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in class today, we learned how to solve for 2 unknowns.I was just wondering how you would solve for 3 unknowns.
thanks

In the strictest algebraic sense, you need 2 equations to solve for 2 variables. Similarily, you need 3 equations for 3 variables, and so on. Some systems one can solve that have more variables than equations, but you usually need number theory tactics for that.

Are you talking about Systems of equations?

If you have 3 unknowns, x, y and z, get the equations in standard form:

$ Ax + By + Cz = D \newline Ex + Fy + Gz = H \newline Ix + Jy + Kz = L \newline$

The capital letters are constants.

Take the coefficents and put them in a 3x3 matrix like so:
\[ P = \[ \begin{pmatrix} A & B & C \\
E & F & G \\
I & J & K \end{pmatrix}
\]
Take what each equation equals and put them in a 3x1 matrix like so:

\[ \begin{pmatrix} D \\
H \\
L \end{pmatrix}
\]

Then multiply the first matrix inverse by the second like so:

$ P^{( - 1)} \cdot Q$

The result will be a 3x1 matrix with the values of each variable

\[ \begin{pmatrix} x \\
y \\
z \end{pmatrix}
\]

whoa.. you may want to fix up your latex.

You might wanna fix the LaTeX  :wink: (EDIT: darn buzzer got it).

But to do a 3x3 system of equations on a calculator just input all the coefficients in a 3x4 matrix (yes it is a 3x[b]4[/b] matrix). Simply find "rref( " under the Matrix menu and input the matrix name. This will give you what each variable equals on the last column.

Yea, sorry about the latex, but I cant figure out what the heck is wrong with it.

[color=darkblue]Moderator Edit: I've fixed your LaTeX; you shouldn't couple $ \$$ tags with the center tags.[/color]

Here are a few ways to solve a $ 3^{rd}$-order system of equations:
Algebraically
Gaussian Elimination
Gauss-Jordan Elimination
Matrix Equation
Cramer's Rule

Post another message if you want me to explain any of these methods.

how is gaussian elimination different from linear combination, or are they the same thing with different names. if they are they same, i really wouldnt suggest it unless you have something factorable thats obvious like a+b=4 b+c=5 a+c=6. it might not work. i just kinda made those numbers on the spot

[url=http://mathworld.wolfram.com/GaussianElimination.html]Gaussian elimination[/url] is specifically done in matrix notation to make it more compact, but it's not really different.  And you don't mean "factorable"; you probably mean "symmetric."


If the price of an article is increased by the fraction $p$, where $p$ is inbetween 0 and 1, then the corresponding decrease in sales must not exceed the fraction $d$ of the original sales, again, $d$ is inbetween 0 and 1. In order to yield the same income, what is the value of $d$ in terms of $p$?

[hide] err... isn't it just 1/p?[/hide]

edit: wait ignore that i misread it

[hide]so if the item's original price is a, then it is increased to $a(1+p)$, right?  and if the original amount of sales is b, then the amount of sales becomes $bd$.  so the original income is just multiplying the price by the amount of sales, so it is $ab$.  and the new income is $a(1+p)bd$.  setting these equal, we get
$ab=a(1+p)bd$

$1=(1+p)d$ because a and b don't equal 0

$d=\frac{1}{1+p}$[/hide]


Let $J$ and $I$ be respectively the Gergonne and incenter of $ABC$. Show that
  $\frac{JA}{IA} + \frac{JB}{IB} + \frac{JC}{IC} < 4$

Excuse me , what is Gergonne point, $treegoner$?

The Gergonne point $J$ is the concurrent point of $AA', BB', CC'$ where $A', B',C'$ are points of tangency of the incircles the sides $BC, CA, AB$.

[hide=" Solution"]

We have $\frac{AJ}{AA'} \geq \frac{AJ}{AI+r} = \frac{1}{1+sin \frac{A}{2}}.\frac{AJ}{AI} > \frac{1}{2}. \frac{AJ}{AI}$.

Hence $\frac{AJ}{AI} + \frac{BJ}{BI} + \frac{CJ}{CI} < 2(\frac{AJ}{AA'} + \frac{BJ}{BB'}+ \frac{CJ}{CC'}) = 4$
[/hide]


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[b]
Express the $ \frac {sin(x)}{x} \equal{} \prod\limits_{i \equal{} 1}^\infty {(1 \minus{} \frac {{x^2 }}{{{i^2}\pi ^2 }})}$ , where $ x$ not equals $ 0$.

Thanx. :)
[/b]

This is a complex analysis problem. Essentially you just need to prove the product converges and then apply Hadamard's Factorization Theorem.

thx Xevarion ,

I just stuck to prove the product is converges ...And I see you think like me about it.

Could you help me plz to prove its convergent.

Thx again  :)

I wasn't having much luck finding good references on this site - my (somewhat incomplete) remarks [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2075729361&t=175841]here[/url] are a place to start.

There must be other places where this principle has been discussed.

Thx Kent :

I'm gonna try to take good informations carry the helpful for me.{from the link you put}

But I'm still looking to ....

Thx Guys


[quote="cdymdcool"]I was wondering if any of the stanford's EPGY math courses will help me prepare for the AIME?  I was about to take aops classes, but it doesn't give a grade so my school won't accept it....[/quote]

What do you mean your school won't accept it? That doesn't make sense to me.

Im guessing for school credit?

But dont worry about that.  Im pretty sure AoPS classes will prepare you a lot better than Stanford's EPGY classes.  I always thought that program was for more non-competition math.

Actually, there is a way to arrange with the instructors of an AoPS class to grade you to attempt to get school credit. (You'd have to arrange this with your school as well.) I think you have to inform the instructors of the class beforehand so you can be evaluated on your performance. (And of course, you must inform your school and get confirmation.)

EPGY has classes explicitly geared toward AMC prep (e.g.: http://epgy.stanford.edu/courses/math/MC12/). I've never taken it, but I'm sure it would help you with AIME. Still, I'd rather take an AoPS class.

Also, I'm not sure why your school would award credit for a math competition prep class -- if you took private piano lessons to help you in a music competition, would they award credit for that?

P.S.: I was just thinking of how, if you responded something along the lines of "Math is more serious and academic-y," I'd respond "But math competitions are very much play. It may be brainy play that's taken very seriously, but so's chess." I then remembered that my local high school actually did offer credit for a summer chess class (but only for people who did some project on how chess is "The game of life" in some way).

Math is generally considered an academic subject. Chess is not - that's the difference.

[quote="rrusczyk"]We can provide a grade if your school needs it.  If a school administrator needs to talk to us about our curriculum, we are happy to do so.

I believe our classes are much more highly designed for problem solving that EPGY's.  For example, there are many AMC/AIME/Mandelbrot/HMMT problems in our books.  I don't think you'll find that in the EPGY curriculum.[/quote]

Mr. Rusczyk,

Do you know who should my academic dean contact if he wants to talk to aops instructors?

Personally, I think that a school should not only accept an AoPS class as a grade but give extra credit as well to any student passionate enough to go beyond "school math."

[quote="hunter34"]Having taking one of the EPGY AMC classes. What it does is assign a certain AMC test, then go over it in the class, and going over a bit of ideas. This differs from AoPS's AMC class because  AoPS focuses on a specific idea, then gives AMC problems on it.[/quote]

How would you compare it to an AoPS class?


algebra
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induction
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Let n be a positive integer. Starting with the sequence 1, 1/2, 1/3, ..., 1/n, form a new sequence of n-1 entries 3/4, 5/12, ..., (2n-1)/(2n(2n-1)), by taking the averates of two consecutive entries in the first sequence. Repeat the averaging of neighbors on the second sequence to obtain a third sequence of n-2 entries and continue until the final sequence produced consists of a single number x_n. Show that x_n<2/n.

This one annoyed me immensely -- at first, I found xn but couldn't show it was less than 2/n.  Then, by the time I figured that out, I didn't have enough time to write up an explanation of the value of xn.

This was my favorite question of the bunch, possibly because it was one of the few that I actually solved. x_n=2/n-1/(n*2^(n-1)). It works out nicely with binomial coefficients.

I meant I found the summation form for it . . . which ones do you think you got?  I'm feeling confident about A1, A2, B1, B2, B4, and B5.  I also submitted complete nonsense for A3 and A6 -- interestingly, the answer for A6 ends up looking very much like a splitting of things mod 12, so I spent a whole lot of time writing that up only to find later that I was completely off-track.

Oh wow, you got a lot of questions. I wanted to have done that well. I only got A1, B2, and B3.

A6 is really awesome. Some of the UCSB people wanted to spend the whole 3 hours on A6 because it was such a neat problem. The solution is painfully simple too.

I was sitting next to Kaloyan somebody at lunch, between the halves of the Putnam.  He was on the Bulgarian IMO team this past year, getting a 41 out of 42.  We had just been talking about how the answer makes very much sense after someone has told you what it is, but I kept talking about how I had no idea how it would pop into anyone's head.  So we gave the problem to Kaloyan, and he looked thoughtful, and within two seconds said, [hide]"Well, what if you take the numbers with odd ones in binary expansion . . ."[/hide] (don't read the spoiler if you want to try the problem).  I didn't stop laughing for ages.  (Aparently, he had seen questions much like this before.)

Which one was B3?  I don't remember it, just that it looked hard and I didn't really try to do it very much as a result.  And I spent a lot of time on brute-force solutions to 5 and 1.  Is there a nice way to do B1?

The question B1, by the way, for those of you interested:

If a, b, c, and d are polynomial functions, is it possible to write

1 + xy + x2y2 = a(x)c(y) + b(x)d(y).  I could only do it with massive equation-bashing.

That's incredible! I really don't know why I didn't think of it. [hide]After all, when we are breaking a set down into two sets, shouldn't binary be an obvious thing to consider?[/hide] (Again, don't look at the spoiler if you want to solve the problem.)

B3 was the product for n!. I knew I'd be able to do it, even if it was going to be really messy. (It wasn't particularly nice.) I wanted to use the partition diagram thing for it, but I couldn't remember if it was called the Ferrers Diagram or the Ferrars Diagram, so I just wrote around it. (I looked it up on MathWorld after, and it turns out both are acceptable.)

I know of no nice solution to B1. The ``nicest'' solution anyone has given me involves solving nine simultaneous equations. I thought I would need twelve.

Yeah, I used 9 -- I thought it highly unpleasant, but I had some time left and I didn't think I was going to get anything else, so I  just kept writing it out.  Blech.  Oh, the product for n! -- I started writing something and then realized I had no idea what I was talking about.  How did you go about it?  It looked like one might want to apply induction, but I wasn't sure how.

I just looked at each prime p <= n and counted the number of factors of p on each side. They agree. They agree because if you write it out as a partition and make it into the Ferrers Diagram, then you interchange rows and columns, then you get another partition. They count the factors in that sort of way.

Okay, having no idea what you're talking about, I shall go look up this Ferrer's Diagram.

By the way, as someone who [i]never[/i] would have thought of the answer to A6, I'm going to ignore your earlier comment about it :).

You definitely could have gotten B4, by the way -- why didn't you go for it?

I thought B6 was interesting -- if it were a number 2, I would have thought I could do it without too much trouble, but as a number 6 I didn't even think about it.

Hey cool, I know Kaloyan! I think I met him initially in Russia when I went to a tournament of the towns conference thingy, then again a few times at IMO. Real cool guy.

Yeah, I really don't know why I didn't do B4. Perhaps it was because I was so sure B1 had a simple and elegant solution that I would find.

I would never have thought of that solution for A6 either. That doesn't mean that I shouldn't have thought of it though.

I think after my analysis final on Tuesday, or possibly tomorrow, I'll ask my professor if he can show me how to solve B6. I should turn it into measurable functions though.

I don't know how anyone could come up with A6 immediately, but I think there is a systematic approach available to us mere mortals.  Namely, put 0 in A, which forces 1 to go into a certain set, which forces 2, ....  Do that for a while, and then look for the pattern.  Well, it's still not easy to deduce the pattern, but at least it's in the realm of possibility.

Is this the greedy algorithm or is that something different...

Simon how often do you use the Ferrer's diagram thing? I saw it used once in someone's MOP notes I think but didn't really take the time to examine all its applications...(so i know the theory but don't know how to use it really)

Bob -- It's not so much that it's a greedy algorithm. It's just that there's only one place for each number to go, otherwise it fails. You can easily show by induction that, up to switching the sets, there can only be at most one way of partitioning them.

As for the Ferrers Diagram, I was staring at my work for a while, and I had written down some cases and was trying to see why two things should be equal. Then I just thought it would be interesting to try the Ferrers Diagram to see what would happen, and I saw that it worked.

I tried your method Ravi -- it's clear that if you do it, only one thing can happen.  I was decieved because it looked very much like the partition was based on mod 12.  I guess you can put that down to my inexperience, or some such thing.  The problem is that if you start listing, there isn't anything in particular that would ever make you think of base 2.

