Super cool!!! I'm definitely going to show up now.

Simon did a goode job too, regardless of the fact that i was only there for 10 minutes. but i did read the transcript, so just as goode

[quote="Fierytycoon"][ComplexZeta] ran today's Math Jam.[/quote]

That was a very nice math jam--the [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=21]transcript[/url] is well worth reading. I must be one of the few forum members here who first "knew" ComplexZeta as the author of his Amazon Listmania list. Only later did I learn about this forum. It's great to see the forum participants really get their juices flowing. I'll have to read some of the (lower difficulty level) jam transcripts with my son to show him what he could learn to do.

That listmania definitely needs to be updated. I'm glad people enjoyed the Math Jam. I enjoyed running it too.


quadratics
number theory unsolved
number theory

Prove that:
For an odd prime $ p$, there are exactly $ \frac{1}{4}(p\minus{}4\minus{}(\minus{}1)^{(p\minus{}1)/2})$ consecutive quadratic residue pairs.

For example, {1,3,4,9,10,12} are the quadratic residues of 13, and there are exactly 2 consecutive quadratic residue pairs mod 13, namely {3,4} and {9,10}.

see [url=http://www.mathlinks.ro/viewtopic.php?t=150499]here[/url].
also u can try to find the sum
$ \sum_{i\equal{}1}^{p\minus{}2} \left(\frac{i(i\plus{}1)}{p} \right)$
and then try to prove that if
$ N\equal{}\{n: 1\le n \le p\minus{}2, \left(\frac{n}{p} \right)\equal{}\left(\frac{n\plus{}1}{p} \right)\equal{}1 \}$
then we'll have
$ |N|\equal{}\frac{1}{4} \left( \sum_{i\equal{}1}^{p\minus{}2} \left(1\plus{}\left(\frac{i}{p} \right) \right) \left(1\plus{}\left(\frac{i\plus{}1}{p} \right) \right) \right)$
then the result follows...


Let $ a,b,c$ be nonnegative numbers.Prove that
\[ \frac {a^2}{\sqrt {(a^2 \plus{} ac \plus{} c^2)(a^2 \plus{} ab \plus{} b^2)}} \plus{} \frac {b^2}{\sqrt {(b^2 \plus{} bc \plus{} c^2)(b^2 \plus{} ab \plus{} a^2)}} \plus{} \frac {c^2}{\sqrt {(c^2 \plus{} ac \plus{} a^2)(c^2 \plus{} cb \plus{} b^2)}} \le \frac {2}{\sqrt 3}
\]
Notice that the equality holds when $ a \equal{} b;c \equal{} 0$
P/s:For you

[quote="Allnames"]Let $ a,b,c$ are nonnegative numbers.Prove that
\[ \frac {a^2}{\sqrt {(a^2 \plus{} ac \plus{} c^2)(a^2 \plus{} ab \plus{} b^2)}} \plus{} \frac {b^2}{\sqrt {(b^2 \plus{} bc \plus{} c^2)(b^2 \plus{} ab \plus{} a^2)}} \plus{} \frac {a^2}{\sqrt {(c^2 \plus{} ac \plus{} a^2)(c^2 \plus{} cb \plus{} b^2)}} \le \frac {2}{\sqrt 3}
\]
Notice that the equality holds when $ a \equal{} b;c \equal{} 0$
P?s:For you[/quote]
 It's wrong .  $ a\equal{}b\equal{}1 ; c\equal{}0$ ;)

I have just edited ,sorry!!
This ineuality is also true
$ \frac {a^2}{\sqrt [3] {(a^2 \plus{} ac \plus{} c^2)(a^2 \plus{} ab \plus{} b^2)}} \plus{} \frac {b^2}{\sqrt [3] {(b^2 \plus{} bc \plus{} c^2)(b^2 \plus{} ab \plus{} a^2)}} \plus{} \frac {c^2}{\sqrt [3] {(c^2 \plus{} ac \plus{} a^2)(c^2 \plus{} cb \plus{} b^2)}} \le \frac {2}{\sqrt[3] 3}$
But I prefer the first to the second one

My proof
\[ \frac {a^2}{\sqrt {(a^2 \plus{} ac \plus{} c^2)(a^2 \plus{} ab \plus{} b^2)}} \plus{} \frac {b^2}{\sqrt {(b^2 \plus{} bc \plus{} c^2)(b^2 \plus{} ab \plus{} a^2)}} \plus{} \frac {c^2}{\sqrt {(c^2 \plus{} ac \plus{} a^2)(c^2 \plus{} cb \plus{} b^2)}} \le \frac {2}{\sqrt 3}
\]
Apply Cauchy schwarz inequality,we have
$ LHS^2\le (a \plus{} b \plus{} c)( \frac {a^3}{(a^2 \plus{} ac \plus{} c^2)(a^2 \plus{} ab \plus{} b^2)} \plus{} \frac {b^3}{(b^2 \plus{} bc \plus{} c^2)(b^2 \plus{} ab \plus{} a^2)} \plus{} \frac {c^3}{(c^2 \plus{} ac \plus{} a^2)(c^2 \plus{} cb \plus{} b^2)}) \equal{} T$
We only need to prove
\[ T \le \frac {4}{ 3}
\]
But It is equivalent to
\[ 3(a \plus{} b \plus{} c)(\sum a^3(b^2 \plus{} c^2) \plus{} abc(a^2 \plus{} b^2 \plus{} c^2))\le 4(a^2 \plus{} ac \plus{} c^2)(a^2 \plus{} ab \plus{} b^2) (b^2 \plus{} bc \plus{} c^2)
\]
Apter expanding
\[ \leftrightarrow 2 (a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3) \le abc(a^3 \plus{} b^3 \plus{} c^3) \plus{} 12a^2b^2c^2 \plus{} \sum a^2b^2(a^2 \plus{} b^2)
\]

\[ \leftrightarrow \sum a^2b^2(a \minus{} b)^2 \plus{} abc(a^3 \plus{} b^3 \plus{} c^3) \plus{} 12a^2b^2c^2\ge 0
\]
Which is true since $ a,b,c\ge 0$
We have done.The equality holds when $ a \equal{} b;c \equal{} 0$

This inequality  is easy if you think in terms of Fermat Point of a triangle by moving on to geometry.
I am not posting ht e complete solution but the idea is simple  because you have many $ a^2\plus{}ac\plus{}c^2$  type expressions as well as you know that things like $ \frac{4}{\sqrt3}$ arer involved.


geometry
3D geometry
quadratics
calculus
integration
modular arithmetic
number theory unsolved

Find all $ (m,n)$ natural number such that $ m\minus{}n$ has four positive factor, $ 4|m\plus{}n$, and $ mn$ perfect square

Write $ m \equal{} ka^2, n \equal{} kb^2$ where $ \gcd(a, b) \equal{} 1$.  If $ m \minus{} n \equal{} k(a \plus{} b)(a \minus{} b)$ has four positive factors, then:

[b]Case:[/b]  $ m \minus{} n$ is a product of two primes.  If $ k \equal{} 1$ let $ a \plus{} b \equal{} p, a \minus{} b \equal{} q$ be two odd primes.  Then $ m \equal{} \left( \frac{p\plus{}q}{2} \right)^2, n \equal{} \left( \frac{p\minus{}q}{2} \right)^2$.  Their sum is divisible by $ 4$ if and only if they are both divisible by $ 4$, so $ p \plus{} q \equiv p \minus{} q \equiv 0 \bmod 4$, contradiction.  If $ a \minus{} b \equal{} 1$ then $ a \plus{} b \equal{} 2b \plus{} 1$.  Let $ k \equal{} p, 2b\plus{}1 \equal{} q$ be two primes (the second of which is odd).  Then $ m \equal{} p(b\plus{}1)^2, n \equal{} pb^2, m \plus{} n \equal{} p(2b^2 \plus{} 2b \plus{} 1)$.  The second factor is odd, so this again gives a contradiction.

[b]Case:[/b]  $ m \minus{} n$ is the cube of a prime.  If $ k \equal{} 1$ let $ a \plus{} b \equal{} p^2, a \minus{} b \equal{} p$ where $ p$ is an odd prime; we run into the same problem as before.  If $ a \minus{} b \equal{} 1$ let $ k \equal{} p, 2b\plus{}1 \equal{} p^2$ or $ k \equal{} p^2, 2b\plus{}1 \equal{} p$; we run into the same problem as before.

Hence no such $ (m, n)$ exist.

the only numbers that have 4 factors are those of the form $ p^3$ or $ pq$, where $ p,q$ are distinct primes.

Case 1. $ m \minus{} n \equal{} p^3$.

note that since $ 4|m \plus{} n$, $ m$ and $ n$ have the same parity, and thus $ m \minus{} n$ is even, and so $ p \equal{} 2$. then $ m \equal{} n \plus{} 8$, so $ 4|m \plus{} n \equal{} 2n \plus{} 8 \equal{} 2(n \plus{} 4)$, which means $ n$ is even: set $ n \equal{} 2k$. then $ mn \equal{} 2k(2k \plus{} 8) \equal{} 4k(k \plus{} 4)$. for this to be a square, we need $ k^2 \plus{} 4k \equal{} s^2$ for some integer $ s$. the discriminant of this quadratic in $ k$ is $ 16 \plus{} 4s^2$, so we need $ 16 \plus{} 4s^2 \equal{} t^2$ for some integer $ t$. this means $ 16 \equal{} (t \minus{} 2s)(t \plus{} 2s)$, whose only solution has $ s \equal{} 0$. this gives $ k \equal{} 0, \minus{} 4$, so there are no positive integral solutions in this case.

Case 2. $ m \minus{} n \equal{} pq$.

again, $ m \minus{} n$ is even, so we may assume WLOG that $ p \equal{} 2$. now we need $ mn \equal{} n(n \plus{} 2q) \equal{} s^2$ for some integer $ s$, i.e. $ n^2 \plus{} 2qn \minus{} s^2 \equal{} 0$. the discriminant of this quadratic in $ n$ is $ 4q^2 \plus{} 4s^2$, so we need $ q^2 \plus{} s^2 \equal{} t^2$ for some integer $ t$. in other words, $ q^2 \equal{} (t \minus{} s)(t \plus{} s)$. the only solutions (up to sign) of this equation are $ (t,s) \equal{} (\frac {q^2 \plus{} 1}{2},\frac {q^2 \minus{} 1}{2})$ and $ (t,s) \equal{} (q,0)$. the first gives us $ n \equal{} \frac { \minus{} 2q \plus{} q^2 \plus{} 1}{2} \equal{} \frac {(q \minus{} 1)^2}{2}$ and thus $ m \equal{} \frac {(q \plus{} 1)^2}{2}$, while the second gives us $ n \equal{} \frac { \minus{} 2q \plus{} 2q}{2} \equal{} 0$, which we immediately disregard. in the former case, however, $ m \plus{} n \equal{} q^2 \plus{} 1\equiv 2\pmod{4}$. so again we have no solutions.

poor problem has no solutions :(

edit: i see t0rajir0u got to this problem before me and confirmed the disappointment

If $ m \plus{} n$ is divisible by 4, they must be both odd or both even. But if both odd, they must be of the form $ m \equal{} 4k_1\pm1, n \equal{} 4k_2\mp1$ for some $ k_1,k_2\in\mathbb{Z}$; then $ mn\equiv \minus{} 1\pmod{4}$ which can’t be a perfect square. Hence $ m$ and $ n$ must be even, and $ m\equiv n\equiv0\pmod{4}$ or $ 2\pmod{4}$. In either case, $ m \minus{} n$ is also divisible by 4; hence the only way for it to have exactly four factors is for $ m \minus{} n \equal{} 8$. It shouldn’t be difficult to check that $ mn$ can’t then be a perfect square.

Here’s the proof. We can write $ n \equal{} 2k \minus{} 4$ and $ m \equal{} 2k \plus{} 4,\ k\geq3$. Then $ mn \equal{} 4(k^2 \minus{} 4)$. In order for this to be a perfect square, $ k^2 \minus{} 4$ must be a perfect square, say $ k^2 \minus{} 4 \equal{} a^2$, so $ k^2 \minus{} a^2 \equal{} 4$. The difference between the two perfect squares must be at least $ k^2 \minus{} (k \minus{} 1)^2 \equal{} 2k \minus{} 1$. But $ k\geq3\Rightarrow2k \minus{} 1\geq5$. So we have the contradiction $ 4\geq5$. Hence $ mn$ can’t be a perfect square.


It's obviously trues because by Am-Gm, we are easily to see that:
LHS $ \geq\ \sqrt{a\plus{}b\plus{}c} \geq$ RHS

You have a mistaken typo. It should be $ a\plus{}b\plus{}c$. Anyway, your hint is so true


linear algebra
matrix
algebra
polynomial
vector
modular arithmetic
function

[b]Theoretical construction:[/b]

If $ \bar{e}_i,i\equal{}1,\ldots,n$ denotes the standard basis of $ K^n$, define the permutation matrix $ P\in GL_n(K)$ via $ P\bar{e}_i\equal{}\bar{e}_{i\minus{}1}$ (indexes are considered modulo $ n$). Note, that $ P^{\minus{}1}\equal{}P^T,P^n\equal{}I$ to be more precise the minimal polynomial of $ P$ is given by $ m_P(x)\equal{}x^n\minus{}1$. We can describe matrices of the form as stated in the problem spec either

in row form $ X_r(\bar{u}): \equal{}\left(\begin{array}{c}\bar{u}^T\\(P\bar{u})^T\\ \vdots \\(P^{n\minus{}1}\bar{u})^T\end{array}\right)$ or in column form $ X_c(\bar{v}): \equal{}\left(\bar{v},P\bar{v},\ldots,P^{(n\minus{}1)}\bar{v}\right)$

For arbitrary $ \bar{a}\in K^n$ with $ X_r(\bar{a})$ invertible choose an (orthogonal) vector $ \bar{0}\not\equal{}\bar{b}\in <P\bar{a},\ldots,P^{n\minus{}1}\bar{a}>^{\perp}$. Direct calculation yields for the coefficient $ [X_r(\bar{a})X_c(\bar{b})]_{ij}\equal{}(P^{i\minus{}1}\bar{a})^TP^{j\minus{}1}\bar{b}\equal{}(P^{i\minus{}j}\bar{a})^T\bar{b}$ from which we conclude $ X_r(\bar{a})X_c(\bar{b})\equal{}\bar{a}^T\bar{b}\cdot I$.

