matrix
Coloring
combinatorics
IMO
IMO 1997
induction

(a)  If we list all the elements in the rows followed by all the elements in the columns, then we have listed every element in the array twice, so each number in S must appear an even number of times. But considering the ith row with the ith column, we have also given n complete copies of S together with an additional copy of the numbers on the diagonal. If n is odd, then each of the 2n-1 numbers appears an odd number of times in the n complete copies, and at most n numbers can have this converted to an even number by an appearance on the diagonal. So there are no silver matrices for n odd. In particular, there is no silver matrix for n = 1997.

(b)  Let Ai,j be an n x n silver matrix with 1s down the main diagonal. Define the 2n x 2n matrix Bi,j with 1s down the main diagonal as follows: Bi,j = Ai,j; Bi+n,j+n = Ai,j; Bi,j+n = 2n + Ai,j; Bi+n,j = 2n + Ai,j for i not equal j and Bi+n,i = 2n. We show that Bi,j is silver. Suppose i ≤ n. Then the first half of the ith row is the ith row of Ai,j, and the top half of the ith column is the ith column of Ai,j, so between them those two parts comprise the numbers from 1 to 2n - 1. The second half of the ith row is the ith row of Ai,j with each element increased by 2n, and the bottom half of the ith column is the ith column of Ai,j with each element increased by 2n, so between them they give the numbers from 2n + 1 to 4n - 1. The only exception is that Ai+n,i = 2n instead of 2n + Ai,i. We still get 2n + Ai,i because it was in the second half of the ith row (these two parts do not have an element in common). The 2n fills the gap so that in all we get all the numbers from 1 to 4n - 1.

An exactly similar argument works for i > n. This time the second half of the row and the second half of the column (which overlap by one element) give us the numbers from 1 to 2n - 1, and the first halves (which do not overlap) give us 2n to 4n - 1. So Bi,j is silver. Hence there are an infinite number of silver matrices.

Well, it is possible to prove that $n \times n$ silver matrices exists for any [b]even[/b] $n$.

a) $ (2k\plus{}1)^2 < (k\plus{}1)(4k\plus{}1)$ for $ k>0$, so if $ n\equal{}2k\plus{}1$ there exists a number from $ S$ which only is used at most $ k$ times, so it can only occur in at most $ 2k$ columns or rows. So there must be an $ i: 1\leq i \leq 2k\plus{}1$ for which it isn't in the $ i$th row or column.

b) For n even just do it. Let $ n\equal{}2m$. It's fairly obvious that we can partition the set of unordered pairs $ \{i,j\}$ from $ \{1,2,....,2m\}$  into $ 2m\minus{}1$ disjoint sets $ T_1, T_2,....,T_{2m\minus{}1}$ of size $ m$ such that each $ i$ is found precisely once in each set. Then take a matrix $ (a_{xy})$ with major diagonal filled with 1s and for each pair $ \{i,j\}, i<j$ in $ T_k$, set $ a_{ij} \equal{} 2k, a_{ji}\equal{}2k\plus{}1$. This gives a silver n by n matrix.

Sorry for reviving a really old topic but I know, in official solution and @Lagrangia 's one, they had just proven that $n$ is power of $2$ so what about $n$ is even, can anyone help me?, @llthigore 's solution doesn't satisfy me at all!

There is a nice recursive construction:
If you have a silver matrix of size $n$, you can construct a silver matrix of size $2n$ by taking two copies of the original matrix on the diagonal and adding [url=https://en.wikipedia.org/wiki/Latin_square]Latin Squares[/url] in the two other corners
But you can also obtain a silver matrix of size $2n-2$ by taking the original matrix in the upper left corner, reflecting it over the diagonal and filling the rest appropriately (this reflection works because you can assume that there are only 1's on the diagonal).
And starting from $2$ with the steps $n \to 2n$ and $n \to 2n-2$ you can obtain any even number.

[quote=Tintarn]
But you can also obtain a silver matrix of size $2n-2$ by taking the original matrix in the [color=#f00]upper left corner, reflecting it over the diagonal [/color] and filling the rest appropriately (this reflection works because you can assume that there are only 1's on the diagonal).
And starting from $2$ with the steps $n \to 2n$ and $n \to 2n-2$ you can obtain any even number.[/quote]

Yes, this is exactly what I need but the red part, I don't know how you reflected, can you clarify, thanks!

I'll show you the example $4 \to 6$. You start with the construction for $n=4$ which canonically looks like
\[\begin{pmatrix} 1 & 2 & 4 & 5\\3 & 1 & 5 & 4\\6 & 7 & 1 & 2\\7 & 6 & 3 & 1 \end{pmatrix}\]
Now, you do the reflection over the diagonal:
\[\begin{pmatrix} 1 & 2 & 4 &  & & \\3 & 1 & 5 &  & & \\6 & 7 & 1 & 2 &  & \\ &  & 3 & 1  & 5 &4\\& &  & 7 & 1 & 2\\ & &  & 6 & 3 & 1\end{pmatrix}\]
(Here of course you need to delete some entries.) And then fill in the gaps:
\[\begin{pmatrix} 1 & 2 & 4 & 8 & 9 & 5 \\3 & 1 & 5 &  9 & 4 & 8 \\6 & 7 & 1 & 2 & 8 & 9\\10 & 11 & 3 & 1  & 5 &4\\11 & 6 & 10 & 7 & 1 & 2\\7 &10 &11  & 6 & 3 & 1\end{pmatrix}\]
Of course there some technicalities to describe the general procedure, but hopefully the idea should be clear.:)

If the silver matrix $a(i,j)$ of size $2^k$ is defined, using the elements $\{-2^k+1, ..., 0, ..., 2^k-1\}$, then you can get a silver matrix of size $2^{k+1}$ by setting

$a(i,j+2^k) = a(i,j)+2^k$ if $a(i,j) \geq 0$

$a(i,j+2^k) = a(i,j)-2^k$ if $a(i,j) < 0$

$a(i+2^k,j) = a(i,j) - 2^k$ if $a(i,j) \leq 0$

$a(i+2^k,j) = a(i,j) + 2^k$ if $a(i,j) > 0$

$a(i+2^k, j+2^k) = a(i,j)$.

It has a lot of nice symmetries, such as $a(i,j) = - a(i,j)$ for all $i,j$.

I really enjoyed this problem :)
[quote=iandrei]An $ n \times n$ matrix whose entries come from the set $ S \equal{} \{1, 2, \ldots , 2n \minus{} 1\}$ is called a [i]silver matrix[/i] if, for each $ i \equal{} 1, 2, \ldots , n$, the $ i$-th row and the $ i$-th column together contain all elements of $ S$. Show that:

(a)  there is no silver matrix for $ n \equal{} 1997$;

(b)  silver matrices exist for infinitely many values of $ n$.[/quote]
[b](a)[/b] We will show that there exists no silver matrix for an odd $n$.
Call the union of the $ i$-th row and the $ i$-th column as the $i$-th [i]block[/i]. Hence, each block must contain all the elements from $1$ to $2n-1$, and clearly, each element in a block must be different from the others (in the same block).
Note that any element in the matrix contributes to at most $2$ blocks. Hence, each element $i$ appears at least $\left \lceil \frac{n}{2} \right \rceil =\frac{n+1}{2}$ times in the matrix. But since there are $2n-1$ elements, hence the total number of elements in the matrix is at least $\frac{(2n-1)(n+1)}{2} >n^2$, a contradiction. $\blacksquare$

[b](b)[/b] We will prove by induction on $n$ that for any $n \geq 1$, there exists a $2^n \times 2^n$ silver matrix. A $2 \times 2$ silver matrix is shown as the base case.
$$\begin{pmatrix}
    1 & 2 \\
    3 & 1 \\
\end{pmatrix}$$
Now, asume that we have found a silver matrix $\{a_{ij}\}$ where $1 \le i, j \le 2^k$ for some $n=k$. For simplicity, set $m=2^{k}$. Then the silver matrix for $n=k+1$ is
$$\begin{pmatrix}
    a_{11} & a_{12} & \dots  & a_{1m} & a_{11}+2m & a_{12}+2m & \dots  & a_{1m}+2m \\
    a_{21} & a_{22} & \dots  & a_{2m} & a_{21}+2m & a_{22}+2m & \dots  & a_{2m}+2m \\
    \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\
    a_{m1} & a_{m2} &  \dots  & a_{mm} & a_{m1}+2m & a_{m2}+2m &  \dots  & a_{mm}+2m \\

    2m & a_{12}+2m & \dots  & a_{1m}+2m & a_{11} & a_{12} & \dots  & a_{1m} \\
    a_{21}+2m & 2m & \dots  & a_{2m}+2m & a_{21} & a_{22} & \dots  & a_{2m} \\
    \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots  \\
    a_{m1}+2m & a_{m2} +2m&  \dots  & 2m & a_{m1} & a_{m2} &  \dots  & a_{mm} \\
\end{pmatrix}$$
The reason why this works is easy to see once you [hide=try it out yourself]
My joblessness is overflowing within, so I decided to add this
$$\begin{pmatrix}
    1 & 2 \\
    3 & 1 \\
\end{pmatrix}    \mapsto \begin{pmatrix}
    1 & 2 & 5 & 6\\
    3 & 1 & 7 & 5 \\
    4 & 6 & 1 & 2 \\
    7 & 4 & 3 & 1 \\
\end{pmatrix} \mapsto \begin{pmatrix}
    1 & 2 & 5 & 6 & 9 & 10 & 11 & 12\\
    3 & 1 & 7 & 5 & 13 & 9 & 12 & 11\\
    4 & 6 & 1 & 2 & 14 & 15 & 9 & 10\\
    7 & 4 & 3 & 1 & 15 & 14 & 13 & 9\\
    8 & 10 & 11 & 12 & 1 & 2 & 5 & 6\\
    11 & 8 & 12 & 13 & 3 & 1 & 7 & 5 \\
    14 & 15 & 8 & 10 & 4 & 6 & 1 & 2 \\
    15 & 14 & 13 & 8 & 7 & 4 & 3 & 1 \\
\end{pmatrix} $$

[/hide] building from the case $n=1$ to $n=2$ to $n=3$, but just for completeness, I will provide a formal explanation as to why this works.
[hide=Why does this work?] (I will show that this is true for the $(m+1)$-th block, and we can then similarly do it for the $(m+2), (m+3) \cdots 2m$-th blocks, and the case for the first $m$ blocks follows by rotating the matrix $180^o$ and using the work mentioned above)

Note that in all these matrices, $a_{ii}=1$ for all $i$, i.e. the diagonal from the top left to the right bottom consists only of $1$s.

Clearly, in any block, the numbers should be $\{1, 2, \cdots 2^{k+2}-1\}$. But here, $\{1, 2 \cdots 2^{k+1}-1\} \equiv \{a_{11}, a_{12} \cdots a_{1m}, a_{21}, a_{31}, \cdots a_{m1}\}$, by the induction hypothesis. Hence, we only need to show $$\{a_{11}+2m, a_{21}+2m \cdots a_{m1}+2m, 2m, a_{12}+2m, \cdots a_{1m}+2m\} \equiv \{ 2^{k+1}, 2^{k+1}+1, \cdots 2^{k+2}-1\}$$ Note that these two sets have the same number of elements, and the minimum of these is $2m=2^{k+1}$ and the maximum possible is (by the induction hypothesis) $2^{k+1}-1+2m=2^{k+1}-1+2^{k+1}=2^{k+2}-1$. Thus, it suffices to show that the elements are distinct.

Clearly, $a_{i1}+2m \ne a_{j1}+2m$ and $a_{i1}+2m \ne a_{1i}+2m \ne a_{j1}+2m$ for any $i \ne j$, by the induction hypothesis.
Also, $2m < 2m+a_{i1}$ and $2m < 2m+a_{1i}$ for all $i$, and hence we are done $\blacksquare$
[/hide]
Hence, we are done by induction. $\blacksquare$

Very nice problem! And a good practice of my proof-writing skills...

[b]a)[/b] Let $n=2k+1$. Call an $i$th row and $i$th column together a $i$-[i]rolumn[/i]. Suppose a silver matrix does exist for $n$. Consider any number $m\in\{1,2,\dots, 4k+1\}$. Say it is present in the $a$th row and $b$th column of the matrix. Then it is part of the $a$-rolumn and the $b$-rolumn. Thus one instance of an entry in the matrix only satisfies being part of two rolumns. Thus to be part of $2k+1$ total rolumns, each of $\{1,2,\dots, 4k+1\}$ must appear at least $\left\lceil\frac{2k+1}{2}\right\rceil=k+1$ times. So there will be at least a total of $(k+1)(4k+1)=4k^2+5k+1$ entries in the matrix. However we only have $(2k+1)^2<4k^2+5k+1$ slots for the numbers, and we have a contradiction. Thus there are no silver matrices for odd $n$, and specifically for $n=1997$. $\square$

[b]b)[/b] We will show that there exists a silver matrix for $n=2^k$, $k\geq 1$.
We will prove this by induction on $k$, where $2^k\times 2^k$ is the dimension of the matrix.
For $k=1$, consider the matrix $\left(\begin{array}{cc} 1&2\\3&1\end{array}\right)$. We assume that a silver matrix exists for $n=2^k$. Call it $A$. Now we wish to prove that a silver matrix exists for $2^{k+1}=2n$. We divide the $2n\times 2n$ matrix into four $n\times n$ matrices. We call them $B_1,B_2,B_3,B_4$ clockwise. Let $B_1=B_3=A$. Then $A$ contains the numbers $1,2,\dots, 2n-1$. The rest of the numbers $2n,2n+1,\dots, 4n-1$ must be filled in $B_2$ and $B_4$. Let $X=\{2n,2n+1,\dots, 3n-1\}$ and $Y=\{3n,3n+1,\dots, 4n-1\}$. So $|X|=|Y|=n$.