Yeah, I'm not saying the question was easy.  (I didn't take the test, so this is just Monday morning quarterbacking on my part.)  Perhaps you need to try many numbers, past 16, to see the pattern more clearly.

By the way, it looks like you did really well on the test; 6 questions is impressive.  When I was a kid, I got about 5 and a half, which was good enough for top 30.

Well, one of the 6 was definitely not written up in full . . . all in all, though, I felt pretty good about it.

Thanks.

[quote="JBL"]Is there a nice way to do B1?

The question B1, by the way, for those of you interested:

If a, b, c, and d are polynomial functions, is it possible to write
1 + xy + x2y2 = a(x)c(y) + b(x)d(y).  I could only do it with massive equation-bashing.[/quote]

I have an alternate approach to B1 that I think works, but I haven't worked it all the way through. Try representing the polynomial multiplications by matrix multiplication. Represent each polynomial by the matrix who's rows are the coefficents of terms containing specific powers of x, and who's columns are the coefficients of terms containing specific powers of y. Then you are looking at trying to solve AC + BD = I, where A represents a(x) and so on..

But I don't think this is possible due to eigenvalue considerations. det (I - BD) = det (AC) = 0, which means the eigenvalues of both AC and BD must be 0, 0, and 1. But that doesn't seem consistent looking at the trace of both sides of AC + BD = 1.

I thought that might be possible -- it looked a lot like there were 2x2 determinants and things.  But I'm just now learning lin. alg., so I didn't really know how to apply it.

I have a friend at my college who did it that way, I think. I guess it's a lot nicer. He does nearly all the problems we give him with linear algebra.

"Quick, how do you integrate this function?"

"Well, first we need to figure out the rank of . . ."

Hmm, that wasn't as funny as I thought it was going to be.  Oh, well.

For B3 (the one about factorials), ComplexZeta said that Ferrers Diagrams work well.  Also, induction on n works well.  In going from n - 1 to n, the only terms on the right that change are those where n/i is a prime power, and it's not hard to calculate the change.

JBL wrote:The question B1, by the way, for those of you interested:

If a, b, c, and d are polynomial functions, is it possible to write
1 + xy + x2y2 = a(x)c(y) + b(x)d(y)

There is a nice solution:   [hide]you can turn it into a linear algebra problem by identifying with a polynomial p(x,y) the matrix whose entry in the ith row and jth column is the coefficient of x^i*y^j in p(x,y).  Then of course 1 + xy + x^2*y^2 becomes our good old identity matrix.  But also, the null spaces for the matrices of a(x)c(y) and b(x)d(y) each have dimension 2, and so they interesect.  This means that the null space of  the matrix of a(x)c(y)+b(x)d(y) must be nontrivial, and so that matrix is definitely not the identity.  QED[/hide]


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number theory unsolved
number theory

Let A be a set of $ n$ positive integers. Show that every sequence of $ 2^n$ numbers taken from A contains a consecutive block of numbers whose product is a square. (For instance, 2,5,3,2,5,2,3,5 contains the block 5,3,2,5,2,3).

Cute little problem. Moreover, $ 2^n$ is the least length that guarantees the result.
For whom is still interested, I will post a solution a week from now ...  :)

I am really interested in this problem. Could you please post a solution (or a sketch of it) sooner...
Thanks in advance.

A first step is to translate it into a vector space (or module) setting:

Given a $ \mathbb Z / 2 \mathbb Z$ vector space $ V$ and $ n$ elements $ a_1, ... , a_n \in V$. Then if $ b_i$ is a sequence of length $ 2^n$ taken from the $ a_k$, two of the sums $ c_m : \equal{} \sum_{i\equal{}1}^m b_i$ for $ 0 \leq m \leq 2^n$ are equal.

Now note that we may assume $ V$ to be $ n$-dimensional, thus having only $ 2^n$ elements, and conclude.

Thank you for answering ZetaX, but please could you provide a little bit more detailed proof (since I am not a very clever person, I find it hard to understand what was done).

it has nothing to do with being clever, but only with having enough experience to follow someone else's logic fluidly...
don't go too harsh on yourself :)

Here's a vector space to fit it in: the space of functions from the set of primes to the integers mod $ 2$. What function goes with a number $ k$? The exponents in its prime factorization.

OK, that's an infinite-dimensional space. What's that about "we may assume $ V$ to be $ n$-dimensional"? Well, only the span of the $ a_i$ matters. Everything else is unreachable.

Let $ a_1,a_2,...,a_n$ be our numbers, and let $ x_1,x_2,..,x_{2^n}$ be the numbers form our sequence.

Define $ s_1,s_2,...,s_{2^n}$ such that $ s_i= x_1 x_2...x_i$ and let $ g(i,j)$ be the number of times element $ a_i$ appears in sequence from $ x_1$ to $ x_j$.

So we can write $ s_i$ as $ s_i= a_1^{g(1,i)}*a_2^{g(2,i)}*...*a_n^{g(n,i)}$

Now let $ f$ be a function from the first $ 2^n$ numbers to the set $ \{ 0,1 \} ^n$ such that

$ f(i)=(g(1,i) \mod 2,g(2,i) \mod 2,....,g(n,i) \mod 2)$

If $ f(i)=(0,0,0,0,...0)$ for some $ i$ then $ s_i$ is a perfect square (this follows from $ s_i= a_1^{g(1,i)}*a_2^{g(2,i)}*...*a_n^{g(n,i)}$).

Otherwise we have $ f(i)=f(j)$ for some $ i<j$. It is easy to see that now $ \frac{s_j}{s_i}$ is a perferct square.
($ \frac{s_j}{s_i}= a_1^{g(1,j)-g(1,i)}*a_2^{g(2,j)-g(2,i)}*...*a_n^{g(n,j)-g(n,i)}$ all exsponents are even since $ g(k,j) \equiv g(k,i) \mod 2$ for every $ k$)

I rest my case :)

Since the integers could be distinct primes, this is equivalent to proving the stricter problem of, given a finite alphabet $ A$ of $ n$ letters $ a_1,a_2,\ldots,a_n$, and a word $ w$ of length $ 2^n$ on this alphabet, to show it contains a nonempty contiguous subword $ x$ ($ w \equal{} \overline{uxv}$, where $ u,v$ could be the empty word), in which each letter appears at an even number of times.

Let us, further on, identify the letter $ a_k$ with the element $ {\bf e}_k \in \mathbb{Z}_2^n$ given by $ {\bf e}_k \equal{} (0,\ldots,0,1,0,\ldots,0)$, where the $ 1$ is at the $ k^{\textrm{th}}$ position. Now the requirement is to find a subword such that the sum of its elements is $ {\bf 0} \equal{} (0,0,\ldots,0)$.

We will apply a classical idea of Erdos. Let $ w \equal{} \overline{x_1x_2\ldots x_{2^n}}$ be the word, with $ x_i \in \{{\bf e}_1,{\bf e}_2,\ldots,{\bf e}_n\}$ for all $ i\equal{}1,2,\ldots,2^n$. Define $ \displaystyle \sigma_k \equal{} \sum_{i\equal{}1}^k x_i$ for all $ k\equal{}1,2,\ldots,2^n$. If any of $ \sigma_k \equal{} {\bf 0}$, we are done, since $ \overline{x_1\ldots x_k}$ can be taken as the subword; otherwise there must exist $ 1\leq p < q\leq 2^n$ such that $ \sigma_p \equal{} \sigma_q$, so then $ \displaystyle {\bf 0} \equal{} \sigma_q \minus{} \sigma_p \equal{} \sum_{i\equal{}1}^{q\minus{}p} x_{p\plus{}i}$, and we can take the subword $ \overline{x_{p\plus{}1}\ldots x_{q}}$.

Let us notice that the result is tight: there exist words of length \framebox{$ 2^n \minus{} 1$} without this property. We can build them inductively. Take $ w_1 \equal{} \overline{a_1}$, and build $ w_{n\plus{}1} \equal{} \overline{w_na_{n\plus{}1}w_n}$ for all $ n\geq 1$.


Morgan has 3 hockey shirts, 2 football shirts and 7 baseball shirts in
her closet. If she randomly selects one of these shirts, what is the
probability that it will not be a baseball shirt? Express your answer
as a common fraction.

Totally, she has 12 shirts: 7 baseball shirts and 5 non-baseball shirts. So, the probability of choosing a shirt that is not a baseball shirt is 5/12.


let $ f: N\minus{}\minus{}>N$ injective ,prove that $ lim[\frac{f(1)}{1^2}\plus{}....\frac{f(n)}{n^2}]\equal{}\plus{}00$

lemma:
[hide]if $ f$  is injective then $ \frac {f(1)}{1^2} \plus{} ....\frac {f(n)}{n^2}\ge1 \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} .....\frac {1}{n}$[/hide]

But:

$ \sum \frac{1}{i}$ converges  :wink:

[quote="FOURRIER"]But:

$ \sum \frac {1}{i}$ converges  :wink:[/quote]
sorry , but it's wrong . it diverge to $ \plus{}00$

Let $ a_n \equal{} \sum\frac {1}{n}$

If $ a_n$ converges but it is an increasing sequence then it has a limit $ \lambda\Leftrightarrow$ there exists $ n_0$ so that :

$ |a_n \minus{} \lambda| < \epsilon,\forall n\ge n_0$

Choose $ \epsilon < 1$ then :

$ |a_{n_0 \plus{} l} \minus{} \lambda| \equal{} \lambda \minus{} a_{n_0} \minus{} \sum_{i\equal{}1}^l \frac {1}{n_0 \plus{}i} < \epsilon \minus{} \frac {l}{n_0 \plus{} l}$

Choose $ l\to\infty$ then $ \lim\frac{l}{n_0\plus{}l}\equal{}1$ then

$ |a_{n_0 \plus{} l} \minus{} \lambda|<0$

 contracdiction  :wink:

quangpbc,
What do you denote by $ a$?
Make it more clear
Thanks  :)

it is known as the harmonic series,take a look at:
[url]http://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29[/url]

[quote="quangpbc"]Let $ a_n \equal{} \sum\frac {1}{n}$

If $ a_n$ converges but it is an increasing sequence then it has a limit $ \lambda\Leftrightarrow$ there exists $ n_0$ so that :

$ |a_n \minus{} \lambda| < \epsilon,\forall n\ge n_0$

Choose $ \epsilon < \frac {1}{n_0 \plus{} 1}$ then :

$ |a_{n_0 \plus{} 1} \minus{} a| \equal{} a \minus{} a_{n_0} \minus{} \frac {1}{n_0 \plus{} 1} < \epsilon \minus{} \frac {1}{n_0 \plus{} 1} < 0$ contracdiction  :wink:[/quote]
This solution is wrong because $ n_0$ depend on $ \varepsilon$
You can use $ a_{2n}\minus{}a_n\geq \frac{1}{2}$
It is an simple solution and special case of Rieman sum.

No, TTsphn,  because $ |a_n \minus{} \lambda|\le\epsilon,\forall \epsilon > 0$, then i can choose $ \epsilon$  :(

[quote="TTsphn"]No you are wrong. :wink:
Because it not true with the defintion of limit of a sequence.[/quote]

I am sorry, i misstake. I have edited this post :)

[quote="FOURRIER"]How do you prove it?

Thanks  :)[/quote]
[u]Proof.[/u] Assume to the contrary that $ \sum_{i\equal{}1}^{\infty}\frac{1}{i}$ diverges.  Then it follows that
\[ \sum_{i\equal{}1}^{\infty}\frac{1}{i}>\frac{1}{2}\plus{}\frac{1}{2}\plus{}\frac{1}{4}\plus{}\frac{1}{4}\plus{}\hdots\equal{}\sum_{i\equal{}1}^{\infty}\frac{1}{i}.\]
Contradiction! $ \Box$

See (in Romanian language) http://mateforum.ro/viewtopic.php?t=102&highlight=

[quote="FOURRIER"][quote="tchebytchev"]lemma:
if $ f$  is injective then $ \frac {f(1)}{1^2} \plus{} ....\frac {f(n)}{n^2}\ge1 \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} .....\frac {1}{n}$[/quote]

Proof Or Hint please  :wink:[/quote]
we suppose that $ \{f(1),f(2),...,f(n)\} \equal{} \{a_1,a_2,...,a_n\}$ such that $ a_n > a_{n \minus{} 1} > ... > a_2 > a_1$ so ($ a_i\ge i$)
^^reordonnment^^ gives $ \frac {f(1)}{1^2} \plus{} ....\frac {f(n)}{n^2}\ge a_1 \plus{} \frac {a_2}{2^2} \plus{} \frac {a_3}{3^2} \plus{} .....\frac {a_n}{n^2}\ge 1 \plus{} \frac {2}{2^2} \plus{} \frac {3}{3^2} \plus{} .....\frac {n}{n^2}$ and we have

Oh right...The IMO 72

Reordonement in english is rearangement  :wink:

Thank you aviateurp  :D

[quote="FOURRIER"]But:

$ \sum \frac {1}{i}$ converges  :wink:[/quote]
take $ U_n\equal{}\sum_{k\equal{}1}^{n}\frac{1}{k}$
we have $ U_{2^n\minus{}1}\equal{}\sum_{i\equal{}0}^{n\minus{}1}(\sum_{k\equal{}2^{i}}^{2^{i\plus{}1}\minus{}1}\frac{1}{k})$
and we have $ \sum_{k\equal{}2^{i}}^{2^{i\plus{}1}\minus{}1}\frac{1}{k}\ge \frac{(2^{i\plus{}1}\minus{}1)\minus{}2^i\plus{}1}{2^{i\plus{}1}}\equal{}\frac{1}{2}$
then $ U_{2^{n}\minus{}1}\ge \frac{n}{2}$
so $ \lim_{n\to \plus{}\infty}U_{2^{n}\minus{}1}\equal{}\plus{}\infty$ so $ (U_n)$ diverge


Could you please help me solve this problem?   Find all nonnegative integers x, y >1 so that 2^x - 3^y =7

I am stuck at the end... I get that x=0 (mod 4) and y=2 (mod 4) but I don't know how to get a contradiction or another solution than x=4 y y=2.