If $ \bar{a}^T\bar{b}\equal{}0$, then $ \bar{b}$ is orthogonal to $ <\bar{a},P\bar{a},\ldots,P^{n\minus{}1}\bar{a}>$, a contradiction, since the dimension of this space equals the rank of $ X_r(\bar{a})$. Therefore, we have $ \bar{a}^T\bar{b}\not\equal{}0$ and by potentially normalizing $ \bar{b}$ such that $ \bar{a}^T\bar{b}\equal{}1$ holds, we can assume, that $ X_r(\bar{a})X_c(\bar{b})\equal{}I$ is valid.

[b]Exact construction:[/b]

Although a pointless exercise (I did not try :D ) apply Gram-Schmidt, i.e. recursively determine vectors $ \bar{c}^{(1)},\ldots,\bar{c}^{(n)}$ via

$ \bar{c}^{(k)}\equal{}P^k\bar{a}\minus{}\sum_{i\equal{}1}^{k\minus{}1}\frac{\bar{a}^TP^{\minus{}k}\bar{c}^{(i)}}{\parallel{}\bar{c}^{(i)}\parallel{}^2}\,P^i\bar{a}$ with the usual convention $ \sum_{\emptyset}\ldots \equal{}0$. Then choose $ \bar{b}$ as an appropriate multiple of $ \bar{c}^{(n)}$.

The [url=http://en.wikipedia.org/wiki/Circulant_matrix]circulant matrices[/url] over $ R$ form a commutative algebra isomorphic to $ R[x]/(x^n \minus{} 1)$ (which you can also see if you diagonalize them), so all you have to do is determine the inverse of $ a_1 \plus{} a_2 x \plus{} ...$ in this ring, which is some multiple of the product of its conjugates.

[quote="t0rajir0u"](which you can also see if you diagonalize them)[/quote]
As a Fourier analyst, I like to claim circulant matrices as part of my territory.

Notation: In an $ N\times N$ circulant matrix, it's a little nicer to let the indices run from $ 0$ to $ N\minus{}1$ rather than $ 1$ to $ N.$ There is arithmetic to be performed on the indices, and that arithmetic is always performed $ \pmod{N}.$

Notation: There's a fairly obvious one-to-one correspondence between complex circulant matrices and vectors in $ \mathbb{C}^N.$ After all, any single row or single column completely determines the circulant matrix. In turn, I would prefer in this instance to refer to $ \mathbb{C}^N$ as the set of all complex valued functions with domain $ \mathbb{Z}/(N\mathbb{Z}).$ (Harmonic analysts prefer to deal with functions defined on groups.)

With that notation understood, the matrix multiplication of circulant matrices is the same as the [i]convolution[/i] of functions on $ \mathbb{Z}/(N\mathbb{Z}).$

And since the relationship between convolution and the Fourier transform is always that convolution on one side of the transform is equivalent to pointwise multiplication on the other side of the transform (possibly with a scale factor thrown in), we get that the discrete Fourier transform diagonalizes all circulant matrices.

That is: Let $ \zeta\equal{}e^{2\pi i/N}$ be the "first" $ N$-th root of unity. Define a matrix $ F$ so that the $ (j,k)$ entry of $ F$ is $ \zeta^{jk}\equal{}e^{2\pi ijk/N}.$ Here, I'm sticking to that first notational convention above and having $ 0\le j,k\le N\minus{}1.$ Note that that product $ jk$ in the exponent can be computed $ \pmod{N}.$

$ F$ is "almost" a unitary matrix. Its columns are orthogonal but a constant multiple away from being orthonormal. To be specific, $ F^*F\equal{}NI.$

Then if $ C$ is a circulant matrix, we have that $ \frac1NF^*CF\equal{}D$ is diagonal. Or interchange the roles of $ F$ and $ F^*$ if you want. Essentially, the eigenvalues of $ C$ are the Fourier coefficients of the vector that determines $ C.$

In the framework I've just set up, we understand sunlight's original question this way: he is asking how to find the inverse of a complex $ N$-vector with respect to convolution. (The identity element for convolution is the Kronecker $ \delta$ - the function which has a $ 1$ in position $ 0$ and a $ 0$ everywhere else. Of course, as a circulant, this is simply the identity matrix.)

In order for such a vector (function on $ \mathbb{Z}/(N\mathbb{Z})$) to have an inverse, its Fourier transform must have no zeros. That is, we must have that

$ \sum_{j\equal{}0}^{N\minus{}1}a_j\zeta^{\minus{}jk}\ne0$ for all $ k\in \mathbb{Z}/(N\mathbb{Z}).$

A particular case for sunlight's setup. (Think of $ n$ complex numbers which form the vertices of an $ n$-gon, and think of the operation of forming a new $ n$-gon by connecting together the centers.)

Let $ B\equal{}\frac12\begin{pmatrix}0&1&0&\dots&1\\ 1&0&1&\dots&0\\ 0&1&0&\dots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&0&0&\dots&0\end{pmatrix}.$

Is $ B$ invertible?

[quote="sunlight"]find the inverse matrix of A
$ A \equal{} \begin{pmatrix} a_1 & a_2 & a_3 & ... & a_n \\
a_n & a_1 & a_2 & ... & a_{n \minus{} 1} \\
a_{n \minus{} 1} & a_n & a_1 & ... & a_{n \minus{} 1} \\
... \\
a_2 & a_3 & a_4 & ... & a_1 \end{pmatrix}$
 :D  :D[/quote]
 Let $ E \equal{} \begin{pmatrix} 0 & 1 & 0 & 0 & ... & 0 \\
0 & 0 & 1 & 0 & ... & 0 \\
0 & 0 & 0 & 1 & ... & 0 \\
... \\
0 & 0 & 0 & ... & 0 & 1 \\
1 & 0 & 0 & ... & 0 & 0 \end{pmatrix}$
Then we have $ A \equal{} a_1I_n \plus{} a_2E \plus{} a_3E^2 \plus{} .... \plus{} a_nE^{n \minus{} 1} \equal{} f(E)$
 Equation of $ E$ is $ ( \minus{} 1)^n(\lambda^{n} \minus{} 1) \equal{} 0$  then $ \lambda_k \equal{} cos(\frac {2k\pi}{n}) \plus{} isin(\frac {2k\pi}{n}) ,\forall k \in$ {$ 0,2,...,n \minus{} 1$} then $ f(E) \equal{} Tdiag(f(\lambda_0),f(\lambda_1),...,f(\lambda_{n \minus{} 1}))T^{ \minus{} 1}$
If $ A \in GL_n(C)$ then $ f(\lambda_k) \neq 0 ,\forall k \in$ {$ 0,1,...,n \minus{} 1$}
and $ A^{ \minus{} 1} \equal{} Tdiag(f^{ \minus{} 1}(\lambda_0),f^{ \minus{} 1}(\lambda_1),...,f^{ \minus{} 1}(\lambda_{n \minus{} 1}))T^{ \minus{} 1}$
we have $ \det(A)\equal{}\prod_{k\equal{}0}^{n\minus{}1}{f(\lambda_k)}$
:roll:  :roll:

[b]As Mr Kent Merryfield [/b]: $ 2B \equal{} E \plus{} E^{n \minus{} 1}$ if take $ n \geq 3$
      $ 2B \equal{} E(I_n \plus{} E^{n \minus{} 2})$
 We have :
   $ n \equal{} 2$ then $ B \in GL_2(C)$
   $ n \geq 3$ then
if $ n$ is even and  $ n\equiv 0 (\mod4)$ then the exist $ t \in${$ 0,1,..,n \minus{} 1$}, $ \lambda^{2}_t \plus{} 1 \equal{} 0$ and since $ det(2B) \equal{} \det(E)\prod_{k \equal{} 0}^{n \minus{} 1}{(1 \plus{} \lambda_k^{2})\lambda_k^{n \minus{} 2}} \equal{} 0$
If $ n$ is odd or $ n \equal{} 4s \plus{} t$ ,$ t \in$ {1,2,3} then $ E$ don't have eigenvalues are $ 0$ or $ \plus{} i, \minus{} i$ then $ \det(B) \neq 0$
 Answer : $ \det(B) \neq 0$ if $ n$ is odd  or $ n \equal{} 4s \plus{} t$ ,$ t \in$ {1,2,3}
               $ \det(B) \equal{} 0$ if $ n \equiv 0 (\mod4)$  !!!
[b]Sorry if I am  wrong[/b] !!!  :blush:

[quote="Kent Merryfield"][quote="sunlight"]B is invertible. (det (2B) =2)
:lol:[/quote]
Not so fast. I don't accept this as the complete answer.[/quote]
oh, i'm sorry. i see this matrix in the solution of one problem in imc2007 day 2 (case n is odd):D

can you help me with this?
The particular for the problem i posted:
$ A\equal{} \begin{pmatrix} 1 & 2 & 3 & 4&... & n \\n & 1 & 2 & 3& ... & n\minus{}1 \\n\minus{}1 & n & 1 & 2&... & n\minus{}2 \\n\minus{}2 & n\minus{}1 & n & 1&... & n\minus{}3 \\...\\2 & 3 & 4 & 5& ... & 1 \end{pmatrix}$
$ B\equal{} \begin{pmatrix} 1 & 2 & 3 & 4& ... & n \\2 & 3 & 4 &  5 &... & 1 \\3 & 4 & 5 &6& ... & 2 \\4 & 5 & 6 &7& ... & 3 \\...\\n & 1 & 2 &3& ... & n\minus{}1 \end{pmatrix}$
find $ A^{\minus{}1}$ and $ B^{\minus{}1}$

[b]The fist question for you  prove that[/b]
$ A ,B \in GL_n{(\mathbb{N})},\forall n \in N^{*}$
[b] Oke , I will answer the question of you !!! [/b]

QuyBac: the actual answer to my problem is that $ B$ is invertible if $ n$ is odd and not invertible if $ n$ is even. (Other than the $ n\equal{}2$ case which - for geometric reasons - I consider degenerate and undefined.)

The eigenvalues, also known as the Fourier coefficients, are what helps decide this.

[quote="Kent Merryfield"]QuyBac: the actual answer to my problem is that $ B$ is invertible if $ n$ is odd and not invertible if $ n$ is even. (Other than the $ n \equal{} 2$ case which - for geometric reasons - I consider degenerate and undefined.)
The eigenvalues, also known as the Fourier coefficients, are what helps decide this.[/quote]
 Please Mr Kent Merryfield  explain to go wrong of me with $ n\equal{}4s\plus{}2,s \in N^{*}$ !
Please to have a try at  $ n\equal{}6,10$  :( Thanks !)

Label the matrices so their entries run from $ 0$ to $ n\minus{}1.$ We have $ B_{jk}\equal{}\frac12$ if $ j\minus{}k\equiv \pm 1\pmod{n}$ and $ 0$ otherwise. Let $ F_{jk}\equal{}\zeta^{jk}$ where $ \zeta\equal{}e^{2\pi i/n}.$

Let's compute the $ jk$ element of $ BF.$ It equals

$ \frac12\left(\zeta^{(j\minus{}1)k}\plus{}\zeta^{(j\plus{}1)k}\right)\equal{}\zeta^{jk}\left(\frac12\left(\zeta^{\minus{}k}\plus{}\zeta^k\right)\right)\equal{}\cos(2\pi k/n)\zeta^{jk}.$

Now left-multiply by $ \frac1nF^*,$ which is the inverse of $ F.$ The $ jk$ element of this product is

$ \frac1n\cos(2\pi k/n)\sum_{m\equal{}0}^{n\minus{}1}\zeta^{\minus{}jm}\zeta^{mk}$

But $ \frac1n\sum_{m\equal{}0}^{n\minus{}1}\zeta^{m(k\minus{}j)}\equal{}\delta_{kj},$ so our matrix element becomes $ \cos(2\pi k/n)$ for $ j\equal{}k$ and zero otherwise.

We have diagonalized $ B$ and its eigenvalues are $ \cos\left(\frac{2\pi k}n\right)$ for $ 0\le k\le n\minus{}1.$

That makes the question of whether $ B$ is invertible the same as the question of whether $ \cos\left(\frac{2\pi k}n\right)$ ever equals $ 0.$  That is, we check to see if we can have $ \frac{2\pi k}n\equal{}\frac{\pi}2.$ This happens when $ k\equal{}\frac{n}4.$

Quy Bac is correct, and I now realize my mistake - this wasn't the problem I intended. I was thinking of this matrix:

$ C\equal{}\frac1n\begin{pmatrix}1&1&0&\dots&0\\ 0&1&1&\dots&0\\ 0&0&1&\dots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&0&0&\dots&1\end{pmatrix}.$

[color=blue]I have a mistake here that if $ \lambda \neq 1$eigenvalues of $ A$ is $ \frac {n\lambda^n_k}{\lambda_k \minus{} 1} \equal{} \frac {n}{\lambda_k \minus{} 1}$ by $ \lambda^n_k \equal{} 1,\forall k \in$ {12,..,n-1}[/color]
Here $ A \equal{} f(E) \equal{} I_n \plus{} 2E \plus{} 3E^2 \plus{} ... \plus{} nE^{n \minus{} 1}$ from that we have  $ \det(A) \equal{} f(1)\prod_{k \equal{} 1}^{n \minus{} 1}{f(\lambda_k)} \equal{} \frac {n(n \plus{} 1)}{2}\prod_{k \equal{} 1}^{n \minus{} 1}{\frac {n}{\lambda_k \minus{} 1}}$ , Here $ \lambda_k \equal{} cos(\frac {2k\pi}{n}) \plus{} isin(\frac {2k\pi}{n}) ,\forall k \in$ {$ 1,2,..,n \minus{} 1$}
And $ \det(E_1) \equal{} \minus{} ( \minus{} 1)^{\frac {n(n \plus{} 1)}{2}}$ then we have :
 $ \det(B) \equal{} \begin{pmatrix} 1 & 2 & 3 & 4 & ... & n \\
2 & 3 & 4 & 5 & ... & 1 \\
3 & 4 & 5 & 6 & ... & 2 \\
4 & 5 & 6 & 7 & ... & 3 \\
... \\
n & 1 & 2 & 3 & ... & n \minus{} 1 \end{pmatrix} \equal{} \det(E_1)\det(A) \equal{} \minus{} ( \minus{} 1)^{\frac {n(n \plus{} 1)}{2}}\frac {n(n \plus{} 1)}{2}n^{n \minus{} 1}\prod_{k \equal{} 1}^{n \minus{} 1}{\frac {1}{\lambda_k \minus{} 1}}$
-------------------------------------------------------
[color=red]If can for me edit here that ! Thanks! [/color]

Let me jump in this discussion with another question.
Let's say I have a Circulant matrix A generated by the vector  (a_0, a_1, ..., a_{-n1})
and  \sum_{i=0}^{n-1}  |a_i| < infinity
Assume also that the circulant matrix has an inverse,  A^{-1}.
Are the elements of the inverse of A also absolutely summable ?

