence A and the integer numbers B and P.

You know there is a solution if gcd(X1,X2,..XN) divides P. So just choose a suitable X1 that allows a solution for X2, X3, .., XN, and repeat for X2, X3, etc.


function
calculus
derivative
integration
real analysis
real analysis unsolved

This problem is from my homework so please don't post a full solution.

I've got some function $f \in L^2(R)$ and some conditions on its Fourier transform:

1. $|\widehat{f}(\theta)| \le \frac{C}{|\theta|^{1+\alpha}}$ for $|\theta| \ge 1$, some constant $C$, and $\alpha > 0$.
2. $|\widehat{f}(\theta)| \le C$ for $|\theta| \le 1$.

I need to prove that $f$ satisfies a Holder condition of order $\alpha$.

I really have no idea how to approach this. Any suggestions? Again, please don't tell me a full solution because this is homework; just a pointer would be good. It would be nice if I could take the derivative and apply Parseval's identity with that and the characteristic function, but I don't think that's legit because $\theta\widehat{f}(\theta) \not\in L^1$...

Well, if we note that $\widehat{f}(\theta) \in L^1$, we can stick it in the reverse transform formula and look at $|f(b) - f(a)|$. Then we can split it into three integrals ($(-\infty, -1]$, $[-1, 1]$, $[1, \infty)$), and the middle one vanishes and we can change variables (switch sign of $\theta$) in the first one to get

$\int_1^\infty \frac{2C}{\theta^{1+\alpha}}(\cos(2\pi\theta b) - \cos(2\pi \theta a)) d\theta$

(unless I've made a mistake). But I don't know where to go from here. How are the $b$ and $a$ going to get an $\alpha$ power on them? How would I even estimate this integral?

Okay I finally figured this out. In case anyone cares, you just have to notice that $\frac{1}{|\theta|^{\alpha + 1}} \le 1$ for $|\theta| \le 1$, and then use that approximation for the function over the whole real line (except a small ball around 0). Then use inverse Fourier transform and a change of variables, and things come out nicely.


AMC AIME MATHCOUNTS HMMT USA(J)MO USAMO geometry

Let's try this again. Everyone was posting in the games section, so I am making a topic for this.

Do not post your name, just post what scores you got on various stuff and where you went these last years.

Grade: 7
AMC 12 2007/2008: 111/126 highest score
AIME 1 2007/2008: 3/5
MathCounts chapter 2007/2008: 41/45
MathCounts state 2007/2008: 36/33(huh.)
MathCounts nationals 2007: 28
NYSML 2007/2008: 7/7
HMMT 2008: 0 combinatorics 3 geometry(taking general for the rest of my life!!!)

I went to CTY last year, and I didn't learn much and I was a sniper target(not literally).

wow did 33 get you to nats? i got a 33 and 15th in texas!!!

Well, it's just to introduce your how "skilled" you are at math...

Grade: 9
AMC 10 2006/2007/2008: 106.5/123/135
AIME 2007/2008: 4/8
USAMO 2008: Something <= 4
MC 2007: 22nd Chapter, 29th State CA

Pretty noob scores...
I'm from the San Francisco Bay Area, Lynbrook High School
I don't know anybody that's also going, so I hope I can meet some new people... :gathering:

Grade 6:

Last Year: 97 AMC 12

This Year: AMC 10 144, AIME 9, USAMO, hopefully 17, did 1, 4, and most of 5,
MathCounts: Chapter 43, State 38, Southern CA, did not make national

Grade: 8
State: TX

Last year:
Mathcounts - 1st chapter (??), 6th state (38)

This year:
Mathcounts - 1st chapter (45), 8th state (35-40?) :(
AMC 8 - 24 (darn you lemmings)
AMC 10A - 132
AIME - 8, qualifying floor/index
USAMO - $ n$ such that $ 7 \leq n \leq 15$, depending on how badly my solutions were written
ARML - made team, not occurred yet

stupid beastly people.

Grade: 12
State (commonwealth): Puerto Rico
Last year: International Spanish Competition - 10th place.
... xD...

Grade: 8
State: IN

Last year: 2nd MC chapter (44).  4th State (37) -__-
27th MC nationals (33)... -_-....
AMC 8 + AMC 10 + AIME fail

ARML: 5

This year
AMC 8 - 25
AMC 10 - 150
AIME - 9
USAMO - err ?

MC Chapter - 2nd (44) -_-
MC State - 1st (XD)

ack. you people are too good at math.

Grade 8, NYC.

Last Year:
MATHCOUNTS: I failed so utterly at this that I got a 32 on the school round and didn't make the team.
AMC 8: 21
AMC10A: 96-ish

This Year:
AMC 8: 23.
MATHCOUNTS: 3rd chapter (44), 6th state (31).
AMC 10A: 120
AMC 12B: 109.5
AIME: 7

I guess I'll follow the trend...

6th grade (last year):

MATHCOUNTS: Didn't make top 10 in chapter  :(
AMC8: 24  :D
AMC10A: 103.5 -_-

7th grade (this year):

MATHCOUNTS:  6th chapter, 7th state and 3rd CD
AMC8: 24  :mad:
AMC10A: 144
AMC12B: 100.5 (heh)
AIME II: 5  :dry: (coulda gotten 7 and made USAMO but whatever.)

Non-math Introduction:

I'm a 7th grade girl from VA.  I like skiing, tchoukball, Linkin Park, ramen, parmesan & garlic Cheezits, pen-spinning, Monopoly, Apples to Apples, singing and math (no duh).

im not going, but what the heck.

MC school:2006-07(7th grade)/07-08(8th grade) = 3rd (36 :blush: )/1st(rather not say)
MC chapter: 06-07/07-08 = 4th (FAIL)/3rd(45)
MC state = rather not talk about it

AMC 8: 06-07/07-08 = 21/23

AMC 10: 06-07/07-08 = 117/126

AIME: 2/5

USAMO: 06-07/07-08 = 31/22<--o.O  KIDDING

Grade: 8
State: TX

This year:
Mathcounts - 12th chapter (39), 15th state (33)
AMC 8 - 23
AMC 10A - 120 (yeah, i know.)
AIME - 5
USAMO - im 2 lousy.
ARML - im going.

I could boast about scores on random competitions...except for the fact that the majority of the other faculty at AMSP have been intimately involved with the IMO and I'm not :D

I'm Alex [Insert last name that would be here except DPatrick was like OMG no names so whatever its on this site somewhere if you care]

Supported by a positive attitude and a tenacious love of learning, I value a level of diversity in my life. I am an avid runner: I was a varsity cross country runner and I frequently play frisbee or touch football with my friends. Constrained by the endless time demands of school, I rarely get to savour the ideas of great writers such as: Emerson, Shakespeare, and Antoine de Saint Exupery. Yet, I make time to experience the wisdom of those authors who can articulate the invisible or otherwise enigmatic facets of life. I may visit China for a few weeks over the summer.

That being said, my chief interest is in mathematics and science. I suspect that I will continue to do Neuroscience research over the summer along with other mathematical projects. I appreciate the elegance of a challenging math problem. I study geometry and algebraic inequalities and create my own problems for enjoyment. As a former AMSP student, I am excited to come back this summer as a counselor. I love the community: I've maintained friendships with numerous students from camp last year. I hope that my observations from the perspective of a learner can make me a better counselor. I'd love to share my insights into geometry and other problem solving topics. This summer should be great! Don't let the competition focus deter you from the fact that Awesomemath is fun math. I hope you can appreciate some of the math that I enjoy when the camp is over.

I won't say my name for security reasons.

I'm also too tired to type something like Altheman did.

So I'll just say that I like soccer, math, and competitive reading.

Wait.

There's something called "competitive reading"?

It makes more sense than competitive cup stacking, I guess :roll:

[quote="diophantient"]Really now, scores do not make friends...[/quote]

totally agreed.  xD

soph from Irvine CA
NO SCORES.
well i do, but i'd rather not post any cuz i'm not smart like all of you guys are.
i pretty much suck at math. dont know how i got in O.o...

Oh man I get a bit ridiculous when I set out to write something. Writing can be more fun that I thought...

I think a more important part of the camp is a desire to learn. Even if you aren't the best at math, you can learn a lot if you try. If you really have no idea what is going on, speak up.