I start by writing... 2^x-3^y= (2^2) + 3  //  2^x-3^y=(3^2) - 2

And I draw conclusions out from there by taking common factors out and studying divisibility. But I'm stuck at trying to find a contradiction in the end.

Thx

well... i solved this problem a couple of days ago, in a small competition :)
you have that both x and y are even, so x = 2m, y = 2n..
you can say 2^x - 3^y = (2^m + 3^n)(2^m - 3^n)...
so, being 7 prime, you have that 2^m + 3^n = 7 and 2^m - 3^n = 1, because 2^m + 3^n is strictly positive, and 2^m - 3^n < 2^m + 3^n..
and there you are ;)


algebra
polynomial
modular arithmetic
blogs

If $a, b,c$ are odd integers, prove that $ax^{2}+bx+c = 0$ does not have a rational number solution.

[hide="hint"] Let $x = \frac{p}{q}$ with $p$ and $q$ odd by the rational root theorem. [/hide]

[hide="Or..."] Consider the polynomial $\bmod 2$. [/hide]

[hide]Since $b^{2}\equiv 1 \pmod 8$ and $4ac\equiv 4\pmod 8$, $b^{2}-4ac\equiv 5\pmod 8$.  It can't be a perfect square because $n^{2}\equiv 0,1,4\pmod 8$, so the discriminant isn't a perfect square, hence the roots are not rational.[/hide]

The method that either paladin8 or I gave extends to any polynomial with an odd number of odd coefficients (provided that the constant term is one of them).

I get paladin's method (that's how I prove it), but not yours.

[hide="Discussion"] First, refer to [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=128871]an old blog post of mine[/url] about modular division.  If a rational solution $\frac{p}{q}$ exists, then $\bmod 2$ it corresponds to a residue.  But $\bmod 2$ our polynomial is $x^{2}+x+1$, which has no solutions $\bmod 2$.

The correspondence fails if $q$ is even.  But if $\frac{p}{q}$ is a root of $ax^{2}+bx+c$ then $\frac{q}{p}$ is a root of $cx^{2}+ax+b$, so that's not an issue. [/hide]


Let a < b be real numbers, and f: [a,b] -> R be a differentiable function. Write C = {x in [a,b] | f ' (x) = 0} for the set of critical points of f.

Two questions:

1. Suppose that f is continuously differentiable. Prove that f(C), the set of "critical values" of f, is a set of measure zero in R. (Given epsilon > 0, how can I use uniform continuity of f ' to find a partition P: a = x_o < x_1 < ... < x_n = b such that whenever an interval [x_j-1 , x_j] contains a point of C, its image f([x_j-1 , x_j]) must have diameter less than epsilon/(1 + b - a).

2. Prove that without the hypothesis that f ' is continuous, it still is true that f(C) is a set of measure zero. (Vitali's lemma can be put to good use here.)

by uniform continuity, given $\epsilon>0$, $\exists \delta >0$ such that
$\mid x-y\mid$ <$\delta$ $\Rightarrow$                                              $\mid f'(x)-f'(y)\mid$<$\frac{\epsilon}{(1+b-a)}$. now take a partition of $[a,b]$ with such that the length of each sub-interval is less than $\delta$.


function
LaTeX
quadratics
algebra
polynomial
Rational Root Theorem

1. How do you solve for $x$: $4x^3-6x=-2$

2. Given that $f$ is the function defined by $f(x) = \frac{x^3-x}{x^3-4x}$, what are the zeros of $f(x)$?

[color=green][size=75]edited by mod...i hope i understood you correctly ... remember order of operations when you're writing without latex[/size][/color]

EDIT- Where's your original question? :? Did you delete it?

oops sorry i wrote down the wrong problem =/

[quote="woohooxd"]how do you solve for x: 4x^3-6x=-2[/quote]

[hide]$4x^3-6x+2=0$.
Dividing by 2 yields
$2x^3-3x+1=0$.
Rational root theorem tells that possible rational zeros include $\pm\frac{1}{2}, \pm1$.
Trying the possible rational zeros, we find that $(x-1)$ works. Dividing by it gives us $2x^2+2x-1$. One solution is $x=1$. The other solutions are given by the quadratic shown. [/hide]

[quote="woohooxd"]2. Given that $f$ is the function defined by $f(x) = \frac{x^3-x}{x^3-4x}$, what are the zeros of $f(x)$?

[color=green][size=75]edited by mod...i hope i understood you correctly ... remember order of operations when you're writing without latex[/size][/color][/quote]
[hide]
First, factor the numerator and the denominator.
$f(x)=\frac{x(x+1)(x-1)}{x(x+2)(x-2)}$

$f(x)=0$ when the numerator is $0$, so that gives three possibilities: $x=0,1,-1$.

However, we can't have $x=0$ since that makes the fraction undefined ($\because$ the denominator becomes $0$).

Therefore, the zeros are $1$ and $-1$.[/hide]


geometry
perpendicular bisector
geometry unsolved

$l$ is a fixed line . $A, B$ are two fixed point . Find $M \in l$ such that : $max(MA,MB)$ gets the minimum value .

Let $P_a$ be the foot of the perpendicular to $\ell$ from $A$. Let $P_b$ be the foot of the perpendicular to $\ell$ from $B$. Let $O$ be the intersection of the perpendicular bisector of $AB$ with $\ell$.

The point we seek is the one of $P_a,P_b,O$ that is between the other two.
(If the perpendicular bisector of $AB$ is parallel to $\ell$, the desired point is $P_a=P_b$)

Case 1: $O$ is between $P_a$ and $P_b$. Then $AP_a<AO<AP_b$ and $BP_b<BO<BP_b$, so $AP_b=\max(AP_b,BP_b)>AO=max(AO,BO)=BO<BP_a=\max(AP_a,BP_a)$. For all $M\neq O$, either $M$ is on the same side as $P_a$ and $MB>OB$ or $M$ is on the same side as $P_b$ and $MA>OA$.

Case 2: $P_a$ is between $O$ and $P_B$. Then $AP_a^2=AO^2-OP_a^2$ and $BP_a^2=BO^2+P_aP_b^2-OP_b^2$, so
$AP_a^2-BP_a^2=OP_b^2-OP_a^2-P_aP_b^2=(OP_a+P_aP_b)^2-OP_a^2-P_aP_b^2$
$= 2OP_a\cdot P_aP_b>0$
and $\max(AP_a,BP_a)=AP_a\le AM\le\max(AM,BM)$ for all $M$.

Case 3: $P_b$ is between $O$ and $P_a$. This is identical to case 2 except for a swap of letters.


calculus
trigonometry
limit
logarithms
calculus computations

How do you find the $ lim_{x\rightarrow0} \frac{\sin{100x}}{x}$ without a calculator?

How about $ lim_{n\rightarrow0} (1+x)^{\frac{1}{x}}$?

Informally, if $ y$ is very near zero, then $ \sin y$ is very nearly the same as $ y.$

The formal way to say this is that $ \lim_{y\to 0}\frac {\sin y}{y} \equal{} 1.$

Hence as $ x\to 0,\ 100x$ eventually becomes very small, and $ \sin 100x$ looks just like $ 100x.$ The limit is the same as the limit of $ \frac {100x}{x},$ namely $ 100.$

That can be done more formally. (It's very likely that some other poster will do so.) But this is the core idea.

For the second one, if $ L \equal{} \lim_{x\to0}(1 \plus{} x)^{1/x},$ then $ \ln L \equal{} \lim_{x\to0}\frac {\ln(1 \plus{} x)}{x}.$

Does that help?

For the first one, instead of doing it intuitively like Kent Merryfield, mathematically just multiply the top and bottom by $ 100$ and using his identity, $ \lim_{y\to0} \frac{\sin(y)}{y} \equal{} 1$
we get $ \lim_{x \to 0} {\frac{100 \cdot \sin{100x}}{100x}} \equal{} 100 \cdot 1 \equal{} 100$

$ \lim_{x \to 0}(1\plus{}x)^{1/x} \equal{} \lim_{u \to \infty}\left(1 \plus{} \frac{1}{u}\right)^u$

[b][i][u]here are the solutions:[/code][/u][/i][/b]

The reasoning here isn't correct. You seem to use the fact that $ 100x\to 0$ and $ x\to 0$, so the only conclusion you should draw is that $ \lim_{x\to 0}\frac{\sin(100x)}{x}\equal{}\frac{0}{0}$.

The notation is a little strange as well, since before the equal sign you are taking a limit, then after the equal sign you still have the $ x$ but the limit is gone.

[quote="Fermat's Little Turtle"]For the first one, instead of doing it intuitively like Kent Merryfield, mathematically just multiply the top and bottom by $ 100$ and using his identity, $ \lim_{y\to0} \frac {\sin(y)}{y} \equal{} 1$
we get $ \lim_{x \to 0} {\frac {100 \cdot \sin{100x}}{100x}} \equal{} 100 \cdot 1 \equal{} 100$[/quote]

Yeh, you are right! :wink:


Find all f:R\{0} ->R such that:
\[xf(x+\frac{1}{y})+yf(y)+\frac{y}{x}=yf(y+\frac{1}{x})+xf(x)+\frac{x}{y}\]

@[b]nguyenhoangvietdeptrai[/b]:Is k fixed in your solution?

I asked that question because if $k$ is fixed then $h(x)=h(y)$ only for $x=\frac{k}{y}$ and not for all $x,y$.
If $k$  is not fixed then what is your $h(x)$?
Can anybody explain me, what is my mistake?

sorrie, I have a mistake
  Well, with each k, we havehave numberless couples (x,y), but not for all x,y. So with each k we have one solution (h(x)=const).
  Thus for any k, we all have such that solution that h(x)= const.
  But the problem is g(x). I will fix that mistake later.


AMC
AIME
AMC 12
function
trigonometry
USA(J)MO
USAMO

Okay, I'm a soon-to-be-freshman, and I am going to be in Algebra 2.  Here's my Summer Plan:

1) June = Go through Intro to C+P and NT, and also work efficiently through an old 1970's Algebra 2 Textbook that is way better than the dumbed down ones today.

2) July-rest of summer = Study AoPS Chapters, do some algebra 2 practice problems, AMC's and AIME's in the school year.

I don't know what else to do. Any thoughts?  :D

If you're intending to do either the AIME or AMC12, you'll need a good understanding of trigonometric functions ($\sin, \cos, \tan, \sec, \csc, \cot$), unless you do them in algebra II (which I did not). I don't know where you're going to get that personally, but keep that in mind.

I am intending to take the AMC 10A/12B, then AIME if I make it (114 twice on AMC 10 this year). I was just thinking of learning some things covered in the AMCs and the AIMEs, and I also need to learn Algebra 2 along with it, as I haven't learned much yet, just the basics. Also, I was thinking that learning Algebra 2 school math would help before I started any Aops Volume 2 chapters, hence my plan. We have some Algebra 2 tournaments during the school year, 4 or 5, so algebra 2 would prove even more useful there.

 I really don't know if this (my plan) will work, for the most part, hence the question.  :o

Hmmm I don't know, but here's my schedule for next year.

June:
First half: Read AoPS 1 and do all the problems.
Second half: Read AoPS 2 and do all the problems.

July:
First half: Possibly finish up problems in AoPS 2, and get started on Problem Solving Strategies by Engel.
Second half: Continue reading Engel. Finish at least the first three chapters of the book. Take a full-length practice USAMO 1999. I'm going to start the practice USAMOs here... for each one I'll keep working on the problems I don't get after time is up.

August:
First half: Continue Engel. Start reading Geometry Revisited, finishing at least one fourth of the book.
Second half: Complete reading at least one half of Engel. Take USAMO 2000.

September:
First half: Go back through the first half of Engel, and do every problem. Continue Geometry Revisited on the side.
Second half: Complete at least one half of Geometry Revisited, moving ahead in Engel on the side. Take USAMO 2001.

