Here's a link to the problem:
http://www.topcoder.com/stat?c=problem_statement&pm=6540

Could someone please help me with the solution? I read the editorial, but I can't really understand how to code it. Could someone either place pseudo-code of the solution or explain the solutions of the contestants cited in the editorial?

What exactly don't you understand? Seems to be DP to me.

To elaborate, let's start out with a greedy solution and see what happens there. Namely, we keep adding the largest remaining block and space out blocks evenly. Then the remaining space between any blocks, as well as on the sides, after placing the first $n$ blocks is
\[S = \frac{L-\sum_{i=1}^{n}l_{i}}{n+1}\]
And we are done if $S < l_{0}$, where $l_{0}$ is the smallest remaining unused block. Is it ever possible that there is a better arrangement? If so, we would have to have $l_{0}\leq S$, yet be able to alter the arrangement in some manner to make it impossible to place an additional block. This arrangement definitely must use $l_{0}$, or else we would still have the same smallest block size, yet have decreased the total amount of space that our used blocks have taken up.

By the same logic, we can show that the used blocks must be in two contiguous chains (in terms of their lengths, not spatially) -- one should contain blocks with consecutive lengths starting with the smallest block, and the other should do the same starting with the largest. For, if not, simply replace all of the blocks not in a contiguous chain with the smallest block with the largest possible blocks that are not yet used. Then there is even less space, but the smallest available block has not changed.

So we can just loop through the number of smallest blocks we use, as well as the number of largest blocks. Since there are only 30 blocks, our algorithm runs in plenty of time. However, there is the problem that we might overstep the bounds of the problem, that is, the total length of blocks we placed might exceed $L$. To get around this, we do the same loop for the contiguous chain at the beginning, i.e. we loop through possibilities for the smallest unused block. Then we need to determine the least number of blocks larger than it that we can use to ensure that not enough space is left. We can do this by keeping track of whether it is possible to get a total length of $l$ using $m$ blocks. This can be solved recurisvely by keeping a two dimensional boolean array that tells us this, or an array of linked lists keeping track of all legal values of $m$ for a given $L_{0}$ (the former seems easier to me, so I will go with that one).

Initially, set all values in the array to false, except for $l=0$, $m=0$. Then, update the array for each block as follows -- loop through the entire boolean array, and if $arr[l][m]$ is true, and the block has length $l_{i}$, then set $arr[l+l_{i}][m+1]$ to be true. At the end, we just need to loop through all values $l$ and $m$, determine which ones take up enough space to disallow us from placing the smallest block, and then among those determine which has the smallest value of $m$.


$1^{\circ}\blacktriangleright x\in C,\ m\in N,\ n\in \overline {0,m}\Longrightarrow \sum_{k=0}^n (1+x)^kC^{n-k}_{m-k}=\sum_{k=0}^n x^kC^{n-k}_{m+1}\ .$
$2^{\circ}\blacktriangleright \{m,n,p\}\subset N^*,\ 0\le p\le n\le m\Longrightarrow \sum_{k=p}^n C^p_k\cdot C^{m-n}_{m-k}=C^{n-k}_{m+1}\ .$


calculus
derivative
function
inequalities
algorithm
inequalities proposed

Here is a nice one
$ a,b,c>0$,Prove that
$ \sum \frac{a}{2}\ge\sum \frac{bc}{\sqrt{(a\plus{}b)(a\plus{}c)}}\ge\frac{\sqrt{3\sum ab}}{2}$
 :)

$ x,y,z>0$
\[ \sum\sqrt{2x^2\plus{}xz}\ge\sqrt{2\sum x^2\plus{}5\sum xy}\]

[quote="math10"][quote="shfdfzhjj"]$ x,y,z > 0$
\[ \sum\sqrt {2x^2 \plus{} xz}\ge\sqrt {2\sum x^2 \plus{} 5\sum xy}
\]
[/quote]
I think you had wrong.Please check[/quote]