[b]Lemma 1:[/b] Any $n\times n$ matrix can be filled with the distinct numbers $a_1,a_2,\dots, a_n$, so that each row and each column contains all the numbers $a_1,a_2,\dots, a_n$.

[b]Proof:[/b] Simply consider the matrix $\begin{pmatrix}a_1&a_2&a_3&\dots&a_n\\a_n&a_1&a_2&\dots&a_{n-1}\\a_{n-1}&a_n&a_1&\dots&a_{n-2}\\.&.&.&\dots&.\\.&.&.&\dots&.\\.&.&.&\dots&.\\a_2&a_3&a_4&\dots&a_1 \end{pmatrix}$, where the $k$th row is formed by shifting the $(k-1)$th row to the right by one entry, and the last entry in the $(k-1)$th row is the first entry in the $k$th row. $\square$

Now going back to the main problem and using Lemma 1, we fill $B_2$ with the elements of $X=\{2n,2n+1,\dots, 3n-1\}$ so that each row and each column contains all the elements of $X$, and $B_4$ with the elements of $Y=\{3n,3n+1,\dots, 4n-1\}$ in a similar fasion.
Then our $2n\times 2n$ ($n=2^k$) matrix is $$\begin{array}{cc}A&B_2\\B_4&A\end{array}.$$ Note that if $1\leq i\leq 2^k$, then the $i$th row contains the elements of the $i$th rows of the upper left corner $A$ and the $i$th row of $B_2$. The $i$th column contains the elements of the $i$th column of the upper left corner $A$ and the $i$th column  of $B_4$. We know that the "$A$ elements" contain all of ${1,2,\dots,2n-1}$ by the induction hypothesis. The "$B_2$ elements" contain all of $2n,2n+1,\dots, 3n-1$ and the "$B_4$ elements" contain all of $3n,3n+1,\dots, 4n-1$. A similar argument goes for $2^k<i\leq 2^{k+1}$. Thus the $i$th row and the $i$th column of the $2n\times 2n$ matrix contains all the numbers $1,2,\dots, 4n-1$, and is thus a silver matrix.

Thus we are done by induction. $\blacksquare$

In the first step we define a machine. our machine is called a matrix permutation :
if we give the matrix A to this , our machine add $+x$ to all elements of matrix $A$ except left large diameter elements ( this diameter $\begin{bmatrix} *&  \\  &* \end{bmatrix}$ ) and fill in the left large diameter elements with  $y$ and $z$ respectively.
we show the new matrix $A$ in $A^{+x}_{y,z}$ form. for example :
$$A : \begin{pmatrix} \bold{1}&2&3&4 \\ 5&\bold{6}&7&8 \\ 9&10&\bold{11}&12 \\ 13&14&15&\bold{16} \end{pmatrix} \rightarrow  A^{+2}_{4,5} : \begin{pmatrix} \bold{4}&4&5&6 \\ 7&\bold{5}&9&10 \\ 11&12&\bold{4}&14 \\ 15&16&17&\bold{5} \end{pmatrix}$$
now,  for the for machine a new silver matrix, it's sufficient to convert the silver matrix $A$ in this way :
$$ \begin{bmatrix} A&A^{+2n}_{n,n+1} \\ A^{+2n}_{n+1,n}&A \end{bmatrix} $$
that's clearly gives the silver matrix because matrix $A$  makes numbers from $1$ to $2n-1$ and matrix permutation  makes number for $2n$ to $4n-1$.
the proof for moving $n$ and $n+1$ in matrix is row $i$_th and column $i$_th $(n < i)$ are same as $+2n$ and they can completely cover numbers $2n$ to $4n-1$ because $A$ was a silver matrix. And if it was in the form of $A^{+2n}_{n,n}$, makes numbers $2n+1$ to $4n-1$ that doesn't give us silver matrix.
$\begin{pmatrix} 1&2 \\ 3&1 \end{pmatrix} \rightarrow \begin{pmatrix} 1&2&4&6 \\ 3&1&7&5 \\ 5&6&1&2 \\ 7&4&3&1 \end{pmatrix} \rightarrow
 \begin{pmatrix} 1&2&4&6&8&10&12&14 \\ 3&1&7&5&11&9&15&13 \\ 5&6&1&2&13&14&8&10 \\ 7&4&3&1&15&12&11&9 \\ 9&10&12&14&1&2&4&6 \\ 11&8&15&13&3&1&7&5 \\ 13&14&9&10&5&6&1&2 \\ 15&12&11&8&7&4&3&1 \end{pmatrix}$

$(\text a)$ FTSOC, let $M=(a_{i,j})$ be a $n\times n$ silver matrix. For odd $n$, we have:
$$2\sum_{i=1}^n\sum_{j=1}^na_{i,j}=\sum_{i=1}^n\left(\sum_{j=1}^n(a_{i,j}+a_{j,i})-a_{i,i}\right)>\sum_{i=1}^n\sum_{j=1}^{2n-1}j=n^2(2n-1),$$
which is a contradiction since $n^2(2n-1)$ is odd. Since $1997$ is odd, there is no such silver matrix.

Got the minor hint to do induction for $(b)$.

$(\text b)$ We claim that there exists a silver matrix for all powers of $2$.

Proceed by induction.
[b]Base case:[/b] $2\times2$
We will use the matrix $\begin{pmatrix}1&2\\3&1\end{pmatrix}$.

[b]Induction step:[/b] $n\times n\Rightarrow2n\times2n$
For any silver $2^n\times2^n$ matrix $M$, we consider the transformation
$$M\mapsto\begin{pmatrix}M&A\\B&M\end{pmatrix},$$
where $A=M+2^{n+1}J_{2^n}$ and $B$ is the matrix obtain by replacing instances of $2^{n+1}+1$ with $2^{n+1}$ in $A$.

For instance:
$$\begin{pmatrix}1&2\\3&1\end{pmatrix}\to\begin{pmatrix}1&2&5&7\\3&1&6&5\\4&7&1&2\\6&4&3&1\end{pmatrix}\to\begin{pmatrix}1&2&5&7&9&10&13&15\\3&1&6&5&11&9&14&13\\4&7&1&2&12&15&9&10\\6&4&3&1&14&12&11&9\\8&10&13&15&1&2&5&7\\11&8&14&13&3&1&6&5\\12&15&8&10&4&7&1&2\\14&12&11&8&6&4&3&1\end{pmatrix}$$

Firstly, we can show that this process, iterated from $\begin{pmatrix}1&2\\3&1\end{pmatrix}$, always produces a matrix with all ones on the main diagonal. Then the main diagonal of $A$ is replaced when we form $B$. Now we can easily show that the resulting matrix is silver.

[hide=Solution (b)]We claim that matrices exist if $n=2^k$ where $k\ge 0.$ We proceed by induction.

[i]Base Case $k=0.$[/i] $\begin{pmatrix}1\end{pmatrix}$ is such a matrix.

[i]Inductive Step.[/i] Consider the matrix
$$M_k=\begin{pmatrix}
x_1 & x_2 & \dots & x_{2^k}\\
x_{2^k+1} & x_{2^k+2} & \dots & x_{2\cdot2^k}\\
\vdots & \vdots & \ddots & \vdots\\
x_{(2^k-1)2^k+1} & x_{(2^k-1)2^k+2} & \dots & x_{2^{k^2}}
\end{pmatrix}$$
We construct the matrix $M_{k+1}=\begin{pmatrix}a&b\\ c&a\end{pmatrix}.$ Note that $M_{k+1}$ is a $2^{k+1}$ by $2^{k+1}$ matrix. Let $a=M_k.$ Let $b$ be the matrix such that $2^{k+1}$ is added to each element of $M_k.$ Let $c$ be the matrix such that $2^{k+1}$ is added to each element of $M_k$ except we only add $2^{k+1}-1$ to the diagonal that spans from upper-left to lower-right.

We claim that $M_{k+1}$ is a silver matrix if $M_k$ is. We consider the $m$th column and $m$th row of $M_{k+1}.$ We see that it will contain row/column $m$ of $a,b,$ and $c.$ Since the smallest terms in $b,c$ are $2^{k+1}$ are the largest term in $a$ is $2^{k+1}-1,$ it suffices to show that column $m$ of $c$ shares no elements with row $m$ or $b$ (or vice versa). This is true because the rows/columns are just $2^{k+1}$ more that those of $a,$ and by the inductive hypothesis are distinct. The only "overlapping" term is also different because we are subtracting one from the "overlapping" term in $c.$

Thus, our induction is complete and we are done.
[/hide]

From [i]Twitch Solves ISL[/i]:

For (a), define a [i]cross[/i] to be the union of the $i$th row and $i$th column. Every cell of the matrix not on the diagonal is contained in exactly two crosses, while each cell on the diagonal is contained in one cross.
On the other hand, if a silver matrix existed for $n=1997$, then each element of $S$ is in all $1997$ crosses, so it must appear at least once on the diagonal since $1997$ is odd. However, $|S| = 3993$ while there are only $1997$ diagonal cells. This is a contradiction.

For (b), we construct a silver matrix $M_e$ for $n = 2^e$ for each $e \ge 1$. We write the first three explicitly for concreteness: \begin{align*} 	M_1 &= \begin{bmatrix} 		1 & 2 \\ 3 & 1 	\end{bmatrix} \\ 	M_2 &= \begin{bmatrix} 		{\color{red}1} & {\color{red}2} & 4 & 5 \\ 		{\color{red}3} & {\color{red}1} & 6 & 7 \\ 		7 & 5 & {\color{red}1} & {\color{red}2} \\ 		6 & 4 & {\color{red}3} & {\color{red}1} 	\end{bmatrix} \\ 	M_3 &= \begin{bmatrix} 		{\color{red}1} & {\color{red}2} & {\color{red}4} & {\color{red}5} & 8 & 9 & 11 & 12\\ 		{\color{red}3} & {\color{red}1} & {\color{red}6} & {\color{red}7} & 10 & 15 & 13 & 14 \\ 		{\color{red}7} & {\color{red}5} & {\color{red}1} & {\color{red}2} & 14 & 12 & 8 & 9 \\ 		{\color{red}6} & {\color{red}4} & {\color{red}3} & {\color{red}1} & 13 & 11 & 10 & 15 \\ 		15 & 9 & 11 & 12 & {\color{red}1} & {\color{red}2} & {\color{red}4} 		& {\color{red}5} \\ 		10 & 8 & 13 & 14 & {\color{red}3} & {\color{red}1} & {\color{red}6} 		& {\color{red}7} \\ 		14 & 12 & 15 & 9 & {\color{red}7} & {\color{red}5} & {\color{red}1} 		& {\color{red}2} \\ 		13 & 11 & 10 & 8 & {\color{red}6} & {\color{red}4} & {\color{red}3} 		& {\color{red}1} \\ 	\end{bmatrix} \end{align*} The construction is described recursively as follows. Let \[ 	M_e' = \left[ 	\begin{array}{c|c} 		{\color{red}M_{e-1}} & M_{e-1} + (2^e-1) \\ \hline 		M_{e-1} + (2^e-1) & {\color{red}M_{e-1}} \\ 	\end{array} 	\right]. \] Then to get from $M_e'$ to $M_e$, replace half of the $2^e$'s with $2^{e+1}-1$: in the northeast quadrant, the even-indexed ones, and in the southwest quadrant, the odd-indexed ones.

I found an alternative construction for part b on the competition itself.

On the main diagonal place all $1$s. Divide the remaining numbers into $n-1$ pairs. Let an $i$-diagonal be all the fields of the form $(x,x+i)$ modulo $n$. Thus, for example, the main diagonal is the $0$-diagonal (though we discount it from further analysis). Within an $i$-diagonal, $i=1,2,\dots,n-1$, let us define the relation of connectedness if two fields of the same $i$-diagonal belong to a common cross ($k$-th row+$k$-th column for some $k$). Note that each cross intersects each $i$-diagonal in exactly two fields. One easily proves that each field is connected to exactly two others, decomposing the $i$-diagonal into cycles. Due to translational symmetry, these cycles are of equal length. If $n=2^m$ for some $m$, these cycles will be of even length. Thus, for each $i$-diagonal we can select a pair of numbers and place them alternately along a cycle, ensuring that for each cross both numbers appear in it. This completes the proof.

Is this a well-written solution for part b?

(a) Let $n=2k+1$, so $S = \{1, 2, \dots, 4k+1\}$.  Call the $i$-th row combined with the $i$-th column the $i$-th [i]grass[/i].  Notice that each element in the matrix is in at most $2$ [i]grasses[/i].  Therefore, to be in $2k+1$ [i]grasses[/i], there must be at least $\left\lceil\frac{2k+1}2\right\rceil = k+1$ copies of each element of $S$ in the matrix.  However, this means we have at least $(4k+1)(k+1) > (2k+1)^2$ entries in the matrix, which is clearly impossible.  Since $1997$ is odd, there is no [i]silver[/i] matrix for $n = 1997$.