November:
First half: Finish half of Geometry Revisited and three fourths of Engel. Do at least the easier problems in Engel while reading through.
Second half: Begin working on Thomas Mildorf's inequalities packet. Learn all the basic inequalities. Take USAMO 2002.

December:
First half: Solve at least 15 problems from Mildorf's inequalities packet, and finish reading Engel. Continue reading Geometry Revisited on the side.
Second half: Solve at least 10 more problems from the inequalities packet, and finish at least three fourths of Geometry Revisited. Take USAMO 2003.

January:
First half: Do around 10 practice AMC tests. Finish Geometry Revisited. Start Mathematical Olympiad Challenges.
Second half: Do all the mock AIMEs for 2007-2008, and go back and do all the problems in Engel. Take USAMO 2004.

February:
First half: Take the AMC 12! I'll try to take the AMC 10 as well. Two practice AIMEs, while doing some problems in Mathematical Olympiad Challenges.
Second half: Three more practice AIMEs, and finish at least one fourth of Mathematical Olympiad Challenges. Take USAMO 2005.

March:
First half: Finish at least half of Mathematical Olympiad Challenges. As many practice AIMEs as possible.
Second half: Take AIME! After that, finish at least three fourths of Mathematical Olympiad Challenges. Take USAMO 2006.

April:
First half: Finish Mathematical Olympiad Challenges. Work through at least two IMO shortlists.
Second half: Take the USAMO!

That schedule is ridiculous.

Take the above statement to mean whatever you want.

Out of curiosity, how could I obtain "Mildorf's inequality packet"

[quote="jrshoch"]That schedule is ridiculous.
[/quote]

Wait, would you care to explain why?

I mean... I thought having a schedule would be a pretty good idea to help me get better since I only got a 9 on USAMO this year.

Do you have some specific feedback?

Maybe other books I should read?  Zeitz?  The introductory AoPS series?  Yeah I only included 2 or 3 books, but maybe I should forget about doing the problems and just get a wider range of exposure.

No no, a schedule is very useful.  It probably will keep you dedicated to your studying too.  I was just startled by the amount of material, it is very thorough.

It was a compliment :)

Sigh, ambition is pretty scary stuff.

You're not going to be able to do all of the problems in AoPS 1 and 2. Thats just not worthwhile. Its a good idea to do a good portion of the questions, including the difficult ones in the books. You are not going to be able to finish allll the problems in a month. Even that in itself would be the equivalent of 40 hour a week job?

That being said, PSS isn't a horrible way to really get into Olympiad style questions. Setting exact chapter numbers.. and amounts of the book is a bad idea.

You want to enjoy what you are doing. This is a recipe for burnout if you actually follow this schedule.

Taking practice USAMOs is only useful for the exercise of the 4.5 hour sitting.

Just doing random Olympiad problems is kinda more useful.

2 weeks is not that long.

I dunno, do what you feel like you can. Don't force yourself to try to do too much.

[quote]Yeah I only included 2 or 3 books, but maybe I should forget about doing the problems and just get a wider range of exposure.[/quote]

That would be a huge mistake.

For all the time that new material ("exposure") is given in AoPS classes/MOP and the like, it doesn't actually help that much on American competitions (except for geometry, which you *must* study). Most problems are approachable if you know the basics, and few problems are going to be downed by some sort of silver bullet theorem.

Solving problems is the only thing that brings significant performance gains (maybe this changes at the extreme upper levels, I don't know), so you should spend almost all of your time doing that. That's why the meat of most of the classes and books on problem solving is actually the problems. Make sure that you do get a wide variety of problems in, though.

All of this is IMX, of course.

Also, I don't think you appreciate exactly how much work you've outlined in your schedule. One of your goals is to completely 500 problems from Engel by the end of September. I don't know if that would be possible even if you were Brian Lawrence and spent eight hours a day on it. And you've got yourself working through like three other problem books simultaneously.

But hey, if it works out, more power to you.  :D

I don't see why you need to do anything more, A2 is plenty enough to get you to Red MOP assuming you're taking the 10.

No pre-calc concepts are allowed to be on the 10, usually only the latter 5 questions or so on the AIME require precalc which gets you to USAMO.

Now this depends on the questions but usually they're mostly ingenuity especially #1 and #4 which are specifically designed to be approachable meaning where you are in your math education isn't really that important at all as long as you know the basics.

That schedule seems a bit ridicolous, just do a couple practice AMC10s, maybe 2 or so the week before and do the ones you don't get. Maybe 3 or so AIMEs two weeks before AIME and finally do a lot of USAMO #1s/#4s and if those are going well, #2s/#5s for USAMO but make sure you know how to write a proof. Making Red MOP is 90% writing a proof, 10% solving a problem.

i think your best bet is to get the basics down quickly...then do problems...the amc contests are really just a set of contrived problems, and practice is the best way to learn the tricks

aspiringmathlete's schedule does look extreme and ridiculous to me. Here's what I did as contest preparation in my time:
Before 9th grade: The only contest I knew about was Mathcounts. There were some team practices, but that's it.
9th grade year: Other than participating in the actual contests, all I did was one practice AHSME. I think that was the famously easy 1994 test, too. I did get a lot out of both the AHSME/AIME/USAMO series and the Pomona talent search.
10th grade year: Again, nothing but the actual contests. I made HM and went to MOP that summer.
11th grade year: I looked back at MOP problems; it took months for everything to sink in. The library of tricks I learned from MOP made an enormous difference on the AIME.
12th grade year: In addition to looking back on MOP materials, I participated in a sequence of mock olympiads along with the others who had a good shot at the IMO team. I think the problems were based on the IMO shortlist.

As for books: I picked up Zeitz at some point, and liked it. I didn't do much of anything with other problem solving books.

What I did, I did because I liked solving problems. The fun is far more important than the prestige.

Completing AoPS II in two weeks? Ridiculous.
I don't think you should force yourself to adhere to a rigid schedule. Sure, you must be disciplined, but if you force yourself too hard, you won't remember a thing and burnout. I, for one, attempted to blast my way through Chapter 12 (analytic geo) of AoPS II in about an hour, and it just didn't work; I couldn't remember three-quarters of what I had read the next day. (I have since reread the chapter, taking time to understand the concepts, etc., and now I'm fine with it.)

I would also recommend Zeitz between AoPS and Engel.

And the heart of these books (AoPS, Zeitz, Engel) are the problems. Do not, in any case, just read the explanations and skip the problems. Just understanding the concepts is no use; you have to be able to apply them. After all, no AIME/USAMO problem (or any good math competition problem, for that matter) is only going to involve applying the concepts straightforwardly; borrowing an  idea from Zeitz, they are not exercises, they are [i]problems[/i].

Well I see what you guys are saying, so I guess I'll just stick to my original schedule and do all the problems in AoPS I, AoPS II, Engel, Math Olympiad Challenges, Geometry Revisited, etc.  Maybe I'll pick up Zeitz, too.

I must say, though, that I think I maybe don't need as much practice "applying concepts" as other people.  I mean... I taught myself calculus just by memorizing the formulas.  I did at most one or two problems in every chapter, and I aced the AP test yesterday.

Also: what I meant by a wider range of exposure is basically that I should memorize the types of problems in each book (and how to approach them), right?  This year I was caught totally unprepared for the USAMO.  But if I know the way to solve all the types of problems, then I'll do much better.

Well, AP tests...you can get by with rote memorization. I think the problems on there are much less creative. But if you can get by in math competitions with simple memorization of formulas, well, more power to you, I guess. It's not the way I would do it, but we're all different.

[quote="aspiringmathlete"]Well I see what you guys are saying, so I guess I'll just stick to my original schedule and do all the problems in AoPS I, AoPS II, Engel, Math Olympiad Challenges, Geometry Revisited, etc.  Maybe I'll pick up Zeitz, too. [/quote]
I still don't think this is a good idea.  It is unreasonable to try to do all the problems in AoPS II.  You're probably ready to move on to more challenging material before you're halfway there.  The same goes for AoPS I, although since the problems are easier, an accomplished problem solver might be able to solve them all with less difficulty—although I don't know why, unless for pedagogy.

[quote]I must say, though, that I think I maybe don't need as much practice "applying concepts" as other people.  I mean... I taught myself calculus just by memorizing the formulas.  I did at most one or two problems in every chapter, and I aced the AP test yesterday.[/quote]
Fine.  However, AP calculus tests do not come close to AIME problems in scope and difficulty.

[quote]Also: what I meant by a wider range of exposure is basically that I should memorize the types of problems in each book (and how to approach them), right?  This year I was caught totally unprepared for the USAMO.  But if I know the way to solve all the types of problems, then I'll do much better.[/quote]
No, no, no, no, no, no!  Each problem is different.  You can't really classify problems as Type G1(b)VII6iii or something.  To be sure, there are a number of steps that are common to most problems.  But these are usually small steps.  It is applying them, stringing them together into a meaningful solution, where the beauty comes.  Poetry is written with words that are (by and large) the same, but by putting words in certain places, one arrives at a beautiful poem.  Memorizing each word in the English language and all its rhymes will probably not make you a much better poet.

Jason :  I think by "Mildorf's inequality packet", they mean [url=http://www.artofproblemsolving.com/Resources/Papers/MildorfInequalities.pdf]this[/url].

[quote="Boy Soprano II"]It is unreasonable to try to do all the problems in AoPS II.[/quote]
I did 80% of the problems in AoPS 2 in two and a half weeks for my AIME preparation. It just takes a lonnnnng time.

Edit: Well, tried to do. I may have been a bit quick on giving up on some.

thanks miles, but the link didn't work.  I'm 99% sure that it's my computer having difficulty with pdf files for some reason though.....

also, nice explanation of problem-solving.  I don't think it's really possible to study for the USAMO.  Knowledge is useful, but most of the time the actual action of writing proofs and working through practice problems is much more useful than cramming information... but everyone's already said that.

sleep is important too ;)

What do you all think of my plan? Aspiringmathlete's schedule is being discussed although I started the topic.  :(

Yours is very focused, but if you're preparing for the AMC and AIME i don't know if NT is as important as the AoPS books.

I'm not good at categorizing stuff though

I would hesitate before using that Algebra II textbook. If your Algebra II class is anything like mine than it is repetitive and boring to begin with and would become completely unbearable if you had already learned everything in depth. I would cut out the Algebra II this summer, hit the AoPS chapters earlier to give yourself more time to do problems in there, and use the old Algebra II book as an extension of your class during the year.

Also, was there a reason that you chose NT and C+P? Unless you are particularly weak in those areas, I would just use the AoPS books to get a mix of things.

I haven't learned everything in depth much. I chose NT And C+P because those are my biggest weaknesses apart from not knowing how to do Above-Basic Algebra II.

I was thinking of using AoPS 2 after June, as I don't have AoPS 1 for the summer.

Maybe I should use AoPS 2 first?

I would personally skip the Algebra 2 textbook completely.

Well, for the AoPS textbook it depends on what level you are at and what you have learned look through the AoPS textbook if you can do almost none of the chapter problems consider looking at Volume 1.

As for the Algebra II textbook you might wannaa sort of skim through it? iono i dont suggest doing hardcore algebra II

First of all, old textbooks I find are even simpler and would be much more "dumbed-down" than today's are. Instead, try using Algebra and Trigonometry by Paul A. Foerster published by Addison-Wesley. ISBN is 0201324601. The book has all the basic A2 stuff, but more importantly explains WHY and offers two or three examples after every chapter showing you exactly how it applied each concept. While formal proofs are a bit sparse, proving them yourself should be an extremely worthwhile task (although you should be a good judge of which problems might be actually feasible) and the book helping you give a glimpse of why is a good start of what to do as well as a measure of difficulty. It also has an "Applications" section at the end of most chapters where you can do some simple word problems.
Note: I am only recommending this because you do not seem to be going through the AoPS books which can function as an Algebra II/Pre-Calc (less precalc) book with emphasis on contest mathematics.


induction
modular arithmetic
algorithm
number theory
relatively prime
Euclidean algorithm

I've read the Wiki definition of it, but I don't really understand.
Could someone explain it to me?
Thanks.

Basically if m and n are relatively prime then for any integers p q there is a solution to the simutaneous congruences  x = p (mod m) and x = q (mod n). And whats more is the solution is unique mod pq.

So for instance if you want know if there is a number that is 1 mod 3 and 2 mod 4, then by the chinese remainder theorem you are guaranteed to find a solution mod 12 (namely 10 mod 12).

Proof is rather simple, since m and n are relatively prime m generates Z_n. So namely there exists a k st mk = (q-p) (mod n) so let x = p+mk and you are done.

The theorem can be generalized to say n different pairwise relatively prime moduli by induction where the above n=2 case is your base case.

Or if you know a bit more you can state it as if $ (m,n)\equal{}1$

Then $ \mathbb{Z}_n  \mathbb{Z}_m$ is isomorphic to $ \mathbb{Z}_{nm}$

Soyy i cant get the cross symbol in between the first 2.