Grade 6:
State:(see location)
Last Year: Nothing
This year: 11th MC Chapter(41),31st MC State(28 PHAIL)

Next Year...hopefully make all my goals

Name: REMOVED DUE TO SECURITY AND PRIVACY REASONS

Is everyone here going to AMSP? I am.

Name is Dai Yang, 8th grade.
COULD have gotten into USAMO, but I guess I'll just wait til next year.
California.
I personally know sooozyy and pythag011, both have posts above mine.
No saying my scores, but I'm not satisfied with them  :( .

Now that I know I'm actually going, I'll join in.

Name: Some know, other's don't. I'm probably not supposed to put it, though.
Residence: Dallas, TX, so I know basically everywhere around the area.
Previous AMP Experience: Track 4 last year
Previous AMP Awards: Karaoke Superstar (along with oh gosh I forgot her username... mysticsomething? she has a lot of posts, but I just can't remember)
Hopefully, I'll be taking these courses: Euclidean Geo or whatever and either Advanced combo or Advanced NT
Interests and Whatever: Running (and I'm actually going to do that this year - Altheman, you up for it?), Music, Pop-Tarts, Swimming, Making stupid mistakes on everything imaginable, Winning, Dallas sports teams, and last but certainly not least, game shows, my favorite of which is Jeopardy

[hide="Scores if you want to see them"]
2006 AMC 10A: 130
2006 AIME I: 6

2007 AMC 10A: 144
2007 AIME II: 5
ARML: 4

2008 AMC 10A: 138
2008 AMC 12B: 132
2008 AIME II: 9
2008 USAMO: 8[/hide]


algebra polynomial function LaTeX Asymptote AMC

What is the maximum number of points of intersect of the graphs of two different fourth degree polynomial functions $ y\equal{}p(x)$ and $ y\equal{}q(x)$, each with leading coefficient $ 1$?

$ \textbf{(A)} \ 1 \qquad \textbf{(B)} \ 2 \qquad \textbf{(C)} \ 3 \qquad \textbf{(D)} \ 4 \qquad \textbf{(E)} \ 8$

[hide]The answer key says that the answer is $ 3$, but from the diagram below, I got $ 4$.  Could someone please tell me why I'm wrong?  Thanks![/hide]

Let $ p(x) = x^4 + a_3x^3 + a_2x^2 + a_1x + a_0$ and let $ q(x) = x^4 + b_3x^3 + b_2x^2 + b_1x + b_0$.
The intersect any time that:
$ p(x) = q(x)$
$ x^4 + a_3x^3 + a_2x^2 + a_1x + a_0 = x^4 + b_3x^3 + b_2x^2 + b_1x + b_0$
$ (a_3 - b_3)x^3 + (a_2 - b_2)x^2 + (a_1 - b_1)x + (a_0 - b_0) = 0$.
Since this is a cubic, there are a maximum of $ 3$ points where $ p(x)$ and $ q(x)$ intersect. If the leading coefficients weren't equal, then there could be a $ 4$th point of intersection.

BTW: What are the equations of those quartics in the picture? If both quartics have a leading coefficient of $ 1$ then the steepness of the the two quartics as $ x \to \pm \infty$ should be about the same, but the black quartic is much steeper as $ x \to \infty$ than the green quartic.

Edit: fixed $ \text{\LaTeX}$.

[quote="Xantos C. Guin"]Let $ p(x) \equal{} x^4 \plus{} a_3x^3 \plus{} a_2x^2 \plus{} a_1x \plus{} a_0$ and let $ q(x) \equal{} x^4 \plus{} b_3x^3 \plus{} b_2x^2 \plus{} b_1x \plus{} b_0$.
The intersect any time that:
$ p(x) \equal{} q(x)$
$ x^4 \plus{} a_3x^3 \plus{} a_2x^2 \plus{} a_1x \plus{} a_0 \equal{} x^4 \plus{} b_3x^3 \plus{} b_2x^2 \plus{} b_1x \plus{} b_0$
$ (a_3 \minus{} b_3)x^3 \plus{} (a_2 \minus{} b_2)x^2 \plus{} (a_1 \minus{} b_1)_x \plus{} (a_0 \minus{} b_0) \equal{} 0$.
Since this is a cubic, there are a maximum of $ 3$ points where $ p(x)$ and $ q(x)$ intersect. If the leading coefficients weren't equal, then there could be a $ 4$th point of intersection.

BTW: What are the equations of those quartics in the picture? If both quartics have a leading coefficient of $ 1$ then the steepness of the the two quartics as $ x \to \pm \infty$ should be about the same, but the black quartic is much steeper as $ x \to \infty$ than the green quartic.[/quote]

I don't know what the equations are.  I didn't use Asymptote or a math graphics program to draw these quartics.  I just know that quartics with a positive leading coefficient generally look like a $ W$.

Also, I don't get:

$ x^4 \plus{} a_3x^3 \plus{} a_2x^2 \plus{} a_1x \plus{} a_0 \equal{} x^4 \plus{} b_3x^3 \plus{} b_2x^2 \plus{} b_1x \plus{} b_0$
$ (a_3 \minus{} b_3)x^3 \plus{} (a_2 \minus{} b_2)x^2 \plus{} (a_1 \minus{} b_1)_x \plus{} (a_0 \minus{} b_0) \equal{} 0$.

Isn't the problem talking about a quartic?  I don't get how the cubic was gotten.

Thanks!

The cubic was gotten from subtracting one quartic from the other, in this case, $ p(x)\minus{}q(x)$.  Since $ p(x)\equal{}q(x)$, you have $ p(x)\minus{}q(x)\equal{}0$.  When you subtract these two quartics, the $ x^4$ term cancels out and you're left with the cubic.

There aren't two cubics.  The cubic equation describes the distance [i]between[/i] the quartics.  This distance can be zero at most three times.


i was looking through the forum and couldn't really find approximations for qualifying scores for state and nationals. Approximately what type of score would you need in a fairly competitive chapter to get to state? And approximately how much would you need to get from state to nationals in a state which usually gets within the top 5 teams?

It depends on the chapter.  If it's in a competitive state, the qualifying score is probably 37-45.

For state, for these competitive states, you'd probably need:
Indiana-43
Virginia-43
California-45
Texas- 46(maybe 45)
Missouri - 42
Michigan-41

what about washington? because im trying to figure out what to aim for in practice. but thanks for the chapter and other state amounts.

The winner of nats 2 years ago is in WA. I don't know the scores though.

Mathcounts is very unrepresentative of the top kids at nationals. If you live in Kansas, you can easily make nats with AoPS, but if you live in California.....good luck

For Michigan, which is fairly competitive (15 place at nats last year), at our most competitive chapter, 41 was the cutoff for CD (well actually a bunch of kids got 41, so it was 41 with some tiebreakers). But that might be overstating michigan a bit, since at our other chapters, a 41 would place you in the top 4.

At states, I think a 38ish got you in the top 4, while a 33ish got you in CD.

But if your talking about top 5 states, thats a WHOLE different ballpark. I think you would need to get a 45 at states in order to even be considered for nats in a top 5 state. Since your from washington, which placed 3rd last year........good luck. You should probably be aiming for a 45.

Cutoff at Washington was either 43 or 44 last year; I know for a fact it was not 45 or 46.

ok, thanks everyone... i now know that I'm not practicing hard enough :(. is there any way to practice for mathcounts without ordering all or most of the competition booklets? and AIME15, do you mean cutoff to get into nationals or into state?

yeah, go to the "mathcounts resources here" thread to get and trade previous MC tests with other people. Also, take a few previous MC tests, find your weakness, and buy the corresponding AOPS book (like if youre bad at counting, get "intro to counting and probability", which is really good)

[quote="henrypickle"]and AIME15, do you mean cutoff to get into nationals or into state?[/quote]

Nationals.

Other than California and Texas, the cutoff would be 43 TOPS.

Why? You guys are all overlooking that it DEPENDS on the DIFFICULTY of the TEST and COMPETITION, in Indiana this year there was a 45 in 2008 a 41 was 1st place and the NAts cutoff was 37

i know it does depend on difficulty and other factors, but isn't it true that the mathcounts state and nationals competitions usually similar cutoffs during the year? Or does it fluctuate largely (like 5 points)?