        (b) (stolen from WOOT induction handout).  We prove that there is a silver matrix for $\boxed{n = 2^k}$ for all positive integers $k$.  In particular, we will prove the case where there is a diagonal of $1$s going from $a_{11}$ to $a_{nn}$, which we'll call $M_n$.  We proceed with induction. For the base case, notice that
        \[
            \begin{bmatrix}
                1 & 2 \\
                3 & 1
            \end{bmatrix}
        \]
        works for $k=1$.  Now suppose $M_k$ is a valid [i]silver[/i] matrix with $1$s on the diagonal.  Define $A_k$ as the matrix $M_k$ with $2^{k+1}$ added to each entry, and $B_k$ as $A_k$ but with all instances of $2^{k+1}+1$ with $2^{k+1}$.  Then we can construct $M_{k+1}$ as follows:
        \[
            M_{k+1} =
            \begin{bmatrix}
                M_k & A_k \\
                B_k & M_k
            \end{bmatrix}
        \]
        Now each [i]grass[/i] contains all integers $1, 2, \dots, 2n-1$ (since each [i]grass[/i] contains a [i]grass[/i] of $M_k$, which is [i]silver[/i]), along with $2^{k+1}+1, 2^{k+1}+2, \dots, 2^{k+1}+2\cdot 2^k-1=2\cdot (2^{k+1}) - 1$ and $2^{k+1}$.  Thus, each [i]grass[/i] has all elements in $S$, and we are done.

For (a), we get a parity issue for $n$ odd.

For (b), we prove that powers of $2$ work. Proceed with induction, with $\left|\begin{matrix}1&2\\3&1\end{matrix}\right|$ as the base case for $2$ by $2$. Let $M_k$ be the silver matrix at $n=2^k$. Consider a new matrix $A_k$, split into four quadrants $NW_k, NE_k, SW_k, SE_k$ with the labeling describing the location. We define $NW_k=SE_k=M_{k-1}$ and $SW_k=NE_k=M_{k-1}+2^k-1$. Transform $A_k$ into $M_k$ by replacing half of the squares in $NE_k, SW_k$ by the indexes' parity.

[b]Remark.[/b] Why did this feel so hard.. I got like 10000 hints on it including basically being given the construction.

Let a \textbf{structure} be the combined row and column of each $i$-th row and $i$-th column cell and the $i$-th row and $i$-th column cells be called the \textbf{$i$-th  center}

Part (a)

Note that there must be at least $n-1$ numbers that aren't on the diagonal of \textbf{centers} because the maximum number of distinct numbers on the diagonal of \textbf{centers} is $n$ and there are $2n-1$ numbers in total. Now note that when a number isn't on the diagonal, it must be in $2$ \textbf{structures} and it must appear in each \textbf{structure}. A number can't have two instances in one structure but there are an odd number of \textbf{structures} because $1997$ is odd. Thus, we have a contradiction because the parities don't match up.

Part (b)

We claim that every $n = 2^{k}$ works. We wish for every number to be in a structure. We induct on $k$. First, note that
\[\begin{tabular}{|c|c|}
    \hline
    1 & 3 \\
    \hline
    2 & 1 \\
    \hline
\end{tabular}\]
works for $k = 1$. Assume that $n = 2^{k}$ has a working construction. For $2^{k+1}$, we have two of the $2^{k}$ constructions on the diagonal of \textbf{centers}. Then, we replace each number in the $2^{k}$ construction with a new number, keeping which numbers on which spaces are equal to each other. \\
We place this new construction on one of the other $2$ places to put a $2^{k}$. Then for the last place to put a $2^{k}$, we just copy the previous construction but change the number on the diagonal of \textbf{centers} to be the only number that hasn't been used. This construction works, which thus concludes the proof.

[hide=remarks]this was so nice. The construction was pretty cool also. [/hide]

For any matrix, the numbers on the diagonal are referred to as [i]rich[/i] numbers and the numbers not on the diagonal are [i]poor[/i].
(a) We show that there is no silver matrix for any odd $n$. A $i$ group has the the row $i$ and column $i$ numbers.
Suppose for contradiction, there is a silver matrix for odd $n$. Any number on $(i,j)$ occupies the row $i$ column $i$ and row $j$ column $j$ group. So, if $i\ne j$ then this number is in exactly two groups. Suppose this number is poor. Then every occurence of this number is in two groups. Since we need a total of $n$ groups, we need exactly $\frac n2$ occurences of each poor number, which is absurd as $n$ is odd. This gives a contradiction. $\blacksquare$

(b) The consturctions for $n=4$ is:
\[\begin{pmatrix} 1 & 4 & 2 & 7\\5 & 1 & 6 & 3\\3 & 7 & 1 & 5\\6 & 2 & 4 & 1 \end{pmatrix}\]
Suppose $n$ is even, so $n=2k$. Construct a $4k-2\times k$ table with entries as ordered pairs $(i,j)$ with $1\leq i,j\leq n$ such that each entry is distinct, and in any row, for any two entries, we have four distinct numbers; i.e. $(i,j)$ and $(i,k)$ must be in different rows. It is possible to construct such a table using a greedy algorithm. Then, each entry in the matrix on $(i,j)$ hosts the row number of $(i,j)$ in our table. It is easy to check that this satisfies all conditions required. $\blacksquare$

Define the $\textit{i-th cross}$ as the set of entries in either the $i$-th row or $i$-th column.

a. We will show that there is no silver matrix for odd numbers $n$. Let $n=2k+1$. Each element is in at most $2$ crosses so there must be at least $k+1$ copies of every element in order to be in $n$ crosses. This implies that the number of elements in the matrix is

\[(4k+1)(k+1)>(2k+1)^2,\]

a contradiction. Thus, for $n=1997$ there cannot be a silver matrix. $\square$

b. For $n=2^k$ there will always be a silver matrix. To see this, we define a recursive sequence:

$$
M_1 = \begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix}
$$
$$
M_k' = \left[ \begin{array}{c|c} M_{k-1} & M_{k-1} + (2^k-1) \\ \hline M_{k-1} + (2^k-1) & M_{k-1} \\ \end{array} \right].
$$

Then to get from $M_k'$ to $M_k$, replace half of the $2^k$'s with $2^{k+1}-1$: in the northeast quadrant, the even-indexed ones, and in the southwest quadrant, the odd-indexed ones. $\square$

[hide=solution]
Call a diagonal point any point that is in the same row and column. If a number isn't on a diagonal point, then it will appear twice when we combine the $i$th row and the $i$th column together. Otherwise, it will appear once. If $n$ is odd, then each number from the set $S$ must appear an odd number of times when we combine the $i$th row and the $i$th column together.So, each number must appear on the diagonal at least once, which isn't possible, as there ae $n<2n-1$ diagonal points. Now, we claim that a silver matrix occurs for all $n = 2^k.$ We will prove this by induction. This is clearly true for $k=1.$ If a silver matrix exists for $n = 2^k,$ let quadrant $2$ be a $n\times n$ silver matrix. Let quadrant $4$ be the same silver matrix. Let qudrant $1$ and $3$ be $n\times n$ silver matricies except with the diagonals replaced with alternating numbers instead of constant and reflected across $x=y.$
[/hide]

$\textbf{(a)}$ We use a bounding argument for odd $n>1$. Note that each integer $1, \ldots, 2n+1$ appears in at most two $\text{row } i \cup \text{col } i$ sets, so we require at least $\left \lceil \frac n2 \right \rceil = \frac{n+1}{2}$ of them. Thus the total number of integers needed is at least
\[(2n-1)\left(\frac{n+1}{2}\right) = \frac{2n^2+n-1}{2},\]

which is greater than $n^2$, contradiction. ${\color{blue} \Box}$

$\textbf{(b)}$ We construct matrices for $n=2^k$ inductively, where we assume $A_{k-1}$ satisfies the given conditions to show $A_k$ does too. Our recursion can be expressed as
\[A_0 = [1], \quad A_1 = \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix}, \quad A_{i+1} = \begin{bmatrix} A_i & B_i \\ C_i & A_i \end{bmatrix},\]

where $B_i$ represents the matrix $A_i$ with each cell increased by $2^{i+1}$ and $C_i$ represents the matrix $B_i$ with each cell of value $2^{i+1}+1$ with $2^{i+1}$. Our inductive assumption tells us this matrix indeed works. $\blacksquare$

(a) To show that there is no silver matrix for $n = 1997$, consider the number of distinct elements that must appear in the rows and columns. For an $n \times n$ matrix where the entries are from $S = {1, 2, \ldots, 2n - 1}$, there are $2n - 1$ distinct elements.

In a silver matrix, each row and each column must contain all elements of $S$. However, in a $1997 \times 1997$ matrix, there are only $1997$ rows and columns. Therefore, it's impossible to cover all $2 \times 1997 - 1 = 3993$ distinct elements of $S$ in just $1997$ rows and columns. Hence, no silver matrix exists for $n = 1997$.

(b) To show that silver matrices exist for infinitely many values of $n$, we can construct a silver matrix for any $n$ such that $n = 2k$ for some positive integer $k$.

Consider an $n \times n$ matrix where the rows and columns are labeled $1, 2, \ldots, n$. For each row $i$, fill the entries of the $i$th row as follows:
$$i,i+1,i+2,\dots,2n-1,1,2,\dots,i-1$$
This arrangement ensures that each row contains all elements of $S = {1, 2, \ldots, 2n - 1}$.

Now, consider each column. For the $j$th column, fill the entries as follows:
$$j,j+1,j+2,\dots,2n-1,1,2,\dots,j-1$$
This arrangement ensures that each column also contains all elements of $S$.

Therefore, for any $n = 2k$, we can construct a silver matrix using this method, ensuring that each row and each column contains all elements of $S$. Since there are infinitely many values of $n = 2k$, silver matrices exist for infinitely many values of $n$.


Prove that the geometric mean of a set of positive real numbers $S$ is equal to the geometric mean of the geometric means of every non-empty subset of $S$.

[hide]
Let $P_{j}$ denote the product of the geometric means of all the subsets with $j$ elements. Then break it up into different cases:

Subsets with one element: Each element appears just once as its own subset
\[P_{1}= a_{1}a_{2}a_{3}\cdots a_{n}\]
Two elements: For each $a_{i}$ term, there are $n-1$ other terms to pair it up with to form a subset, so each $a_{i}$ term appears $n-1$ times in $P_{2}$
\[P_{2}=(a_{1}a_{2}a_{3}\cdots a_{n})^{\frac{{n-1\choose 1}}{2}}\]
For three elements, we choose two terms, and for four elements, we choose three, etc...
\[P_{3}=(a_{1}a_{2}a_{3}\cdots a_{n})^{\frac{{n-1\choose 2}}{3}}\]
Continuing until we get
\[P_{n}=(a_{1}a_{2}a_{3}\cdots a_{n})^{\frac{{n-1\choose n-1}}{n}}\]
Taking the product of all the $P_{j}$ terms, and writing $1$ as $\frac{{n-1\choose 0}}{1}$
\[P_{1}P_{2}P_{3}\cdots P_{n}= (a_{1}a_{2}a_{3}\cdots a_{n})^{ \left(\frac{{n-1\choose 0}}{1}+\frac{{n-1\choose 1}}{2}+\frac{{n-1\choose 2}}{3}+\cdots+\frac{{n-1\choose n-1}}{n}\right)}\]
Using the relationship
\[(i+1)\binom{n}{i+1}= n\binom{n-1}{i}\Longrightarrow \frac{{n-1\choose i}}{i+1}= \frac{{n\choose i+1}}{n}\]
we can prove the equation
\[\sum_{k=0}^{n-1}\frac{{n-1\choose k}}{k+1}= \frac{2^{n}-1}{n}\]
So the product of the geometric means is equal to
\[(a_{1}a_{2}a_{3}\cdots a_{n})^{\frac{2^{n}-1}{n}}\]
The total number of subsets is equal to $2^{n}-1$, since the empty set is not included. So to get the geometric mean of the geometric means, we raise their product to the $\frac{1}{2^{n}-1}$ power.
\[P_{1}P_{2}P_{3}\cdots P_{n}^{(\frac{1}{2^{n}-1})}= (a_{1}a_{2}a_{3}\cdots a_{n})^\frac{1}{n}\]
which is the geometric mean of the set $\{a_{1}, a_{2}, a_{3}, \cdots, a_{n}\}$, as we wanted.
[/hide]

When I solved this problem I used a little more formal notation :D, but I followed exactly the same idea than you :)