$ \sum_{i\equal{}0}^{n\minus{}1} |a_i| < \infty\ ?$

That's a finite sum. As long as each of the $ |a_i|$ is a complex number, the sum is going to be finite. And if the inverse exists, then each of the entries of the inverse is a complex number, that sum is also finite.

There's really nothing to see here - I don't understand the purpose of the question.

[quote="Kent Merryfield"]$ \sum_{i \equal{} 0}^{n \minus{} 1} |a_i| < \infty\ ?$

That's a finite sum. As long as each of the $ |a_i|$ is a complex number, the sum is going to be finite. And if the inverse exists, then each of the entries of the inverse is a complex number, that sum is also finite.

There's really nothing to see here - I don't understand the purpose of the question.[/quote]

Kent, you are right.
I actually meant :
$ \lim_n \sum_{i \equal{} 0}^{n \minus{} 1} |a_i| < \infty$.
 So, for increasing n, the row sums of the original matrix A are bounded in absolute value. What can we say about the row sums of the inverse of A when n tends to infinity ?
What your post means is that, in the general case, the sum of the absolute values of the rows of the inverse of A is of order O(n). What I am looking for is conditions on the original vector "a" that would insure the sum of absolute values of the vector that generates the inverse of A is also bounded. Any idea ?


calculus
integration
graph theory
number theory
RSI
research

A year ago I took a precalculus class and did very well.  My teacher told me that I should seriously consider majoring in math.  So I continued on to calculus, and somtimes when my teacher (the same teacher from pre calc) was not feeling good he would let me go to the board.  I would often go to his office and talk about wanting to do math research.  He even gave me an extra book to go over during the summer.  Now I'm in calculus 2 (different teacher) and I have continued to do very well.  My teacher told me to do more challenging problems after I corrected an improper integral problem she did completely wrong in class.  Then I ended up getting a 97 on the exam.  The point I'm trying to make is that I do math all of the time.  That's all I think about and the only reason why I'm saying this is because my real name is covered.  I feel like it's a waste of time to spend long lengths of time at a university and then start doing research when I'm 30 something.  I spend so much time learning on my own because I hardly learn anything from teachers.  What is the point in spending a whole bunch of time and money at a university when I feel like I really do have something to offer mathematics and I can do it on my own?  I have found mathematics to be a very lonsome thing.  None of my friends understand my passion for math, in fact they all hate it!  I made really good friends with my teacher that I had for a year but it turns out that he can't keep talking to me because he's either busy or doesn't think it's appropriate. But it's so frustrating because he was the only person who understood me.  I don't want to just be another student, I want to do research!  So next semester I decided I'm going to buy the books for some of the higher up classes and go ahead and do it on my own.  But, can someone please tell me how far I have to go?

It's good that you have a desire to do these things but if you want to do research in a university setting, you're going to have to stay all the way through because you're going to need that Ph.D., and the guidance of an experienced pro in the field.

As I understand it, research typically begins at the graduate level.  Some of the higher-tier colleges have undergraduate research programs that allow you, with the guidance of a professor, to begin research as early as your freshman year of college.  There are also summer programs and other opportunities to do research at the [i]high school[/i] level - consider all the math projects submitted to competitions like [url=http://www.siemens-foundation.org/en/competition.htm]Siemens[/url] or [url=http://www.societyforscience.org/sts/]Intel[/url].  In other words, it's never too early to start doing research.

On the other hand, the ease of doing research alone (and research is rarely done alone) depends heavily on 1) your mathematical background and experience, and 2) the nature of the specific field of mathematics you wish to work in.  Some fields like (certain branches of) combinatorics and graph theory do not require too many prerequisites, whereas if you wish to do research in a heavy field like algebraic number theory you will need years of background and experience to properly understand the current state of the field.  Your best bet is to seek out a professor at a local university and ask for guidance; you'll have a very tough time doing any research without a mentor to guide you.

Consider the following:  (real) calculus in its entirety was more or less worked out by the 19th century.  So what exactly have mathematicians been up to for the last two centuries?

On the third hand, you might find [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php]this article[/url] interesting.

I understand what you said about research not being done alone.  It's not good to do it alone because you can make a lot of mistakes.  I looked on my universities website and they have some professors who assist undergraduates in research.  so I am going to try to get involved in that.  I am just going to try to get as far as I can with the time I have.  Thanks for your replies.

[quote="improper_integral"]It's not good to do it alone because you can make a lot of mistakes.[/quote]

Guidance can be of enormous help just in determining what to work on.  Deciding on a question to answer is half the struggle.  Without direction, you could waste years working on a problem that a professor would've told you just don't have the background for (say, the Riemann Hypothesis).  Again, this depends heavily on the field; some problems are like this and some problems aren't.

As far as my own experience goes, at the Research Science Institute last year my mentor was new and wanted to give me a research problem that fit my application (where I talked about some pretty difficult problems) without being enormously difficult.  In hindsight, the problem we landed on (in the field of elliptic curves) may have been a little difficult, but I learned a lot in the process and was able to write a decent paper anyway.

Also, if you are interested in finding some current research problems, you may wish to check out Open Problem Garden: http://garden.irmacs.sfu.ca/

[quote="JRav"]When I entered college last year, I asked about research opportunities.  As t0rajir0u pointed out though, there aren't too many research opportunities for undergrads because NSF tends to fund graduate students.  However, there are several summer programs that one can apply to.[/quote]

Can you tell us some of these programs?

I see that you are from Greece.  Do you attend school in the US because that makes a big difference.

Several universities host Research Experience for Undergraduates (REU's) in which small groups of students work with a professor on some research topic (usually in the advisor's field).  These are pretty competitive, but most programs will not admit international students.

Oh... I'm student at the university of Athens, Greece.  :(

Some programs will allow international applicants, but many won't (usually because their insurance won't cover you or something).  Also, for this summer, most of the deadlines have already passed. :(


How many games do i have to play on FTW before getting to chat?

I'm a new user so i haven't played that many games yet.

The FAQ indicates it's generally less than 60 games.

Usually it is about 40-55ish but it changes. I think you also have to have a couple of posts and a post rating that is under 2.
Also this belongs in the FTW forum.
And Also use the search button before posting a question it can save you time.

[quote="monkeygirl13"]a post rating that is under 2.
[/quote]

Over?

I've played between 20-30 games,
yet I have 800< posts and have a 3< post rating.
And I chat :).

I could chat after about 15 posts and 20 games with a "Not rated yet".

It generally varies from person to person, and the amount of games played is currently "top secret", although everybody generally has a different amount of posts/games.

It was 60 games for me. Shouldn't this be on the FTW thread?

In a sense, yes.

In another sense, considering the cornucopia of FTW threads saying the exact same thing, this thread shouldn't happen at all. Moral of the day: [color=red]stare[/color] [color=green]at[/color] [color=violet]that[/color] [color=orange]pretty[/color] [color=red]"[/color][color=orange]S[/color][color=#CCCC00]e[/color][color=green]a[/color][color=blue]r[/color][color=darkblue]c[/color][color=indigo]h[/color][color=violet]"[/color] [color=olive]button[/color] and you'll [color=blue]eventually[/color] be [color=orange]enlightened[/color] by the [color=violet]usefulness[/color] of it.


analytic geometry
graphing lines
slope
calculus
derivative
quadratics
linear algebra

Find the equation of the tangent line to the curve x^2 + y^2 = 169 at the point (5,-12).

Differentiate implicity, then $ y'$ will be the slope of the tangent line. Plug the coordinates into the derivative to get the value of the aforesaid slope.

Apply the point slope formula and you are done.

2x+2y(dy/dx) = 0
(dy/dx) = (-2x)/2y

now what do i do?

As said earlier, the idea behind these problems is first to differentiate, use the chain rule somewhere, solving for a variable which you can then write a formula for.

In your case, you've got $ x^2 \plus{} y^2 \equal{} 169$.
Differentiate WRT x.

$ 2x \plus{} 2y\cdot \frac {dY}{dX} \equal{} 0$
Solve for $ \frac {dY}{dX}$ to get a slope. (it turns out to be $ \minus{} \frac {x}{y}$)

Remember from algebra where they taught you that any line can be represented in "point slope form"?

All that means is this: If you have a point on a line (oh, say, $ (y_1, x_1)$), and you have the slope (let's call it $ m$), then the equation of the line is $ Y \minus{} y_1 \equal{} m(X \minus{} x_1)$

So, we've got our slope at any given point ($ \minus{} \frac {x}{y}$) and we have our coordinates: $ (5, \minus{} 12)$

So plug those values in:
$ Y \minus{} ( \minus{} 12) \equal{} \minus{} \frac{5}{ \minus{} 12} \cdot (X \minus{} 5)$
$ Y \plus{} 12 \equal{} \frac {5}{12}X \minus{} \frac{25}{12}$
$ Y \equal{} \frac {5}{12}X \minus{} \frac {169}{12}$

The equation of the tangent line to the curve $ x^2 \plus{} y^2 \equal{} r^2$ at the point $ (x_0,\ y_0)$ is $ x_0x \plus{} y_0y \equal{} r^2$.
Isn't this a formula? Using scalar product gives a solution directly.

The equation of the tangent line to the quadratic form $ \mathbf{XAX}^T \equal{} 0$ (where $ \mathbf{X} \equal{} \left( x, y, 1 \right)$ and $ \mathbf{A}$ is a symmetric $ 3 \times 3$ matrix) at a point $ \mathbf{P}$ is given by

$ \mathbf{PAX}^T \equal{} 0$.

This is called the [i]polarization[/i] of the quadratic form $ \mathbf{XAX}^T \equal{} 0$ (and is valid in any number of variables).

In this case, $ \mathbf{A} \equal{} \left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & \minus{}r^2 \end{array} \right)$.  Substituting produces the same result as kunny.  One way to see this result is to view such forms as Taylor series in two or more variables, and to center them at $ \mathbf{P}$; then we derive the equation of the tangent line (plane, hyperplane) simply by discarding any terms of higher than linear order.


[quote="Math Geek"]Hmm, this question seems annoying :mad::

The equation MATH=COU+NTS is true if each of the letters is replaced by a different digit and the equation is seen as the sum of two three-digit numbers resulting in a four-digit number.  If C=3, a solution can be found in which one of the three-digit integers is a multiple of 23.  In that case, what four-digit integer does MATH represent?

PS: I have the answer key, so include a solution with your answer.[/quote]
+COU
+NTS
MATH
Well the only three digit number is COU and NTS so first, we'll try have COU be the multiple of 23.
322
345
368
391
However, we can use logic to find that unless COU is 391, you cannot obtain a T while adding a T. This also means that U and S must be > 9.
However, since U would be 1, and the only way to get > 9 would be to add 9, and we have already used 9, it will not work.
Therefore, COU cannot be the multiple of 23.
Now we can start substituting numbers.
+3OU
+NTS
MATH
Usable digits: 1 2 4 5 6 7 8 9 0
You will immediately notice that N will need to be either 6, 7, 8, or 9.
Also, O will have to be either 0 or 9.
If N is 7, then O will have to be 0, but then A cannot be 0, so N cannot equal 7.
If N is 8, then O will need to be 9. Also, by checking the multiples of 23, we can find that NTS needs to be 874, because the other multiples don't work out.
+39U
+874
127H
Usable digits: 4 5 6 0
U+4 needs to be greater than 9, meaning that U needs to be 6.
Therefore, the final equation becomes.
396+874=1270
So MATH is 1270.

Nice logic.  Yeah, that's the answer.  Thanx!


There are 10 people, A, B, C, D, E, F, G, H, I, and J.
They will sit in a row containing 10 chairs inside a cinema.
A, B, C, and D aren't allowed to sit next to each other.
E, F, G aren't allowed to sit next to each other.
H and I aren't allowed to sit next to each other.
One such possible arrangement is:
A-B-D-F-C-G-I-J-H-E
And this arrangement is not possible:
A-B-F-G-E-D-H-C-J-I
How many seating arrangements are possible?


integration
logarithms
probability and stats

If $ P$ and $ Q$ are two prob. measures on a measure space, we define $ I(P\|Q) \equal{} \int \log\left( \frac {dP}{dQ}\right) dP \equal{} \int \frac {dP} {dQ} \log\left(\frac {dP}{dQ}\right)$   when $ P < < Q$.
Question is, when would $ I(P\|Q)$ be $ \infty$?

check: http://en.wikipedia.org/wiki/Kullback-Leibler_divergence
there are some mistakes in your definitions ..

There is one $ dQ$ missing. It is $ I(P\|Q) \equal{} \int\log\left(\frac {dP}{dQ}\right) dP \equal{} \int\frac {dP}{dQ}\log\left(\frac {dP}{dQ}\right)dQ$ when $ P$is absolutely continuous w.r.to  $ Q$
Apart from that I don't think there is any mistake.


function
algebra
polynomial
induction
algebra unsolved

Find the functions satisfying f(x+y) = f(x) + f(y) +xy

[hide="thoughts and hint from book"]asking for all real functions
the solution suggests a method by assuming that f(x) is a polynomial[/hide]

Let $ f(x) \equal{} \frac{x^2}{2} \plus{} g(x)$, then $ g(x \plus{} y) \equal{} g(x) \plus{} g(y)$.