No,and the stronger is true,too.
\[ \sum\sqrt {2x^2 \plus{} xz}\ge\sqrt {2\sum x^2 \plus{} 7\sum xy}
\]
 :lol:

[quote="shfdfzhjj"]

No,and the stronger is true,too.
\[ \sum\sqrt {2x^2 \plus{} xz}\ge\sqrt {2\sum x^2 \plus{} 7\sum xy}
\]
:lol:[/quote]
No, it's wrong already. Try $ x\equal{}y\equal{}1$ and $ z\equal{}0.$ :wink:

[quote="arqady"][quote="shfdfzhjj"]

No,and the stronger is true,too.
\[ \sum\sqrt {2x^2 \plus{} xz}\ge\sqrt {2\sum x^2 \plus{} 7\sum xy}
\]
:lol:[/quote]
No, it's wrong already. Try $ x \equal{} y \equal{} 1$ and $ z \equal{} 0.$ :wink:[/quote]
Yes,I thought to ineq true we need to find a number larger
But when ineq not nice.

[quote="shfdfzhjj"][quote="arqady"][quote="shfdfzhjj"]

No,and the stronger is true,too.
\[ \sum\sqrt {2x^2 \plus{} xz}\ge\sqrt {2\sum x^2 \plus{} 7\sum xy}
\]
:lol:[/quote]
No, it's wrong already. Try $ x \equal{} y \equal{} 1$ and $ z \equal{} 0.$ :wink:[/quote]
Sorry,it should be
\[ \sum\sqrt {2x^2 \plus{} xz}\le\sqrt {2\sum x^2 \plus{} 7\sum xy}
\]
[/quote]
No,it is wrong (with x=8,y=6,z=5)

[quote="shfdfzhjj"]$ \sum_{i \equal{} 1}^n{a_i} \equal{} 1,\sum_{i \equal{} 1}^n{b_i} \equal{} 1$
\[ \sum_{1\le i < j\le n}{a_ia_j} \plus{} \sum_{1\le i < j\le n}{b_ib_j} \plus{} \sum_{i \equal{} 1}^n{a_ib_i}\le1
\]
[/quote]
$ \Leftrightarrow LHS\le(\sum_{i\equal{}1}^n{a_i})(\sum_{i\equal{}1}^n{b_i})$
$ \Leftrightarrow \sum_{1\le i < j\le n}{a_ia_j} \plus{} \sum_{1\le i < j\le n}{b_ib_j} \le \sum_{1\le i<j\le n}({a_i b_j \plus{} a_j b_i})$
$ \Leftrightarrow \sum_{1\le i<j\le n}{(a_i \minus{}b_i)(a_j\minus{}b_j)}\le 0$
Let $ x_i\equal{}a_i\minus{}b_i$, we have $ \sum_{i\equal{}1}^n x_i\equal{}0$.
So we have
$ 0\equal{}(\sum_{i\equal{}1}^n x_i)^2\equal{}\sum_{i\equal{}1}^n x_i^2 \plus{}2\sum_{1\le i<j\le n}{x_i x_j}\ge 2\sum_{1\le i<j\le n}{x_i x_j}$.
Therefore
$ 0\ge\sum_{1\le i<j\le n}{x_i x_j}\equal{}\sum_{1\le i<j\le n}{(a_i \minus{}b_i)(a_j\minus{}b_j)}$.
Done.

[quote="shfdfzhjj"]$ \sum_{i \equal{} 1}^n{a_i} \equal{} 1,\sum_{i \equal{} 1}^n{b_i} \equal{} 1$
\[ \sum_{1\le i < j\le n}{a_ia_j} \plus{} \sum_{1\le i < j\le n}{b_ib_j} \plus{} \sum_{i \equal{} 1}^n{a_ib_i}\le1
\]
[/quote]
I found another solution which is better than first one.
$ LHS\le\sum_{1\le i < j\le n}{a_ia_j} \plus{} \sum_{1\le i < j\le n}{b_ib_j} \plus{} \frac{\sum_{i \equal{} 1}^n{(a_i^2\plus{}b_i^2})}{2}$
$ \equal{}\frac{(\sum_{i\equal{}1}^n a_i)^2\plus{}(\sum_{i\equal{}1}^n b_i)^2}{2}\equal{}1$.