Consider the set $S=\{a_{1},\ a_{2},\cdots,\ a_{n}\}$. We wish to count how many times appears $a_{i}$ in the $r$-subsets of $S$, $1\leq r <n$. It's easy to see that this number is $\dbinom{n-1}{r-1}$. So, if $G$ is the geometric mean of the geometric means of the non-empty subsets of $S$, we have \[G=\left[\prod_{i=1}^{n}\prod_{r=1}^{n}a_{i}^{\frac{1}{r}\binom{n-1}{r-1}}\right]^{\frac{1}{2^{n}-1}}=\left[\left(\prod_{r=1}^{n}a_{i}\right)^{\sum_{r=1}^{n}\frac{1}{r}\binom{n-1}{r-1}}\right]^{\frac{1}{2^{n}-1}}\] But observe that \[\sum_{r=1}^{n}\frac{1}{r}\binom{n-1}{r-1}=\frac{1}{n}\sum_{r=1}^{n}\frac{n}{r}\binom{n-1}{r-1}=\frac{1}{n}\sum_{r=1}^{n}\binom{n}{r}=\frac{2^{n}-1}{n}\] Hence
\[G=\left[\left(\prod_{r=1}^{n}a_{i}\right)^{\frac{2^{n}-1}{n}}\right]^{\frac{1}{2^{n}-1}}=\left(\prod_{r=1}^{n}a_{i}\right)^\frac{1}{n}\] that is, $G$ is the geometric mean of the elements of $S$, as we wanted to show.


function
limit
calculus
calculus computations

Let $ f: [a,\infty) \to \mathbb{R}$ be a functions such that $ f_{| (a,\infty)}$ has an antiderivate.If $ f$ is continuous at $ x\equal{}a$, is it true that $ f$ has an antiderivate ? I can't find an argument.:blush:

Let $ F$ be the antiderivate.
It is enough to show that $ \lim_{x \downarrow a} F(x)$ exists,
or that if $ x_1, x_2, \cdots$ is a sequence in $ (a,\infty)$ and $ \lim_{n \to \infty} x_n \equal{} a$, then $ \lim_{n \to \infty} F(x_n)$ exists.
We know that $ \lim_{n \to \infty} F'(x_n)$ exists (say $ \equal{}c$), because $ f$ is continuous in $ a$.
Then (intermediate value theorem) $ F(x_n) \minus{} F(x_m) \equal{} F'(\zeta_{m,n})(x_n\minus{}x_m)$ for all $ m.n$.
This shows that $ F(x_1),F(x_2),\cdots$ is a Cauchy sequence, and consequently convergent.


modular arithmetic
invariant
geometry
geometric transformation
reflection
calculus
integration

Given any set of $9$ points in the plane such that there is no $3$ of them collinear, show that for each point $P$ of the set, the number of triangles with its vertices on the other $8$ points and that contain $P$ on its interior is even.

Let $P=\{P_0,P_1,\ldots,P_8\}$ be the set of the $9$ points in the plane.  Fix an arbitrary $P_0$.

Now let $C$ be the number of triangles with vertices on the other $8$ points which include $P_0$ in their interior.  We must show $C\equiv 0 \pmod{2}$.

Now, fix some arbitrary $P_i$, $1\le i\le 8$.  Now, draw line $P_0P_i$, and consider the pairs of the other $7$ points.  To solve this problem, we will show that $C\equiv 0\pmod{2}$ remains invariant upon reflecting $P_i$ through $P_0$ (in other words, constructing $P_i'$ with the rule that the midpoint of $P_iP_i'$ is $P_0$).  By doing so, we can reflect the right series of points to have all $8$ on the same half of the plane, and therefore since $C=0$ at this time, we have $C\equiv 0\pmod{2}$ for all formations of the $8$ points and arbitrary $P_0$.

Now, if a pair of points $(P_x,P_y)$ lie on the same side of $P_0P_i$, then $\bigtriangleup P_iP_xP_y$ does not contain $P_0$ and neither does $\bigtriangleup P_i'P_xP_y$.  So we can focus on only the pairs of points that have one on each side of line $P_0P_i$.

Let $P_x$ and $P_y$ be on opposite sides of line $P_0P_i$. Now, notice that if $\bigtriangleup P_iP_xP_y$ contains $P_0$, then $\bigtriangleup P_i'P_xP_y$ does not.  The converse of this statement is also true.  Thus, let there be $a$ points on one side of line $P_0P_i$ and thus $7-a$ points on the other side of $P_0P_i$.  Then let $B$ be the number of triangles with the above conditions that contain $P_0$.  So, we have $a(7-a) - B$ triangles that do not contain $P_0$.  But since

$\displaystyle B\equiv a(7-a) - B\pmod{2}$ for all integral $a,B\ge 0$, then $C$'s parity is invariant.  So we are done.

Assume that the set has $n$ points.  Consider a point $P$ that is inside $k\geq1$ triangles whose vertices are in the set, and let $ABC$ be any such triangle.  For each triangle $ABC$, and for each point in the set $Q\neq A,B,C,P$, either $A,B,C,Q$ are the vertices of a convex quadrilateral, and $P$ is inside exactly two of the four triangles formed by these vertices, one of which is $ABC$ (each one of the two diagonals of the quadrilateral divide it into two disjoint triangles, inside exactly one of both must be $P$), and if $A,B,C,Q$ are the vertices of a concave quadrilateral, then one of the triangles formed by these three vertices contains the fourth inside it, hence also $P$, while the other three possible triangles are disjoint and cover the previous triangle, hence exactly one of them has $P$ inside it, again exactly two of the four triangles formed by these vertices have $P$ inside it.  Since this happens for each one of the $n-4$ points $Q$ that are not $A,B,C,P$, we conclude that, for each triangle $ABC$, there are exactly $n-4$ triangles that share exactly one side with $ABC$ and have $P$ inside it, plus $ABC$ itself, and no other triangle shares one side with $ABC$.  Note that, if we count each edge $AB,BC,CA$ once for each triangle containing $P$ which admits it as a side, and add over the three of them, we obtain $n-4+3=n-1$.  Adding over the $k$ triangles that have $P$ inside them, we obtain $k(n-1)$.  Now, given any segment whose ends are points in the set other than $P$, let $u$ be the number of triangles that have this segment as one of its sides, and have $P$ inside them.  For each one of these $u$ triangles, the segment is counted $u$ times in $k(n-1)$, hence $\sum u^2=k(n-1)$.  But if we count the $3k$ edges of the $k$ triangles, we are counting each one of these segments $u$ times (as many times as it appears in a triangle), hence $\sum u=3k$.  We conclude that
\[(n-4)k=\sum u^2-\sum u=\sum u(u-1),\]
where the sum is carried out over all possible segments formed by two points of the set other than $P$, and each term in the sum is clearly even, hence $(n-4)k$ is necessarily even.  If $n$ is odd (e.g. $n=9$), then $k$ is even, and we are done.

[i]Note:[/i] One cannot say anything about the case when $n$ is even, since clearly $k=1$ would be odd if we consider a triangle with $P$ inside it for $n=4$, but in a regular $n$-gon with $n$ even, each point $P$ is inside $0$ (even) triangles.

Consider each of  the $\binom{8}{3}$ triangles that may or may not contain $P$ as vertices of a graph.  We draw a "graph edge" between two of these vertices (that is, triangles) if both triangles contain $P$ and they share a "geometrical edge" when considered as triangles and not vertices of the graph we are constructing (LOL).

If a triangle $ABC$ contains $P$ on it's interior, draw the lines $AP,BP,CP$, and then convince yourself that this triangle will have 5 "graph edges" when considered as a vertex of the graph. If it does not contain $P$, then it will have 0 "graph edges".

The number of vertices with odd degree on a graph is even.
(I considered writing this without the edge distinction, for pure uninteligible awesomeness)

Is there something wrong?

Of course, we're gonna change 9 for an arbitrary $n>1$ odd number. Let $A=\{P_1, P_2, \dots P_{n-1} \}$ be $n-1$ points on the plane. Our mission is to prove that, no matter where we put the last point $P_n$, it will be inside of an even number of triangles.

We define the function $f: \mathcal{P} \backslash \{\overline{P_i P_j}, i, j \in \{ 1, 2, \dots n-1\} \} \rightarrow \mathbb{N}_0$ with $f(P) = \text{number of triangles with vertices in}\ A\ \text{which contain P}$. If we prove that $f$ is always even, we're done.

Take the $n-1$ points and draw all the segments $\overline{P_iP_j}, i, j \in \{ 1, 2, \dots n-1\}$. They divide the plane in regions. If I have a region $\mathcal{R}$ and two points $A, B \in \mathcal{R}$, we have $f(A)=f(B)$, because we can move A to B without crossing any segment, so It wont cross any triangle boundary.

Now, if we have two adyacent regions $\mathcal{R}, \mathcal{S}$, and two points $A \in \mathcal{R}, B \in \mathcal{S}$, the difference $f(A)-f(B)$ is even. Indeed, when we cross their common boundary, we´re changing the state (in-out) on a lot of triangles. In fact, we have $n-3$ triangles we're changing state, because we have two fixed points (the ones who delimite the boundary). So, if $r = \text{number of triangles which have common points with}\ \mathcal{R}$ and $s = \text{number of triangles which have common points with}\ \mathcal{S}$, it's pretty clear that $r+s=n-3$ and $f(A)-f(B)=r-s=r+s-2s=(n-3)-2s$ which is an even number.

Now we're ready to prove the problem. If we start with a point $P$ very far (on the exterior region), it's clear that $f(P)=0$. Now, we can get to every region crossing some boundaries, and using the statement we prove, f(P) doesn't change it's parity, i.e, it's always even.


Let:$ a;b;c>0$ and $ abc\equal{}1$.Prove that:
$ \sum \frac{2a^3\plus{}3}{a(b\plus{}c)^2}\plus{}\frac{4}{\sum \frac{a\plus{}b}{c}\plus{}2} \ge \frac{17}{4}$

[quote="caubetoanhoc94"]Let:$ a;b;c > 0$ and $ abc \equal{} 1$.Prove that:
$ \sum \frac {2a^3 \plus{} 3}{a(b \plus{} c)^2} \plus{} \frac {4}{\sum \frac {a \plus{} b}{c} \plus{} 2} \ge \frac {17}{4}$[/quote]
The following stronger inequality is true too.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab\plus{}ac\plus{}bc\neq0.$ Prove that
\[ \frac{2a^2\plus{}bc}{(b\plus{}c)^2}\plus{}\frac{2b^2\plus{}ac}{(a\plus{}c)^2}\plus{}\frac{2c^2\plus{}ab}{(a\plus{}b)^2}\plus{}\frac{16abc}{(a\plus{}b)(a\plus{}c)(b\plus{}c)}\geq\frac{17}{4}\]


inequalities
logarithms
function
algebra
domain
linear algebra
matrix

Given $5n$ real numbers $r_i, s_i, t_i, u_i, v_i \geq 1 (1 \leq i \leq n)$, let $R = \frac {1}{n} \sum_{i=1}^{n} r_i$, $S = \frac {1}{n} \sum_{i=1}^{n} s_i$, $T = \frac {1}{n} \sum_{i=1}^{n} t_i$, $U = \frac {1}{n} \sum_{i=1}^{n} u_i$, $V = \frac {1}{n} \sum_{i=1}^{n} v_i$.  Prove that $\prod_{i=1}^{n}\frac {r_i s_i
t_i u_i v_i + 1}{r_i s_i t_i u_i v_i - 1} \geq \left(\frac {RSTUV +1}{RSTUV - 1}\right)^n$.

Applying jensen's ineq for five multivariables!!!

can you please explain the multivariable jensen?  i cant find it anywhere

Let $X=(x_1,x_2,x_3,x_4,x_5)$, $F(X)=\ln \frac{x_1x_2x_3x_4x_5+1}{x_1x_2x_3x_4x_5-1}$. Then $F(x)$ is a convex function in a domain $(0,+\infty)^5$. For proving this, we must assure that Hessian $H(X)=(\partial^2 F/\partial x_i\partial x_j)_{i,j=1}^5$ is non-negative defined in every point $(x_1,x_2,x_3,x_4,x_5)$. We have
$\partial^2 F/\partial x_i\partial x_j=\frac{2p(p^2+1)}{x_ix_j(p^2-1)^2}, i\ne j$
and
$\partial^2 F/\partial x_i^2=\frac{4p^3}{x_i^2(p^2-1)^2}.$

(here $p=x_1x_2x_3x_4x_5$) So, we have to prove that the matrix with diagonal entries $2p^2$ and not-diagonal elements $p^2+1$ is not-negative defined (we multipled rows and columns by corresponding $x_i$
and then multiply by some constant). But it is clear: matrix with equal entries is non-negative defined, and our matrix differs from it by a positive scalar matrix.

So, just apply Jensen for this function.


algebra
polynomial
complex analysis
complex analysis unsolved

Let $ P$ be a polynomial of degree $ n$  with complex coefficients and $ P(0)\equal{}0$. Then the polynomial has constant term equal to 0?

I say yes, since $ P(0)$ always equals the constant term of the polynomial, all other terms vanishing, whether the coefficients are real or complex.

I am trying to solve a problem on polynomials and I want to be sure for this step. :lol:


function
induction
inequalities
algebra
functional equation

Let $f$ be a function from the set of non-negative integers into itself such that for all $n\geq 0$ we have that \[\left(f(2n+1)\right)^{2}-\left(f(2n)\right)^{2}= 6f(n)+1\ \ \text{and}\ \ f(2n) \geq f(n).\] How many numbers less than $2003$ are there in the image of $f$?