[quote="seamusoboyle"]Or if you know a bit more you can state it as if $ (m,n) \equal{} 1$

Then $ \mathbb{Z}_n \mathbb{Z}_m$ is isomorphic to $ \mathbb{Z}_{nm}$

Soyy i cant get the cross symbol in between the first 2.[/quote]

try

[code]\times[/code]

to get $ \times$

Back to the topic, the Chinese Remainder Theorem is a way to find a solution to a system of simultaneous congruences. Notice that the modulos need to be pairwise relatively prime, otherwise the system might have no solution.

Example. $ x\equiv 1\pmod{2}$ and  $ x\equiv 2\pmod{4}$
[hide]
This has no solution! The first equation implies $ x$ is odd and the second implies that $ x$ is even. The reason is because $ 2$ and $ 4$ are not relatively prime.[/hide]

Example. $ x\equiv 13\pmod{21}$ and  $ x\equiv 2\pmod{17}$
[hide]
Here is how you do it: the first equation implies that $ x \equal{} 13 \plus{} 21 k$, where $ k$ is a positive integer. Plugging in to the second equation,

$ 13 \plus{} 21 k\equiv 2\pmod{17}$

Reducing,

$ 4 k\equiv 2 \minus{} 13\equiv 6\pmod{17}$

$ 2k\equiv 3\pmod{17}$ (we're allowed to divide because 2 is relatively prime to 17)

$ 2k\equiv 20\pmod{17}$

$ k\equiv 10\pmod{17}$

Since $ x \equal{} 13 \plus{} 21 k$,

$ x\equiv 13 \plus{} 210 \equal{} 223\pmod{357}$
[/hide]
If there are more congruences, just repeat this procedure over and over again. (Try it, with modulos $ 15,17,21$ or another set that is pairwise relatively prime)

EDIT: Why did I think that 20/2=3?

[quote="vishalarul"]
Example. $ x\equiv 13\pmod{21}$ and  $ x\equiv 2\pmod{17}$

Here is how you do it: the first equation implies that $ x = 13 + 21 k$, where $ k$ is a positive integer. Plugging in to the second equation,

$ 13 + 21 k\equiv 2\pmod{17}$

Reducing,

$ 4 k\equiv 2 - 13\equiv 6\pmod{17}$

$ 2k\equiv 3\pmod{17}$ (we're allowed to divide because 2 is relatively prime to 17)

$ 2k\equiv 20\pmod{17}$

$ k\equiv 3\pmod{17}$

Since $ x = 13 + 21 k$,

$ x\equiv 13 + 63 = 76\pmod{21\cdot 17}$
)[/quote]
If I am not wrong i think you yourself have solved the set of eqns wrongly
Is 76 congruent to 2 mod 17 ???

I think he messed up in going from $ 2k\equiv{20}\pmod17$ to k being 3 mod 17. It should be 10, giving an answer of 223 as the smallest.

Thanks everyone!

Is there a formula or algorithm to solve the congruence
$ ax\equiv b \pmod c$
if given $ \gcd(a,c) \equal{} 1$?

The algorithm is as follows. Suppose we want to calculate $ ax \equiv b \pmod{c}$ or $ x \equiv a^{ - 1}b \pmod{c}$; we need to find $ a^{ - 1}$. So for now consider the simpler problem that $ ax \equiv 1 \pmod{c}$. We can apply the Euclidean Algorithm as $ gcd(a,c) = 1 \Longrightarrow am + cn = 1$ for some integers $ m,n$; in particular we want to find the value of $ m$.

[hide="Applying The Euclidean Algorithm"]
(sorry my explanation isn't very good) In this case, we  can apply the Euclidean algorithm as follows:

Let $ c = q_1 \cdot a + r_1$, where $ q_1$ is the quotient and $ r_1$ the remainer upon division of $ c / a$. Then repeatedly compute the $ r$s and $ q$s that satisfy:
$ a = q_2 \cdot r_1 + r_2$
$ r_1 = q_3 \cdot r_2 + r_3$
$ r_2 = q_4 \cdot r_3 + r_4$
Continue this until we finally reach some $ r_k = 1$. Then, we work back upwards:
\begin{align*}r_k = 1 & = r_{k - 2} - q_k \cdot r_{k - 1} \\
& = r_{k - 2} - q_k (r_{k - 3} - q_{k - 1} \cdot r_{k - 2}) \\
& = (1 + q_kq_{k - 1}) \cdot r_{k - 2} - q_k \cdot r_{k - 3} \\
& ... \end{align*}
On the right hand side we keep substituting the $ r$ with the greatest subscript with the next two lower terms and recursively work back upwards. At the end we will be left with $ 1 =$ some constant $ \times\ a +$ some constant $ \times\ c$, and the answer we are looking for, $ m$, is that constant that is the coefficient of $ a$.[/hide]

Now $ a^{ - 1} \equiv m \pmod{c}$, and returning the original problem we would have $ x \equiv bm \pmod{c}$.

The above method will get you the right answer, but be careful using $ a^{\minus{}1}$ or inverses in general when the modulus ($ c$ in the above case) is not prime.


superior algebra
superior algebra unsolved

Let $ E/K$ and $ L/K$ be finite normal field extensions. Show that there exists a K-homomorphism $ \sigma : E  \rightarrow L$ if and only if there exist  $ f,g\in K[x]$ satisfying the following conditions:
1.$ g$ divides $ f$
2.$ L$ is a splitting field of $ f$ over $ K$ and $ E$ is splitting field of $ g$ over $ K$.

What have you done so far? Can you reduce to the case that the normal extensions are simple?


ARML
projective geometry
videos
trigonometry
Science Olympiad
Pascal\u0027s Triangle

Any thoughts on ARML t-shirt designs? Let's ignore [color=pink]color[/color] for now :| .

Really, as long as the design isn't extremely dorky, the color [i]is[/i] the most important factor.

I wasn't there, but I've heard nothing but good things about the pink shirts...

Umm but it is trivial for an individual to determine his color preference, so it takes about a day of discussion to finalize the color. The design is much more difficult. And I for one think it is at least as important.

Someone suggested Desargues (the proof thereof, involving extension into 3 dimensions). It is a possibility.

hm just throw in some awesome pictures like the ones at http://www.algebraicsurface.net/
http://www.freigeist.cc/gallery.html

Focusing on something a little more wild than lines and points would make me very happy.

color is the most important thing

I find t-shirt designs that try to be overtly mathematical usually just lame and nerdy anyway, so they all equate to a coolness factor of around 0

so all the coolness thus derives from the color

[quote="MysticTerminator"]I find t-shirt designs that try to be overtly mathematical usually just lame and nerdy anyway, so they all equate to a coolness factor of around 0[/quote]

Agreed. And this means that design is important.

eh. if you want. BUT color is obviously important. and hot pink is obviously the best color.

o_o

I think that white is by far the best color (although I probably won't even go to the ARML...)

OffTopic:
Our Science Olympiad shirts for Ligon were pink.... .. ... ..  ....  .....

[hide="only open if you know what 1+2 is"]

Congrats to Vitek at Science Olympiad.. Got like half the medals that enloe got.

I myself, got half the medals that Ligon got... but that was 2 medals...

[/hide]

unfortunately you are wrong it has been proven that hot pink is the best

How about something from this video?

[url]http://youtube.com/watch?v=MiMWJ1xBo8w[/url]

(45,000,000,001?, maybe)

Or a list of all the numbers n such that a regular n-gon is constructible with compass and straightedge?

Or a Pascal's Triangle with 25-ish rows?

Or a bunch of differently sized, differently colored dots on the back, and the simple NC ARML logo on the front?

(I especially like the dots idea.  :) )

Write the contestant's names on them (too bad not_trig isn't going =()

Oh man bots can be pretty awesome sometimes. Especially when two posts in a row are bot posts: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=200585[/url].

[quote="PI-Dimension"]

Congrats to Vitek at Science Olympiad.. Got like half the medals that enloe got.

I myself, got half the medals that Ligon got... but that was 2 medals...

[/quote]

Uh...

I got more medals than you...

uh...

dude dnkywin way to go completely off-topic

also i did not get like half of enloe's medals, maybe like 1/3 at most and i think less

also i do not remember you being at science olympiad states hrmmmm unlike berman and jane

dude next year's t-shirts are going to be awesome

As in the [color=violet]color[/color]?

Seriously, pink shirts win championships.


geometry
geometric transformation
homothety

Two circles radii R(1)  and R(2)  centres O(1) and O(2)  lie inside a circle centre O.

They touch each other at A and the larger circle at A(1) and A(2)

Prove that    O(2)A(1),   O(1)A(2)      and  OA  are concurrent

The tangency point of 2 touching circles is collinear with their centers. Thus the point tripples $(A, O_1, O_2)$, $(A_1, O_1, O)$ and $(A_2, O_2, O)$ are collinear. The tangency points $A, A_1, A_2$ are homothety centers of the circle pairs $(O_1), (O_2);\ (O_1), (O)$ and $(O_2), (O)$, respectively, with homothety coefficients $h, h_1, h_2$:

$h = \frac{\overline{O_1A}}{\overline{O_2A}} = -\frac{R_1}{R_2} < 0$

$h_1 = \frac{\overline{O_1A_1}}{\overline{OA_1}} = \frac{R_1}{R} > 0$

$h_2 = \frac{\overline{O_2A_2}}{\overline{OA_2}} = \frac{R_2}{R} > 0$

where the segments $\overline{O_1A}, \overline{O_2A}, \overline{O_1A_1}, \overline{O_2A_2}, \overline{OA_1}, \overline{OA_2}$ are considered to be oriented. Consider the triangle $\triangle OO_1O_2$ with the points $A_1, A, A_2$ of the side lines $OO_1, O_1O_2, O_2O$.

$\frac{\overline{OA_1}}{\overline{O_1A_1}} \cdot \frac{\overline{O_1A}}{\overline{O_2A}} \cdot \frac{\overline{O_2A_2}}{\overline{OA_2}} =  -\frac{R}{R_1} \cdot \frac{R_1}{R_2} \cdot \frac{R_2}{R} = -1$

By Ceva's theorem for oriented segments, the cevians $OA, O_1A_2, O_2A_1$ are concurrent. It is of no consequence that the concurrency point is not inside of the triangle $\triangle OO_1O_2$.


function
inequalities
inequalities proposed

Prove that if $ a,b,c\geq0$, then
\[ \frac{1}{a^{2}\plus{}b^{2}\plus{}c^{2}}\leq\sum_{\rm cyclic}\frac {a}{a^{3}\plus{}(b\plus{}c)^{3}}\leq\frac{1}{ab\plus{}bc\plus{}ca}.\]

[quote="Erken"][quote="stephen38"][quote="zaya_yc"]$ 1\plus{}A^{3}\equal{} (1\plus{}A)(1\minus{}A\plus{}A^{2})\geq(\frac{(1\plus{}A)\plus{}(1\minus{}A\plus{}A^{2})}{2})^{2}\equal{} (1\plus{}\frac{A^{2}}{2})$[/quote]
and [quote="zaya_yc"]$ \frac{1}{2}(B\plus{}C)^{2}\geq B^{2}\plus{}C^{2}$[/quote]
 --->

I don't think those are true. :maybe:[/quote]
Yes,he made a mistake but anyway his solution is correct. :)[/quote]
yes,indeed it is a nice one. :)

The second part is very easy too:
Denote $ \alpha_{a}\equal{}\frac{a}{a\plus{}b\plus{}c}$,$ \alpha_{b}\equal{}\frac{b}{a\plus{}b\plus{}c}$,$ \alpha_{c}\equal{}\frac{c}{a\plus{}b\plus{}c}$.
Then we should prove that $ S \equal{}\sum\alpha_{a}\frac{1}{\sum a^{2}\plus{}2bc\minus{}ac\minus{}ab}\leq\frac{1}{ab\plus{}bc\plus{}ca}$.
Since $ \sum\alpha_{a}\equal{} 1$ and function $ f(x) \equal{}\frac{1}{x}$ is concave so we can use Jensen.If we denote $ x \equal{} a\plus{}b\plus{}c,y \equal{} ab\plus{}bc\plus{}ca,z \equal{} abc$,then after using Jensen inequality:
$ S\leq\frac{x}{x^{3}\minus{}3xy\plus{}9z}\leq\frac{1}{y}$.The last inequality follows from Schur:$ x^{3}\plus{}9z\geq 4xy$.Thus we are done. :wink:
It is very nice problem,pvthuan. :lol:

[quote="Erken"]The second part is very easy too:
Denote $ \alpha_{a}\equal{}\frac{a}{a\plus{}b\plus{}c}$,$ \alpha_{b}\equal{}\frac{b}{a\plus{}b\plus{}c}$,$ \alpha_{c}\equal{}\frac{c}{a\plus{}b\plus{}c}$.
Then we should prove that $ S \equal{}\sum\alpha_{a}\frac{1}{\sum a^{2}\plus{}2bc\minus{}ac\minus{}ab}\leq\frac{1}{ab\plus{}bc\plus{}ca}$.
Since $ \sum\alpha_{a}\equal{} 1$ and function $ f(x) \equal{}\frac{1}{x}$ is concave so we can use Jensen.Then if we use the ABC theorem inequality will be obvious. :wink: if someone want i could write it with more details.Equality occurs $ \Longleftrightarrow a \equal{} b > 0,c \equal{} 0$.
[/quote]
yes,erken.Could you write your sol more details?? :)

To stephen38:
Look above,i've already edit it.Do you need in more details anymore? :wink:

[quote="stephen38"]OK. :) Dear erken,could you post some problems using ABC method???I heard that it was a powerful method,but i have never seen the application of this method.[/quote]
Look for my post in "Inequalities Theorems and Facts" about ABC theorem.Maybe you didn't see,but the previous problem was solved by me with using Schur too.
You could try to solve for example Iran TST 1996,just use consequences,that i've already posted in "ABC theorem".