[hide="(Incomplete?) Solution"]
[b]Lemma.[/b] $ f(x\plus{}n)\minus{}f(x)\equal{}nf(1)\plus{}nx\plus{}\frac{(n\minus{}1)n}{2}$ for any integer $ n$.
[hide="Proof"]In the given condition, let $ y\equal{}1$ to get
\[ f(x\plus{}1)\minus{}f(x)\equal{}f(1)\plus{}x\]
Thus, the lemma holds for a base case $ n\equal{}1$.

Now assume that what we wish to prove holds for some integer $ m$.
That is, $ f(x\plus{}m)\minus{}f(x)\equal{}mf(1)\plus{}mx\plus{}\frac{(m\minus{}1)m}{2}$.
Simply plug in $ m\equal{}m\plus{}1$ to get:
$ f(x\plus{}m\plus{}1)\minus{}f(x)\equal{}(m\plus{}1)f(1)\plus{}(m\plus{}1)x\plus{}\frac{m(m\plus{}1)}{2}$,
which takes care of cases where $ n>1$.
Plugging in $ m\equal{}m\minus{}1$ takes care of cases where $ n<1$.

So by induction, the lemma holds. $ \Box$[/hide]

Now apply this by letting $ n\equal{}\minus{}y$ (Note: this is weird because $ n$ is an integer but $ y$ is any real, but...) and we get
\[ f(0)\minus{}f(y)\equal{}\minus{}yf(1)\minus{}y^2\minus{}\frac{y(\minus{}y\minus{}1)}{2}.\]
But plugging in $ x\equal{}y\equal{}0$ to the given identity shows that $ f(0)\equal{}2f(0)\implies f(0)\equal{}0$. We may also assume $ f(1)$ to be some constant $ k$. Substituting this in, and negating both sides:
\[ f(y)\equal{}y^2\plus{}yk\minus{}\frac{y(y\plus{}1)}{2}\equal{}\boxed{\frac{y}{2}(y\plus{}2k\minus{}1)},\]
for any constant $ k$.

Checking, this function satisfies the problem, so we are done.
[/hide]

[quote="grn_trtle"](Note: this is weird because $ n$ is an integer but $ y$ is any real, but...)[/quote]
Weird indeed; it means that your definition only works for positive integers. The post before yours shows that this is equivalent to the Cauchy equation and thus has infinitely many solutions, not all continuous.

But plugging the result into the given condition shows that it works out, no matter what you sub in for $ y$...  :huh:

Sorry, allow me to rephrase. You have only shown what the values on positive integers must be. Your solution says little about what the function does on nonintegral values, so you have missed a large family of solutions.

Hi :)

the solution is the function: $ f(x)\equal{} \frac{x^2}{2} \plus{}ax$

we can prove that $ f$ are continue and derivabl function than $ \frac{f(x\plus{}y)\minus{}f(x)}{y} \equal{} \frac{f(y)}{y} \plus{} x$

 $ y \rightarrow 0$ ===> $ f'(x)\equal{}  f'(0) \plus{} x$ (because f(0)=0).

than $ f(x) \equal{} f'(0)x \plus{} \frac{x^2}{2} \plus{} c$

$ f(0)\equal{}0$ ===> $ f(x)\equal{} \frac{x^2}{2} \plus{} ax \ \ / \ with \ a\equal{}f'(0) \in \mathbb{R}$


solve this equation in  R

$\sqrt[3]{x}+\sqrt[3]{1-x}=2$

Since $f(x)=x^{3}$ is a bijection in $\mathbb{R}$, cubing an equation at any point doesn't lead to the loss of a solution. So

$x+1-x+3\sqrt[3]{x(1-x)}(\sqrt[3]{x}+\sqrt[3]{1-x})=8$

The last pair of parentheses is equal to $2$, hence

$1+6\sqrt[3]{x(1-x)}=8\iff\sqrt[3]{x(1-x)}={7\over 6}\iff x(1-x)={7^{3}\over 6^{3}}$

That quadratic, however, has no real solution since its discriminant is negative.


geometry
3D geometry
linear algebra
matrix
linear algebra unsolved

[quote="Pascual2005"]Let $ A,B$ be two matrices with integer entries such that $ \det A \equal{} 0$ and $ \det B \equal{} 0$, and such that $ AB \equal{} BA$. Prove that $ \det A^3 \plus{} B^3$ is a perfect cube.

A friend gave this problem to me but I strongly suspect that it is only treu for matrices of dimensions $ 2\times 2$.[/quote]
It is true for all dimension $ n\geqslant 1$, and you have $ \det (A^3\plus{}B^3)\equal{}0$.

Hi JC , are you awaked?
The asked result is true only if $ n\equal{}2$.
$ det(A^3\plus{}B^3)\equal{}(trace(A)trace(B))^3$ or $ 0$.

Hi loup blanc, I am indeed.
0 is a perfect cube isn't it :wink:
Hence, the asked result is true for all $ n>0$.

[quote="loup blanc"]
The asked result is true only if $ n \equal{} 2$.
$ det(A^3 \plus{} B^3) \equal{} (trace(A)trace(B))^3$ or $ 0$.[/quote]
By the way, I'm not sure your formula is correct
Take $ A\equal{}B\equal{}\begin{pmatrix}  0 &0 \\
0&1
\end{pmatrix}$.
$ \det(A^3 \plus{}B^3) \equal{} 0 \neq (trace(A)trace(B))^3 \equal{} 1$

Hi JC,
1) the result is false for $ n\equal{}3$.
Take $ A\equal{}diag(0,1,1),B\equal{}diag(1,0,1)$ : $ det(A^3\plus{}B^3)\equal{}2$.
2) If $ n\equal{}2$ , $ A,B$ are simultaneously triangularizable over the reals and $ det(A^3\plus{}B^3)\equal{}$ EITHER $ (trace(A)trace(B))^3$ OR $ 0$.

loup blanc,
[i]Au temps pour moi[/i], you are right.


calculus conics ellipse algebra polynomial Vieta MATHCOUNTS

[quote="archimedes1"]Who is going?
What are you participating in?
Discuss![/quote]
Me!!! Algebra 2 for the Gunn team woot. I hope I don't fail this, I can't fail 4 competitions in 1 week, that's just too much. I already failed 3, so I need to unfail this one.

Open division for Harker (dangit i hate the 3 person restriction)

im in open too
hey another chance to see u all in real life!

I had to do alg2 since everything else got taken by more pro people at gunn :( I hate being a noob.

[quote="xscapezaer"]I'm not positive. Maybe calculus, but maybe leapfrog if vishalarul is willing...

Homestead is going to fail tomorrow; we only have 4 people going total as of right now. :|

[quote="archimedes1"]Judging by [url]http://faculty.harker.org/MisaelF/SCVMA/2007SCVMA/2007problems/2007Algebra2MadHatter.pdf[/url], I;m not sure how excited I am about the Algebra II contest...[/quote]

Go look at the results for that one... :P

Wait, I shouldn't be bragging about that, I failed... oops[/quote]

DUDE 2nd IS NOT FAILING. WHY does everyone have an Ubeyish attitude nowadays?

[hide="rant"]Did Homestead even register yet? [size=200]WHERE ARE WE EVER GOING TO GET THE MONEY?!!?![/size][/hide]

Hey guess what? You got FIRST!!!!!!!!!!!!!!

AAAAAAAAAAAAAARRRRRRRRRRRRRRRRRGGGGGGGGHHHHHHHHHHHHHHHHHHHH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[/misery]

Anyway, undoubtedly we'll figure something out.

Wait a sec...

/me does a double take

/me does not have a Ubeyish attitude!!!

I think...

And one more post and you'll be 1337. :rotfl:

[quote="xscapezaer"]I'm not positive. Maybe calculus, but maybe leapfrog if vishalarul is willing...

Homestead is going to fail tomorrow; we only have 4 people going total as of right now. :|

[quote="archimedes1"]Judging by [url]http://faculty.harker.org/MisaelF/SCVMA/2007SCVMA/2007problems/2007Algebra2MadHatter.pdf[/url], I;m not sure how excited I am about the Algebra II contest...[/quote]

Go look at the results for that one... :P

Wait, I shouldn't be bragging about that, I failed... oops[/quote]

I feel your pain
I had to take the Geo test AND i failed on it because i didn't know a thing about ellipses

[quote="archimedes1"]Who is going?
What are you participating in?
Discuss![/quote]
What is SCVMA?

Wow, that was a weird weird...
[quote="The Algebra II Contest Writer"]
If $ a \ne 0$ and $ r$ and $ s$ are the roots of $ a_0 \plus{} a_1 x \plus{} a_2x^2 \equal{} 0$, then the equality
\[ a_0 \plus{} a_1 x \plus{} a_2 x^2 \equal{} a_0 \left( 1 \minus{} \frac {x}{r} \right) \left( 1 \minus{} \frac {x}{s} \right) \text{holds...}
\]
A. for all values of $ x, a \ne 0$
B. for all values of $ x$
C. only when $ x \equal{} 0$
D. only when $ x \equal{} r$ or $ x \equal{} s$
E. only when $ x \equal{} r$ or $ x \equal{} s,a \ge 0$
[/quote]
  :|  Is there any particular reason the polynomial is in backwards standard form but the coefficients are forward, other than to trip me up?  How are $ D$ and $ E$ different if we are already given $ a \ne 0$? What [i]is[/i] $ a$?

A for problem 36, why does Alpha negotiate with inanimate objects such as steps on a moving staircase?

In addition, calculators should not have been allowed.  Some of the problems were just asking to be TI-bashed.  Also, the multiple choice was a nice feature, which allowed some of the last problems to be solved in under a minute.[/quote]

[quote="junggi"][quote="serialk11r"][quote="CircleSquared"]I got a 41 (failage) while only guessing on like 4.
The 1st placer, who was from my school (yeaaaah) told me she guessed on lots of the first 25, and got a 46

 :wallbash: im the unluckiest guy in the world.[/quote]
WOW a 41 is fail? Holy crap our Alg2 section sucked then. Quatto and I each got something like 40/50...[/quote]
wow major fail guys[/quote]

no seriously it was screwingly easy

it didnt even involve precalc
it was just like alg2+bit harder

[quote="CircleSquared"][/quote][quote="junggi"][quote="serialk11r"][quote="CircleSquared"]I got a 41 (failage) while only guessing on like 4.
The 1st placer, who was from my school (yeaaaah) told me she guessed on lots of the first 25, and got a 46

 :wallbash: im the unluckiest guy in the world.[/quote]
WOW a 41 is fail? Holy crap our Alg2 section sucked then. Quatto and I each got something like 40/50...[/quote]
wow major fail guys[/quote][quote="CircleSquared"]

no seriously it was screwingly easy

it didnt even involve precalc
it was just like alg2+bit harder[/quote]
yeah so
wow major fail guys

[quote="CircleSquared"][/quote][quote="junggi"][quote="serialk11r"][quote="CircleSquared"]I got a 41 (failage) while only guessing on like 4.
The 1st placer, who was from my school (yeaaaah) told me she guessed on lots of the first 25, and got a 46

 :wallbash: im the unluckiest guy in the world.[/quote]
WOW a 41 is fail? Holy crap our Alg2 section sucked then. Quatto and I each got something like 40/50...[/quote]
wow major fail guys[/quote][quote="CircleSquared"]

no seriously it was screwingly easy

it didnt even involve precalc
it was just like alg2+bit harder[/quote]
Well duh it was easy, that's exactly why I was like holy crap we sucked. The Alg2 test was basically the stuff we're learning in class right now..."find population of blah", "find p and q such that blah" etc. A lot of the problems were identical to the crap we do in math class...even identical in form to the stuff on tests.

[quote="serialk11r"][/quote][quote="CircleSquared"][/quote][quote="junggi"][quote="serialk11r"][quote="CircleSquared"]I got a 41 (failage) while only guessing on like 4.
The 1st placer, who was from my school (yeaaaah) told me she guessed on lots of the first 25, and got a 46

 :wallbash: im the unluckiest guy in the world.[/quote]
WOW a 41 is fail? Holy crap our Alg2 section sucked then. Quatto and I each got something like 40/50...[/quote]
wow major fail guys[/quote][quote="CircleSquared"]

no seriously it was screwingly easy

it didnt even involve precalc
it was just like alg2+bit harder[/quote][quote="serialk11r"]
Well duh it was easy, that's exactly why I was like holy crap we sucked. The Alg2 test was basically the stuff we're learning in class right now..."find population of blah", "find p and q such that blah" etc. A lot of the problems were identical to the crap we do in math class...even identical in form to the stuff on tests.[/quote]

scary

>hijacks topic<

someone make sure the answer to the problem with m, n, and 33 i posted earlier
is NOT 2 like archimedes said
because i can come up with 34 pairs of the top of my head that work

please

[quote="bowei"][quote="archimedes1"]Oh, sorry, I thought it was there  :oops:

[quote="CircleSquared"]dont remember

but take a look at this

$ m^3 \plus{} n^3 \plus{} 99mn \equal{} 33^3$
how many ordered pairs (m,n) are there

a. 2
b. 3
c. 33
d. 35
e. 99

i think its 34 for some reason[/quote]
Multiple choice ftw!
First of all, note that $ (33,0)$ is a solution, so $ m,n$ are at most $ 33$.
Note that for every $ (m,n)$ that works, $ (n,m)$ works too.  Hence our number of solutions is even.  HOWEVER, what if $ m \equal{} n$, in which case there will not be a corresponding solution?  Then we are looking for $ 2m^3 \plus{} 99m^2 \equal{} 33^3$, or $ m^2(2m \plus{} 99) \equal{} 3^3 \cdot 11^3$.  From this, we see that either $ 3 | 2m \plus{} 99$ or $ 3 | m^2$; hence, $ 3 | m$.  Similarly, $ 11 | m$.  Hence, the only possible $ m \le 33$ are $ m \equal{} 0$ or $ m \equal{} 33$, and neither of these work.  Hence, there are an even number of solutions, so there must be A: 2, solutions, namely $ (33,0)$ and $ (0,33)$.[/quote]
okay let me kill this.
note that (33,0), (0, 33), and (-33, -33) are solutions.
all the rest can be paired and therefore # of sltns is odd.
consider equation mod 3
m^3+n^3=0
residues of x^3 mod 3 are: 0, 1, 2.
therefore, 3|m+n.
consider equation mod 11:
m^3+n^3=0
residues of x^3 mod 11 are: 0, 1, 8, 5, ..., 6, 3, 10.
for two to add up to 0, $ m \equiv 11 \minus{} n$
therefore, 11|m+n.
therefore, 33|m+n.
if m >0, n>0 because mn >0.
therefore m, n <= 33.
therefore m+n=33.
substuting this in, we see taht the equation holds.

now do something cool with negative numbers: check that if m+n=-33 or -66, then m=n=-33.
then we have 35 solutions.
qed.[/quote]

Well I think Bowei's solution makes sense so the answer is neither 2 nor 34 but actually 35?