The triangle $ ABC$ is inscribed in a circle with radius $ 12$. If $ \overline{AB}\equal{}8$ and $ \overline{AC}\equal{}9$, find altitude relative to the side $ BC$.

[hide]The length is equal to (AB)sinB, and we know AC/sinB = 2R by law of sines. So the length of the altitude is (AB)(AC)/2R = 3[/hide]


limit integration logarithms inequalities function calculus derivative

If $\mu(X)=1$ and $f\in L^q$ for some $q>0$.  Prove that
\[ \lim_{p\to 0}\|f\|_p=\exp\left(\int_X \log |f|d\mu\right)  \]
----

This is what I have.  We can assume that $|f|\not=0$ a.e.

1) $\log\|f\|_t\geq \int_X\log|f|d\mu$ for all $0<t<q$, by Jensen's inequality applied to the concave function $\log(x)$

2) The quotients $\frac{\int_X |f|^td\mu-1}{t}$ are $\geq\log\|f\|_t$ for $0<t<q$, by the elementary inequality $x\geq \log x+1$.

3) As $t\to 0$, $\frac{\int_X |f|^td\mu-1}{t}\to\int_X\log|f|d\mu$.  This is where I'm a bit shaky.  Since the derivative w.r.t $t$ of the integrand is $|f|^t\log|f|$, we should be able to apply dominated convergence and differentiate under the integral sign to get the result, since we've assumed that $|f|\not=0$ a.e.  (we can make this assumption because if $|f|=0$ on some set of positive measure, then it makes a contribution of zero to both sides.)

4) Take $\log$s.  Then
\[ \begin{eqnarray*} \lim_{p\to 0}\|f\|_p&=&\exp\left(\lim_{p\to 0} \frac{1}{p}\log\left(\int_X |f|^pd\mu\right)\right) \\ &\leq&\exp\left(\lim_{p\to 0} \frac{\int_X |f|^pd\mu-1}{p}\right) \\ &=&\exp\left(\int_X \log |f|d\mu\right) \\  \]
where we use $2)$ and $3)$ in the middle step.  For the other half,
\[ \begin{eqnarray*} \lim_{p\to 0}\|f\|_p&=&\exp\left(\lim_{p\to 0} \frac{1}{p}\log\left(\int_X |f|^pd\mu\right)\right) \\ &\geq&\exp\left(\int_X \log |f|d\mu\right) \\  \]
by $1)$

Putting it all together yields the result.  Does this work?

[quote="blahblahblah"]
3) As $t\to 0$, $\frac{\int_X |f|^td\mu-1}{t}\to\int_X\log|f|d\mu$.  This is where I'm a bit shaky.  Since the derivative w.r.t $t$ of the integrand is $|f|^t\log|f|$, we should be able to apply dominated convergence and differentiate under the integral sign to get the result, since we've assumed that $|f|\not=0$ a.e.  (we can make this assumption because if $|f|=0$ on some set of positive measure, then it makes a contribution of zero to both sides.)
[/quote]
Firstly, it makes contribution of $-\infty$ to the right hand side and $\frac{-\text{const}}{t}$ to the left hand one. Secondly, I'm not sure the dominated convergence theorem is applicable here: what function do you propose for the integrable majorant? There are several ways out, of course, but, as written, the argument cannot be called satisfactory at this point.

I was hoping to use $|f|^q$ as the majorant, which I don't think works.  What I meant about making a contribution of zero to both sides was that if $|f|=0$ on some set of positive measure, then when we integrate over it, we get zero, and when we take the exponential of the integral of the logarithm, we get $\exp(-\infty)$, which is zero.

Anyways, do you have any suggestions on how to proceed?  It seems like we need to differentiate under the integral somewhere, unless I'm missing another approach.

[quote="blahblahblah"] What I meant about making a contribution of zero to both sides was that if $|f|=0$ on some set of positive measure, then when we integrate over it, we get zero, and when we take the exponential of the integral of the logarithm, we get $\exp(-\infty)$, which is zero.[/quote]
But it is not an additive $0$ on either side!

[quote="blahblahblah"]
Anyways, do you have any suggestions on how to proceed?  It seems like we need to differentiate under the integral somewhere, unless I'm missing another approach.[/quote]
You can differendiate under the integral sign if the function is separated away from $0$ (then $|f|^q$ can, indeed, be used as a majorant). After that one can run the usual approximation argument. Replace $f$ by $f_\delta$ where $f_\delta(x)=f(x)$ if $|f(x)|>\delta$ and $\delta$ otherwise. Then $\|f\|_t\le \|f_\delta\|_t$, so $\limsup_{t\to 0+}\|f\|_t\le\exp\left\{\int_X \log |f_\delta| \right\}$. Then one can use the monotone convergence theorem to show that the right hand side converges to $\exp\left\{\int_X \log |f| \right\}$ as $\delta\to 0$.

That's a neat idea, and one I hadn't seen before.  Thank you :)

Okay, I think I might have another workaround for 3)

Say $f=0$ on a set of positive measure, call it $T$.  Then
\[ \lim_{p\to 0} \|f\|_p=\lim_{p\to 0} \mu(X\setminus T)^{\frac 1p} = 0  \]
Similarly,
\[ \exp\left(\int_X \log|f|\right)=\exp(-\infty)  \]
We have to have $\int_X \log |f|=-\infty$ in that case, because otherwise $\log|f|$ isn't integrable.  So WLOG $\mu(f^{-1}(\{0\}))=0$.

Next, consider
\[ \int_X \left(\frac{|f|^t-1}{t}\right)d\mu  \]
If $|f|\leq 1$, then the quotient $\frac{|f|^t-1}{t}$ is montonically increasing in $t$, with pointwise limit $\log|f|$.  So we can apply the monotone convergence theorem.

Conversely, if $|f|\geq 1$, that same quotient is monotonically decreasing in $t$, so we take our majorant to be
\[ \frac{|f|^q-1}{q}  \]
since $f\in L^q$, and then apply the dominated convergence theorem to this half of the integral.

[quote="blahblahblah"]It seems like we need to differentiate under the integral somewhere, unless I'm missing another approach.[/quote]

I know this is an old topic, but in my edition of the book the problem is in chapter 3, and differentiability is only discussed in chapter 7 so I think we should only use theorems from the first 3 chapters.

We can assume $ f\geq 0$. From Holder's Inequality we get that $ \|f\|_r<\|f\|_s$ if $ r<s$ so the limit $ l=\lim_{p \to 0}\|f\|_p$ exists. From Jensen's Inequality we get that $ l\geq \exp(\int_X \log(f)d\mu)$.

If we take $ f_n: X\to [0,\infty)$ simple measurable functions such that $ f_n\leq f_{n+1}\leqq$ and $ f_n(x)\to f(x)$ for all $ x$, we have that $ \|f-f_n\|_p \to 0$ as $ n \to \infty$ for every $ p$. So if we take $ \varepsilon>0$ there exists $ n_0$ such that for all $ n\geq n_0$ we have $ \|f-f_n\|_{p_0}<\varepsilon$. Then $ \big| \|f\|_p-\|f_n\|_p\big|\leq \|f-f_n\|_p<\varepsilon$ for all $ p\leq p_0$ and $ n\geq n_0$. So, $ l\leq \lim_{p\to 0} \|f_n\|_p + \varepsilon$.

But, because $ f_n$ are simple functions and $ \mu(X)=1$, by taking logarithms and l'Hospital rule we have $ \|f_n\|_p \to \exp( \int_X \log(f_n)d\mu  )\leq \exp( \int_X \log(f)d\mu  )$

So, $ \exp(\int_X \log(f)d\mu)\leq l\leq \exp(\int_X \log(f)d\mu)+\varepsilon$ which gives the conclusion.

Is it correct?

[quote="Thor"]I know this is an old topic, but in my edition of the book the problem is in chapter 3, and differentiability is only discussed in chapter 7 so I think we should only use theorems from the first 3 chapters.[/quote]

But the differentiability discussed in chapter 7 is the [i]differentiability of measures[/i].  here we are only using the 'usual' notion of differentiability of functions, e.g., the notion that one might find in Rudin's [i]Principles of Mathematical Analysis[/i].  I think that my proof (with the modified step 3)) is fine, although the step 4) in the first post should have liminfs and limsups in place of the limits :).

Then again, there could be mistakes; I just skimmed it and it has been nearly two years.

[quote="blahblahblah"]But the differentiability discussed in chapter 7 is the [i]differentiability of measures[/i].  here we are only using the 'usual' notion of differentiability of functions, e.g., the notion that one might find in Rudin's [i]Principles of Mathematical Analysis[/i]. [/quote]

 :oops:
I see. I know differentiability of functions(here it is in the high school curriculum) but I don't know how to use it for the Lebesgue integral. But now I see that you don't actually use it in your final solution, so it's ok  :) .


linear algebra
matrix
geometry
rectangle
superior algebra
superior algebra solved

we consider the matrix A=(a_ij), i=1,..n and j=1,..,(n+1) with complex elements with the properties that \sum{j=1,n+1}a_ij=0,for any i=1,...,n and rank(A)=n.
If exists X=(x_1,x_2,...,x_(n+1))\in M_1,n+1(C) and AX^t=O_n,1 prove that x_1=x_2=...=x_(n+1).
X^t is the transpose of the matrix X.

[quote="Kuba"]we consider the matrix A=(a_ij), i=1,..n and j=1,..,(n+1) \in M_n(C) with the properties that \sum{j=1,n+1}a_ij=0,for any i=1,...,n [/quote]

kuba can you check the above two lines
because you wrote A \in Mn(C) so A is square matrix, but for for the subscript j you say 1,...,n+1

Maybe you mean A is rectangle matrix n row, n+1 columns ?

yes.. my fault... i want to abridge the sentence "with complex elements" :D
sorry again

A has rank n.
Assume that the minor formed by the first n columns is invertible. Denote it by B.
Let tU=(1,...,1) in M1,n(R).

Then ai,n+1=-\sum {j=1..n} ai,j, so A=[B;-BU] (B is the square block and -BU is the last column).

Let Y=(x1,...,xn).
AX=0 <=> [B;-BU] [Y;xn+1]t=0 <=>
B(tY)-xn+1BU=B(tY-xn+1U)=0. Since B is invertible, this means tY=xn+1U, which means xn+1=xi for all i=1..n, hence the conclusion.


A fitness center charges a membership fee plus a fixed amount per day for each day the center is used. If Claire paid $ \$105$ for 30 days of use, and Adrian paid $ \$123$ for 39 days of use, how many dollars is the membership fee?

To see how much $ 9$ days of use cost, subtract $ \$105$ ($ 30$ days of use charge) from $ \$123$ ($ 39$ days of use charge). We find that $ 9$ days of use costs $ \$18$. Thus, each day of use costs $ \$2$. Subtracting $ 30(\$2)\equal{}\$60$ from $ \$105$, we get $ \fbox{\$45}$, the membership fee.

r15s11z55y89w21 is right but  apparently, the answer says $55


abstract algebra
superior algebra
superior algebra unsolved

Can we describe the localization at prime ideals $ \mathfrak{p}$ in $ \mathbb{Z}_n$, the ring of integers module $ n$, nicely?

Yes: By the Chinese remainder theorem, the representation $ n\equal{}p_1^{k_1}\cdot\ldots\cdot p_r^{k_r}$ implies $ \mathbb Z/n\mathbb Z\equal{}\prod \mathbb Z/p_i^{k_i}\mathbb Z$. Thus $ (\mathbb Z/n\mathbb Z)_\wp\equal{}\mathbb Z/p_i^{k_i}\mathbb Z$ if $ \wp\equal{}p_i\mathbb Z$ and the 0-ring otherwise.


let $ a, b, c, d$ be real numbers such that: $ abcd > a^2\plus{}b^2\plus{}c^2\plus{}d^2$. Prove that: \[ abcd>a\plus{}b\plus{}c\plus{}d\plus{}8\]

$ \frac{(a\plus{}b\plus{}c\plus{}d)^2}{4}\leq a^2\plus{}b^2\plus{}c^2\plus{}d^2<abcd\leq \left(\frac{a\plus{}b\plus{}c\plus{}d}{4}\right)^4$
therefor: $ \left(\frac{a\plus{}b\plus{}c\plus{}d}{4}\right)^4 > \frac{(a\plus{}b\plus{}c\plus{}d)^2}{4}$
$ \Leftrightarrow a\plus{}b\plus{}c\plus{}d>8 \Leftrightarrow 2(a\plus{}b\plus{}c\plus{}d)>8\plus{}a\plus{}b\plus{}c\plus{}d$
then:
$ abcd>a^2\plus{}b^2\plus{}c^2\plus{}d^2\geq\frac{(a\plus{}b\plus{}c\plus{}d)^2}{4}>2(a\plus{}b\plus{}c\plus{}d)>a\plus{}b\plus{}c\plus{}d\plus{}8$

[quote="peine"]$ \frac {(a \plus{} b \plus{} c \plus{} d)^2}{4}\leq a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 < abcd\leq \left(\frac {a \plus{} b \plus{} c \plus{} d}{4}\right)^4$
therefor: $ \left(\frac {a \plus{} b \plus{} c \plus{} d}{4}\right)^4 > \frac {(a \plus{} b \plus{} c \plus{} d)^2}{4}$
$ \Leftrightarrow a \plus{} b \plus{} c \plus{} d > 8 \Leftrightarrow 2(a \plus{} b \plus{} c \plus{} d) > 8 \plus{} a \plus{} b \plus{} c \plus{} d$
then:
$ abcd > a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2\geq\frac {(a \plus{} b \plus{} c \plus{} d)^2}{4} > 2(a \plus{} b \plus{} c \plus{} d) > a \plus{} b \plus{} c \plus{} d \plus{} 8$[/quote]
Thanks

peine wrote that abcd [u]<[/u]((a+b+c+d)/4)^4,but it's not true for every real number.