I have one question connected with Erken's solution.

Maybe I'm stupid but You have used Jensen's inequality in reversed sign because(in my opinion )$ f(x)\equal{}1/x$ is not concave but convex for nonnegative arguments.

sorry if it is spam....

It's my old and easy problem. See the proofs here:  :)
http://www.mathlinks.ro/Forum/viewtopic.php?t=225558


In an $ n \times n$ square of lattice points, compute the number of distinct distance between two lattice points.

[hide="Stuff"]Each distance corresponds to the distance from the bottom-left corner to some other lattice point, which in turn corresponds with its square.

We know the square distances range from 1 to 2(n-1)^2 (smallest to largest).

So the problem is equivalent to the following:

Find the number of $ a,b$ satisfying:

$ 1 \leq a^2\plus{}b^2 \leq 2 (n\minus{}1)^2$
$ 0 \leq a,b \leq n\minus{}1$

How one would go about that is beyond me, especially with the second condition.[/hide]

@TZF

I think it's not to find the number of $ a,b$, but the number of $ a^2\plus{}b^2$,

so that $ 3^2\plus{}4^2$ and $ 5^2\plus{}0^2$ aren't counted as two.

There's also the fact that as $ n$ gets larger and larger there begin to exist $ 0<c\leq n\minus{}1$ such that $ c^2$ can be represented as the sum of two positive squares in more than one way. *coughbritishflagtheoremcough*


Let $A$ and $B$ be disjoints sets whose union is the set of natural numbers. Show that for every natural number $n$ there exist distinct $a,b>n$ such that \[ \{a,b,a+b\} \subseteq A \quad \text {or} \quad \{a,b,a+b\} \subseteq B. \]

i think i posted once a solution for this one use the search function

Ah, I found it. ( http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=disjoint+sets&t=43383 ) Thanks.


Find all value of m such that :
$ 9x^2 \plus{} 20y^2 \plus{} 4z^2 \minus{} 12xy \plus{} 6xz \plus{} myz > 0$
with x,y,z not simultaneous equal 0

why nobody solve this problem ?
I need some comments.... :|

I had an approach, but I think it led to a dead-end.

Here it is, anyway.

I tried writing it as the sum of squares, i.e., $ (ax \plus{} by)^2 \plus{} (cy \plus{} dz)^2 \plus{} (ez \plus{} fx)^2$ where $ a^2 \plus{} f^2 \equal{} 9$, $ b^2 \plus{} c^2 \equal{} 20$, $ d^2 \plus{} e^2 \equal{} 4$, $ ac \equal{} \minus{}6$, $ ef \equal{} 3$, and $ cd \equal{} m/2$. I then added them up and tried to write the sum as a sum of squares, but I didn't really get anywhere.

Well, those are my comments...

[quote="Zhero"]I had an approach, but I think it led to a dead-end.

Here it is, anyway.

I tried writing it as the sum of squares, i.e., $ (ax \plus{} by)^2 \plus{} (cy \plus{} dz)^2 \plus{} (ez \plus{} fx)^2$ where $ a^2 \plus{} f^2 \equal{} 9$, $ b^2 \plus{} c^2 \equal{} 20$, $ d^2 \plus{} e^2 \equal{} 4$, $ ac \equal{} \minus{} 6$, $ ef \equal{} 3$, and $ cd \equal{} m/2$. I then added them up and tried to write the sum as a sum of squares, but I didn't really get anywhere.

Well, those are my comments...[/quote]
thank you for your comment,I also try to do as you and I obtain$ (3x \plus{} z \minus{} 2y)^2  \plus{} 2(2y \plus{} z)^2  \plus{} (2y \minus{} z)^2  \plus{} 4y^2  \plus{} myz > 0$,so I need to find m such that $ 4y^2  \plus{} myz > 0\,....$

[quote="facis"]I think that the truth of this statement depends on both $ m$, $ y$ and $ z$. if you complete the square,

$ 4y^2 \plus{} myz \plus{} m^2 z^2 / 16 > m^2 z^2 / 16$
$ (2y \plus{} mz/4)^2 > (mz/4)^2$

which is true when

$ 2y \plus{} mz/4 > mz/4 \implies y > 0$
or
$ 2y \plus{} mz/4 < \minus{} mz/4 \implies y < \minus{} mz/4$

So the values that $ m$ can take on depend on the values of $ y$ and $ z$.  If we are going to get better bounds, we cannot just ignore the perfect squares in the original inequality.[/quote]
I don't think so because we need to find m with all value of x,y,z (meaning x,y,z always changing).

the thing is, the bounds I found imply the given inequality, but they do not mean that the inequality is false if they are violated.

Anyhow, lets try making it the square of a single trinomial:

$ (3x \minus{} 2\sqrt {5}y \plus{} 2z)^2 \equal{} 9x^2 \plus{} 20 y^2 \plus{} 4z^2 \minus{} 12\sqrt {5}xy \minus{} 8 \sqrt {5} yz \plus{} 12 xz$

So we have $ (3x \minus{} 2\sqrt {5}y \plus{} 2z)^2 \plus{} (12 sqrt{5} \minus{} 12)xy \plus{} (8\sqrt {5} \plus{} m)yz \minus{} 6xz > 0$

...not sure if that was useful, but it would certainly simplify the inequality if $ m \equal{} \minus{} 8 \sqrt {5}$

Edit:
Actually, it may help to play with the plusses/minuses in the squaring, and see what inequalities result.  Anyway, I'm too tired to try it now; if no one else gets around to it first, I'll give it a shot later.

[hide="Here is my idea"]$ (3x \minus{} 2y \plus{} z)^2 \equal{} 9x^2 \plus{} 4y^2 \plus{} z^2 \minus{} 12xy \plus{} 6xz \minus{} 4yz \ge 0 \implies \\
16y^2 \plus{} 3z^2 \plus{} (m \plus{} 4)yz > 0$

Let $ yz \ge 0$

$ (4y \minus{} z\sqrt3)^2 \equal{} 16y^2 \plus{} 3z^2 \minus{} 8\sqrt {3}yz \ge 0 \implies \\
m \plus{} 4 > \minus{} 8\sqrt3 \implies m > \minus{} 4(2\sqrt3 \plus{} 1)$

Now let $ yz < 0$

$ (4y \plus{} z\sqrt3)^2 \equal{} 16y^2 \plus{} 3z^2 \plus{} 8\sqrt {3}yz \ge 0 \implies \\
m \plus{} 4 < 8\sqrt3 \implies m < 4(2\sqrt3 \minus{} 1)$

so $ \boxed{m \in \langle\minus{} 4(2\sqrt3 \plus{} 1), 4(2\sqrt3 \minus{} 1)\rangle}$
[/hide]

[quote="Flame"]$ (3x - 2y + z)^2 = 9x^2 + 4y^2 + z^2 - 12xy + 6xz - 4yz \ge 0 \implies \\
16y^2 + 3z^2 + (m + 4)yz > 0$
Let $ yz \ge 0$
$ (4y - z\sqrt3)^2 = 16y^2 + 3z^2 - 8\sqrt {3}yz \ge 0 \implies \\
m + 4 \ge - 8\sqrt3 \implies m \ge - 4(2\sqrt3 + 1)$
Now let $ yz < 0$
$ (4y + z\sqrt3)^2 = 16y^2 + 3z^2 + 8\sqrt {3}yz \ge 0 \implies \\
m + 4 \le 8\sqrt3 \implies m \le 4(2\sqrt3 - 1)$
so $ \boxed{m \in %Error. "display" is a bad command.
( - 4(2\sqrt3 + 1), 4(2\sqrt3 - 1)%Error. "display" is a bad command.
)}$
note how it can be equal since $ y$ and $ z$ can't simultaneously equal $ 0$[/quote]

OH ! your resulf semilar to my.... I obtain a good solution for this problem :
$ \begin{array}{l}9x^2 + 2y^2 + 4z^2 - 12xy + 6xz + myz > 0 \\
\Leftrightarrow 9x^2 + 2x(3z - y) + 20y^2 + 4z^2 + myz > 0 \\
\Leftrightarrow \Delta ^' _x = - 144y^2 - 27z^2 - 36yz - 9myz < 0 \\
\Leftrightarrow \Delta ^' _y = (36z + 9mz)^2 < 4.144.27.z^2 \\
\Leftrightarrow (4 + m)^2 < 192 \\
\Leftrightarrow - 4 - 8\sqrt 3 < m < 8\sqrt 3 - 4 \\
\end{array}$
[hide="theorem"]Given:$ y = {\rm{ax}}^2 + bx + c(a \ne 0)$.
+If $ \Delta < 0$,then $ ya > 0$ with all value of x.
+If $ \Delta = 0$,then $ ya > 0$  with $ \forall x \ne - \frac {b}{2a}$.
+If $ \Delta > 0$,($ x_1$,$ x_2$ be a two roots of equation ;suppose$ x_1$<$ x_2$ )then $ ya > 0$  with $ \left[ \begin{array}{l}x > x_2 \\
x < x_1 \\
\end{array} \right.$ and $ ya < 0$ with $ x_1 < x < x_2$[/hide]
thank for all who have contributed ideas in this topic. :lol:

I noticed it's open interval instead of closed and I corrected it so it's same solution  :)

Flame: nice solution.

MR.129 - very nice.  I especially like your technique, since it is easy to apply, and takes no great amount of insight to use.  but be more careful about typos, your solution is a little confusing because of them.

btw, in your theorem, don't you mean $ x \neq \minus{}b/(2a)$?

[quote="facis"]Flame: nice solution.

MR.129 - very nice.  I especially like your technique, since it is easy to apply, and takes no great amount of insight to use.  but be more careful about typos, your solution is a little confusing because of them.

btw, in your theorem, don't you mean $ x \neq \minus{} b/(2a)$?[/quote]
OH!Thank you for your prompt, I have revised. :P


Yes there is a formula, but I'm not really in a mood to TeX it  :wink: (rather long and ugly; it's more interesting to find it yourself).
There are some algorithms that use very similar moves in each step (in fact, mostly commutators with small differences), but you have to use google to find them.

[quote="neurosurgeon1118"]Does anyone know how to solve them more quickly without having to use a lot of algorithms (meaning under 5 if possible). My regular times are about 12 minutes for revenge and 18 for the professor),also is their a formula thats simple to calculate how many permutations the cubes have.

:wallbash_red: Thx in advance[/quote]

Yes, there are 2 (or 5) new ones to learn (at least for me) which are based off of 2nd layer algorithms (1-4) (pairing the dedges) and the other one is self-intuitive (getting the middles)
This doesn't solve the parity cases but who cares about those


$ 1.)$ Find the solution :
\[ \sqrt {2x^2 \minus{} 5x \plus{} 2} \minus{} \sqrt {x^2 \minus{} x \minus{} 2 } \equal{} \sqrt {x^2 \minus{} 3x \plus{} 2}
\]

$ 2.)$ Solve the equation :

\[ \sqrt[3]{x\plus{}7} \minus{} \sqrt[3]{2x\minus{}1} \equal{} \sqrt[3]{x}\] .

[hide="Hint for #1"]
Factor each quadratic[/hide]
[hide="Solution for #1"]
$ \sqrt {2x^2 \minus{} 5x \plus{} 2} \minus{} \sqrt {x^2 \minus{} x \minus{} 2 } \equal{} \sqrt {x^2 \minus{} 3x \plus{} 2}$
$ \equal{} \sqrt {(x \minus{} 2)(2x \minus{} 1)} \minus{} \sqrt {(x \minus{} 2)(x \plus{} 1)} \equal{} \sqrt {(x \minus{} 2)(x \minus{} 1)}$
And $ \sqrt {0} \minus{} \sqrt {0} \equal{} 0$ so a solution is $ \boxed{x \equal{} 2}$.
[/hide]

$ \sqrt {2x^2 \minus{} 5x \plus{} 2} \minus{} \sqrt {x^2 \minus{} x \minus{} 2 } \equal{} \sqrt {x^2 \minus{} 3x \plus{} 2}
\equal{} \sqrt {(x \minus{} 2)(2x \minus{} 1)} \minus{} \sqrt {(x \minus{} 2)(x \plus{} 1)} \equal{} \sqrt {(x \minus{} 2)(x \minus{} 1)}$ so $ x \in (\minus{} \infty ; \minus{}1] \cup [2 ; \plus{} \infty)$ so , if $ x \in [2 ; \plus{} \infty)$ the problem is very simple .