[quote="CircleSquared"][/quote][quote="serialk11r"][/quote][quote="CircleSquared"][/quote][quote="junggi"][quote="serialk11r"][quote="CircleSquared"]I got a 41 (failage) while only guessing on like 4.
The 1st placer, who was from my school (yeaaaah) told me she guessed on lots of the first 25, and got a 46

 :wallbash: im the unluckiest guy in the world.[/quote]
WOW a 41 is fail? Holy crap our Alg2 section sucked then. Quatto and I each got something like 40/50...[/quote]
wow major fail guys[/quote][quote="CircleSquared"]

no seriously it was screwingly easy

it didnt even involve precalc
it was just like alg2+bit harder[/quote][quote="serialk11r"]
Well duh it was easy, that's exactly why I was like holy crap we sucked. The Alg2 test was basically the stuff we're learning in class right now..."find population of blah", "find p and q such that blah" etc. A lot of the problems were identical to the crap we do in math class...even identical in form to the stuff on tests.[/quote][quote="CircleSquared"]

scary

>hijacks topic<

someone make sure the answer to the problem with m, n, and 33 i posted earlier
is NOT 2 like archimedes said
because i can come up with 34 pairs of the top of my head that work

please[/quote]
Yeah, I know  :)   I was making the point that the answers must be wrong.  Did you plug in $ x$ for $ m$ and $ 33 - x$ for $ n$ like I said previously?  That would give 34.

[quote="archimedes1"][/quote][quote="CircleSquared"][/quote][quote="serialk11r"][/quote][quote="CircleSquared"][/quote][quote="junggi"][quote="serialk11r"][quote="CircleSquared"]I got a 41 (failage) while only guessing on like 4.
The 1st placer, who was from my school (yeaaaah) told me she guessed on lots of the first 25, and got a 46

 :wallbash: im the unluckiest guy in the world.[/quote]
WOW a 41 is fail? Holy crap our Alg2 section sucked then. Quatto and I each got something like 40/50...[/quote]
wow major fail guys[/quote][quote="CircleSquared"]

no seriously it was screwingly easy

it didnt even involve precalc
it was just like alg2+bit harder[/quote][quote="serialk11r"]
Well duh it was easy, that's exactly why I was like holy crap we sucked. The Alg2 test was basically the stuff we're learning in class right now..."find population of blah", "find p and q such that blah" etc. A lot of the problems were identical to the crap we do in math class...even identical in form to the stuff on tests.[/quote][quote="CircleSquared"]

scary

>hijacks topic<

someone make sure the answer to the problem with m, n, and 33 i posted earlier
is NOT 2 like archimedes said
because i can come up with 34 pairs of the top of my head that work

please[/quote][quote="archimedes1"]
Yeah, I know  :)   I was making the point that the answers must be wrong.  Did you plug in $ x$ for $ m$ and $ 33 - x$ for $ n$ like I said previously?  That would give 34.[/quote]

yeah i was kinda wondering why i got an answer that wasn't one of the choices
i was also wondering why such an easy q was a #50

...well now i know

by the way, did anyone get the function one i posted

Err I tried plugging in $ 1$ and $ 0$ (that's the *only* thing I know to do with stuff like that), but it kinda did nothing at all... And plugging in $ 0$ made it blow up :( .

[quote="sonny"]
Well I think Bowei's solution makes sense so the answer is neither 2 nor 34 but actually 35?[/quote]
Bowei's solutoin uses (-33,-33) which violates $ m, n$ $ \ge$ $ 0$.

For the function problem:
Just remember that $ g(x) \equal{} 1/x$ is cyclic.  Basically, plug in $ 1/x$ for $ x$, then you can solve and get $ f(x) \equal{} 2/x \minus{} x$.
As in,
$ f(1/x)\plus{}2f(x)\equal{}3/x$
$ f(x)\plus{}2f(1/x)\equal{}3x$
This technique is in Aops vol. II.

[quote="archimedes1"][quote="sonny"]
Well I think Bowei's solution makes sense so the answer is neither 2 nor 34 but actually 35?[/quote]
Bowei's solutoin uses (-33,-33) which violates $ m, n$ $ \ge$ $ 0$.

For the function problem:
Just remember that $ g(x) \equal{} 1/x$ is cyclic.  Basically, plug in $ 1/x$ for $ x$, then you can solve and get $ f(x) \equal{} 2/x \minus{} x$.
As in,
$ f(1/x) \plus{} 2f(x) \equal{} 3/x$
$ f(x) \plus{} 2f(1/x) \equal{} 3x$
This technique is in Aops vol. II.[/quote]

Um well I didn't take the test but if the condition is mn>0 then -33,-33 actually work.

[quote="bowei"]no, all that is required is that $ mn \geq 0$.[/quote]
Oh, so they can both be negative.  in that case, you are correct.

*Pops in*
Uh guys, a few problems seem to be stolen from the 1999 AIME, like that weird mn>=0 one...
*Pops out*


inequalities
trigonometry
geometry unsolved
geometry

ABC is not an obtuse triangle.
prove that:
$ \frac{sinAsinB}{sinC}\plus{}\frac{sinAsinC}{sinB}\plus{}\frac{sinCsinB}{sinA}\geq \frac{5}{2}$

[quote="tuandokim"]ABC is not an obtuse triangle.
prove that:
$ \frac {sinAsinB}{sinC} \plus{} \frac {sinAsinC}{sinB} \plus{} \frac {sinCsinB}{sinA}\geq \frac {5}{2}$[/quote]

This inequality is equivalent : $ \boxed{\frac {ab}{c} \plus{} \frac {bc}{a} \plus{} \frac {ca}{b} \ge 5R}$

$ a^2 \plus{} b^2 \equal{} c^2 \plus{} 2ab\cos\gamma$ and  $ ab \equal{} \frac {2S}{\sin\gamma}$ , $ R \equal{} \frac {c}{2\sin\gamma}$

So we can write : $ \frac {ab}{c} \plus{} \frac {bc}{a} \plus{} \frac {ca}{b} \equal{} \frac {ab}{c} \plus{} \frac {c}{ab}(c^2 \plus{} 2ab\cos\gamma) \equal{} \frac {2S}{c\sin\gamma} \plus{} \frac {c^3\sin\gamma}{2S} \plus{} 2c\cos\gamma \ge \frac {5c}{2\sin\gamma}$

$ \frac {2S}{c^2\sin\gamma} \plus{} \frac {c^2\sin\gamma}{2S}\ge \frac {5}{2\sin\gamma} \minus{} 2\cos\gamma$

Let $ t \equal{} \frac {2S}{c^2\sin\gamma}$  and we must prove : $ t \plus{} \frac {1}{t} \ge \frac {5}{2\sin\gamma} \minus{} 2\cos\gamma$

We can assume :$ \boxed{\frac {\pi}{2} \ge \gamma \ge \frac {\pi}{3}}$

$ S \le \frac {c^2\sin\gamma}{8\sin^2\gamma} \equal{} > t \le \frac {1}{4\sin^2(\frac {\gamma}{2})} \le 1$

$ f(t) \equal{} t \plus{} \frac {1}{t} \ge f(\frac {1}{4\sin^2 (\frac {\gamma}{2})} \equal{} \frac {1}{4\sin^2(\frac {\gamma}{2})} \plus{} 4\sin^2(\frac {\gamma}{2}) \ge \frac {5}{2\sin\gamma} \minus{} 2\cos\gamma < \equal{} >$  $ \frac {1}{4\sin^2(\frac {\gamma}{2})} \ge \frac {5}{2\sin\gamma} \minus{} 2$

Let $ f \equal{} ctg (\frac {\gamma}{2}) \ge 1$

Then :$ < \equal{} > f^3 \minus{} 5f^2 \plus{} 9f \minus{} 5 \equal{} (f \minus{} 1)(f^2 \minus{} 4f \plus{} 5)\ge 0$

Equality will be for $ a \equal{} b$ and $ \gamma \equal{} \frac {\pi}{2}$ :)


How can I prove that we can find rational and irrational numbers in any interval?

[hide="Rational"]
If $ I\equal{}[a,b]$, set $ n>\frac 1{b\minus{}a}\Rightarrow \frac 1n < b\minus{}a$ and suppose there is no integer m with $ \frac mn \in I.$ Getting a contradiction out of this is fairly easy.
[/hide]

[hide="Irrational"]
Set $ I\equal{}[a,b]$ as above and look at the interval $ I'\equal{}\left[\frac a{\sqrt{2}},\frac b{\sqrt{2}}\right]$ which contains at least one rational number $ \frac mn$ so $ x\equal{}\frac mn \sqrt 2 \in I$ and $ x\not\in\mathbb{Q}.$
[/hide]

[quote="bertram"][hide="Rational"]
If $ I \equal{} [a,b]$, set $ n > \frac 1{b \minus{} a}\Rightarrow \frac 1n < b \minus{} a$ and suppose there is no integer m with $ \frac mn \in I.$ Getting a contradiction out of this is fairly easy.
[/hide][/quote]

[hide="Personally..."] I like to construct the rational explicitly.  In particular, suppose the binary expansions of $ a, b$ first differ at digit $ d$ so that the $ d^{th}$ digit of $ a$ is $ 0$ but the $ d^{th}$ digit of $ b$ is $ 1$ (which we can assume because $ a < b$).  Replace the subsequent digits of $ a$ with $ 1$.  The resulting number is strictly greater than $ a$ and strictly less than $ b$, and rational. [/hide]
[hide="Alternately..."] Without loss of generality, $ a, b > 0$.  Consider the portion of the first quadrant of the Euclidean plane defined by the inequalities $ ax < y < bx$.  Since its area is clearly infinite, it contains an infinite number of lattice points, whence a point $ (m, n), m, n \in \mathbb{N}$ such that $ an < m < bn \Leftrightarrow a < \frac{m}{n} < b$. [/hide]

I [i]must[/i] learn to use $ \in$ correctly. $ a\in\mathbb{Z}$ translates to $ a$ lies in $ \mathbb{Z}$ right? I tried to look it up but the actual denotation of it got drowned out in jargon about rings etc.

This may help:
http://www.mathlinks.ro/Forum/viewtopic.php?t=144939

[quote="nateharman1234"]Heres one for irrational based of the result for rational:

WLOG a, b>0

Consider the interval [ln(a), ln(b)], by the first part we know that there is some rational number r in this interval.  This puts e^r in [a,b], and e to a rational power must be irrational[/quote]

This method is quite complicated compared to the method I gave (scale the interval down by some arbitrary irrational number and use the fact that there will be a rational number in there). Also the proof doesn't hold for $ a,b<0$ (that case is obvious; regardless, it has to be addressed :))


geometry
3D geometry
tetrahedron
circumcircle

Let ABCD be a tetrahedron whose 4 altitudes are concurrent and let R be the radius of its circumcircle (sphere). Let $ h_1;h_2;h_3;h_4$ be the lengths of the altitudes corresponding A,B,C,D and $ R_1;R_2;R_3;R_4$ be the circumcircle's radius of the opposite faces of A,B,C,D. Prove that
$ \frac{1}{h_1\plus{}2\sqrt{2}R_1}\plus{}\frac{1}{h_2\plus{}2\sqrt{2}R_2}\plus{}\frac{1}{h_3\plus{}2\sqrt{2}R_3}\plus{}\frac{1}{h_4\plus{}2\sqrt{2}R_4} \geq \frac{1}{R}$.
When does equality occurs?

To mod: Please lock this topic because it is a contest problem in Mathematics And Youth Magazine of Viet Nam. The dealine is 15-04-2008!


Given cos(3x)/cos(x) = 1/3, determine the value of sin(3x)/sin(x).

Any ideas?

Well I did manage to solve it, finally.  Here's how I did it:

[hide]cos(x) = 3cos(3x)

cos(x) = 3cos(x + 2x)

cos(x) = 3(cos(x)cos(2x) - sin(x)sin(2x))

1 = 3cos(2x) - 6sin(x)sin(x)cos(x)

1 = 3cos(2x) - 6sin^2(x)

1 = 3 - 6sin^2(x) - 6sin^2(x)

-2 = -12sin^2(x)

1/6 = sin^2(x)

Then basically do the same sum identity trick on the other side to convert them all to sin^2 s. [/hide]

I'm pretty sure there is a shorter way though.


Cuáles son los equipos?
Qué equipo iberoamericano se lo va a ganar?

Este es el Equipo colombiano Definitivo:

1. Santiago Cuellar
2. Miguel Acosta
3. Jorge Olarte
4. Gustavo Nicolás Páez
5. Haiden Liu Weng
6. Carlos Carvajal

La mitad del equipo es repetido, así que pue ser una buena competencia...
Por favor, los que puedan, suban sus pruebas de selcción, para verlas  :lol:

Hi!!
Peruvian team:
[b]PER1.-JESUS RAMOS
PER2.-FERNANDO MANRIQUE
PER3.-CESAR CUENCA
$ PER4.\minus{}$AMILCAR VELEZ
PER5.-TOMAS ANGLES
PER6.-IVAN MUÑOZ[/b]--
Respecto a los selectivos, los de Perú los puedes encontrar en http://selectivos-peru.blogspot.com
y los selectivos de colombia?..onde los encontramos Skuge...
Chaufas!!