[quote="NextPeace"][quote="shfdfzhjj"]$ \sum_{i \equal{} 1}^n{a_i} \equal{} 1,\sum_{i \equal{} 1}^n{b_i} \equal{} 1$
\[ \sum_{1\le i < j\le n}{a_ia_j} \plus{} \sum_{1\le i < j\le n}{b_ib_j} \plus{} \sum_{i \equal{} 1}^n{a_ib_i}\le1
\]
[/quote]
I found another solution which is better than first one.
$ LHS\le\sum_{1\le i < j\le n}{a_ia_j} \plus{} \sum_{1\le i < j\le n}{b_ib_j} \plus{} \frac {\sum_{i \equal{} 1}^n{(a_i^2 \plus{} b_i^2})}{2}$
$ \equal{} \frac {(\sum_{i \equal{} 1}^n a_i)^2 \plus{} (\sum_{i \equal{} 1}^n b_i)^2}{2} \equal{} 1$.[/quote]

Thank you for your interesting,one of my friends found this proof,too.

[quote="shfdfzhjj"][quote="math10"][quote="shfdfzhjj"]$ x,y,z > 0$
\[ \sum\sqrt {2x^2 \plus{} xz}\ge\sqrt {2\sum x^2 \plus{} 5\sum xy}\]
[/quote]
I think you had wrong.Please check[/quote]
This one is really true,I have a simple proof. :lol:[/quote]
Yes, it's obvious after squaring.

[quote="arqady"][/quote][quote="shfdfzhjj"][quote="math10"][quote="shfdfzhjj"]$ x,y,z > 0$
\[ \sum\sqrt {2x^2 + xz}\ge\sqrt {2\sum x^2 + 5\sum xy}\]
[/quote]
I think you had wrong.Please check[/quote]
This one is really true,I have a simple proof. :lol:[/quote][quote="arqady"]
Yes, it's obvious after squaring.[/quote]

How about others?

[quote="shfdfzhjj"]$ a,b,c,d,e\ge0$
$ (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1)(d^2 \plus{} 1)(e^2 \plus{} 1)\ge(a \plus{} b \plus{} c \plus{} d \plus{} e \minus{} 1)^2$[/quote]

WLOG $ a\ge b\ge c\ge d\ge e\ge 0$. Clearly the LHS is greater than or equal to $ 1$. Let $ s\equal{}a\plus{}b\plus{}c\plus{}d\plus{}e$. Then it suffices to consider when $ 1\le (x\minus{}1)^2\iff (x)(x\minus{}2)\ge 0\implies x\ge 2$. Now $ 2\le a\plus{}b\plus{}c\plus{}d\plus{}e\le \frac{1}{4}(a\plus{}b\plus{}c\plus{}d)\plus{}(a\plus{}b\plus{}c\plus{}d)\equal{}\frac{5}{4}(a\plus{}b\plus{}c\plus{}d)$ so $ a\plus{}b\plus{}c\plus{}d\ge \frac{8}{5}\ge 1$.

Let $ f(a,b,c,d,e)\equal{}$ LHS minus RHS

Then the derivative of $ f$ wrt $ e$ is $ \frac{(a\plus{}b\plus{}c\plus{}d\minus{}1)}{(a^2\plus{}1)(b^2\plus{}1)(c^2\plus{}1)(d^2\plus{}1)\minus{}1}\ge 0$. So $ f$ as a function of $ e$ is smallest when $ e\equal{}0$.

Now we are left with $ a\ge b\ge c\ge d\ge 0$ and we can continue.

This obviously generalizes to $ n$ variables by induction.