I can't solve the equation if $ x \in (\minus{} \infty ; \minus{}1]$ . Maybe someone has a solution .

Well the problem above can be simplified to:
[hide]$ \sqrt {(x \minus{} 2)(2x \minus{} 1)} \minus{} \sqrt {(x \minus{} 2)(x \plus{} 1)} \equal{} \sqrt {(x \minus{} 2)(x \minus{} 1)}$
$ \sqrt {2x \minus{} 1} \minus{} \sqrt {x \plus{} 1} \equal{} \sqrt {x \minus{} 1}$
[/hide]
However, after rewriting and squaring it, I got to this scenario
[hide]$ \minus{} \frac {1}{2} \equal{} \sqrt {x^2 \minus{} 1}$ which I have a feeling can be proven there is no solution by proving a scenario like this:
[list]Prove or disprove there is no solution for $ a\in\mathbb{C}$ such that
$ \sqrt {a} \equal{} \minus{} 1$[/list][/hide]

But I am uncertain about this and not sure if I approached it correctly.

\sqrt {(x - 2)(2x - 1)} - \sqrt {(x - 2)(x + 1)} = \sqrt {(x - 2)(x - 1)}
from this step
i guess you can assume that if x-2=0, and x=2 is one of the solutions
and assume that if x-2 does not equal 0
you can divide \sqrt {(x - 2)}  on both sides of the equation
\sqrt {(2x - 1)} - \sqrt {(x + 1)} = \sqrt {(x - 1)}
you will get 4x^2 + 8x - 5 = 0 at the end (if i did the math right)
and factor it into (2x + 5)(2x - 1)
x = \frac {-5}{2}  and [code] x = \frac {1}{2}

[quote="Igd0612"] you will get 4x^2 + 8x - 5 = 0[/quote]
Did you mean $ 4x^2\minus{}5\equal{}0$?
After checking, $ x\not\equal{}\pm\frac{\sqrt{5}}{2}$ either.

yep, i just checked
these 2 roots doesnt work either......
maybe its because i squared the equation twice

[quote="lgd0612"]\sqrt {(x - 2)(2x - 1)} - \sqrt {(x - 2)(x + 1)} = \sqrt {(x - 2)(x - 1)}
from this step
i guess you can assume that if x-2=0, and x=2 is one of the solutions
and assume that if x-2 does not equal 0
[b][color=darkred]you can divide \sqrt {(x - 2)}  on both sides of the equation
\sqrt {(2x - 1)} - \sqrt {(x + 1)} = \sqrt {(x - 1)} [/b][/color]
you will get 4x^2 + 8x - 5 = 0 at the end (if i did the math right)
and factor it into (2x + 5)(2x - 1)
x = \frac {-5}{2}  and [code] x = \frac {1}{2}[/quote]

You can divide if and only iff $ 2x \minus{} 1 > 0$ , $ x \plus{} 1 > 0$ and $ x \minus{} 1 > 0$ . Otherwise the $ \sqrt {\text{ }}$`s do not exist .

can't use [i]i[/i]?

for example
what if u got 5i-2i=3i, does that work?
i am not sure either...

x can be a real number even when 2x-1<0, x+1<0, and x-1<0

for example:
 \sqrt {(2x - 1)} - \sqrt {(x + 1)} = \sqrt {(x - 1)},
if 2x-1<0, x+1<0, x-1<0
lets say at the end the equation is going to be 5i-2i=3i   the squareroot of a negative # is not a real number
but the solution for x can be a real number and also works

If you can't divide by $ \sqrt {x \minus{} 2}$ and $ x \neq 2$ then you can divide by $ \sqrt {2 \minus{} x}$.  :)

[quote="linker"]No , I forgot to specific that $ x \in \mathbb{R}$ .[/quote]
The critical part of this problem is the the loss of equivalence when you square the both sides. A simple example is:
$ x\equal{}1$ implies $ x^2\equal{}1$. But $ x^2\equal{}1$ doesn't imply $ x\equal{}1$, since $ x$ can also be $ \minus{}1$.
In this case, the usual method is to solve and check. But sometimes it will get too complicated in the calculation, so you need to carefully record the bounds on the variable before squaring the equation.

[quote="linker"]$ 2.)$ Solve the equation :
\[ \sqrt [3]{x \plus{} 7} \minus{} \sqrt [3]{2x \minus{} 1} \equal{} \sqrt [3]{x}
\]
.[/quote]

[color=darkred] How about this one ? ..Has somebody a solution ?. [/color]

One solution I immediately recognized is $ \boxed{x \equal{} 1}$, but I'm not sure how else to tackle this.


[hide]
Subtract $\displaystyle\frac{3}{2}$ from both sides. Then, divide both sides by $\displaystyle\frac{4}{3}$ to get $x = -\frac{13}{18}*\frac{3}{4}$. $x = \boxed{-\frac{13}{24}}$[/hide]

[quote="hellyeah"]Solve for x . Write your answer as a fraction in simplest form.
4/3x + 3/2 = 7/9

How would i find out what x is :?:
[color=red]text[/color][/quote]

$\frac{4}{3}x=\frac{7}{9}-\frac{3}{2}=-\frac{13}{18}$
$x=\frac{-\frac{13}{18}}{\frac{4}{3}}$
$x=\boxed {-\frac{13}{24}}$

[quote="manifold"][hide]You both seem to have made the same mistake- you solved for 1/x, so the answer should be the reciprocal of that.[/hide][/quote]

That's only if the problem is 4/(3x) but I believe it to be 4x/3 which is what TRAN THAI HUNG put which means he got the wrong answer for his equation.

[quote="hellyeah"]Hey thanks yall for the help. :lol:[/quote]

You're welcome


A rectangular prism has sides 3, 4, and 5.  What is the volume of the region within the prism that within 1/2 unit of the prism's surface?

Am I missing any cases in the following solution?

[hide]The region in the prism not within 1/2 unit of the surface is another prism with dimensions (3-1), (4-1), and (5-1), so the area of that prism is 2*3*4=24, so the volume of the region within 1/2 unit is 60-24=36.[/hide]

[hide]the new dimentions are 2,3,4 i think

2*3*4=24[/hide]


did anyone go to the invitational mu alpha theta competition this past weekend, Nov 13, in Tallahassee, FL at Chiles High School?

I did.

I got 4th on both Ciphering and Individual in the Algebra II division. Overall, I liked the tournament, though there were quite a number of careless mistakes in the given answers in just my division alone.

thats coo, good jobl...luckily there wasnt a single mistake in my division of geometry....i got 2nd in individual and 5th in ciphering

I didn't go, considered it a waste of time quite frankly.  I'm going to the regional competitions though.


When I seem to log in to my username (its [b]chess[/b], ID [b]6939[/b]), it always requires me to log in twice before I can view my status. Is this fixable?

Heh, this isn't an isolated incident. Happens to me as well. I personally thought it was just my computer.

The problem is that when you go tp usamts.org without the www, they send you to http://www.usamts.org after you login, so you'll have to sign in again because they're recognized as 2 separate sites.  Should be easy enough to fix.


x.y .z
integers such that
${\frac{24}{5}=x+\frac{1}{y+\frac{1}{z+1}}}$

Write LHS as mixed number a multiple of times to get (x,y,z)=(4,1,3).


[quote="chickendude"][hide]from the original problem, we see that $x > 0$, restricting our solution set even further to
$0 < x < 27$ and $x \neq 3$
[/hide][/quote]
Except $3^{-1-\sqrt{5}}> 0$, so your original interval was correct.


assume that P(x) is polynomial with integers coefficient. show that :
Q(x)=P(x^4)*P(x^3)*P(x^2)P(x)+1
has no integer root.

degP=n
Assume Q[x] have x_0 is an integer root
=> P(x_0^4).P(x_0^3).P(x_0^2).P(x_0) =-1
but P(x_0^k) is integer so we have
there are three -1 one 1 or three 1 one -1
so we easily have x is different 0,1,-1
=> we have k in {1,2} so that P(x_0^k) different P(x_0^(k+2))
=>lx_0^k-x_0^(k+2)l is divisble by lP(x_0^k)-P(x_0^(k+2))l
=>lx_0*(x_0^2-1)l is divisble by is divisble by 2
=>x_0=0,1,-1(contracdition)

So Q[x] doesn't have an integer root
Q.E.D

Modulo 3 we have x4 = x2 and x3 = x so P(x4) = P(x2) and P(x3) = P(x).
Hence Q(x) = P(x2)2P(x)2 + 1 (mod 3).
If x' is an integer root of Q(x) then P(x'2)2P(x')2 = -1 (mod 3).
But this is impossible as no perfect square is equal to -1 (mod 3).


Does the infinite sequence

$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+......$

End up equaling one?

all the terms in the sequence is in the form $\displaystyle\frac{1}{x(x+1)}=\frac{x+1-x}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}$.
so
$\displaystyle\left(\frac11-\frac12\right)+\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)+...+\left(\frac{1}{x-1}-\frac{1}{x}\right)+\left(\frac{1}{x}-\frac{1}{x+1}\right)$
everything cancels out except for $\displaystyle\left(\frac11-\frac{1}{x+1}\right)=\frac{x}{x+1}$

since $x=\infty$, $\frac{\infty}{\infty+1}$=$\frac{\infty}{\infty}=1$
I don't know if the last part is right.

I would be careful about how you wrote things.  For example, $\frac{\infty}{\infty}$ doesn't make any sense.  Infinity is not a number that can be manipulated in that way.  Instead, saying that $\lim_{x\to \infty}\frac{x}{x+1}=1$ is a much better way of saying it.

You can't manipulate $\infty$ like a number.

To find the limit as x approaches infinity you divide by the highest power in the denominator and then evaluate the results realizing that x is getting huge.

$\lim_{x\to \infty}\frac{x}{x+1}$
$\lim_{x\to \infty}\frac{1}{1+\frac{1}{x}}$
Since x is getting really really big, 1/x is getting really really small so we can let it be 0.
$\frac{1}{1+0} = 1$
[/hide]

It's actually easier to see if you don't make the last recombination step; when you have $1 - \frac{1}{x+1}$, it's fairly clear that as x gets big, this expression goes to 1.  (Also, an argument by "substitution of infinity," which as people have pointed out is not rigorous and will often fail you, becomes clearer when you write it like this because you needn't divide infinity by infinity.)


inequalities
floor function
linear algebra
linear algebra unsolved

Let $n \in \mathbb N$, $n \geq 2$.

(a) Give an example of two matrices $A,B \in \mathcal M_n \left( \mathbb C \right)$ such that \[ \textrm{rank} \left( AB \right) - \textrm{rank} \left( BA \right) = \left\lfloor \frac{n}{2} \right\rfloor .  \]

(b) Prove that for all matrices $X,Y \in \mathcal M_n \left( \mathbb C \right)$ we have \[ \textrm{rank} \left( XY \right) - \textrm{rank} \left( YX \right) \leq \left\lfloor \frac{n}{2} \right\rfloor .  \]

[i]Ion Savu[/i]

I really don't know how to do this problem... I tried time and time again, but I can't find the right idea.

 :blush:

the second part was solved here http://www.mathlinks.ro/Forum/viewtopic.php?highlight=sylvester+rank&t=6544

For the first part we can take
$\ A_{2k}=\left(\begin{array}{cc} I_k & O_k \\ O_k & O_k \end{array}\right)$ $\ B_{2k}=\left(\begin{array}{cc} O_k & I_k \\ O_k & O_k\end{array}\right)$ and $\ A_{2k+1}=\left(\begin{array}{cc} A_{2k} & 0 \\ 0 & 0\end{array}\right)$ $\ B_{2k+1}=\left(\begin{array}{cc} B_{2k} & 0 \\ 0 & 0\end{array}\right)$

Very interesting

[url= https://artofproblemsolving.com/community/c7h2576955_matrix_and_rank] see [/url]
https://artofproblemsolving.com/community/c7h2576955_matrix_and_rank

[quote=pezzeepenguinstalker]I swear... most of your most recent College Math posts are literally post farming with two word sentences... maybe think about contributing something helpful?[/quote]

who are you talking to, the OP?

It was referencing #5 who revived a 15 year thread just to say it was interesting

pezzeepenguinsta , honestly I don't see the point of your remark. When someone reports an old result that he considers interesting, I find it very good and, moreover, very common in a forum like ours. If that doesn't interest you then, please, don't disgust others.
In conclusion, keep your strength to perform more constructive acts.