Let $ x,y,z\geq 0$ and $ x+y+z+xyz=4$ . Prove that

$ x+y+z\geq xy+yz+zx$

See here (my last post) http://www.mathlinks.ro/Forum/viewtopic.php?t=127956


function
calculus
derivative
logarithms
integration
calculus computations

The digamma function [denoted $\psi (z)$] is the logarithmic derivative of the gamma function, that is $\psi (z) = \frac{d}{dz}\log \Gamma (z)$.

Prove that for $\Re z>0,$  $\psi (z) = \sum_{n=0}^{\infty}\frac{1}{n+1}\sum_{k=0}^{n}(-1)^{k}\left(\begin{array}{c}n\\k\end{array}\right) \log (z+k)$

Hint: differentiate  [url=http://mathworld.wolfram.com/MalmstensFormula.html]Malmstén's Formula[/url].

Use the formula $\log z = \int_{0}^{\infty}\left( e^{-t}-e^{-zt}\right) \frac{dt}{t}$ for $\Re z>0$


geometry
parallelogram
geometry unsolved

In parallelogram $ ABCD$, with center $ O$ and $ \angle AOB > 90$, $ B_1$ is on the extension of $ DB$ through $ B$ and $ A_1$ is on the extension of $ CA$ through $ A$ such that $ A_1B_1 \parallel AB$ and $ 2\angle A_1B_1C \equal{} \angle ABC$. Prove that $ A_1D \perp B_1C$.

Let $ E \equiv A_1D \cap B_1C$ and $ F$ the intersection of the inner bisector of $ \angle ABC$ with $ DC.$ We shall prove that $ \triangle EA_1B_1$ is necessarily right. Since $ \angle A_1B_1C \equal{} \frac {_1}{^2}\angle ABC$ and $ A_1B_1 \parallel AB,$ it follows that $ BF \parallel CB_1.$ Let $ M \equiv EO \cap A_1B_1.$ From $ DC \parallel A_1B_1,$ it follows that cevians $ A_1C$ and $ B_1D$ of $ \triangle EA_1B_1,$ meet on its E-median. Thus, $ EO$ passes through the midpoints of $ A_1B_1$ and $ AB$ $ \Longrightarrow$ $ EO$ is the midline of the parallels $ AD,BC.$  Therefore, $ \triangle MB_1E$ is homothetic to the isosceles $ \triangle CBF,$ due to $ BF \parallel CB_1,$ $ DC \parallel A_1B_1 ,$ $ EM \parallel CB$ $ \Longrightarrow$ $ \triangle EMB_1$ is isosceles with legs $ ME\equal{}MB_1$ and $ M$ is the midpoint of $ A_1B_1$ $ \Longrightarrow$ $ \triangle EA_1B_1$ is right.

Similar like Luis you can prove, using Ceva, that M is the midpoint... blah blah blah
Is easy note that $ \angle A_1M'E \equal{} \angle ABC\equal{}2\angle A_1B_1C\equal{}\angle A_1B_1C \plus{} \angle M'EB_1$, and then M'FE is isosceles and finish.
But I have a Question: ¿Do you think there is a solution without using Ceva?
I say that because a lot of problem don't have solution (not that I know) without this idea.

[quote="Immanuel Bonfils"]Please, where is $ M'$ ?[/quote]

Sorry I should have said M, my M' is the M of Luis


algebra
polynomial
integration
trigonometry
real analysis
real analysis unsolved

Let $ P\in \mathbb R\ [X]$ be a polynominal with degree $ n\ge 2$.
Choose $ F\in\int P(x)\sin x\ \mathrm {dx}$, $ G\in\int P(x)\cos x\ \mathrm {dx}$.
Prove that the system $ \|\begin{array}{c}
F(x)=0\\\
G(x)=0\end{array}$ has a finite number of solutions.

A method:

$ \int P(x)\cos(x)dx\equal{}f(x)\sin(x)\plus{}g(x)\cos(x)\equal{}0$.

$ \int P(x)\sin(x)dx\equal{}g(x)\sin(x)\minus{}f(x)\cos(x)\equal{}0$,

where $ f$ and $ g$ are some polynomials. Square the two equations and add them up,

and you get that $ f^{2}(x)\plus{}g^{2}(x)\equal{}0$.

Thus, $ x$ is a root of some polynomial, i.e., $ f^{2}\plus{}g^{2}$ and hence,

the system has a finite number of solutions.


Consider the statements: I: $ (\sqrt { \minus{} 4})(\sqrt { \minus{} 16}) \equal{} \sqrt {( \minus{} 4)( \minus{} 16)}$, II: $ \sqrt {( \minus{} 4)( \minus{} 16)} \equal{} \sqrt {64}$, and $ \sqrt {64} \equal{} 8$.  Of these the following are [u]incorrect[/u].

$ \textbf{(A)}\ \text{none} \qquad \textbf{(B)}\ \text{I only} \qquad \textbf{(C)}\ \text{II only} \qquad \textbf{(D)}\ \text{III only} \qquad \textbf{(E)}\ \text{I and III only}$

$ (\sqrt{\minus{}4})(\sqrt{\minus{}16})\equal{}(2i)(4i)\equal{}\minus{}8$, $ \sqrt{(\minus{}4)(\minus{}16)}\equal{}\sqrt{(64)}\equal{}8$. So statement I is incorrect.
$ \sqrt{(\minus{}4)(\minus{}16)}\equal{}8$ so $ \sqrt{(64)}\equal{}\sqrt{(\minus{}4)(\minus{}16)}\equal{}8$ is correct.
$ \sqrt{64}\equal{}8$ is obviously correct.
So the only incorrect statement is I $ \Rightarrow \boxed{B}$.


find all positive integers x and y such that
 $x +y^2 +z^3 =xyz$  , where z is the greatest common divisor of x nad y

Look at the 249th post in this thread:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=40038&postorder=asc&start=240[/url]


LaTeX AoPSwiki articles Asymptote Support \/closed

I am encountering a problem that I have been ignoring for some time because it was pretty insignificant: at times, some latex does not show up on a particular post. Initially I read posts that appeared like "let  be the intersection of $ AB$ and $ CD$" so I had to click the Quote button to see that the poster had meant "let $ E$ be the intersection of $ AB$ and $ CD$". I notice that the letter E (by itself, like $ E$) does not show up often.

Also in the AopsWiki article [url=http://www.artofproblemsolving.com/Wiki/index.php/2008_AMC_12A_Problems/Problem_22]2008 AMC 12A #22[/url], the asymptote image shows up as the following:
[img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/e/9/2/e9218358a6d7c5e0c8d92a87a664a6311253a6f8.png[/img]
But when I click on "Edit" and "Preview" without changing any text, I get the intended image:
[img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/5/6/7/5677ea0c1d97fe7a1ce5b4e9ffc2da5990e98e3d.png[/img]
Anyone have similar problems or know what is going on?

This is a problem with parsing on the forum and Wiki so best asked in the Support forum. I am transferring this post to that forum.

It has already been discussed. The problem seems to be that the dollar signs in the Asymptote labelling command get translated into <math> tags before executing the code. It is really irritating and it would be nice to do something about it ASAP because it essentially disables Asymptote on Wiki.

This is also an issue in the PDF downloads - asymptote is very messed up there.

Hey, this was also 2008 AMC 10A Problem #25!

[quote="azjps"]There is a work-around to this problem: [url=http://www.artofproblemsolving.com/Wiki/index.php?title=2008_AMC_12A_Problems/Problem_22&curid=6477&diff=23573&oldid=23496]use \( ... \) instead of math tags[/url]. See this thread by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=183139]Boy Soprano II[/url].[/quote]Not required anymore. You can use [code]$ .. $[/code] within the asy tags in the wiki.

Also the + - issues (them not appearing in the PDFs) are now solved.

All that remains is support for Asymptote (making a separate thread for that tho) images within the generated .pdfs


Find all primes $ p$ for which numbers $ \frac{p^{p}\minus{}1}{p\minus{}1}$  and  $ \frac{p^{p}\plus{}1}{p\plus{}1}$ are primes too.