No es por ser antipaticos, pero los selectivos de Colombia usan problemas que no deben ser publicados hasta despues de la IMO. De todas formas creo que es muy probable que durante las sesiones de entrenamiento hagan casi todos, depronto con un par de excepciones. Si quieren se los podria mandar por mail.

Ah.... tengo entendido q la seleccion de columbia es la misma prueba de la olimpiada bolivariano. es verdad??

No. La olimpiada Bolivariana es realmente la ronda final de la olimpiada nacional Colombiana. Los mejores diez en la ronda final y los cinco mejores del nivel intermedio de la olimpiada nacional presentan el selectivo de la IMO.

Si, los selectivos de Colombia tocará discutirlos después de la olimpiada. Seguro cuando se acabe los subimos.

Dónde están los otros equipos???

Hay muy pocos países en este foro. Los únicos países hispanohablantes que yo sepa que tienen personas que escriben usualmente aquí son Perú, Colombia, Paraguay y tal vez Argentina (aunque no mucho, y casi nunca en el foro de hispanohablantes). Brasil escribe más en español que la mayoría de nosotros :(

Pero si son sólo cuatro :wacko: :(

Hay equipos que aseguran una posición muy baja en una IMO al presentar equipos incompletos :(

Jutaro tiene razón. Este foro se está volviendo aburrido, pues mucha gente que ponía muchas cosas en el foro no pone nada desde hace mucho... :(

Por ejemplo, la encuesta de "Cuál será el mejor equipo en la IMO 2007" fue abierta en Mayo... Y ahora ya estamos a una semana de IMO y nadie parece emocionado por la IMO :(

[quote="Jutaro"]Claro que hay varios países, lo que pasa es que por alguna razón este foro se ha vuelto bastante... ¿aburrido? :( A veces pasan días sin que hayan posts nuevos... Recuerdo que en los tiempos de los retos de la semana había más actividad jajaja :lol:

Por cierto el equipo de El Salvador para esta IMO es el siguiente (no estoy seguro del orden de los estudiantes):

Héctor Enmanuel Alberti Arroyo
Nahomy Jhopselyn Hernández Cruz
Luis Damián Reyes Rodríguez
Otto Tang Chen

¡La IMO ya está cerca! :D[/quote]

Bueno es cierto lo que dices...pero sucede que cuando algunos ponen nuevos posts y nadie postea una solucion y por eso como que mejor es postearlo en ingles.El problema es que tenemos mucha flojera de escribir (me incluyo XD).
JeJe veo que participara este año en la IMO otro chen!!..pero esta ves por El salvador..xD. Bueno aya nos veremos bye bye..

HOla a todos los hispanohablantes!!
Wno Como sabemos la IMo ya esta cerca , a una semana.
alguien sabe de casualidad por algun dato por ahi..si la IMO sera en un hotel o en una universidad....y si nos daran bastantes gaseosas cocacolas gratis (ya q son auspiciadores..) :rotfl:
Bueno bye!..
                 Ahi nos vemos...:P
                    En la madre patria..y ole!!-.-

[quote="chichi"]hola amigos...
por casualidad si tengo esa informacion, nos vamos a estar metidos en un colegio(no es ni hotel ni universidad jejej)... no recuerdo bien como se llama, y los leaderes y detutys estara en un hotel.
la organizadora nos dio esta informacion ya que el consulado de españa no nos queria dar la visa ya q dice q tenemos que tener/saber garantia economica y reservacion de hotel etc. como cualquier turista...[/quote]
Chs...casi le acierto !!.. wno Chichi ahi nos vemos xD ..
Seria una buena idea saber que hispanohablantes que participan en mathlinks..iran a la IMO...bueno safo ..cuidense.

[quote="skuge"]Bueno ya averigué, no es un colegio. Se llama Colegio Mayor Elias Ahuja pero es una universidad. Hoy estuve ahí, está bien, se parece un poco a la de Eslovenia y no está haciendo tanto calor....[/quote]
Perate un momento..el equipo de colombia ya esta en españa??!!!...q buen paseo se daran xD...
Ahi nos vemos...chaufas...
por cierto..cual eres tu  :wink:

Por se acaso viste como es el baño...?
el baño tambien es como en eslovenia q es en un esquina del pasillo?

Sí, estuvimos un día en españa y ahora estamos en Portugal con el equipo de aquí pero no estamos paseando sino entrenando :wink: ... no pude ver los cuartos ni los baños... sólo el edificio y el primer piso el comedor se ve bien.....

[quote="aev5peru"][quote="skuge"]No parece que hayan publicado ninguna información al respecto... nadie ha hablado de eso ni en el foro ni en la página de la olimpiada...[/quote]
Jeje...bueno..creo q me confundi..xD...por la imagen rapida que le di a este sitio http://www.imo-2008.es/sponsors_ESP.html
Bueno ahi nos vemos....ahi dice q cocacola es uno de los otros colaboradores...
Chaufas!![/quote]
[color=blue]Nuestro amigo está ilusionado con que Coca Cola provea bebidas gratis…[/color]

Felicitaciones a los representantes de Perú por los buenos resultados que han obtenido, y si no me equivoco es la primera medalla de oro que gana un peruano, así que doble celebración  :D

:first:  :first:

Yo tambien felicito a todos los participantes peruanos y brasileros por los excelentes resultados que tuvieron. Hay dos equipos latinos en el top 20, muy bien.

http://www.imo-2008.es/paises/CHI_ESP.html

A mucha honra felicito al equipo chileno, se nota que se ha superado, dos medallas y una mencion, quizas algo histórico, pero mas que eso por el esfuerzo de cada uno, necesitabamos algo así.

sdls.


[quote="tim1234133"]Prove that the sequence defined by

$y_{0}=1,$        $y_{n+1}=\frac12(3y_{n}+\sqrt{5y^{2}_{n}-4})$, $(n\geq0)$

consists only of integers.[/quote]
$y_{n+1}=\frac12(3y_{n}+\sqrt{5y^{2}_{n}-4})$, $(n\geq0)$
is the biggest solution of the equation
$x^{2}-3y_{n}x+y_{n}^{2}+1$.
So we want to prove that if $y_{n+1}^{2}-3y_{n}y_{n+1}+y_{n}^{2}+1 = 0$, then $x^{2}-3y_{n+1}x+y_{n+1}^{2}+1$ has integer solutions. But from the first equation one has $y_{n+1}^{2}+1 = 3y_{n}y_{n+1}-y_{n}^{2}= y_{n}(3y_{n+1}-y_{n})$. Substituting in the second equation the result is obvious.

Daniel

That's interesting.  It's very similar to the equation in my post "Fibonacci Generator".

In fact, that equation will generate a sequence of Fibonacci numbers.

[quote="nguyentrang"][quote="Farenhajt"]After some algebraic juggling we get $\sqrt{5y_{n+1}^{2}-4}={5\over 2}y_{n}+{3\over 2}\sqrt{5y_{n}^{2}-4}$. Hence

$y_{n+2}={7\over 2}y_{n}+{3\over 2}\sqrt{5y_{n}^{2}-4}\qquad(*)$

[/quote]

Can you please explain these two equations better? How they come?[/quote]

If $y_{n+1}={1\over 2}(3y_{n}+\sqrt{5y_{n}^{2}-4)}$, then

\begin{eqnarray*}5y_{n+1}^{2}-4 &=&{5\over 4}(9y_{n}^{2}+6y_{n}\sqrt{5y_{n}^{2}-4}+5y_{n}^{2}-4)-4\\ &=&{70\over 4}y_{n}^{2}+{15\over 2}y_{n}\sqrt{5y_{n}^{2}-4}-9\\ &=&{25\over 4}y_{n}^{2}+2\cdot{5\over 2}y_{n}\cdot{3\over 2}\sqrt{5y_{n}^{2}-4}+{9\over 4}(5y_{n}^{2}-4)\\ &=& \left({5\over 2}y_{n}+{3\over 2}\sqrt{5y_{n}^{2}-4}\right)^{2}\end{eqnarray*}

and the second equation is obtained by simple plugging in of the above expression and the expression for $y_{n+1}$


abstract algebra
vector
group theory
superior algebra
superior algebra unsolved

Show $ \mathbb{Q} \otimes_{\mathbb{Q}} \mathbb{Q}$ and $ \mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Q}$ are $ \mathbb{Q}$-module isomorphic.

It's easy to show the first tensor product is a one-dimensional vector space over $ \mathbb{Q}$ spanned by $ 1 \otimes 1$ since it's $ \mathbb{Q}$-bilinear. How can I show the same for the product over $ \mathbb{Z}$?

This is because any two rational numbers are commensurable; any finitely generated subgroup of $ \mathbb{Q}$ is (infinite) cyclic. It the follows that any two elements $ \frac ab\otimes \frac cd$ and $ \frac pq\otimes \frac rs$ are commensurable. Explicitly, $ bdqs\cdot \frac ab\otimes \frac cd\equal{}acrt\cdot \frac pq\otimes \frac rs$. We can't have two linearly independent elements, and the qhole thing is a one-dimensional space isomorphic to $ \mathbb{Q}$.


Anyone ever play this version of Tower Defense:

[url]http://www.onemorelevel.com/games.php?game=415[/url]

What level did you get up to? And what configuration of towers did you use?

I got up to level 71, using water towers placed in the middle of the map, and leaf towers along the ridge that extends along the entrance.

That game takes too long. I did it twice, but I'm too lazy to do it again.

:D

[img]http://img67.imageshack.us/img67/5651/score69gu9.png[/img]

Here's another fun game: http://www.onemorelevel.com/games.php?game=432

I lasted 75 seconds. :ninja:

THIS GAME IS INSANE!!!

not to mention hard... I can't get past level 20. Not to mention 16...

I managed to get to 75 using earth towers

[quote]not to mention hard... I can't get past level 20. Not to mention 16...[/quote]
lol try upgrading towers

yea earth towers are the best, got to level 86 with them :P

I WAS upgrading towers.

P.S. I thought that the picture above were dead towers. Now I see that you can upgrade towers MORE than twice. I got to level 50 with 7 or 6 citezens.

@I_like_pie: I couldn't get to THAT level with that pattern!

P.S. On the ball game, I lasted 78 seconds.

edit: I just got 117 seconds.

10 balls, 240 something seconds, a+, super boring

edit:
[quote="anirudh"]Its edited.[/quote]
your head's edited

[quote="anirudh"]Its edited.[/quote]
My picture? No, it's not. I can't prove it, but you can't prove it's edited either... :roll:

The bottom-most tower is a fully upgraded water tower. The middle one (with the blue ball flying around it) is almost fully upgraded.

The living tower is the highest nature tower.

The ice tower that's in a blur, I don't know.

As high as I can get with fire: century inferno tower.

I just scored 69 on that game!

[quote="Ars"]Highest level was a 71

Highest Earth is Living Tower best damage except for flower

Highest Ice is the blur which has the best range (most expensive)

Highest fire is Firelord Elemental Tower which has similar damage to ice but much worse range (least expensive)[/quote]
I'm not sure what the highest fire tower is called, but it is the most expensive (1500 compared to 1000 and 1100) and it's also animated (it constantly changes from red to orange and back).

I meant total cost
The Flower Tower is definatly worth it in terms of damage it is upgradable but will lose range for damage

towers fire a lot faster when they dont have to rotate, so putting them at the end of long walkways lets them fire super fast.

I got to level 72 using six fully upgraded ice towers and one random termination tower

my bad, i meant termination tower, but i randomly placed it

8 living towers, 2 supernatural, and 2 ice towers (not sure which upgrade).

[img]http://img246.imageshack.us/img246/299/score73bj5.png[/img]

[quote="i_like_pie"]The ones right before living towers. I think they cost 640, and living cost 1100.[/quote]
OHHH living towers are upgraded nature towers? oh....I thought it meant like living as in not destroyed, and I was thinking "huh towers can die...?" So then I thought supernatural was a term for dead towers.... :blush:

[quote="#H34N1"]Its edited.[/quote]
You are one to talk aren't you :mad:

[quote="#H34N1"]THIS GAME IS AMAZING! I got to level 542 with a 3rd time upgraded 3K tower for 15K then Ihad to go do something else but would have probably pwned the game.[/quote]
You could at least lie a little better.

how do you guys get by with out fully upgraded towers? the highest i gotten was a 6X with 2 termanation towers, two or three Highest level ice and a few highest grass

Getting better... :D

[img]http://img182.imageshack.us/img182/1825/score80dr8.png[/img]

Put Ice Towers at the end of all the long passageways and Grass ones at the end of the short, never buy the flower tower, the more you ugprade it the more it sucks.

I definitely agree with the putting ice towers at the end of long passageways, but flower towers are definitely good.  They have the highest attack at the end (but the do fire more slowly).

Are you suggesting all ice towers, Ignite?

[quote="Ignite168"]No, just in 5 of the spots.

Flower Towers have the worst range at the end, they'll only be able to hit something extremely close by and green towers are much more efficient at that stage.[/quote]
the range of the living towers is still enough to hit the majority of most walkways. they are way cheaper than fully upgraded ice towers

I got to level 79 with the idea that they fire faster. See who can get to the highest level with only sorm towers. I tryed it and i got to level 26...


1) A Polynomial f(x) with integer coefficients has f(7)=x and f(9) = 13.

x cannot be:

A. 9
B. 1
C. 3
D. 5
E. NOTA

2) If f(x) is a fifth degree polynomial, and f(0)=3, f(1)=5, f(2)=6, f(3)=10, f(4)=11, and f(5)=12, find f(6).