Since $ abcd \equal{} \mid abcd\mid$ it follows that $ \mid a\mid\plus{}\mid b\mid\plus{}\mid c \mid\plus{}\mid d\mid \geq a\plus{}b\plus{}c\plus{}d$, which makes the ineq. valid for all real a,b,c and d if it is valid for positive reals.

$ 900 > \mid 6\mid \plus{} \mid 6\mid \plus{} \mid \minus{} 5 \mid \plus{} \mid \minus{} 5\mid \plus{} 8 \equal{} 30 \geq 6 \plus{} 6 \plus{} \minus{} 5 \plus{} \minus{} 5 \plus{} 8 \equal{} 10$.  As you see, the inequality is true for your values too. Since the sum of the avsolute values of the variables is greater than the sum without them it is only needed to prove the inequality for positive reals.
(inequalities are transitive, by the way)

the inequality is only true for $ a,b,c,d \in \mathbb{R^\plus{}}$ not  $ a,b,c,d \in \mathbb{R}$


p=4k+3 is a prime number .Suppose
x^2+y^2=z^(p+1/2)
prove that xy is divisible by p.


[quote="star99"]I have to do something about leadership for an extracurricular activity, and I wanted to know what Aops members think about a couple of things.

What is leadership?
What is service?
Why do you do service?

And on a side note, who are your role models?

Thanks.[/quote]

I think that a leader has to be a nice person out of work, but when he/she's working, he/she has to be really cold and rude, and with no tolerance

[quote="José"]I think that a leader has to be a nice person out of work, but when he/she's working, he/she has to be really cold and rude, and with no tolerance[/quote]

Wow, seriously José?  Do you think a leader is able to earn the respect of his followers when he is cold and rude?  I've had several leadership roles and never have I found this to be the case.

Star99, I'll answer each of your questions in turn:

[b]What is leadership?[/b]
Leadership, in short, is getting a group of people to accomplish a goal while having everybody acknowledge that you are in control.  Leadership is needed mostly when everybody has different ideas about how something should get done.  Many hands make light work, but only when united under a common cause, and the leader is the one that either puts his foot down and just says what is going to happen or is the one that takes all the ideas and comes up with a compromise that is suitable for everybody.  One of the most important parts about leadership, and I think this is what José meant to say, is that you cannot back down.  If your followers (for lack of a better word) can get you to change your mind or go back on something you have taken a step back towards the "everybody has an idea and nobody is unifying them" stage.  You must always stay involved in the project, never leave, slack, or take breaks.  Set the example, and do [i]more[/i] work than your followers.  A leader can't have any chinks in his armour and can't pick and choose when to be excellent.  People should leave your project thinking that their idea was the one that worked and most importantly, everybody should have a feeling of satisfaction and overall happiness!

[b]What is service?[/b]
Ok, service and leadership are two different things.  In fact, I'm not quite sure why you are asking about service.  Most of the time service does not involve leadership unless you are the coordinator of some project at school or whatnot.  Service, I guess, is just helping some cause or person, but if leadership is what you're shootin' for then service is not your only/best option.  Service is very good though, just not always for leadership.

[b]Why do you do service?[/b]
I do service for Honor Society hours so I can wear the white thing at graduation, lol.  No, seriously, service is for helping people.  I do service because I enjoy helping people.

[b]Role Model[/b]
My role model is Cal Ripken Jr., baseball player with the most consecutive games ever!

You said you were looking for leadership opportunities.  Here's what I do. . .

Marching Band (HUGE leadership opportunities, HUGE!)
Peer Mentoring, Middle School Programs
Take command of group projects in class (make sure not to be a snot about it, and don't screw up, always stay one step ahead of what's going on and don't let anybody down, be prepared for everything and also expect to end up doing a lot of the work)

Good Luck!

answering, I do think that: when you're cold at work, your low-positioned people (I don't know how to put that) work better, then they realised that you did it for them.... just that... That's why you, as a leader, have to be nice when you're not working

[quote="José"]That's why you, as a leader, have to be nice when you're working[/quote]Wait, I thought you just said that you have to be cold and rude when you're working.

In my opinion, a good leader is easy to work with but fair towards everyone.  Being a good arbiter and being able to make decisions are the two most important qualities.


superior algebra
superior algebra unsolved

How do I show that a group of even order must contain an odd number of elements of order 2?

The sets

$ A \equal{} \{g \in G: g^{2} \equal{} 1\}, \qquad B \equal{} \{g \in G: g^{2} \neq 1\}$

provide us with a partition of [i]the[/i] group.

Show that $ \vert B \vert$ is an even number. It implies that  $ \vert A \vert$ is even, too. You'll be done once you get rid of the $ 1$ in $ A$.

Regards,

[quote="coquitao"]

Show that $ \vert B \vert$ is an even number.

[/quote]

How does one go about proving that?

OK... If $ B \equal{} \emptyset$, you are done. Otherwise, you have to notice that $ g \in B \Rightarrow g^{ \minus{} 1} \in B$. That's it!  :)


Let $ a, b, c$ are positive reals. Find the minimum of $ ab\plus{}bc\plus{}ca$ if $ a\plus{}b\plus{}c \geq 7$ and $ abc \geq 12$.

hello, the searched minimum is $ 16$.
Sonnhard.

[quote="rofler"]Replacing $ max(a,b)$ and $ min(a,b)$ with $ max(a,b) \plus{} k$ and $ min(a,b) \minus{} k$ decreases both $ abc$ and $ ab \plus{} ac \plus{} bc$. We decrease until $ abc \equal{} 12$.

Then $ c \equal{} \frac {12}{ab}$ and we have to minimize $ ab \plus{} \frac {12(a \plus{} b)}{ab}$ given $ a \plus{} b \plus{} \frac {12}{ab}\ge 7$

Reducing $ a \plus{} b$ while keeping $ ab$ fixed will decrease both expressions, we can reduce until either $ a \equal{} b$ or $ a \plus{} b \plus{} \frac {12}{ab} \equal{} 7$

Case i) $ a \equal{} b$

We must minimize $ a^2 \plus{} \frac {24}{a}$ given $ 2a \plus{} \frac {12}{a^2}\ge 7$

Case ii) $ a \plus{} b \plus{} \frac {12}{ab} \equal{} 7$

Then we have $ abc \equal{} 12$ and $ a \plus{} b \plus{} c \equal{} 7$, and we have to minimize $ ab \plus{} ac \plus{} bc$

Both cases seem tedious,but doable, with calculus.[/quote]
If we have $ a\equal{}b$, we can just reduce $ a\plus{}c$ and keep $ ac$ fixed, so we only have to check case ii :)


I have a two-kilobyte PDF file that I would like to add in my tex file. Yes, really in it, so I have one file instead of two.

So instead of the regular [code]\includegraphics[height=3cm]{UGent}[/code] I would like something similar to [code]\includegraphics[height=3cm]{\x00\x30\xc3\x32\xfa\x40\x32\xc5\x99\xfc...}[/code]which stores the file contents in HEX or ASCII or with \char132 or in any other way.

This would have the huge benefit (to me) that I don't need to use a second file (now it makes things much more difficult for my purposes). I attached the file. I believe this should be doable with a 2-kilobyte PDF file (even if it increases my tex file with 20 kb).

Thanks in advance!

I cant imagine what you are trying to do. \includegraphics is for showing an image so how can that image be a sequence of hex characters? If you want to include a file you use \include but then of course you must convert it to a readable text file first.

Well, I currently need 2 files to compile my documents: the tex file and this university logo. I would like to reduce this to 1 file, by hardcoding the logo inside my tex file. Is this still unclear?

Not necessarily through \includgraphics, any other way to give the direct input is fine too. :)

I still don't understand. Whatever version of \include you use you will always need two files, the tex file and the other logo file when compiling. \include means read the other file and then use it for compiling. BUT the resulting compiled pdf file will only be one file and that is the one that you will distribute.

The logo file is a picture so cannot be converted to text, which is what the tex file format uses. The only way to not use \include is to reconstruct the drawing using asymptote, jPicEdit or other drawing package which produces tex commands.

[quote="stevem"]\include means read the other file and then use it for compiling.[/quote] Basicly what LaTeX does (if I understand well) with \input{FileName} is:
FileContents = readFile(FileName)
inputInDocument(FileContents)

So I mainly wondered if there is a way to supply FileContents rather than FileName. But apparently there is not. Then I'll see if I can get it converted to asymptote or similar.

But what is the point of using \input - that needs an external file and you don't want that?
It's simple to supply the FileContents - just open the pdf file in a text editor copy and paste the pdf text into the tex document yourself. BUT it won't of course compile since it doesn't tell TeX what to do - TeX needs commands hence the need for Asymptote or similar.
A PDF file is really a series of instructions to a PDF reader as to how to display the file - those instructions won't mean anything to TeX.


Let $a,\ b$ be positive numbers such that $a+b<\frac{3}{2}.$

Prove that $(1-a)(1-b)<\frac{1}{1+a+b}<1-ab.$

we can assume $a+b=k<3/2$
Then you can easily solve this problem with $ab\leq (k^{2})/4$

umm my little post above is flawed anyway.. sorry


geometry
geometric transformation
number theory
algebra proposed
algebra

Yes, and one can find a solution here :
http://www.animath.fr/tutorat/dossier_02034sol.html

(if someone wants a translation, just ask :))

[quote="Fermat -Euler"]Find all positive integers $a$, $b$ such that $2^a + 1$ is a multiple of $2^b - 1$.[/quote]

Please post Number Theory problems in the section of the Number Theories... ;)  ok?


induction
function
algebra
domain
strong induction

I cant understand inducction can someone help me :blush:  :blush:  :blush:

Can you tell us what exactly confuses you? That way we can help you better.

hello, here are some problems with solutions, i hope this will help you to understand induction:
http://school.maths.uwa.edu.au/~gregg/Academy/1995/inductionprobs.pdf
Sonnhard.

The basic idea of induction is that you can prove something for all integers > n (n is usually 1). You can, of course, use it to prove that it works for all integers < n, as well.

I'll call what you are trying to prove as "prove that f(x) is true", even if it isn't actually a function that you are trying to prove true or false.

So, the first thing you do is prove that f(n) is true. This you can do by substitution Then, you prove that if f(k) is true, for any given k, then f(k+1) [or f(k-1), if you are trying to prove that all integers < n work] is also true. Generally, you take f(k) or whatever, and substitute all of the k's with "k+1" and find some way to reduce it back to f(k) or something that is obviously true based on f(k) being true. You then say that the induction is complete, and Q.E.D.

The theory is that it works for k=n, as proven, so it works for k=n+1, as proven in the second step. Since it works for k=n+1, it works for k=n+2, etc.

Often, you will see induction explained as a "domino effect". The most common domain to prove something over is $ n\ge1$ for integers. ($ n$ is a natural number. Also note that your basis can be at any other value of $ x$, like $ 3$, if you wanted to prove a statement for $ n\ge3$.)

So, imagine that the first domino is proving that the statement is true for $ n=1$. Then, if you prove that if the statement is true for any value $ n=k$, it is then true for $ n=k+1$, that starts the rest of the domino chain. Proving that it is true for $ n=1$ shows that it is true for $ n=2$, which makes it true for $ n=3$, etc.

[url]http://en.wikipedia.org/wiki/Mathematical_induction[/url]

[b][u]Example: Show that, for all natural numbers $ n$,[/u][/b]

\[ \displaystyle\sum_{x=1}^{n}x=\frac{n{(}n+1{)}}{2}\]

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

For $ n=1$:

$ \displaystyle\sum_{x=1}^{1}x=1$

and

$ \frac{1{(}1+1{)}}{2}=\frac{2}{2}=1$

So the statement is true for $ n=1$.

Now, let's assume that

\[ \displaystyle\sum_{x=1}^{k}x=\frac{k{(}k+1{)}}{2}\]

Then,

\[ \displaystyle\sum_{x=1}^{k+1}x=\displaystyle\sum_{x=1}^{k}x+{(}k+1{)}=\frac{k{(}k+1{)}}{2}+{(}k+1{)}=\frac{k{(}k+1{)}+2{(}k+1{)}}{2}=\frac{{(}k+2{)}{(}k+1{)}}{2}=\frac{{(}k+1{)}{(}{(}k+1{)}+1{)}}{2}\]

Therefore,

\[ \displaystyle\sum_{x=1}^{n}x=\frac{n{(}n+1{)}}{2}\]

for $ n\ge1$.

As a small exercise, why don't you try using induction to prove the same thing for $ n\ge0$?

There was a Math Jam about induction:
http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=259

There are two types of induction - strong and regular(lol).

strong induction is basically induction except you use a whole lot of base cases instead of just one. Induction is usually useful when proving something dealing with positive integers.