Only $ p \equal{} 3$. See http://www.mathlinks.ro/viewtopic.php?t=169135 (shows that if $ p \equiv 1 \mod 4$, then $ \frac {p^p \minus{} 1}{p \minus{} 1}$ isn't prime; proving that if $ p \equiv 3 \mod 4$ and $ p > 3$, then $ \frac {p^p \plus{} 1}{p \plus{} 1}$ isn't prime is complete the same).


I am an international student from China.
Now I have got admitted by University of Wisconsin-Madison and University of Minnesota-Twin Cities. Both are math majors.

The math rankings of the two universities are similar but the comprehensive ranking of UWisc is much higher. UMN's advantage is the expense. The tuition for intl students is cheap this year.So the total expense of UMN is about 6000 dollars/year less than that of UWisc.

My family could exatly afford UMN while will be in debt if I attend UWisc.

I prefer UMN but I am always afraid of the grad application four years later. UWisc's prestige and ranking are better so if I go to UMN will I be in disadvantage regarding the "name" when applying to grad schools?

I am good at math and owned the first prize of Chinese national math competiton.But because of my very bad SAT score(English is always not my strongpoint) and financial ability I could not apply for better universities or colleges.
So I place my hopes on the grad application. I want to go to top schools such as MIT,Harvard in the future.

Thanks!

I would be interested to hear the views of others about this. I think the OP knows that I am an alumnus (NOT a math major) of the U of MN. Students in Minnesota or Wisconsin can attend either university with in-state tuition reciprocity.

I wonder that why umn's comprehensive ranking is so low while its individual program's ranking is often within the top 30?
Even if the comprehensive ranking is for undergrad but these rankings differ too much.
Does it mean that umn does not give much attention to undergraduates?

I fell that umn is underrated........

Minnesota historically got a lot of the top students in Minnesota, a state of moderate population (above the median population of U.S. states) but UW Madison took fewer low students, compared to Minnesota Twin Cities, because the UW system is divided into more distinct campuses and the overall state population is spread around more geographically. So I think what drags down the overall ratings of Minnesota Twin Cities, despite the known presence of some strong students in the toughest programs, is the equally known presence of some mediocre students in many of the easier programs.

I know almost nothing about these two schools, but I'll still share my thoughts:

US News undergrad rankings do NOT evaluate the quality of the undergrad education.  If you want to know what they're evaluating, just look at the information used to produce the ranking.  Since you are picking between only two schools, it's easy enough to look at the raw numbers for those two schools and compare them based on the factors that matter to *you*.

As you mentioned, the math depts of the two schools are comparable, so it stands to reason (but is not a given) that the quality of your mathematical education and your mathematical opportunities at the two schools is comparable.  (In particular, I don't think Minnesota would hurt you for grad apps.  But DO improve your English skills.  People care about that.)

In fact, if I were you and ALL I cared about was my math education (and my parents' finances), I'd go with the cheaper school.  However, other things tend to be important to undergrads (such as geography, social environment, etc).

If you might not go into math (always possible), there are other important concerns, such as the strength of other departments.  And related to the issue of prestige is the question:  "What is my degree worth on the job market?"  In most of the country, and definitely in China, no one will know or care about the difference between Minnesota and Wisconsin.  In the Midwest they might possibly know or care.


sounds like HMMT except its not, and besides how come the other sites don't get to get a copy of mathematica?

[quote="AIME15"]At Las Vegas, the guy stating who got what place said we would all get a copy of Mathematica, and instructions would be on the ARML site...where are they?[/quote]
Yeah I recall them saying that too...

Last year, we got it through our coaches. I suggest talking to them; individually you can't do much about it.

Wait another week or two before you ask your coaches. We don't have the answers yet.

Well it's been a month now... got any updates?

Apparently some people have received instructions by email.  I'm not sure if they came from Mathematica, ARML, or their coaches.  So be patient--you should have them soon.


vector
topology
function
calculus
derivative
analytic geometry
geometry

Here I use definations in "Treatise on Analysis" by Jean A. Dieudonne.

Remember that if  $X$ is submanifold  of differencial manifold $Y$, and $dim_{x}(X)=n$,  $dim_{x}(Y)=N$, then at $x \in X$, there is an open nbh $U$ of $x \in Y$, and a submersion $g: U\to R^{N-n}$, such that $X \cap U=g^{-1}(0).$

And if $Y=R^{N}$, and suppose $0 \in X$, so by implicit function Thm, there is an open nbh $V$ of $0$ in $R^{n}$, such that $X \cap U$ is the set of points $x=(t_{1}, t_{2},...,t_{N})$ such that $(t_{1}, t_{2},...,t_{n}) \in V$ and $t_{n+i}=f_{i}(t_{1},t_{2},...,t_{n})$, $(1\leq j\leq N-n)$
Where $f_{i}$ are unique $C^{\infty}$ functions, which vanish togather with their first derivatives, at the origin.

For each point
$x=(t_{1}, t_{2},...,f_{1}(t_{1}, t_{2},...,t_{n}),...,f_{N-n}(t_{1}, t_{2},...,t_{n}))$
The tangent vector space of X at x, $T_{x}(X)$ satisfy the N-n linear equations: for a point $(p_{1},p_{2},...,p_{N}) \in T_{x}(X)$, (identify $T_{x}(X)$ as canonical subspace of $T_{x}(Y)$)
$p_{n+j}=\sum_{i=1}^{n}D_{i}(f_{j}(t_{1}, t_{2},...,t_{n})) \bullet p_{i}$,      $(1\leq j\leq N-n)$

($D_{i}$ is partial derivative with respect to i's coordinate, it has the usual meaning of a linear mapping)

My question is:
why $T_{x}(X)$ satisfy these linear equations?
( Do I treat $T_{x}(X)$ a submanifold of $T_{x}(Y)=R^{N}$, and use implicit function Thm again, but how come those unique functions are exactly
those linear equations?

[quote="loup blanc"]A $T_{x}(X)$ basis is $(e_{1},...,e_{n})$ where $e_{i}=\frac{\partial{x}}{\partial{t_{i}}}=(0,...,1,...,0,D_{i}(f_{1}),...,D_{i}(f_{N-n}))$.

Loup, you computation is simple and I think this is Dieudoone's book intended to do. However I have questions regarding to the above statement.

(1) I am confused with different definations of $T_{x}(X)$ basis in the literature. Are you using definations from Kobayashi's book "Foundations of Differencial Geometry"? He avoided vector differentials and He defined $T_{x}(X)$ having a basis of "partial defferative operators" at point x:
$e_{i}=\frac{\partial}{\partial{t_{i}}}$.How come you have $e_{i}=\frac{\partial{x}}{\partial{t_{i}}}$?

(2) Dieudonne's book (see my original post), defined $T_{x}(X)$ basis using vector differentials
(derivetives are linear mapping between Banach spaces (here is R^n)). I hope you read Dieudonne before, (since you from France  :) ). I feel intuitively it is same as Kobayashi's defination (which seems to be the convention nowadays), but couldn't figure out exactly.

[quote="loup blanc"]Hi Kalazealot
my "$e_{i}$" isn't the same as that of Kobayashi;$\frac{\partial}{\partial{t_{i}}}: x\rightarrow \frac{\partial{x}}{\partial{t_{i}}}$ is an operator which transforms a function into a directional derivative.
You can also write the following:
If $x(t_{1},\cdots,t_{n})=x(t)=(t_{1},\cdots,t_{N})$ then
   $T_{x(t)}(X) =((Dx)(t))(\mathbb{R}^{n})$ where $Dx$ is the $x$'s derivative:
$(Dx)(t): (p_{1},\cdots,p_{n})\rightarrow(p_{1},\cdots,p_{n},\sum_{i=1}^{n}D_{i}f_{1}(t).p_{i},\cdots)$
You obtain the same result (fortunately).[/quote]
Loup, thanks. Now I totally understand. This expression of x is just the conanical injection of this submanifold $X$ into $R^{N}$, call it $f$. So using Dieudonne's book: $T_{x}(f)=Df$, is the map from $T_{x}(X)\to T_{x}(R^{N})$.
Do you know how to show this by using Kobayashi's defination of tangent space? (bases of tangent space are partial differential operators of $R^{n}$,  not of $R^{N}$ ($dim(X) =n$).
I can't find in his book "canonical injection of submanifold into manifold"...