[quote="Bicameral"]1) A Polynomial f(x) with integer coefficients has f(7)=x and f(9) = 13.

x cannot be:

A. 9
B. 1
C. 3
D. 5
E. NOTA
[/quote]

If $ f(7) \equal{} k$ and $ f(9) \equal{} 13$, then the answer is E; in fact, any [i]linear[/i] polynomial will fit A-D. If $ f(x) \equal{} mx \plus{} b$, then plugging in $ x \equal{} 9$ yields $ 9m \plus{} b \equal{} 13 \implies b \equal{} 13 \minus{} 9m$. Plugging that value of $ b$ for $ f(7)$ yields $ f(7) \equal{} 7m \plus{} 13 \minus{} 9m \equal{} 13 \minus{} 2m$, so $ m$ is clearly is an integer for all odd $ k$.

In general, if $ \frac{f(b)\minus{}f(a)}{b\minus{}a}$ is an integer, then a linear polynomial with integer coefficients exists.

For #2, just let itbe ax^5+bx^4+cx^3+dx^2+ex+f=0 and the plug in your values to get simultaneous equations.

Alternatively, construct a [url=http://mathworld.wolfram.com/FiniteDifference.html]finite differences[/url] chart.

i see this as easier to understand
[hide]
first note that the last 3 given input/output pairs are in linear progression
start with the equation $ f(x)\equal{}x\plus{}7$
but this does not work for x=0,1, or 2
here is where you make adjustments
for x=2, the real f(x) is different than the current f(x)
so you want to make an adjustment that will change the value of f(x) for x=2 but not for x=3,4,5
here is what you do:
the new f(x) is in the form of $ f(x)\equal{}x\plus{}7\plus{}k(x\minus{}3)(x\minus{}4)(x\minus{}5)$
plugging f(x)=6 and x=2 into the equation gives k. (i will not show work here)
the current f(x) is now $ f(x)\equal{}x\plus{}7\plus{}\frac{1}{2}(x\minus{}3)(x\minus{}4)(x\minus{}5)$
but for x=1, f(x) is supposed to equal 5, but it currently does not.
similar to the last adjustment, the new adjustment is
$ f(x)\equal{}x\plus{}7\plus{}\frac{1}{2}(x\minus{}3)(x\minus{}4)(x\minus{}5)\plus{}k(x\minus{}2)(x\minus{}3)(x\minus{}4)(x\minus{}5)$
plugging in f(x)=5 and x=1 gives k
$ f(x)\equal{}x\plus{}7\plus{}\frac{1}{2}(x\minus{}3)(x\minus{}4)(x\minus{}5)\plus{}\frac{3}{8}(x\minus{}2)(x\minus{}3)(x\minus{}4)(x\minus{}5)$
as an exercise to you :D you should make the final adjustment for x=3
you shuold get
[hide]
$ f(x)\equal{}x\plus{}7\plus{}\frac{1}{2}(x\minus{}3)(x\minus{}4)(x\minus{}5)\plus{}\frac{3}{8}(x\minus{}2)(x\minus{}3)(x\minus{}4)(x\minus{}5)\plus{}\frac{19}{120}(x\minus{}1)(x\minus{}2)(x\minus{}3)(x\minus{}4)(x\minus{}5)$
[/hide]
plugging in x=6 gives you the answer, $ \boxed{44}$
[/hide]

IS that easier or harder than a 6-eq. system of equations?


sorry,  There was an error.  I edit it now

$ a,b,n$are natural numbers, $ b > 1$ and $ (b^{n} \minus{} 1)| a$.

prove that ) express $ a$ in "$ b$ notation system". Then $ a$ contains at least $ n$ digits which is $ not$ zero.

(I don't know exact word in English. So I offer you 2 examples below)

ex) 2 notation system:  5= 101 (2) and it contains 2 digits which is not zero.
     3 natation system:   17= 122 (3) and it contains  3 digits which is not zero.

but if we take $ a\equal{}999,b\equal{}10,n\equal{}3$ we have $ b^{n}\minus{}1|a$ but $ a\equal{}\overline{111}^{(10)}$ contains no zero


If $ S(a)$ is the sum of all divisors of a. Prove that if $ a\equal{}(105)^{2004}$ then $ S(a)<\frac{35}{16}a$.

[quote="Elio (n)"]Is there any formula to find the sum of all divisers of a number?[/quote]

of course, this: [quote="SimonM"]$ n \equal{} \prod_{i \equal{} 1}^{r}p_{i}^{a_{i}}$ $ \sigma (n) \equal{} \prod_{i \equal{} 1}^{r} \frac {p_{i}^{(a_{i} \plus{} 1)} \minus{} 1}{p_{i} \minus{} 1}$[/quote]


algebra
polynomial
calculus
function
limit
superior algebra
superior algebra solved

[quote="fredoniahead"]Hello:

Is there a proof which states that a polynomial $\sum_{k-0}^{n}A_{k}x^{n-k}$ must have finite terms?.  I thought perhaps there was a proof.

Thanks[/quote]

I dont understand the question sorry  :?:

Formally a polynomial is the expression:
$\sum_{n=0}^{\infty}a_{n}x^{n}= a_{0}+a_{1}x+a_{2}x^{2}+...$   and $a_{i}\in R$ where $R$ is some ring.

Where [b]almost all[/b] $a_{i}$ are zero.
And $x$ is treated as an [i]intermediate[/i]. Meaning it doesnt represent a number just a symbol (think of it as a matrix). But if we subsitute values for $x$ we get values back. I am just not sure exactly what the question is.

[quote="fredoniahead"]If n tends to infinity, does it just becomes a power series?. This is probably a stupid question, but I just wondered. It always seemed to me a polynomial has finite terms.[/quote]

There is a distintion between a "polynomial" in field theory (much of what this subforum is devoted to) and a "polynomial" in analysis (much of what the Calculus subforum is devoted to).

A "polynomial" in field theory is what exactly I wrote above it is just that infinite thing. But with the condition that almost all coefficients are zeros. Which tells us that there is a highest coefficient that doest have zero in front and everything else is zero. So in that case polynomials are always finite.

A "polynomial (infinite series)" in calculus is a [i]function[/i]. It is something different from the polynomial as above.

Yes they have finite terms by definition. And yes, power series can be seen as some limit (in fact $K[[x]] \cong \lim_{\substack{\leftarrow \\ n}}K[x]/(x^{n})$ if you know projective limits).


Prove that:
 if all of the bisectors of a triangle are less than 1 then the area of that triangle will be less than $ \frac{{\sqrt 3 }}{3}$ :D

[hide="Solution"]
1.If the triangle is an acute angled one(or right angled),let in contrary $ S > \frac {\sqrt {3}}{3}$
Since all of the bisectors of a triangle are less than 1,we have the perprndiculars $ h_a,h_b,h_c \leq 1$
$ a \equal{} \frac {2S}{h_a} > \frac {2\sqrt {3}}{3}....$
Assume $ \angle A \geq \angle B \geq \angle C,90^0 \geq \angle A \geq 60^0$
$ h_b \equal{} c*sinA > \frac {2\sqrt {3}}{3} \frac {\sqrt {3}}{2} \equal{} 1$ that's impossible!
2 It's a obtuse angled one.
It is easy to show the two short sides are less than 1,so $ S \leq \frac {1}{2} < \frac {\sqrt {3}}{3}$
[/hide]

Yeah, that's correct.Do you think we can use trigonometry for this stuff?
 :roll:


algebra
polynomial
algorithm
Rational Root Theorem

$a$ is a real root of $x^{3}+2x^{2}+10x-20 = 0$. Show that $a^{2}$ is irrational.

[quote="t0rajir0u"][hide="Solution"] Rational Root Theorem quickly verifies that there are no rational roots.  In other words, if $a$ is a real root, then this polynomial is its minimal polynomial over the rationals.

Ergo, if $a^{2}= q$ were to be rational, then $x^{2}-q$ would be the minimal polynomial of $a$ over the rationals - contradiction.  (It's a good lemma to prove that the minimal polynomial of a root $a$ must divide any polynomial such that $a$ is its root.) [/hide][/quote]
I was hoping you'd notice RRT trivializes yet another Spanish problem. :rotfl:

Spanish problems are quite funny.

[quote="t0rajir0u"][hide="Solution"] Rational Root Theorem quickly verifies that there are no rational roots.  In other words, if $a$ is a real root, then this polynomial is its minimal polynomial over the rationals.

Ergo, if $a^{2}= q$ were to be rational, then $x^{2}-q$ would be the minimal polynomial of $a$ over the rationals - contradiction.  (It's a good lemma to prove that the minimal polynomial of a root $a$ must divide any polynomial such that $a$ is its root.) [/hide][/quote]
How does one prove the minimal polynomial lemma?

[hide="Solution"] Let the minimal polynomial of some root $a$ over some ring be $m(x)$.  Part of the definition of the minimal polynomial is that $m(x)$ has minimal degree.

Say that there exists $p(x)$ in the same ring such that $p(a) = 0$.  Division algorithm:

$p(x) = m(x) q(x)+r(x)$

Ergo,

$p(a) = 0 = m(a) q(a)+r(a) = r(a)$

Either $r(x) = 0$ or $\text{deg }r < \text{deg }m$; but the latter is contradiction because $m$ is minimal.  Hence $r(x) = 0$ and $m | p$. [/hide]

[quote="Gyan"][quote="Farenhajt"][hide="A bit simplified"]No rational roots $\implies a^{2}=\frac{20-10a}{a+2}\not\in\mathbb{Q}$[/hide][/quote]
[hide]FWIW , True, but may need a little more (may be another step  or rational or explanation ....)  for the skeptics.. eg why irrational/irrational can't be rational in this case... [/hide][/quote]

[hide="Answer"]Just assume ${20-10a\over a+2}=r$ where $r$ is rational. Then $20-10a=ar+2r\iff 20-2r=a(r+10)\iff a={r+10\over 20-2r}\in\mathbb{Q}$ which is impossible.[/hide]


What is the smallest integer multiple of $\ 9997$, other than $\ 9997$ itself, which only contains odd digits?

I once though would it be $\ 99979997$, but then I programed this question but the answer is a lot smaller.  :ninja:

Is that 33339995?

 ;)

9997 * 3335.

I noticed 9997* 4001 contains all odds.

9997*1 or 2 or 3: 9+9 must releases 0, thus impossible for 1001,2001 or 3001 to have all odds.

9997* 4 = 39998.

39988000 + 9997 -> Notice 39988 would be replaced by 39997.

339898+ 99 -> 339997, That's 9997*3401.

But that's not the smallest.

9997* 3330 = 33290010, after dividing each digit, it's possible to have 9997*3330-3401 to have all odds.

And I find 9997*3335 is all odded.


function
geometry
geometric transformation
reflection
limit
complex analysis
algebra

Find all entire functions f such that $|f(z)|=1$ on the unit circle $|z|=1$.

Does the following solution work? It is a bit tangly...

By Schwartz Reflection Principle, from the restriction of $f$ in the unit disk, we can extend it to the outside of the disk since the values of $f(z)$ lie on a circle when $z$ is on the boundary.

Since $f$ is entire, the extension and $f$ itself must be identical. Hence there are no poles, which means that $f$ has no zeroes in $0<|z|<1$. The only possible zero in $|z|<1$ is $z=0$.

If $f(0)\neq 0$ then $\lim_{z\to\infty} f(z)$ exists. So $f$ is bounded. From Liouville's theorem, $f(z)$ is a constant. $f(z)=c, |c|=1$.

If $f(0)=0$ then $\lim_{z\to\infty} f(z)=\infty$. The order of the zero should be finite, say, $m$. Then $g(z) = f(z)/z^m$ is entire, too; and it has no zeroes, $|g(z)|=1$ when $|z|=1$. From the argument above, we see that $g(z)$ is a constant. Hence $f(z) = c z^m$, where $|c|=1$.

Yes your solution is good.  However, you should write down the relation the extension satisfies to better see why f can only be zero at z = 0.  Also the fact that the limit is infinite as z goes to infinity implies that f has a pole at infinity.

We can also do this:

First, we find those functions holomorphic on some domain containing the closed unit disk which satisfy the same hypothesis. We consider $f$ to be defined on an open disk of radius $r$ s.t. all the roots of $f$ have absolute value $<\frac 1r$.

We can perform induction on the number of roots. If $\alpha$ is a root of $f$ with $|\alpha|<1$, then we can apply the induction hypothesis to $\frac f{\varphi_\alpha}$, where for $|t|<1,\ \varphi_t(z)=\frac{z-t}{1-\bar tz}$, and in the end we conclude that $f$ must have the form $\lambda\prod_{i=1}^n\varphi_{\alpha_i}$ for some $|\alpha_i|<1$. Next, since $f$ is entire, we conclude that all the $\alpha_i$'s must, in fact, be $0$.


Hello,

I was wondering if anyone knows how to generate 256 different sequences from 256 numbers [0:255], where no number is repeated within the same row of numbers, at the same time no two consecutive numbers exist more than once in the different 256 sequences.

Can anyone help me with this please ?

here is what I mean by a simple example of fewer numbers.

if n = 4, then we have a matrix of 4 by 4

1 2 3 4
2 4 1 3
3 1 4 2
4 3 2 1

where numbers from 1 to 4 are represented in 4 different sequences where no two consecutive numbers exist more than once in the different sequences.

So I need to do the same thing with n=256 i.e a matrix of 256 by 256.

Thank you very much in advance.

Dina

I'm curious what the source of this question is.

[hide]Here's an analysis of your sequences of length 4: add 0 in front of all your sequences.  Then the first sequence is what you get by adding 1 each time, the second sequence is what you get by adding 2 each time, the third is what you get by adding 3 each time and the fourth is what you get by adding 4 each time (where in all cases you take the numbers you get modulo 5).  You can do the same thing whenever you have $ p \minus{} 1$ numbers for some prime $ p$.  Since 257 is prime, you can do it for 256, too.[/hide]

Does anyone else have a more general construction?  (Or a proof that they don't exist -- e.g., consider the case of permutations of length 3.)

Thank you very much for your reply, I really appreciate it.

The source of the question is that these sequences are the inputs to a circuit and by changing the sequences the power consumption of the circuit differs slightly, but still needs to be known, so the same numbers have to be presented in different sequences without repetition inside the same sequence.