[quote="Altheman"]

This obviously generalizes to $ n$ variables by induction.[/quote]

No,please check your proof carefully.It is wrong when 6 variables. :P

[quote="Vasc"][quote="shfdfzhjj"]$ x,y,z\in R$,then
$ x^4(x \minus{} y)(x \minus{} z) \plus{} y^4(y \minus{} z)(y \minus{} x) \plus{} z^4(z \minus{} x)(z \minus{} y) \ge (y \minus{} z)^2(z \minus{} x)^2(x \minus{} y)^2$
$ x,y,z\ge0$,then
$ x^4(x \minus{} y)(x \minus{} z) \plus{} y^4(y \minus{} z)(y \minus{} x) \plus{} z^4(z \minus{} x)(z \minus{} y) \ge 5(y \minus{} z)^2(z \minus{} x)^2(x \minus{} y)^2$[/quote]
This is a particular case ($ k \equal{} 0$) of my inequality

$ \sum x^2(x \minus{} ky)(x \minus{} kz)\ge (5 \minus{} 3k)(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2$

for $ k\le 4$. See http://www.mathlinks.ro/Forum/viewtopic.php?t=271839[/quote]
Very nice result!I think is should be
$ \sum x^2(x \minus{} ky)(x \minus{} kz)(x\minus{}y)(x\minus{}z)\ge (5 \minus{} 3k)(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2$
But how about
$ x,y,z\in R$,then
$ x^4(x \minus{} y)(x \minus{} z) \plus{} y^4(y \minus{} z)(y \minus{} x) \plus{} z^4(z \minus{} x)(z \minus{} y) \ge (y \minus{} z)^2(z \minus{} x)^2(x \minus{} y)^2$

[quote="Jianxing113725"]How about this one :maybe:
\[ x^5(x \minus{} y)(x \minus{} z) \plus{} y^5(y \minus{} z)(y \minus{} x) \plus{} z^5(z \minus{} x)(z \minus{} y)\]

\[ \geq 3(x \plus{} y \plus{} z)(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2\]
[/quote]
It's true,easy to prove.:D

[quote="shfdfzhjj"] But how about
$ x,y,z\in R$,then
$ x^4(x \minus{} y)(x \minus{} z) \plus{} y^4(y \minus{} z)(y \minus{} x) \plus{} z^4(z \minus{} x)(z \minus{} y) \ge (y \minus{} z)^2(z \minus{} x)^2(x \minus{} y)^2$[/quote]
For $ x,y,z\in R$, I think this is easier

$ \sum x^2(x\plus{}2y)(x\plus{}2z)(x\minus{}y)(x\minus{}z) \plus{}(x\minus{}y)^2(y\minus{}z)^2(z\minus{}x)^2 \ge 0$,

but this is very hard

$ \sum x^2(x\plus{}3y)(x\plus{}3z)(x\minus{}y)(x\minus{}z) \plus{}\frac {133}{64}(x\minus{}y)^2(y\minus{}z)^2(z\minus{}x)^2 \ge 0$.

The coefficient $ \frac {133}{64}$ is the best.

[quote="Vasc"][quote="shfdfzhjj"] But how about
$ x,y,z\in R$,then
$ x^4(x \minus{} y)(x \minus{} z) \plus{} y^4(y \minus{} z)(y \minus{} x) \plus{} z^4(z \minus{} x)(z \minus{} y) \ge (y \minus{} z)^2(z \minus{} x)^2(x \minus{} y)^2$[/quote]
For $ x,y,z\in R$, I think this is easier

$ \sum x^2(x \plus{} 2y)(x \plus{} 2z)(x \minus{} y)(x \minus{} z) \plus{} (x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2 \ge 0$,

but this is very hard

$ \sum x^2(x \plus{} 3y)(x \plus{} 3z)(x \minus{} y)(x \minus{} z) \plus{} \frac {133}{64}(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2 \ge 0$.