Here $X$ is a $n$-dimensional submanifold of $\mathbb{R}^{N}$ and $x: \mathbb{R}^{n}\rightarrow{Y}$ is a $X$'s parameterization. The Kobayashi's definition is the same than that of Dieudonné: a $T_{x}(X)$'s basis is given by the operators:  $\frac{\partial}{\partial{t_{i}}}$,$i=1,\cdots,n$.
  More generally $X$,a $n$-dimensional manifold, can be embedded in $\mathbb{R}^{2n+1}$ (Whitney); let $f$ be this embedding; $Y=f(X)$ is a submanifold of $\mathbb{R}^{2n+1}$;let $y=f(x)$; we know $T_{y}(Y)$ (see what precedes); let $\gamma$ a curve in $X$ s.t. $\gamma(0)=x$; $(f\circ\gamma)'(0)\in{T_{y}(Y)}$ and conversely we may define $T_{y}(Y)$ thus; moreover we may identify $T_{x}(X)$ and $T_{y}(Y)$.


let $n=2k$ for some $k\in\mathbb{Z^+}$.

if we fully expand the following, and combine the similar terms, what is the total number of remaining terms (in terms of n)?

$\left (a+(b+c)\right )^n+\left (a-(b+c)\right )^n$

(a,b,c are different variables)

um, you could apply the binomail theorem twice...


Let $a,b,c,d$ be nonnegative real numbers such that $a+b+c+d=4$. Prove that
\[(a^{2}-a+1)(b^{2}-b+1)(c^{2}-c+1)(d^{2}-d+1)\geq 1.\]

In VI Forum, I proposed the stronger inequality

http://www.ch06k.com/forum/showthread.php?t=188

My solution is only in 5 lines shortened.


Let $U_n$ be the set of n-th roots of unity and consider all possible sums of different elements of $U_n$. How many different numbers do we get?
   I really believe that this is monstruously ( if this word exists) difficult, so I would be glad to see even some estimations for this number.

It came immediately $2^n-1$ as an upper bound. For $p$-prime it came that the answer should be exactly $2^p-1$, otherwise if $n$ is composite the answer is strictly less than $2^n-1$... Indeed seems monstruosly...
By the way, do you know and use some specialised math software for "huge" computations? (like Mathematica or Maple - at least can someone supply a licence for me... :blush:  ?!?)

About Maple and others things, I have no idea. Grobber could help you, i think. Related to the problem, I have managed to prove that if $n=p^k$ then the cardinal is $ (2^p-1)^{p^{k-1}}$. I consider this already non-trivial. For the other values, no idea.


Let $ A\equal{}\{[n\sqrt{2}]\plus{}[n\sqrt{3}]|$ where $ n\equal{}1,2,3\dots\}$.Prove that $ A$ contains infinitely many odd and even integers.

$ a_n\equal{}[\sqrt{2}n]$
$ b_n\equal{}[\sqrt{3}n]$
$ a_{n\plus{}1}\minus{}a_n\in\{1,2\}$ and $ b_{n\plus{}1}\minus{}b_n\in \{1,2\}$
Suppose exist $ N$ such that for all $ n\geq N$ then $ a_n\plus{}b_n$ is an odd number .
Then we have $ a_{n\plus{}1}\minus{}a_n\equal{}b_{n\plus{}1}\minus{}b_{n},\forall n\geq N$
$ \Rightarrow b_{n\plus{}1}\minus{}a_{n\plus{}1}\equal{}b_n\minus{}a_n\equal{}...\equal{}b_{N}\minus{}a_{N}$
Therefore $ \sqrt{3}n\minus{}1\minus{}\sqrt{2}n<b_{N}\minus{}a_{N}$
It gives contradiction when n tends to infinite .
So this sequence contains infinite odd numbers .
Similar ,it contains infinite even numbers .


Will whoever ends up being the mod please sticky this and delete this paragraph?

Hey! Let's work on a list of linguistics-related competitions. Post here if you have one to add.

Of course the origin of this forum was about the NAMCLO: North American Computational Linguistics Olympiad. Its website may be found at namclo.org [suprise!]

[url]http://olympiads.win.tue.nl/ilo/index.html[/url]

That has links to the ILO, Moscow Olympiad, NAMCLO, St. Petersburg Olympiad, and Dutch Olympiad.

New forums may not need moderators; some of the arts/sciences forums still don't have them. But I really would like to moderate this, so the admins may make me moderator if they wish.

Heh, both of the users of the Linguistics forum want to be mods, if the admins wish, of course ;). [Though I don't care if I'm not that much.]

Could someone please explain exactly what linguistics are?  I get that they have something to do with words & logic  :roll:

Thanks.

As an example, try the problem I posted. It shouldn't be too hard. But keep in mind that problems like that are to linguistics as say number theory is to math.

As a different, more realistic example, historical linguistics deals with treating how languages evolve, and thus may be interested in the following correspondences:

Latin - Gothic (extinct language related to German and English)

pes - fotus (foot)
tertius - thridja (third)
canis - hunds (hound)
quod - hwa (what)
baculum - peg (English)
decem - taíhun (ten)
gelu - cold (English)
(Greek) φρατηρ - brothar (brother)
(Greek) θυρα - daúr (door)
(Greek) χην - goose (English)

This information (and of course lots more like it) led to Grimm's Law that says the following consonants in most European languages shift to the consonant to their right in Germanic languages:

bh-b-p-f
dh-d-t-th
gh-g-k-h
gwh-gw(gu)-kw(qu)-hw

Then if we examine the English word "eight" (Old English "eahta") and try to back-track to an earlier language (we cannot back-track directly to any language like Latin, because English and Latin are really only sisters in the same generation), we got "akda". But kd is impossible to pronounce, so "akta" is more likely. Indeed, Latin for eight is "octo".

Oh, I get it; nice explanation!

In my Latin class I learn things such as "The farmer watches the forest." and "Where are the big waves?"

Woah! I just realized that's like a Latin pun. Unde undae magnae sunt (absunt?)?

Where might I find this year's NAMCLO problems?

BTW, solafide, were you going to do some "vocab study"? I still feel pretty lonely on this forum.

Speaking of NAMCLO, when are the results coming out? For internet people? I took it and it was actually a lot of fun, even though I'd never done anything about linguistics before.

[quote="funcia"][quote="Boy Soprano II"][quote="jb05"]I still feel pretty lonely on this forum.[/quote]
Sorry, this week is insane for me.[/quote]
it is for everyone.  :(
Really good Latin pun![/quote]

i don't get it... where's the pun?

pun = play on words
unde undae = play on words

maybe pun was the wrong word...

@wii: Mr. Radev just said about 3 this afternoon.


If a, b, c are the sidelengths of a triangle, then prove the inequalities

$a^3+b^3+c^3+3abc-2b^2a-2c^2b-2a^2c\geq 0$

and

$3a^2b+3b^2c+3c^2a-3abc-2b^2a-2c^2b-2a^2c\geq 0$.

  Darij

Of corse everybody knows the well known inequality $(a+b-c)(b+c-a)(c+a-b)<abc$
By expanding we get : $a^3+b^3+c^3+3abc \geq a^2b+ab^2+b^2c+bc^2+c^2a+ca^2$
Then we can prove that $a^2b+ab^2+b^2c+bc^2+c^2a+ca^2 \geq 2ab^2+2bc^2+2ca^2$
which is equivalent to $ab(a-b)+bc(b-c)+ca(c-a) \geq 0$
And we are done  :lol:

I think you are wrong, the last inequality cannot be valid always. By the way, something tells me that the old Vasc technique works very well for this case of inequalities, of course after the substitution $a=y+z$,..., but I'm too lazy to make the computations! :D  :D

Yes, Erdos' solution in post #2 is wrong. Actually, I have tried to derive the inequalities from some basic abc inequalities like Schur, AM-GM or $\left(b+c-a\right)\left(c-a\right)^2+\left(c+a-b\right)\left(a-b\right)^2+\left(a+b-c\right)\left(b-c\right)^2\geq 0$ (the latter is to use the condition that a, b, c are sides of a triangle), but didn't succeed. I include my original proofs of the inequalities into the spoiler below. I hope that somebody will find better proofs.