Consider a triangle whose sidelengths $a$, $b$, $c$ are integers, and which has the property that one of its altitudes equals the sum of the two others.

Then, prove that $a^2+b^2+c^2$ is a perfect square.

[quote="andyciup"]Denote the altitudes of the triangle by $h_a$, $h_b$, $h_c$, and assume WLOG that $h_a=h_b+h_c$.

Let $S$=the area of the triangle,and a,b,c the sides of the triangle.Then $h_a=\frac{2S}{a}$ and $h_b$ and $h_c$ are written in the same form.Then $\frac{2S}{a}=\frac{2S}{b}+\frac{2S}{c}$ therefore $\frac{1}{a}=\frac{1}{b}+\frac{1}{c}$ therefore $a= \frac{b*c}{b+c}$.Thus we have $a^2+b^2+c^2=(\frac{b*c}{b+c})^2+b^2+c^2=\frac{(b^2+c^2+bc)^2}{(b+c)^2}$ =$(\frac{(b^2+c^2+bc)}{b+c})^2$  QED I guess :)[/quote]

I think you should show that $b+c | b^2+c^2+bc$.

Andyciup's solution is correct: If an integer is the square of a rational number, then it is the square of an integer; thus, since $a^2+b^2+c^2=\left(\frac{b^2+c^2+bc}{b+c}\right)^2$, it follows that $a^2+b^2+c^2$ is the square of an integer.

  darij


This was problem #5 at a high school math competition at my school today, in the contest you have 5 minutes to solve this problem(not written exactly as in the competition because I don't have the paper but that's basically the problem):

5) Given a right triangle $ ABC$, with the right angle on C such that $ AC\equal{}2\minus{}\sqrt{3}\plus{}\dfrac{1}{11}$ and $ BC\equal{}1\minus{}\dfrac{2}{11}\plus{}\dfrac{\sqrt{3}}{11}$. Find $ \angle B$ given that $ \tan{\dfrac{\pi}{12}}\equal{}2\minus{}\sqrt{3}$

Yeah, I had figured it out just a tad bit too late :( .

[hide=" Solution"]Rewrite the sides as the following:

$ AC \equal{} \tan (\frac {\pi}{12}) \plus{} \frac {1}{11}$

$ BC \equal{} 1 \minus{} \frac {1}{11} \tan (\frac {\pi}{12})$

So,

$ \tan \angle B \equal{} \frac {AC}{BC} \equal{} \frac {\tan (\frac {\pi}{12}) \plus{} \frac {1}{11}}{1 \minus{} \frac {1}{11} \tan (\frac {\pi}{12})} \equal{} \tan ( \frac {\pi}{12} \plus{} \tan^{ \minus{} 1} ({\frac {1}{11}}))$

$ \therefore \boxed{\angle B \equal{} \frac {\pi}{12} \plus{} \tan^{ \minus{} 1} ({\frac {1}{11}}})$[/hide]


Let $ABC$ a triangle and $AD$ , $BE$ and $CF$ concurrent cevians . ( $D \in$[$BC$] , $E \in$[$AC$] and $F \in$ [$AB$] .)
  Let $AD \cap BE \cap CF = \{M\}$ . Prove that :

$\frac{AM}{AD}+\frac{BM}{BE}+\frac{CM}{CF}= 2$ .

AM/AD=1-MD/AD=1-[MBC]/[ABC]

result follows


I believe that the HM tournament (the first questions of the GUTS round) has the same level difficulty questions as one would find towards the end of the AMC 12.  The harder questions on the HM tournament seem to have questions with the same difficulty to the AIME.

What do you guys think?

Because of the time restraints, the Harvard-MIT thing can be kind of hard to get perfects on. Nothing is too bad however, not past #5 or so on the AIME I'd say. It's a good competition, but comparing it to the AMC tests involves a lot of factors.

I'm pretty sure you're allowed to bring computers to the guts round.


What is the hardest ap class you've ever taken, or the hardest ap you're taking now?

For me, AP Spanish Lit is killer. Especially since I didn't get to take AP Spanish Lang.

The only one I'm taking is AP Phys C; its quite challenging in fact;
Last year I took AP Bio; the teacher wasn't very good, and we only got through about 3/5th of the book. At my new school, some friends of mine are in AP Bio, and I could imagine how a complete courseload could be extremely time consuming

[quote="neelnal"]Well, you might be asking this to the wrong crowd.

For a non-math/science community, the answer is probably AP Physics C.

But for me, it is AP English Language and Comp. or AP Spanish Language.[/quote]

Actually, the hardest ap I've taken is probably AP Physics C, especially at the beginning of the year because I was so overwhelmed. During the 3rd and 4th advisory I finally caught up because I was doing homework.

The others, with the exception of World Hist have been/are pretty easy.

Calc BC
Calc AB
ENG lang
Eng Lit
French lang
Phys B

I'm having trouble in world hist because I don't do the work in time.

AP EURO IS SO HARD!!!!

apush is so much easier...but then again, i never read a single page in euro...which would explain my 4 :(

[quote="mdk"][quote="Elemennop"]oh god, AP Physics C is so easy at our school. I don't think I've done homework in the last two weeks.[/quote]

Do you have a major projects every week, like a rube goldberg device etc...[/quote]

No, usually I take a nap or eat lunch.

AP English is evil.

AP Bio is a piece of cake. :D

[quote="pieterminate"]AP statistics definitely.  It was really hard to keep awake... :roll:[/quote]

Lawl.
I completely agree.

In retrospect, I concur with pieterminate.
 :rotfl:

[quote="mdk"]Why do you guys think that AP Stats is hard, it is easiest AP ever.
I guess AP's are relative.[/quote]

We were kidding around.

To study for AP, or wait...HA!

Well, I take one or two practice tests for timing. My school does a good job of preparing us.

I would definitely say AP Virgil or AP Music Theory.

Well it seems to me like it would be pretty hard to date any AP.

On a more serious note, AP Physics C seems to be pretty tough in general at our school, probably in part because the test is basically comprised of physics olympiad problems.

[quote="tokenadult"]How is everyone studying AP stats? What kind of textbooks do you use? What do the homework assignments look like?[/quote]

my ap stat teacher really really really sucked. it was his first yr, so no one learned a thing, and most didn't care anyway (90percent of students taking it went to sleep the first 5 minutes)

i basically just read the textbook, do the problems in it and used Princeton review.  these combined should easily give u a 5.

[quote="juicybooty911"]i basically just read the textbook, do the problems in it and used Princeton review.[/quote]

What textbook? What did you think of it?

Get the Barron's AP Stat review book; it's the only thing you'll need. I started studying off that 2 days before the AP and got a 5. Since your kid has a strong background in math, tolken, it should be fine. A lot of it is very intuitive.

[quote="neelnal"]Wait, who is finding AP Stats hard?

It is basically boring common sense and a repitition of everything I learned in 5th and 6th grade.[/quote]

not really, i don't think i learned much of statistics back in 5th grade. but of course, some people take calc in middle school, hehe.

ap stat, in my opinion, is harder than ap calc but then that's not saying much. for almost everyone here, apstat should be a breeze.

i don't exactly remember what book i used, but i will find out tomorrow in school.

Here's an interesting review of statistics textbooks, by a statistician very concerned about education reform:

[url=http://statland.org/MAAFIXED.PDF]Advice to Mathematics Teachers on Evaluating Introductory Statistics Textbooks[/url]

AP English Lit.  It's the only AP I didn't get a 5 on (got a 3).  But of the 11 AP exams I have taken or will take, 9 of them were math or science, so they are just easy for me.  I also took AP Government, but I'm good at that.

Why is it that everyone who takes stat (including me) has the same story: "My teacher sucked and we didn't learn anything, but the course is so easy I just read a textbook and got a 5"?  Do all stat teachers suck?  Does everyone get a 5?

[quote="renatzu"]

Why is it that everyone who takes stat (including me) has the same story: "My teacher sucked and we didn't learn anything, but the course is so easy I just read a textbook and got a 5"?  Do all stat teachers suck?  Does everyone get a 5?[/quote]

Well, i believe most teacher really don't have much of an in depth understanding of stat since most of them didn't take much classes  on the subject. my teacher only took one or two stat classes before he started teaching stat (partly bc he was the only one who can teach at my school)

and everyone gets 5 by reading textbook is pretty common since we are all math superpowers and have good self learning abilities. and this isn't just stat, i slept in my ap physics and read only half of the book and looked over princeton review and got a 4 (parents were definitely not too happy about that score).

[quote="renatzu"]Why is it that everyone who takes stat (including me) has the same story: "My teacher sucked and we didn't learn anything, but the course is so easy I just read a textbook and got a 5"?  Do all stat teachers suck?  Does everyone get a 5?[/quote]

That's the reason I'm not taking a math this year.  Actually, I had we've had two ap stat teachers since I entered high school ('07 !) and both of them I intentionally avoided.

[hide="Dialogue between current AP stat teacher and her hon calc class this year"]
Teacher (after writing a problem on the board and comparing her                      answer to the book's to find she was wrong): Just pretend that answer's right
Class(indignant): If we get a problem on the test wrong can we pretend it's right?
Teacher: No.
Class:But you didn't explain the problem!
Teacher: You get the general idea.  Sometimes in math, there's
             more than one way to approach the same problem,
             and you can get more than one answer and they can
             both be right.
The class was pretty upset.
[/hide]

I'm planning on just going independent study for the exam sometime next semester. I don't know how early I should (or will) start, though.

AP US History was ungodly hard for some reason :'(.  I did horrible

i dont think you can really say that there is a hardest ap exam. this is because different people have different strengths. for instance someone who is good at math might find the ap statistics and calculus exams easy but might have trouble with one on a foreign language, and vise versa.
(this is just an example, i am not trying to imply that math people are bad with language in any way.)

I was under the impression that most people who take AP Stats do it independent study... I guess that's in keeping with the idea that it's easier to learn from the book than from the teacher.

I've taken/am taking a lot of AP classes for my school, but not compared to many people here. Still, the tests seem ridiculously easy to me... Usually, the course covers far more material than the test, and the test only evaluates a vague understanding of the few topics it does include. (World history, biology, and statistics were especially limited.)

Any AP Spanish would be very hard for me because advanced education in foreign languages doesn't exist in my school system, and non-advanced classes don't teach very much AT ALL.

And about AP Physics C... it's probably easy if you're only taking 1/2 of it, hard if you're taking all of it.

For Physics and Chemistry... you can tick off a few basic processes/concepts you have to learn, and then you'll do fine. If you don't master these things, you'll be unsure what to do and thus have difficulty--very simple. For language, by contrast, it depends on how good your intuition is, and it's harder to just study and improve.


First I would like to say hi to everyone it looks like this is a very nice forum. I hope that I placed this in the right forum to. My friend who is very good with math gave me this problem. He keeps telling me that it is very easy and that I should now how to do it but I am a complete math idiot and I was wondering if anyone of you could be so kind and tell me why this problem does not work and why. Also I did not know how to make a square rout in this so I did it like how you would do it in a graphing calculator using the ^ symbol.
a=b Suppose a equals b
a^2 = ab  Multiply by a
a^2 - b^2 = ab-b^2 subtract by b^2
(a-b)(a+b) = b(a-b) factor
a+b=b divide by (a-b)
b+b = b substitute a for b
2b=b add b + b
2 = 1 divide by b

Again he did something wrong and I cant seem to figure it out.
Thanks for any help

[quote="Jacinto"] divide by (a-b)[/quote]

Here, because $a=b$, and dividing by $0$ is only possible in the dark side :starwars:

Pierre.

wait why cant you divde by  (a-b)? what do you mean the darkside?

$a-b=0$ so you cannot divide through $a-b$. The dark side ... hmm, have you ever watched Star Wars? :)


If a cyclic quadrangle ABCD has sides a,b,c,d and semiperimeter u,area K is given by
                      $ K\equal{}\sqrt{(u\minus{}a)(u\minus{}b)(u\minus{}c)(u\minus{}d)}.$ Prove that.
...yildobekir...