The coefficient $ \frac {133}{64}$ is the best.[/quote]

If $ {x,y,z}\in{R},k\ge{2},$ then we have

 $ \sum{x^2(x\plus{}ky)(x\plus{}kz)(x\minus{}y)(x\minus{}z)}$
$ \plus{}\frac{k^4\minus{}20\plus{}4k^2\plus{}12k}{4(k\plus{}1)^2}(x\minus{}y)^2(y\minus{}z)^2(z\minus{}x)^2\ge0.$

The coefficient $ \frac{k^4\minus{}20\plus{}4k^2\plus{}12k}{4(k\plus{}1)^2}$ is the best.

[quote="fjwxcsl"][quote="Vasc"][quote="shfdfzhjj"] But how about
$ x,y,z\in R$,then
$ x^4(x \minus{} y)(x \minus{} z) \plus{} y^4(y \minus{} z)(y \minus{} x) \plus{} z^4(z \minus{} x)(z \minus{} y) \ge (y \minus{} z)^2(z \minus{} x)^2(x \minus{} y)^2$[/quote]
For $ x,y,z\in R$, I think this is easier

$ \sum x^2(x \plus{} 2y)(x \plus{} 2z)(x \minus{} y)(x \minus{} z) \plus{} (x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2 \ge 0$,

but this is very hard

$ \sum x^2(x \plus{} 3y)(x \plus{} 3z)(x \minus{} y)(x \minus{} z) \plus{} \frac {133}{64}(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2 \ge 0$.

The coefficient $ \frac {133}{64}$ is the best.[/quote]

If $ {x,y,z}\in{R},k\ge{2},$ then we have

 $ \sum{x^2(x \plus{} ky)(x \plus{} kz)(x \minus{} y)(x \minus{} z)}$
$ \plus{} \frac {k^4 \minus{} 20 \plus{} 4k^2 \plus{} 12k}{4(k \plus{} 1)^2}(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2\ge0.$

The coefficient $ \frac {k^4 \minus{} 20 \plus{} 4k^2 \plus{} 12k}{4(k \plus{} 1)^2}$ is the best.[/quote]
I have proved it.What's the way you find this beatiful inequality?Schur decomposition or any other new method?

[quote="fjwxcsl"]
If $ {x,y,z}\in{R},k\ge{2},$ then we have

 $ \sum{x^2(x \plus{} ky)(x \plus{} kz)(x \minus{} y)(x \minus{} z)}$
$ \plus{} \frac {k^4 \minus{} 20 \plus{} 4k^2 \plus{} 12k}{4(k \plus{} 1)^2}(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2\ge0.$

The coefficient $ \frac {k^4 \minus{} 20 \plus{} 4k^2 \plus{} 12k}{4(k \plus{} 1)^2}$ is the best.[/quote]

You are right, fjwxcsl, since equality holds for $ 2x \plus{} k(y \plus{} z) \equal{} 0$ and $ y \equal{} z$. :lol: Have you also a proof?
In addition, the inequality holds also for $ k\le \minus{}4$. We posted once the case with nonnegative $ x,y,z$ on this forum.

[quote="Vasc"][quote="fjwxcsl"]
If $ {x,y,z}\in{R},k\ge{2},$ then we have

 $ \sum{x^2(x \plus{} ky)(x \plus{} kz)(x \minus{} y)(x \minus{} z)}$
$ \plus{} \frac {k^4 \minus{} 20 \plus{} 4k^2 \plus{} 12k}{4(k \plus{} 1)^2}(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2\ge0.$

The coefficient $ \frac {k^4 \minus{} 20 \plus{} 4k^2 \plus{} 12k}{4(k \plus{} 1)^2}$ is the best.[/quote]

You are right, fjwxcsl, since equality holds for $ 2x \plus{} k(y \plus{} z) \equal{} 0$ and $ y \equal{} z$. :lol: Have you also a proof?
In addition, the inequality holds also for $ k\le \minus{} 4$. We posted once the case with nonnegative $ x,y,z$ on this forum.[/quote]

Yes, Vasc,  the inequality

$ \sum{x^2(x \plus{} ky)(x \plus{} kz)(x \minus{} y)(x \minus{} z)}$
$ \plus{} \frac {k^4 \minus{} 20 \plus{} 4k^2 \plus{} 12k}{4(k \plus{} 1)^2}(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2\ge0$
 holds also for $ k\le \minus{} 4$.