[hide="Proofs of the inequalities"][quote="darij grinberg"]If a, b, c are the sidelengths of a triangle, then prove the inequalities

$a^3+b^3+c^3+3abc-2b^2a-2c^2b-2a^2c\geq 0$

and

$3a^2b+3b^2c+3c^2a-3abc-2b^2a-2c^2b-2a^2c\geq 0$.[/quote]

[i]Proofs.[/i] We start by showing the first inequality:

$a^3+b^3+c^3+3abc-2b^2a-2c^2b-2a^2c\geq 0$.

Since this inequality is cyclic, we can WLOG assume that $a\geq b$. Then, $a-b\geq 0$. Also, b + c - a > 0 (since b + c > a, because a, b, c are the sidelengths of a triangle), and $a^2+b^2+c^2-bc-ca-ab\geq 0$ (since

$a^2+b^2+c^2-bc-ca-ab=\frac12\left(\left(b-c\right)^2+\left(c-a\right)^2+\left(a-b\right)^2\right)$).

Thus,

$a^3+b^3+c^3+3abc-2b^2a-2c^2b-2a^2c$
$=\left(b+c-a\right)\left(a^2+b^2+c^2-bc-ca-ab\right)+2\left(a-b\right)\left(c-a\right)^2\geq 0$,

and the first inequality is proven.

The second inequality is harder. We start with a lemma:

[color=blue][b]Lemma 1.[/b] If x and y are two nonnegative reals, then $x^3+y^3+2x^2y-3xy^2\geq 0$.[/color]

[i]Proof of Lemma 1.[/i] If $y\leq x$, then $x-y\geq 0$, and thus $x^3+y^3+2x^2y-3xy^2=x\left(x-y\right)\left(x+3y\right)+y^3\geq 0$.

If $x\leq y\leq 2x$, then $y-x\geq 0$ and $2x-y\geq 0$, so that $x^3+y^3+2x^2y-3xy^2=\left(y-x\right)^3+\left(2x-y\right)x^2\geq 0$.

Finally, if $2x\leq y$, then $y-2x\geq 0$ and thus also $y-x\geq 0$ (since y - x = (y - 2x) + x), so that $x^3+y^3+2x^2y-3xy^2=x^3+y\left(y-x\right)\left(y-2x\right)\geq 0$.

Hence, we get $x^3+y^3+2x^2y-3xy^2\geq 0$ in every case, and Lemma 1 is proven.

Now we are ready to establish the second inequality

$3a^2b+3b^2c+3c^2a-3abc-2b^2a-2c^2b-2a^2c\geq 0$.

In fact, since this inequality is cyclic, we can WLOG assume that a is the greatest among the numbers a, b, c. Then, $a\geq c$ and $a\geq b$, so that the numbers a - c and a - b are nonnegative. Hence, we can apply Lemma 1 to these numbers a - c and a - b, and we get

$\left(a-c\right)^3+\left(a-b\right)^3+2\left(a-c\right)^2\left(a-b\right)-3\left(a-c\right)\left(a-b\right)^2\geq 0$.

Also, as we saw above, b + c - a > 0 and $a^2+b^2+c^2-bc-ca-ab\geq 0$.

Thus,

$3a^2b+3b^2c+3c^2a-3abc-2b^2a-2c^2b-2a^2c$
$=\left(b+c-a\right)\left(a^2+b^2+c^2-bc-ca-ab\right)+\left(\left(a-c\right)^3+\left(a-b\right)^3+2\left(a-c\right)^2\left(a-b\right)-3\left(a-c\right)\left(a-b\right)^2\right)$
$\geq 0$.

This proves the second inequality. $\blacksquare$ [/hide]

  Darij

can we prove by let $a=x+y,b=y+z,c=z+x$?

we can prove this in following way

$3a^2b+3b^2c+3c^2a-3abc-2b^2a-2c^2b-2a^2c\geq 0$.

$case$ $1$ $a\geq b\geq c$

$\Leftrightarrow 2(a-b)(b-c)(a-c) + b(a-b)(a-c)+ a(b-c)^2 \geq 0$

$Q.E.D.$

$case$ $2$ $a\geq c\geq b$

$\Leftrightarrow (a-c)(ac+b^2-2bc)+a(c-b)(a-b) \geq 0$

this is true cuz

$a(c-b)(a-b)\geq 0$ and $ ac+ b^2 - 2bc= c(a-b) - b (c-b) $

Here, $|c| \geq |b|$ and $ |a-b| \geq |c-b| $

$Q.E.D.$

[quote="zhaobin"]can we prove by let $ a \equal{} x\plus{}y,b \equal{} y\plus{}z,c \equal{} z\plus{}x$?[/quote]

Of course  :D
And it is very easy ...The first inequality is equivalent to $ 2(xy^{2}\plus{}yz^{2}\plus{}zx^{2})\geq 6xyz$ which is a plain AM-GM

and the second is equivalent to
$ x^{3}\plus{}2xy^{2}\plus{}y^{3}\plus{}2yz^{2}\plus{}z^{3}\plus{}2zx^{2}\geq 3(x^{2}y\plus{}y^{2}z\plus{}z^{2}x)$

which I think is a well known one

Well, according to the theorem shown here

http://www.mathlinks.ro/Forum/viewtopic.php?t=148928

It suffices to consider the case $ a \equal{} b\plus{}c$.

Moreover, for the general problem

Let $ a,b,c$ be three side-lengths of a triangle. The following inequality is true
\[ r(a^{3}\plus{}b^{3}\plus{}c^{3})\plus{}k(a^{2}b\plus{}b^{2}c\plus{}c^{2}a)\plus{}l(ab^{2}\plus{}bc^{2}\plus{}ca^{2})\plus{}sabc\ge 0,\]
if and only if it's true for $ a \equal{} b\plus{}c$.

(r,k,l,s may be negative)


$a,b,c$ are positive real numbers. Prove that :

$\frac{a^{2}+bc}{b+c}+\frac{b^{2}+ac}{a+c}+\frac{c^{2}+ab}{a+b}\geq a+b+c$

Just vonicular schur !  :D
$L-R = \sum \frac{(a-b)(a-c)}{b+c}$

[quote="vietnam2007"]What does it say [b]vonicular schur[/b] ? because I know only [b]shur inequality[/b].[/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?t=49427
Vonicular Schur is a very strong inequality !! :)

Thank you :)  but it's called vornicu-shur inequality  :P

a,b,c are positive real numbers. prove that
$ \frac{a^{2}\plus{}bc}{b\plus{}c}\plus{}\frac{b^{2}\plus{}ac}{a\plus{}c}\plus{}\frac{c^{2}\plus{}ab}{a\plus{}b}\geq a\plus{}b\plus{}c$

it's equal to
$ \frac{a^{2}\plus{}bc}{b\plus{}c}\plus{}\frac{b^{2}\plus{}ac}{a\plus{}c}\plus{}\frac{c^{2}\plus{}ab}{a\plus{}b}\minus{}a\minus{}b\minus{}c \equal{} \frac{(a^{2}\minus{}b^{2})^{2}\plus{}(b^{2}\minus{}c^{2})^{2}\plus{}(c^{2}\minus{}a^{2})^{2}}{2(a\plus{}b)(b\plus{}c)(c\plus{}a)}\geq 0$

oh very easy problem:
$ \sum_{}^{} \frac {a^2}{b \plus{} c} \plus{} \sum_{}^{} \frac {ab}{a \plus{} b} \ge \sum_{}^{}a$   $ \Leftrightarrow$
$ \sum_{}^{} \frac {a^2}{b \plus{} c} \plus{} \sum_{}^{} \frac {ab}{a \plus{} b} \plus{} \sum_{}^{} \frac {a^2}{a \plus{} b} \ge \sum_{}^{} a \plus{} \sum_{}^{} \frac {a^2}{a \plus{} b}$ $ \Leftrightarrow$
$ \sum_{}^{} \frac {a^2}{b \plus{} c} \ge \sum_{}^{}  \frac {a^2}{a \plus{} b}$ which is obvious


Has anyone here attended a boarding school similar to Cate School or one of the Phillips Academy's? What were the classes like? Did a majority of the student population care about their education? Overall, was it a good experience?