The area of a convex cuadrilateral ABCD is given by the formula (Brahmagupta's formula)

$ K \equal{} \sqrt {(u \minus{} a)(u \minus{} b)(u \minus{} c)(u \minus{} d) \minus{} abcd.cos^2\frac {A \plus{} C}{2}}$

When ABCD is cyclic $ A \plus{} C \equal{} 180^o$ then the factor $ abcdcos^2\frac{A\plus{}C}{2}$ equals $ 0$

[hide] $ \widehat{BAD} \equal{} \alpha$ and $ \widehat{BCD} \equal{} 180 \minus{} \alpha.$ $ AB \equal{} a,AD \equal{} d ,BC \equal{} b ,CD \equal{} c.$
$ A(ABCD) \equal{} \frac {1}{2}\cdot ad\cdot \sin \alpha \plus{} \frac {1}{2}\cdot bc\cdot \sin \alpha \equal{} \frac {(ad \plus{} bc)}{2}\cdot \sin \alpha$
By the Law of Cosinus in ABD and BCD triangle;
$ BD^{2} \equal{} a^{2} \plus{} d^{2} \minus{} 2ad\cdot \cos \alpha$
$ BD^{2} \equal{} b^{2} \plus{} c^{2} \plus{} 2cd\cdot \cos \alpha$
$ \rightarrow\cos \alpha \equal{} \frac {a^{2} \plus{} d^{2} \minus{} b^{2} \minus{} c^{2}}{2ad \plus{} 2bc}$
$ \cos^{2} \alpha \equal{} 1 \minus{} \sin^{2} \alpha$
$ \sin^{2} \alpha \equal{} \frac {(2ad \plus{} 2bc)^{2} \minus{} [(a^{2} \plus{} d^{2}) \minus{} (b^{2} \plus{} c^{2})]^{2}}{(2ad \plus{} 2bc)^{2}}$
$ \equal{} \frac {(2ad \plus{} 2bc \minus{} a^{2} \minus{} d^{2} \plus{} b^{2} \plus{} c^{2})(2ad \plus{} 2bc \plus{} a^{2} \plus{} d^{2} \minus{} b^{2} \minus{} c^{2})}{(2ad \plus{} 2bc)^{2}}$
$ \equal{} \frac {[(b \plus{} c)^{2} \minus{} (a \minus{} d)^{2}][(a \plus{} d)^{2} \minus{} (b \minus{} c)^{2}]}{(2ad \plus{} 2bc)^{2}}$
$ \equal{} \frac {[(b \plus{} c \plus{} a \minus{} d)(b \plus{} c \minus{} a \plus{} d)][(a \plus{} d \plus{} c \minus{} b)(a \plus{} d \plus{} b \minus{} c)]}{(2ad \plus{} 2bc)^{2}}$
$ a \plus{} b \plus{} c \plus{} d \equal{} 2u$
$ b \plus{} c \plus{} a \minus{} d \equal{} 2u \minus{} 2d$
$ b \plus{} c \minus{} a \plus{} d \equal{} 2u \minus{} 2a$
$ a \plus{} d \plus{} c \minus{} b \equal{} 2u \minus{} 2b$
$ a \plus{} d \plus{} b \minus{} c \equal{} 2u \minus{} 2c$
$ \equal{} \frac {16(u \minus{} d)(u \minus{} b)(u \minus{} c)(u \minus{} a)}{(2ad \plus{} 2bc)^{2}}$
$ \sin \alpha \equal{} \frac {4\sqrt {(u \minus{} d)(u \minus{} b)(u \minus{} c)(u \minus{} a)}}{2ad \plus{} 2bc}$
$ A(ABCD) \equal{} \frac {(ad \plus{} bc)\cdot 4\sqrt {(u \minus{} d)(u \minus{} b)(u \minus{} c)(u \minus{} a)}}{2\cdot 2\cdot (ad \plus{} bc)^{2}}$
$ A(ABCD) \equal{} \sqrt {(u \minus{} d)(u \minus{} b)(u \minus{} c)(u \minus{} a)}$ [/hide]

$ f\sqrt[u]{n}!^{!}_{!}!\longrightarrow M\alpha T\sqrt[h]{eM\sqrt{a}t_{i}^{c^{S}}}$$ F^{T}W$
...yildobekir...


geometry
topology
probability
calculus
Putnam
real analysis
complex analysis

This is a question for those of you who have already completed your undergrad or really anyone else who has a opinion.

I go to a college where a math degree requires the typical calculus/differential equations sequence, modern algebra sequence, real analysis sequence, and then [i]BAM[/i] 18 hours of math electives. Right now I'm registering for the fall (a bit late I know), and I got to thinking, "Man I really have no idea how to spend these electives".

That leads me to my question: What are some particularly important courses to take? Complex analysis? Partial Differential Equations? Modern Geometry? The college I go to is very large with a wide variety of courses so if you think of something fire it out.  :lol:

Statements like "partial differential equations are more important than algebra" don't make much sense. If you did not have any geometry yet, take some. Same for complex/functional analysis, topology, combinatorics... But the most important thing is not which classes you take - it's what you take from those classes.

For our "general option" math major we require a semester of probability/statistics (really, just probability - random variables). In some places, that isn't even in the Math Dept.

I'll echo what mlok said, and say as well that if there's a particularly exciting, charismatic, and demanding faculty member, it might well be worth taking a course from him or her whether it's the "right" course or not. (The "grapevine" on your campus may have some information about who that might be, but much of the information on said grapevine was put there by students whose interests and ability levels might be quite different from yours.)

Important for what?

If you're not going to graduate school in math (or a related field), then it really doesn't matter what you take.  Sadly, you are unlikely to *use* what you learn in math classes, and you're really learning for learning's sake (which is a good thing).  So just take what you think will be most interesting to you, and take classes from good professors.

If you *are* going to graduate school, then it's true that some courses might be more important than others.  If you're going to do pure math, I recommend perusing the table of contents of Thomas Garrity's  book, All the Mathematics You Missed But Need to Know for Graduate School, to figure out which topics are important.  However, "need to know" is an exaggeration.  No one "needs" to learn all of those topics, but a high quality math education should include most of them.  But even if you're going to graduate school, the basic advice still holds:  Just take what you think will be most interesting to you, and take classes from good professors.

However, if you held a gun to my head, and I had to order you to choose some classes, I'd probably pick complex analysis and topology.  In the unlikely even that your school's curriculum includes a linear algebra course that doesn't overlap with the courses you've already taken, then linear algebra wins hands down.

Thank you all for the advice, and I agree with you that I should just take what interests me. After all, if it's not required why force myself through a semester of misery in a class I know I'll hate? Forget PDE! I'm taking Number Theory  :P

Also might I add that I am absolutely [i]shocked [/i]that an A&M professor posts on this board. Maybe we'll bump into each other in Blocker sometime, mlok.

Since I teach calculus, I do not know much about the upper-level classes at A&M. If I were at A&M as an undergrad, I would look into some of these:

304: Linear Algebra
409: Advanced Calculus I
411: Mathematical Probability
427: Introduction to Number Theory
469: Introduction to Mathematical Biology
470: Communications and Cryptography
   and last, but definitely not least,
490: The Putnam Challenge

Whatever area of math you like, try to get more opportunities to solve problems, as opposed to memorizing theorems and plugging data into them.

Gig 'em  :thumbup:

You're the man, mlok  :thumbup:  Thanks again.


Sandra Day O'Connor retired today, saying she was getting too old and wanted to spend more time with family. Thoughts, anyone? I was really surprised by it.

This was something my civics teacher said that might happen ---- during President Bush's presidency, several (at least 1) Supreme Court judge will retire and President Bush will select the new Supereme Court that shares same viewpoint as him.

Of course, the Congress has to approve him/her first but this is something big you know.. President Bush now has one empty seat on the Supreme Court, and really, selecting a Supreme Court judge that shares same viewpoint as you do (you = President) doesn't happen often. This will give more "power" (anyone can rephrase that word) to Republicans..

[quote="bleumoose"]except... O'Connor's um.... Republican?[/quote]

But she occasionally votes with the liberal judges.  Particularly on explosive issues like abortion.


I don't know if there's a specific reason as to why the "forum" link has disappeared from the sidebar, and I wasn't quite sure where to post this, so I thought I should clarify here. So, uner the community button, where the word forum used to exist, it now ceases too.

You are talking about the AoPS skin. I think that if you push the Community button you get to the Forum, however I also think that it should say Forum there somewhere. We'll fix it.

Richard simplified the sidebar yesterday to make the website easier for newcomers to navigate.  We'll take into advisement that the lack of "Forum" might confuse some people.

it is a bit confusing since some people "like me" have always gone to the home forum page using that link

Yeah, the forum button was easier to click . The Community button is of similar length and starts with the same letter as the classroom button, and I keep on clicking the classroom button. It sorta takes a long time to load too...

you should just replace the community button with the forum button...

Separating the Community and Classroom links would be a great idea :D  I kinda think that Community would be easier for visitors to understand than Forum...

How about My Classes be moved down to under Classroom, where it makes since, and either the top tabbar or the sidebar have all the links? Because right now, there are 2 ways to get to My Classes, and then there's the annoying Blogs tab that takes you to the old blogs...


inequalities
AMC
USA(J)MO
USAMO
inequalities solved

Let a,b,c be real positive numbers. Prove that

 (2a+b+c) 2 /(2a 2 +(b+c) 2 )+ (2b+c+a) 2 /(2b 2 +(c+a) 2 )+(2c+a+b) 2/(2c 2 +(a+b) 2 ) \geq 8

Sorry, you are right. I have found it in a web site and I haven't remembered it was USAMO 2003/5.

Sorry, again.

I think the sign is "<="

anyway a solution can also be found at:

http://www.kalva.demon.co.uk/usa/usoln/usol035.html

cheers! :D :D


this is the first pole i've ever made! i wonder if i did it right...

i personally think Harry is a hot headed  :furious:  idiot,  but you may think differently.

i think i like fred and george the best, and then Peeves,

what's FTW, srry, i am not good w/ abbreviations  :(

thx.

this is obviously not a popular topic.  *sigh*

=P that's ok. it's only been alive for 2-3 days  :D . I voted for Fred & George because they are the funniest.

I voted for Snape... Snape's the man

I like Fred and George because of their attitudes toward life, and I like Snape for his dedication.

fred and george r the funniest, so i voted for them.

Well, I don't have a favorite, but here's my rundown:

Harry-Dan Radcliffe is now a British porn star ([i]Equus[/i]) And he's a stupid rich kid, I watched a commentary. Someone shoot him.
Ron- He's cool
Hermione- her character is annoying, but Emma Watson is hot. (Ahem, Yule Ball) :D
Fred and George-they're hilarious.
Snape- cool character, and Alan Rickman is a fantastic actor.
Viktor Krum-SLAVIC PRIDE! He's Bulgarian, I'm Russian.

[quote="aanjohri"]i personally think Harry is a hot headed  :furious:  idiot,  but you may think differently.

i think i like fred and george the best[/quote]

aghh i voted for snape cuz he's evil/good lol. i agree Harry is hot headed, and i think the name Harry is kinda weird.

i forgot Fred/George was on there *groans*

i'm one of the few, but I voted 4 sirius, b/c he's like this mad prisoner who goes good....i think that's cool...I think Ron is an absolute idiot, and harry goes stupid in the 7th book...seriously!

I think Harry's cool. To some extent anyway.

I think we can do some of bit spolier here.  :blush: So, I hide my next writing.

Please don't open if you has not yet read Harry Potter: The Deathly Hallows
[hide]
Well, I've noted that many peoples hate Snape before 7th Book launched. Now, it seems that peoples turn their mind.
Yeah, Snape had dedicated all his life to his love, Lily Potter that inherited to Lily' son. Though, he hated some parts of Harry (from James), but for Lily, he does protect Harry. It's tragic that he ended with kills from Nagini. Even, he couldn't tell the truth from his own speak to people including Harry.  (He used pensieve). yeah, Snape is one of the bravest man.
He is a hero, a real hero in Harry Potter, I think.
Snape, who had loved Lily since he was 9 years old. But, never get the love from Lily. Well, It's pretty sad to read the chapter Prince's Tale, which recounted Snape's past life. Especially, when he read a letter from Lily in Grimmauld Place, take it's photo, love, and signature.  :(

And, I think James Potter (Dad of Harry) is not too good person.
[/hide]

I think Snape was the coolest guy ever...

[quote="mustafa"]VOLDERMORTINO![/quote]

there's something wrong with you. :rotfl:  :rotfl:  :rotfl:

SNAPE's the MAN! He was the best in the 7th book...

[quote="i_like_pie"]Other FTW![/quote]
uhhh...not really...


Let $ a;b;c>0$ and satisfying $ a\plus{}b\plus{}c\equal{}1$
Prove that:
$ \frac{a\minus{}bc}{a\plus{}bc}\plus{}\frac{b\minus{}ca}{b\plus{}ca}\plus{}\frac{c\minus{}ab}{c\plus{}ab} \leq \frac{3}{2}$

$ \frac{a\minus{}bc}{a\plus{}bc} \equal{} 1 \minus{} \frac{2bc}{a^2\plus{}ab\plus{}ac\plus{}bc} \equal{} 1\minus{} \frac{2bc}{(a\plus{}c)(a\plus{}b)} \equal{}
1\minus{}\frac{2b^2c\plus{}2bc^2}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$
Thus, the inequality is equivalent with
$ \sum \frac{b^2c \plus{} bc^2}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \geq \frac{3}{4}$
clearing the denominators, our inequality is equivalent with
$ \sum b^2c\plus{}bc^2 \geq 6abc$, true by Muirhead.

Our inequality is equivalent with \[ \sum \frac{bc}{a\plus{}bc} \ge \frac{3}{4}.\] By Cauchy Schwarz Inequality, we obtain \[ LHS \ge \frac{(ab\plus{}bc\plus{}ca)^2}{3abc\plus{}a^2b^2\plus{}b^2c^2\plus{}c^2a^2}.\] It suffices to prove that \[ 4(ab\plus{}bc\plus{}ca)^2 \ge 9abc\plus{}3(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)\] or equivalently, \[ a^2b^2\plus{}b^2c^2\plus{}c^2a^2 \ge abc.\] It comes from \[ (ab)^2\plus{}(bc)^2\plus{}(ca)^2 \ge abc(a\plus{}b\plus{}c) \equal{} abc\] as desired.