Best regards,

Dina

[hide="Discussion"] The question can be rephrased in graph-theoretic terms as follows:  can we find $ n$ edge-disjoint [url=http://en.wikipedia.org/wiki/Hamiltonian_path]Hamiltonian paths[/url] in the complete directed graph on $ n$ vertices (that is, the graph on $ n$ vertices with two directed edges in opposite directions between every vertex)?  Unfortunately, the only articles I have been able to find with the keywords "disjoint Hamiltonian paths" are about [url=http://mathworld.wolfram.com/Tournament.html]tournaments[/url], and while it's true that the complete directed graph on $ n$ vertices is a composition of two tournaments, it's not clear that the set of Hamiltonian paths we want can always be decomposed in this way.

I'm firmly convinced that there are no solutions for $ n$ odd but haven't been able to find a proof.  [/hide]
Edit:  JBL's construction can be extended to any $ n$ for which there exists $ m$ such that $ n \equal{} \varphi(m)$ by using multiples of the group of units $ \bmod m$ and then relabeling.  This further suggests that there are no solutions for $ n$ odd.

[quote="t0rajir0u"]JBL's construction can be extended to any $ n$ for which there exists $ m$ such that $ n \equal{} \varphi(m)$ by using multiples of the group of units $ \bmod m$ and then relabeling.  This further suggests that there are no solutions for $ n$ odd.[/quote]
This is actually quite wrong.  I'll keep working.

Here's a construction for even $ n$: take a "good permutation" as in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=199166[/url], then add 1 repeatedly: for example, for $ n \equal{} 8$, take the good permutation 81726354 and successively add 1 mod 8:
81726354
12837465
23148576
34251687
45362718
56473821
67584132
78615243
(t0r, is this the sort of thing you were shooting for with your previous comment?)
Funny, the connections one finds ... still need to settle odd $ n$, though.
Also interesting is that this is a Latin square -- will every such list be a Latin square?  (My guess is no.)

[quote="JBL"](t0r, is this the sort of thing you were shooting for with your previous comment?)[/quote]
Well, although it took me awhile to realize it, the problem with replacing $ \{ 1, 2, ... p \minus{} 1 \}$ with any other set of units (in increasing order) is that it's not possible to control the consecutive differences properly and they end up repeating, and those repetitions turn out to indicate repetition of ordered pairs.

That's interesting, though.  That makes this problem something of a generalization of the good permutation problem in a slightly different direction, and we can now consider the problem of finding disjoint Hamiltonian paths that exhaust the edges in a (directed) [url=http://en.wikipedia.org/wiki/Regular_graph]regular graph[/url].  This in turn suggests that the good permutation problem for finite groups can be phrased in terms of [url=http://en.wikipedia.org/wiki/Cayley_graph]Cayley graphs[/url].  Of course, this is all speculation.

That's an interesting suggestion -- hopefully, I'll remember to give it more thought at some point :)

As an aside, let me answer the question from my previous post: the construction I gave using good permutations gives a Latin square, and permuting the first column so that it matches the first row actually makes the resulting table symmetric.  However, there is at least one table of this form that does not give a Latin square (and as a consequence cannot be made symmetric by rearranging rows) for $ n \equal{} 6$.  (For $ n \equal{} 4$, all such lists of permutations are equivalent to the one given by the O.P.)

There are no tables of this form for $ n \equal{} 5$, further suggesting that they exist only for even $ n$ (or, I suppose, for $ n \equal{} 1$ in the same stupid sense that good permutations exist in this case).  So, I'm quite thoroughly convinced that there are no solutions for $ n$ odd, but I still don't have a proof.  Perhaps a moderator could move this to some place where someone is more likely to come up with one, e.g. either preolympiad or one of the combinatorics fora?

Completely unrelatedly: t0r, was your RSI mentor Ryan Reich?

Yes.  He's mentoring Zane Li now, right?

Okay, cool.  Yes, he worked with Zane and one other student (whose name escapes me, although I could find it easily if you're interested) this year.

Here's the "not nice" table of size 6:

123456
213654
314625
426153
516432
635241

("The" meaning the one I've found and referenced in my previous post -- there may be others.)

Edit: okay, now I can generate all such tables of size 6: there are 736 of them up to symmetries (applying a permutation to the entries, and permuting the rows).

Also, the conjecture that there are no tables for odd $ n$ turns out to be badly false: the first of many for $ n \equal{} 7$ is
1234567
2135476
3146275
4157263
5361742
6524371
7325164

My mathematica program spent 6 or 7 hours to find that there are 466144 fundamentally different tables of size 7, none of which is a Latin square.  So, the new conjecture is that odd tables exist of all sizes 7 or larger (but that none of them are Latin squares?) -- any clever suggestions for a construction?

(Would have added this to the previous post, but I can't edit it any more.)

Edit: and what do you know, we're not the first people to think about this: http://www.research.att.com/~njas/sequences/A004078
These are apparently called "Tuscan squares."

Later edit: There are two Tuscan Latin squares of size 6:
123456
241635
315264
462513
536142
654321

is the one you get using a good permutation as the base and

123456
246135
362514
415263
531642
654321

which arises from performing shuffles starting with 123456 and 654321 (and probably the same construction works for every even $ n$?)

If you can follow the pattern in the image
the formula C(r,c)= (c-r) - n*Floor((c-r)/(n+1))+1
corresponds to every row (r) and column (c)

p.s * means multiplication

Here is the image i was unable to post because of size

to Torajir0u:I think Peter Zhang would be the other student
to JBL: Hello! I'm Bob's Korean friend. I think we met once at the presentations.

Hi ywl!  I hope you and Bob are enjoying the end of summer.

Can anyone else not view ManuelS's image?


I think the problem had better reduce as following:
Hypothesis: $a^{3}+b^{3}+c^{3}=\frac{1}{2}$
Conclusion: $\frac{4}{3}+5abc\leq(a+b+c)(a^{2}+b^{2}+c^{2})\leq 2+3abc$

But now i have no idea to kill it


probability
symmetry
number theory
relatively prime
AMC

Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumilated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B$. The probability that team $A$ finishes with more points than team $B$ is $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

[hide="Solution"]Each team will play $5$ more games, and Team A must win at least as many of their remaining games as Team B.  Let $P(n)=$ the probability that Team A wins $n$ of their last $5$ games, or the probability that Team B wins $n$ of their last $5$ games.  Then we have

$P(0)=P(5)=\frac{1}{32}$
$P(1)=P(4)=\frac{5}{32}$
$P(2)=P(3)=\frac{10}{32}$

The probability that Team A will finish with more points than Team B is

$P(0)P(0)+P(1)(P(0)+P(1))+P(2)(P(0)+P(1)+P(2))\ldots$

After some simple calculations, we find that the answer is $\boxed{831}$.[/hide]

Solafidefarms, you should change 50% to $50\%$ for the LaTeXed PDF file. ;)

I believe I have a slightly nicer solution, props to Mark Hou for pointing this out.

[hide]There are 5 remaining games. Find the probability that team A wins the same number of games as team B:

Number of ways to win same number:
(5 0)(5 0) + (5 1)(5 1) + (5 2)(5 2) + (5 3)(5 3) + (5 4)(5 4) + (5 5)(5 5) = (10 5)

Number of total configurations of games: 2^10

P = (10 5) / 2^10 = 63/256

So the probabilty that they don't win the same number of games is 1 - P = 193/256.

But by symmetry the number of ways to win more games is the same as the number of ways to lose more games, so the probability that A wins more games than B is:

(193/256)/2 = 193/512.

A wins more total games than B if they win the same number or if A wins more, so the total probaiblity is

193/256 + 126/512 = 319/512, so the answer is 831.[/hide]

A generalization: Prove that, if there are $n$ teams participating, and $A$ beats $B$ in their game, the probability of $A$ beating $B$ in the final ranking is $\frac{1}{2}+\frac{\binom{2n-4}{n-2}}{2^{2n-5}}.$

[hide="Hint"]
$\sum_{i=1}^{n}\binom{n}{i}^{2}= \binom{2n}{n}.$
[/hide]

[quote="Karth"]A generalization: Prove that, if there are $ n$ teams participating, and $ A$ beats $ B$ in their game, the probability of $ A$ beating $ B$ in the final ranking is $ \frac {1}{2} \plus{} \frac {\binom{2n \minus{} 4}{n \minus{} 2}}{2^{2n \minus{} 5}}.$

[hide="Hint"]
$ \sum_{i \equal{} 1}^{n}\binom{n}{i}^{2} \equal{} \binom{2n}{n}.$
[/hide][/quote]
what is n stand for? and Would anyone prove it? Seem to me, it is not right.

[hide=So long "soccer"s]
There would be $2^{21}$ total possible outcomes, but the first game's outcome is already determined by the problem, thus the total is simply $2^{20}$ ways.

Observe that the maximum points any team could have is $6$ points, since they all play $6$ games.
We could divide the desirable outcomes into cases. (btw I let the teams be $A_1$,$A_2$ and so on)

Case 1: $A_1>A_2$ with $A_1=6$
$2^5*2^{10}.$
Case 2: $A_1=5$
$\binom{5}{4}*[2^5-1]*2^{10}.$
Case 3: $A_1=4$
$\binom{5}{3}*[2^5-\binom{5}{4}-\binom{5}{5}]*2^{10}.$
Case 4: $A_1=3$
$\binom{5}{2}*[2^5-\binom{5}{5}-\binom{5}{4}-\binom{5}{3}]*2^{10}.$
Case 5: $A_1=2$
$\binom{5}{1}*[2^5-\binom{5}{5}-\binom{5}{4}-\binom{5}{3}-\binom{5}{2}]*2^{10}.$
Case 6: $A_1=1$
$\binom{5}{0}*[2^5-\binom{5}{5}-\binom{5}{4}-\binom{5}{3}-\binom{5}{2}-\binom{5}{1}]*2^{10}.$

Add all these, and divide it by $2^{21}.$
You get $\dfrac{2^{10}*638}{2^{21}*2^{11}}.$
This simplifies to $\dfrac{319}{512}.$
$319+512=831.$
[/hide]

Is there a better or faster way than this?

[hide=Solution]
Team $A$ finishes with more points than $B$ if and only if $A$ scores at least as many points as $B$ in the remaining five games for each of them. There are $2^{10}=1024$ possible outcomes of these ten games. The number of outcomes in which $A$ scores [i]exactly[/i] as many points as $B$ is $\tbinom{5}{0}^2 + \tbinom{5}{1}^2 + \ldots + \tbinom{5}{5}^2 = 252$ (since if both $A$ and $B$ score $k$ points then there are $\tbinom{5}{k}^2$ possible outcomes). Then there are $1024-252=772$ outcomes in which $A$ and $B$ score an unequal number of points. By symmetry, there are $\tfrac12 \cdot 772 = 386$ outcomes in which $A$ scores [i]strictly more[/i] points than $B$. In total, we have $386+252=638$ outcomes in which $A$ scores at least as many points as $B$, so the probability is $\tfrac{638}{1024}=\tfrac{319}{512}$. The answer is $319+512=831$.
[/hide]


modular arithmetic
calculus
integration
parameterization

Someone posted a question some days ago that what are the last two digits of the number 1234^23.
I have got the answer 24. Is it correct? Also mention the solution so that I could get to know better solutions.

[hide="ignore this"]list out the first few powers of 4. They repeat in the tens digits and the units digits every 10.

$4^1=04$
$4^2=16$
$4^3=64$
$4^4=56$
$4^5=24$
$4^6=96$
$4^7=84$
$4^8=36$
$4^9=44$
$4^{10}=76$
$4^{11}=04$
$4^{12}=16$

We can see that $4^{23}$ has the same last two digits as $4^3$ because it has the same remainder when divided by 10, and the last two digits are 64.

This is the middle school way  :blush:[/hide]

You're both wrong :P

You don't just need the powers of 4; you need the powers of 34.  You'll find that [hide]$34^{10 n + 3} \equiv 04 \pmod {100}$, so $1234^23$ ends in 04. [/hide]

[hide]

Suppose $1234^{23}=x$ then applying mod and FLT we have

$x\equiv 0 \mod 4$
$x\equiv 34^{23}\equiv 9^{23}\equiv 9^3\equiv 4 \mod 25$

Now letting $x=4p$ and $x=25q+4$ for some integer $p,q$ .

Solving $4p=25q+4$ for integral solution $p,q$ , we can see that $(p,q)=(26,4)$ satisfy it . Hence the parameter for $p,q$ is $p=26+25k$ , $q=4+4k$ for all integer $k$ .

Hence $x=4p=4(26+25k)=104+100k$ which w can see it leaves a remainder of $4$ when divided by $100$[/hide]

[quote="varun"]the last two digits of the number 1234^23.[/quote]

[hide]
$1234^{23} \equiv 34^{23} \pmod {100}$
We can find $34^{5} \equiv 24 \pmod {100} \rightarrow 34^{20} \equiv 76 \rightarrow 34^{23} \equiv 04 \mod {100}$
[/hide]


inequalities
geometry
circumcircle
geometry solved

$1.\blacktriangleright$ Let $M$ be a point inside a triangle $ABC; R$ is the circumradius of $\triangle ABC$.
Prove that: $\sum\frac{MA}{a^{2}}\geq\frac{1}{R}$
$2.\blacktriangleright$  Let triangle $ABC$. Prove that: $\sum\frac{MA}{a}\geq\sqrt{3}$
with any point $M$ on the plane.

1. http://www.mathlinks.ro/Forum/viewtopic.php?t=6073
2. http://www.mathlinks.ro/Forum/viewtopic.php?t=21016

  darij


[quote="Ahiles"]Let $ P$ be a point of triangle $ ABC$. Draw a line through $ P$ which divides area of triangle in two parts with ratio $ 3: 4$.[/quote]
Is $ P$ on the sides or inside the triangle?

[quote="Ahiles"]I've got the same question.... In romanian it is written [i]"...punctul $ P$ al triunghiului $ ABC$..."[/i]. I think it is on the sides....[/quote]Thank you for your answer but both cases should be valid...