He's saying that simply saying a problem is interesting and no other remark is basically post farming, and I have to agree. At least make some comment regarding how to solve the problem or its origin or something similar. Also, his username isn't 'pezzeepenguinsta'


Hi,
  I am an engineering student, want to brush up on basic math.
 I am particularly weak in geometry. So, i am in a dilemma about which of the following two books to buy
Can somebody point out the adv and disadv of either of them. I am not looking to score high in olympiad or anything. It should provide a clear understanding of basic concepts and should have stimulating problems.
I have done much advanced stuff in math.. like i have taken real analysis and pretty good at complex no.s or trigonometry.. but feel very weak in the basics and on how to approach  challenging problems in general

Books:
A) Intro to Geometry by Aops
B) plane geometry or planimetry by Kiselev(not sure about the title, but heard that it is very good)

If you are looking to just brush up, I know that the AoPS will provide what you need. I like the format of those books. It basically goes concept 1->example->simple question->thinking question. I haven't heard of the other book, but I by no means have researched geometry books.


vector
linear algebra
matrix
algebra
polynomial
linear algebra unsolved

[color=purple] Let $ V$ be a vector space over a field $ \mathbb{Z}/p$. Prove that : the endomorphism $ f : V \rightarrow V$ is diagnalizable iff $ f^p \equal{} f$
 :D   :)
[/color]

You didn't say that $ V$ was finite dimensional, but I'm going to answer this on the assumption that that's what you meant to say - in other words, that we're talking about an $ n\times n$ matrix $ A$ for some $ n.$

Suppose $ A$ is diagonalizable. Then diagonalize it, which means we might as well assume that it's diagonal in the first place. But since $ \lambda^p \equal{} \lambda$ for all $ \lambda,$ we have $ [\text{diag}(\lambda_1,\dots,\lambda_n)]^p \equal{} \text{diag}(\lambda_1^p,\dots,\lambda_n^p) \equal{} (\lambda_1,\dots,\lambda_n).$

Hence $ A \equal{} PDP^{ \minus{} 1}$ and $ A^p \equal{} PD^pP^{ \minus{} 1} \equal{} PDP^{ \minus{} 1} \equal{} A.$

Now for the converse. There's a theorem that a matrix is diagonalizable if and only if it's minimal polynomial splits into [i]distinct[/i] linear factors. But we have that $ A^p \equal{} A,$ which means that the minimal polynomial for $ A$ divides $ x^p \minus{} x.$

$ x^p \minus{} x \equal{} x(x \minus{} 1)(x \minus{} 2)\cdots(x \minus{} p \plus{} 1)$ splits in to distinct linear factors, so any divisor of that polynomial must also split into distinct linear factors. Hence $ A$ is diagonalizable.


MATHCOUNTS FTW AMC USA(J)MO USAMO geometry AIME

Blah I am starting a Fantasy MC League, much like with fantasy baseball, etc.

Post if you wish to play. Rules posted later.

Woah I have a list of predicted top 25 on my blog.

The Rules:

There are 3 positions:

Sprint

Target

Team

A draft occurs, and you take a total of 12 people (6 if lots sign up for this game) for your team, and place them in each category. Note, each state counts as a person for the team round.

Your total score is (Sum of sprint socres for team)+2(Sum of Targets)+5(Sum of Team Rounds).

You get 5 points for each person who make countdown, and then 1 point for each cd your team member wins (a bye gets 1 point). You also get 10 points if a team member wins masters round.

Post to join:

Mewto55555
Xpmath

/in.

The message is too small. Please make the message longer before submitting.

And izzy signs up, completing our 6.

In a random order:

Mewto55555
Jonathonchou711
Pythag011
Isabella2296
Math154
Xpmath

This is the order in which you pick.

M,J,P,I,154,X,X,154,I,P,J,M....etc. until every person has 10 things (2 states, 8 students).

I choose [b]Bobby Shen[/b]

Someone pm chou. Pythag, pm chou and tell him its his turn.

Dude why must I pick last, even though I get to pick again right after?

Um. I'm not sure how this works, so please forgive me of my stupidity. Do I choose a person? If I do, then I choose Darryl Wu? :huuh:

[quote="jonathanchou711"]Um. I'm not sure how this works, so please forgive me of my stupidity. :huuh:[/quote]

You are forgiven.  :wink:

I like country music though! jkjkjkjk.  :lol:

[quote="moogra"][quote="AIME15"][quote="serialk11r"]Man a certain organ must be missing from it.[/quote]

THATS WHAT SHE SAID[/quote]

That was horrible.[/quote]

Well there are good ones and bad ones. This one didn't really make sense.

[quote="moogra"][quote="AIME15"][quote="serialk11r"]Man a certain organ must be missing from it.[/quote]

THATS WHAT SHE SAID[/quote]

That was horrible.[/quote]

I would say it, but I'd be insulting myself...

my science teacher once said, "it's small but you can still do it"

female...
but we were talking about something sciencey

[quote="pianoforte"]hmm, something smells fishy here...[/quote]

I have decided to refrain!

[quote="moogra"]
That was horrible.[/quote]

I have also decided to refrain!

i'd say one other person is enough, but that's kinda weird. anyone else wanna join?


What is your advice for getting into RED MOP? Our young mathletes need your help. Thank you.

You should first qualify for the USAMO. After you have done that, you should solve at least one problem on the USAMO and write it up in a reasonable manner. That will be pretty helpful... presumably the hard part is solving a problem on the USAMO, but one problem is pretty manageable if you just work on some older USAMO #1/#4's to help get a sense of the difficulty and then practice proof-writing.

Look through Engel. Last year, this boosted me over the course of one week from 0-1 problems to 1-2 problems regularly. The only other thing I did to make red was practice a bunch of USAMOs under timed conditions. At first, it is hard to force yourself to work for 4.5 hours (okay... it still is for me), so do it in very comfortable conditions and when there's nothing better to do. Without forcing myself to stay for 4.5 hours, I tended at first to just want to get up and stop after half an hour, which isn't very good.

I imagine Andover has a good enough math team that they will have lots of USAMO tips.
EDIT: My bad. Andover only had 1 USAMO qualifier this year, 1 the year before, 0 the year before that. There are definitely schools in NC that beat that.

And look through the AMC forum. This question has been asked over and over again there.


LaTeX
quadratics
algebra
number theory unsolved
number theory

Find all pairs of positive integers $ x$,$ y$ for which $ \frac{x^2\plus{}y^2}{x\minus{}y}$ is an integer
that divides 1995.

Note that 1995=3.5.7.19

Let d=(x,y).x=da, y=db Denote the given expression by f(x,y) = S ( Im poor at LATEX type setting). S=df(a,b).
Let p be a prime that divides f(a,b). this would imply p divides $ a^{2} \plus{} b^{2}$ and hence, -1 is a quadratic residue mod p ( (a,b)=1 ). Since -1 is a quadratic residue mod p iff p=1(mod4), and since p should be equal to one of 3,5,7,19 we have p=5.
This would force f(a,b)=+/- 1 or 5. since a,b are positive integers, we see that only the second case is possible.

$ a^{2} \plus{} b^{2} \equal{} 5(a \minus{} b)$, $ a^{2} \minus{} 5a \plus{} b^{2} \plus{} 5b \equal{} 0$ Considering this as a quadratic equation in a, discriminant=$ 25 \minus{} 4(b^{2} \plus{} 5b)$= a perfect square. so b=1 and a=2 or 3.
f(a,b)=-5 would give the same set of solutions with the variables reversed.

So, possible (a,b) ordered pairs are (1,2) , (1,3) , (2,1) , (3,1)

Given f(a,b)=5, d can have 8 possible values.
So, there are 32 possible ordered pairs (x,y) taking d= 1, 3, 7, 19, 3.7, 7.19, 3.19, 3.7.19


<math> [\sqrt{n}]+[\sqrt{n+1}]=[\sqrt{4n+2}]</math>,where n is an integer.I am a beginner ,but how to show the formula correctly?

um this definetly does not belong in the FF. i think it should be in high school math> idk. Ask a mod

This should go in either Intermediate or Pre-Olympiad.


find the sum of the infinite series \[ 1\plus{}\frac{1}{3}\plus{}\frac{2}{9}\plus{}\frac{1}{9}\plus{}\frac{4}{81}\plus{}\frac{5}{243}...\]

how do you do this??

(btw the answer is $ \frac{7}{4}$

So that 1/9 is 3/27, right?

[hide="hint"] Let the sum be S. Find S/3 and then subtract to get a geometric series.[/hide]

Another method is do to

$ 1\plus{}\left(\frac13\plus{}\frac19\plus{}\frac1{27}\plus{}\cdots\right)\plus{}\left(\frac19\plus{}\frac1{27}\plus{}\frac1{81}\plus{}\cdots\right)\plus{}\cdots$,

i.e. first simplify the geometric series in the parentheses, and then sum the remaining geometric series.


Calculate the de Broglie wavelenght of a 0.05 eV (thermal) neutron.

Making a nonrelativistic calculation,
the answer is  1.28 A.

$\frac{1}{2}mv^2=eV$, thus

${\lambda=\frac{h}{mv}=\frac{h}{m \sqrt{\frac{2eV}{m}}}=\frac{h}{\sqrt{2meV}}}$


geometry
3D geometry
inequalities
geometry unsolved

a triangle ABC has 2 points X and Y on AC and BC respectively.
Z is a point on XY such that

AX / XC = CY / YB = XZ / ZY.

prove that area of ABC is

  [ (cube root of area of AXZ ) + (cube root of area of BYZ ) ] ^3 .

i tried but failed .
 :maybe: if area of AZB is proved double the area of XYC , THE QUESTION WILL BE  CRACKED

This is a very easy problem. It's only direct consequence of $ Yasin Yilmaz's Theorem$ (also known as yayik's lemma).  :wink:

Let $ {\frac{AX}{XC}\equal{}\frac{CY}{YB}\equal{}\frac{XZ}{ZY}\equal{}x}$; and $ {S\triangle ABC\equal{} S}$ SO WE GET :$ {(S \triangle AXZ)^3\plus{}(S \triangle BZY)^3 \equal{}(\frac{x}{1\plus{}x}\plus{}\frac{1}{1\plus{}x})*cubesqrt{S}}$ So the result easily follows.
This problem can be harder :a problem from CGMO 2003 is not hard but nice and it is similar to this. In that problem we have to prove an inequality and we can use AM-GM inequality to solve it , dear mathlinkers , try it!You will enjoy it . :)
view the problem from CGMO 2003here  http://www.mathlinks.ro/viewtopic.php?p=1359429#1359429

cnyd please send me the link to access what you referred to as

Yasin Yilmaz's Theorem (also known as yayik's lemma).

i couldn't find it on mathworld

CCMath1
what was S in {S\triangle ABC= S} .

also, what did you mean by '' cubesqrtS'' .
how did you bring about this :

{(S \triangle AXZ)^3+(S \triangle BZY)^3 =(\frac{x}{1+x}+\frac{1}{1+x})*cubesqrt{S}}

please help me to understand this.


Let $ M$ be any point which lie in $ (O,\frac{EF}{2})$. $ N,P\in EF$ such that $ ON\equal{}OP, MN\cap (O)\equal{}\{A,C\}, MP\cap (O)\equal{}\{B,D\}$,$ OB\cap AC\equal{}\{K\}, OA\cap BD\equal{}\{L\}, CD\cap EF\{Q\}$. Prove that $ K,L,Q$ are collinear.

Corect me if I am wrong:
$ (O)$ is the circle with the center in $ O$ , and from $ (O,\frac{EF}{2})$ I understand that $ EF$ is a diameter in this circle.

[quote="balan razvan"]Corect me if I am wrong:
$ (O)$ is the circle with the center in $ O$ , and from $ (O,\frac {EF}{2})$ I understand that $ EF$ is a diameter in this circle.[/quote]
Yes, you are right. I think my problem is nice. Who can solve it? :D

$ OA, OB$ cut $ (O)$ again at $ X, Y,$ respectively, $ ABXY$ is a rectangle. $ NY \parallel BD$ cuts $ (O)$ again at $ Z,$ $ BDYZ$ is also a rectangle $ \Longrightarrow$ $ O \equiv BY \cap DZ.$ By Pascal theorem for:
(1) cyclic $ YAZBCX,$ intersections $ N \equiv YZ \cap AC,$ $ O \equiv YB \cap AX,$ $ S \equiv ZX \cap CB$ are collinear $ \Longrightarrow$ $ S \in EF.$
(2) cyclic $ CXDYZB,$ intersections $ Q \equiv CD \cap XY,$ $ S \equiv CB \cap XZ,$ $ O \equiv DZ \cap YB$ are collinear $ \Longrightarrow$ $ Q \equiv CD \cap EF.$
(3) cyclic $ ABCXDY,$ intersections $ K \equiv AC \cap BY,$ $ L \equiv AX \cap BD,$ $ Q \equiv CD \cap YX$ are collinear. QED.