I would like to know the answer to this question too. My son is applying to [url=http://www.exeter.edu]Phillips Exeter Academy[/url] and to [url=http://www.andover.edu]Phillips Academy[/url] (Andover, MA) this year for admission as a ninth grader ("junior") next year. AoPS moderator mathfanatic went to Exeter last year--his [url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=1320]formerly sky-high post count[/url] has hardly been added to since he started school there, so I guess he is quite busy during the school year.

Yeah, I'd like to hear from anyone attending any of schools mentioned in this thread.

i'm currently applying to andover and exeter for 11th grade

Hey I'm a 4-year senior at Exeter.  It's intense, but I think we all grow from the experience.  No regrets =)

I'm happy to answer questions until we go back to school in a week.  You can reply here, or email slcampbell@exeter.edu.  Good luck with apps, everyone!

Hi, saracampbellsoup, you're on the Exeter math team, right? How has that been as an experience? What are some of the regional events the team does besides the national competitions?

People at math club are hard-working, but really chill, and we have fun.  I think the Harkness method we use in class also carries into math club.  Much of what we learn is from each other, so we work well in team competitions, because we're used to doing math collaboratively.  We keep each other motivated and we meet up outside of math club to work on problems together.  We send teams to HMMT and ARML.

I thought I'd revive the thread to ask if any other AoPSers are planning to apply to boarding school this year?

Hi tokenadult,

It's certainly been a while, and as you rightly guessed, Exeter has kept me pretty busy.  With respect to math club, I agree with everything Sara said.  We do have a good group of students who are talented at solving problems but who also get along well with each other.  Also, I think students have the chance to do what is normally regarded as "competition math" in math class, especially in T3X, a math course for the brightest of the new 9th and 10th graders.  You should ask about it when your son applies, and do e-mail Mr. Feng (his e-mail is listed the [url=http://www.unl.edu/amc/c-com,spon,dir/usamoc.html]AMC website[/url]) since he teaches the course.

Exeter, however, is not just about math.  The Academy is strong in many areas--I think history and classical languages are usually considered very rigorous, but all the departments are pretty solid.  And then there are all the extracurriculars, from our school newspaper, The Exonian (which I'm partial to since I am involved in it), to more esoteric activities, like fencing club.  Your son (and anyone applying to Exeter) should keep in mind that math will only take up a small part of his time at Exeter.

Anyway, if you have any questions, feel free to e-mail me (I think my e-mail is in my profile).  I'll be on winter break starting next thursday, so I'll be able to get back to you.

Keone

Hi, mathfanatic,

It's good to see you posting here again. My son got a very nice email reply from Mr. Feng a while ago, so we have been communicating about the Exeter math team. Another big extracurricular interest of my son's is soccer, but I have no idea where his skills will rank among the students vying for various prep school varsity, junior varsity, and other soccer teams.

It may be that I first discovered [url=http://math.exeter.edu/dept/materials/index.html]Exeter's phenomenal math problem sets[/url] back when you were thinking out loud about applying. Those are indeed one really good reason to apply to Exeter: there seem to be few schools in the English-speaking world that teach secondary mathematics in quite that way. (That approach is routine for gifted math students in China, but we're not going there in the next few years.) What I particularly like about a curriculum such as Exeter's, as a foreign language major, is the expectation that students take foreign languages throughout high school. Here in Minnesota it is remarkably hard to find a high school that even offers German anymore, much less Chinese and Japanese. And the history and English classes appear to involve rather more reading than most schools here expect, so that is all good too.

Would anyone care to comment on what the science courses are like at Big Red? The Web page with course descriptions mentions what topics are covered by each course, but without details about textbooks used, methods of assessment, and depth expected for each topic. That is one of the areas I would like to know more about as we continue the application process and try to figure out what to do for next year.

I heard of somebody who got mail from Phillips Academy (Andover) based on some kind of recruitment criterion like that. We applied "cold" after a friend of ours told us that anyone who is applying to Exeter ought to apply to Andover too. As you can imagine, that friend is a current Andover student.   :P  It will be interesting to see how the applications turn out. Our interview trip to New England comes in two weeks.

from the start my only goal is andover :D , and i only requested andover package and info. but after reading them and hearing about exeter, i decide i would just try exeter. just thrown in an application u know.:P

[quote="log0"]Has anyone been getting letters from Phillips Academy telling them to apply? The letter said that they saw AMC scores blah blah blah and sent me a whole packet of info.[/quote]
yea, a couple of people in my high school got a letter from Exeter about applying for 12th or even 13th grade, but they just decided to stick with the high school and worry about college applications instead. i guess a boarding school seems like a very strict environment, my friend from [url=http://www.ncssm.edu/]North Carolina school for science and math[/url] told me that she feels it was a lot harder than MIT, but if that's what you want, go for it :)

any know how you send the optional parent statment (recommendation form) for exeter to the school? they didn't provide an envelope.

do i just send it in a plain one? thanks :lol:

im not poor enough to get a scholarship for a boarding school, but im not rich enough to afford one either  :mad:  so in the end im completely dependent on self study and my local college

Some local parents here are lining up some really cool joint classes for their children (homeschoolers), to which my son is invited, so we are losing interest in boarding school here. It all depends on what your other plan is, for sure.

About science at Exeter...

I’ve liked my science courses.  The textbooks are… standard?  The department is up to date on what books are out there and takes student feedback very seriously.  We complained about our chem textbook, so they changed it haha.

For example, AP physics uses this for mechanics:

[url=http://www.amazon.com/gp/product/0534408427]Physics for Scientists and Engineers[/url]

Science grades depend on tests, labs, homework, and projects.  If you work hard and do well on everything else, a few bad tests probably won’t mess up your grade.  And teachers love to give extra help.  I found a huge jump in the level of depth from 8th grade public school to 9th grade at Exeter.  I think most of us get 5’s on AP exams, because we’re used to doing harder things in class.  Also, our labs have lots of cool stuff!


Does anyone have any information about this?

How does DNA account for errors in transmittance?

There are DNA repair enzymes that spot errors in the DNA sequence and repairs them. I forget what they are called though.

The cell will also produce more cell cycle inhibitors so that the error doesn't manifest itself all over the place. If the problem can't be solved, apoptosis will occur.

I just finished studying chapter 16 in Campbell. Apparently DNA Polymerase self-corrects errors it makes.

However, Campbell says that "changes in DNA are usually corrected before they become self-perpetuating mutations. Each cell continuously monitors and repairs its genetic material". How does it know when its genetic material needs repair? Campbell talks about the methods of repair but doesn't go into details about how those errors are detected.

Do any of you know of sources/articles about this?

I believe one of the examples in Campbell's is a thymine dimer...not complete sure if I remember correctly. That would be an example of "correcting" DNA, although it's not really an error in the code. Perhaps it's something like this?

There are several methods for error correction/minimization in DNA.  The posters above mentioned a few.  Here are a few more.

As you know, DNA is composed of 4 nucleotide bases.  A pairs with T and G with C.   These bases have distinct chemical properties and distinctive 3 dimensional  shapes.  During DNA replication, structure of paired bases is important.  An AT pair is 3 dimensionally distinquishable from an erroneously transcribed AG pair.  We have enzymes capable of recognizing these differences and excising the erroneously paired base.  The bond strength between an A and T is also different than the bond strenght of a GC base pair.  This means that transcription of A or T (particularly long chains of AT repeats) is more error prone than transcription of G or C.  It also means it's harder the melt a DNA strand containing a high GC content.

Genomes also have various ratios of coding vs noncoding base pairs.  While non-coding DNA has many purposes (priming replication, preserving 3 dimensional DNA structure, driving geneticists crazy) a mutation in a non-coding region may be less deleterious than a mutation in a coding region.

There is some level of redundancy in our genes.  There is also redundancy in the genetic code itself.  DNA is transcribed in nucleotide triplets.  Three nucleotides code for an amino acid (building blocks of proteins).  So while GCA codes for Alanine, so does GCG.  In fact the third nucleotide in a triplet is often refered to as the "wobble position".  Be careful regarding this, however.  Just becaue GCA and GCG code for the same thing doesn't mean a mutation from an A to a G is silent.  A's and G's are different.  They have different binding affinities and may influence the way DNA is supercoiled even if they don't alter the end protein.   Supercoiling affects how accessible a gene is to transcription.