(The coefficient $ \frac {k^4 \minus{} 20 \plus{} 4k^2 \plus{} 12k}{4(k \plus{} 1)^2}$ is the best.)

[quote="shfdfzhjj"][/quote][quote="fjwxcsl"][quote="Vasc"][quote="shfdfzhjj"] But how about
$ x,y,z\in R$,then
$ x^4(x - y)(x - z) + y^4(y - z)(y - x) + z^4(z - x)(z - y) \ge (y - z)^2(z - x)^2(x - y)^2$[/quote]
For $ x,y,z\in R$, I think this is easier

$ \sum x^2(x + 2y)(x + 2z)(x - y)(x - z) + (x - y)^2(y - z)^2(z - x)^2 \ge 0$,

but this is very hard

$ \sum x^2(x + 3y)(x + 3z)(x - y)(x - z) + \frac {133}{64}(x - y)^2(y - z)^2(z - x)^2 \ge 0$.

The coefficient $ \frac {133}{64}$ is the best.[/quote]

If $ {x,y,z}\in{R},k\ge{2},$ then we have

 $ \sum{x^2(x + ky)(x + kz)(x - y)(x - z)}$
$ + \frac {k^4 - 20 + 4k^2 + 12k}{4(k + 1)^2}(x - y)^2(y - z)^2(z - x)^2\ge0.$

The coefficient $ \frac {k^4 - 20 + 4k^2 + 12k}{4(k + 1)^2}$ is the best.[/quote][quote="shfdfzhjj"]
I have proved it.What's the way you find this beatiful inequality?Schur decomposition or any other new method?[/quote]

An New Inequality Mechanical Proving Algorithm,shfdfzhjj.

[quote="Vasc"]
For $ x,y,z\in R$, I think this is easier

$ \sum x^2(x \plus{} 2y)(x \plus{} 2z)(x \minus{} y)(x \minus{} z) \plus{} (x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2 \ge 0$,

[/quote]
Because
$ \sum_{cyc} x^2(x \plus{} 2y)(x \plus{} 2z)(x \minus{} y)(x \minus{} z)\plus{}\prod_{cyc}(x\minus{}y)^2\equal{}$
$ \equal{}(x\plus{}y\plus{}z)^2\sum_{cyc}(x^4\minus{}x^3y\minus{}x^3z\plus{}x^2yz)\geq0.$

[quote="arqady"][quote="Vasc"]
For $ x,y,z\in R$, I think this is easier

$ \sum x^2(x \plus{} 2y)(x \plus{} 2z)(x \minus{} y)(x \minus{} z) \plus{} (x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2 \ge 0$,

[/quote]
Because
$ \sum_{cyc} x^2(x \plus{} 2y)(x \plus{} 2z)(x \minus{} y)(x \minus{} z) \plus{} \prod_{cyc}(x \minus{} y)^2 \equal{}$
$ \equal{} (x \plus{} y \plus{} z)^2\sum_{cyc}(x^4 \minus{} x^3y \minus{} x^3z \plus{} x^2yz)\geq0.$[/quote]
Yes, Arqady. My proof is the same.