You also have to remember that some coding areas are more important than others.  Maybe a DNA mutation does result in an amino acid change.  The three dimensional location of this change in the completed protein is important as is the amino acid change itself.  Is the new amino acid on a binding surface?  Is the new amino acid a different charge?  Is is bulkier?  Does it change the way the protein is folded?

If you're looking to study coding theory as relates to DNA you can try something like this: http://www.liebertonline.com/doi/abs/10.1089%2F10665270152530818 (not that I'v read the paper-- it was the first marginally relevant one that popped up in my search)

If you're interested in a more detailed discussion of the molecular mechanisms, I suggest you pick up a molecular cell biology text.


AwesomeMath
summer program
Mathcamp
FTW
search
ARML
Duke

awesomemath is a mathcamp that helps you improve ur math skillzzz  for more information, please go to,

awesomemath.org

Only if you aren't looking to increase your mathematical problem solving skills.

[quote="modularmarc101"]Is CTY at Hopkins better or Awesomemath?[/quote]

depends on what you want to do/improve...
my suggestion:
awesomemath ftw!

I did CTY, and it wasn't very tough...

I'll look into awesomemath

oh okay, yea cuz they sent me a broshure and i was like nahhh cty must be better but I guess its not..If only i could find where i put the broshure.... :ninja:

[quote="pinkmuskrat"]Sorry, what's CTY?[/quote]
Center for Talented Youth

CTY offers tests such as the SAT and ACT, but also offers a VERY difficult test for the really and truly smart kids. This test, the spacial battery test I believe(I don't completely remember the name), is extremely challenging and tests the brain(some math involved).

The way CTY works is if you do well on the CTY test that you take, you get to take a special CTY course which is quite helpful. The course is very good, but getting the course is a bit more difficult. Plus the class is not very costly as you have to do well to get in.[/b]

@batteredbutnotdefeated & pinkmuskrat: There's this really helpful thing called a search engine...
http://tinyurl.com/qur4wp
http://tinyurl.com/qp9pwr

I've been to CTY but not Awesomemath, but I do know CTY doesn't really focus on problem-solving math that much, and as 1=2 said, if you only care about improving your problem solving skills, go to Awesomemath (I mean, just look at the instructors there).

As far as I recall, the spatial battery test was more about rotating and memorizing images in your brain and didn't really have that much math at all? On their website [url=http://cty.jhu.edu/ts/stbtiming.html]here[/url], the test components don't really look like they involve math.

nice try andersonw...

I don't get rickrolled that easily! :P

Um, neither of them are rickrolls.

The logic "I saw a tinyurl which was a rickroll so all tinyurls are rickrolls" is flawed.

Maybe some people have finally started to become suspicious about rick rolls because of some people though.

CTY classes don't cost above 100 bucks, I believe... That is if you score high enough...

[quote="not_trig"][quote="1=2"][url=losethegame.com]Gamerolls[/url] are so much more fun.

Actually I think I'm pretty standard when it comes to reading and writing, and somehow I qualified for the reading talent search. The only subject that I see myself talented in is mathematics. IMHO it probably isn't that hard to qualify for CTY talent search.
I'm not talking about the STB, that's another thing altogether.

AwesomeMath gets you to the harder competitions: ARML, IMO, top person at PUMaC, etc. CTY just teaches you new stuff that you're likely to forget (Case in Point: me and the hardest class my local college had to offer; Game Theory). And the classes at CTY are dirt expensive (oxymoron and 3 grand a class), while AwesomeMath does a few hundred dollars more, but that includes room, board, transportation to/from airport, and any trips. Last year all but two people (alexhhmun and I) went to Six Flags for a day. That is EXPENSIVE. CTY has you overstimulated enclosed in a campus (at least that's my experience).

Overall I would suggest you to go to AwesomeMath.[/quote]

awesomemath is actually quite successful at producing moppers (yay zuming)

let's see... from my year, AMP '06, there were 8 '07 red MOPpers and 2 '07 blue MOPpers; the next year, 3 blue 1 red (at least; I don't remember completely :/?), and this year, 1 black 3 blue at least. so yeah. (the 3 blue last year made MOP again XD and they were all on our team contest team. yeahhh.)[/quote]

This year, MOP produced (at least) 2 black and 10 red, though one of the black moppers was a counselor.
Throwing in my opinion, regardless of the cost, AwesomeMath is an incredible experience and you learn an incredible amount of math from it. I went to Duke TIP (the counterpart of CTY) and in my opinion, you will learn a lot more and benefit more, have a better experience at, and meet more people in a positive environment at AwesomeMath.


Solve in reals:
yz=3y+2z-8
zx=4z+3x-8
xy=2x+y-1

hello, your first equation gives $ z\equal{}\frac{3y\minus{}8}{y\minus{}2}$, by substituting this in equation (2) we get
$ \frac{x(3y\minus{}8)}{y\minus{}2}\equal{}\frac{4(3y\minus{}8)}{y\minus{}2}\plus{}3x\minus{}8$ (4). From (3) we get $ y\equal{}\frac{2x\minus{}1}{x\minus{}1}$, inserting this in (4) we have $ \frac{\left(3(\frac{2x\minus{}1}{x\minus{}1})\minus{}8\right)x}{\frac{2x\minus{}1}{x\minus{}1}\minus{}2}\equal{}\frac{4\left(\frac{3(2x\minus{}1)}{x\minus{}1}\minus{}8\right)}{\frac{2x\minus{}1}{x\minus{}1}\minus{}2}\plus{}3x\minus{}8$, simplifying this we get $ \minus{}2(x\minus{}2)(x\minus{}3)\equal{}0$ and so we find our solution with $ x_1\equal{}2,y_1\equal{}3,z_1\equal{}1$ or $ x_2\equal{}3,y_2\equal{}\frac{5}{2},z_2\equal{}\minus{}1$.
remark: if we assume that $ y\equal{}2$ we get the contradiction $ 0\equal{}\minus{}2$ in the first eqaution of our system, if we assume $ x\equal{}1$ we get $ 0\equal{}1$ in your third equation.
Sonnhard.

Systems $ \Longleftrightarrow$ $ \begin {cases} y\equal{}\frac{2z\minus{}8}{z\minus{}3}\\x\equal{}\frac{4z\minus{}8}{z\minus{}3}\\ xy\equal{}2x\plus{}y\minus{}1 \end {cases}$

$ \Longleftrightarrow$ $ \begin {cases} y\equal{}\frac{2z\minus{}8}{z\minus{}3}\\x\equal{}\frac{4z\minus{}8}{z\minus{}3}\\ \frac{8(z\minus{}2)(z\minus{}4)}{(z\minus{}3)^2}\equal{}\frac{8(z\minus{}2)}{z\minus{}3}\plus{}\frac{2(z\minus{}4)}{z\minus{}3}\minus{}1 \end {cases}$

$ \Longleftrightarrow$ $ \begin {cases} y\equal{}\frac{2z\minus{}8}{z\minus{}3}\\x\equal{}\frac{4z\minus{}8}{z\minus{}3}\\ 8(z\minus{}2)(z\minus{}4)\equal{}8(z\minus{}2)(z\minus{}3)\plus{}2(z\minus{}3)(z\minus{}4)\minus{}(z\minus{}3)^2\end {cases}$

$ \Longleftrightarrow$ $ \begin {cases} y\equal{}\frac{2z\minus{}8}{z\minus{}3}\\x\equal{}\frac{4z\minus{}8}{z\minus{}3}\\ z^2\minus{}1\equal{}0\end {cases}$

$ \Longleftrightarrow$ $ \begin {cases} y\equal{}\frac{2z\minus{}8}{z\minus{}3}\\x\equal{}\frac{4z\minus{}8}{z\minus{}3}\\ z\equal{}1\end {cases}$ OR $ \begin {cases} y\equal{}\frac{2z\minus{}8}{z\minus{}3}\\x\equal{}\frac{4z\minus{}8}{z\minus{}3}\\ z\equal{}\minus{}1\end {cases}$

$ \Longleftrightarrow$ $ \begin {cases} x\equal{}2\\y\equal{}3\\ z\equal{}1\end {cases}$ OR $ \begin {cases} x\equal{}3\\y\equal{}\frac{5}{2}\\ z\equal{}\minus{}1\end {cases}$

Thank you very much!

You may try these systems if you are interested:

http://www.mathlinks.ro/viewtopic.php?t=182073