[quote="Vasc"][quote="shfdfzhjj"]$ x,y,z\in R$,then
$ x^4(x \minus{} y)(x \minus{} z) \plus{} y^4(y \minus{} z)(y \minus{} x) \plus{} z^4(z \minus{} x)(z \minus{} y) \ge (y \minus{} z)^2(z \minus{} x)^2(x \minus{} y)^2$
$ x,y,z\ge0$,then
$ x^4(x \minus{} y)(x \minus{} z) \plus{} y^4(y \minus{} z)(y \minus{} x) \plus{} z^4(z \minus{} x)(z \minus{} y) \ge 5(y \minus{} z)^2(z \minus{} x)^2(x \minus{} y)^2$[/quote]
The last is a particular case ($ k \equal{} 0$) of my inequality

$ \sum x^2(x\minus{}y)(x\minus{}z)(x \minus{} ky)(x \minus{} kz)\ge (5 \minus{} 3k)(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2$

for $ k\le 4$ and $ x,y,z\ge 0$. See http://www.mathlinks.ro/Forum/viewtopic.php?t=271839[/quote]

Actually, when $ 1\le k\le 4$, the inequality

$ \sum x^2(x\minus{}y)(x\minus{}z)(x \minus{} ky)(x \minus{} kz)\ge (5 \minus{} 3k)(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2$

holds for any $ x,y,z\in R$.

[quote="Vasc"][quote="shfdfzhjj"] But how about
$ x,y,z\in R$,then
$ x^4(x \minus{} y)(x \minus{} z) \plus{} y^4(y \minus{} z)(y \minus{} x) \plus{} z^4(z \minus{} x)(z \minus{} y) \ge (y \minus{} z)^2(z \minus{} x)^2(x \minus{} y)^2$[/quote]
For $ x,y,z\in R$, I think this is easier

$ \sum x^2(x \plus{} 2y)(x \plus{} 2z)(x \minus{} y)(x \minus{} z) \plus{} (x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2 \ge 0$[/quote]

Actually, when $ \minus{}2\le k\le 1$, the inequality

$ \sum x^2(x\minus{}y)(x\minus{}z)(x \minus{} ky)(x \minus{} kz)\ge (k\plus{}1)(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2$

holds for any $ x,y,z\in R$.


vector
calculus
parameterization
integration
analytic geometry
calculus computations

For a curve $x=y=z^{2}$, the differential length vector $d\bold{l}$ at a point $(x,y,z)$ is $dx\bold{i_{x}}+dy\bold{i_{y}}+2zdz\bold{i_{z}}$.

What is the physical meaning of $d\bold{l}$?

[quote="fedja"]Before talking about physical meaning of $d\bold{l}$, it would be nice to figure out how you construct it for an arbitrary curve and what those $dx$, $dy$ and $dz$ mean. I'm personally quite confused about this notion as you presented it. For instance, what will $d\bold{l}$ be for the curve $x^{3}=y^{3}=z^{6}$? Will it, in your opinion, be $3x^{2}\,dx\bold{i}_{x}+3y^{2}\,dy\bold{i}_{y}+6z^{5}\,dz\bold{i}_{z}$ or something else?[/quote]
Hmmm... I see what you mean. The curves $x=y=z^{2}$ and $x^{3}=y^{3}=z^{6}$ are essentially the same curves, I think. Yet, they have different differential length element $d\bold{l}$!?

Actually, the example I presented was from a book about electromagnetism where they reviewd some of the basic concepts in vector calculus. So, in the end, I don't think I even know the definition of $d\bold{l}$ let alone its physical meaning!

Please, help!

Usually, in vector calculus one takes a curve given parametrically as $x=x(t),y=y(t),z=z(t)$ and puts \[d\bold{l}=[x'(t)\bold{i}_{x}+y'(t)\bold{i}_{y}+z'(t)\bold{i}_{z}]\,dt\,.\] For instance, the curve you had can be given by $(t^{2},t^{2},t)$ and, with this parametrization, you will have \[d\bold{l}=[2t\bold{i}_{x}+2t\bold{i}_{y}+\bold{i}_{z}]\,dt\,.\] Of course, one can choose a different parametrization for the same curve like, say, $(s^{6},s^{6},s^{3})$ and get \[d\bold{l}=[6s^{5}\bold{i}_{x}+6s^{5}\bold{i}_{y}+3s^{2}\bold{i}_{z}]\,ds\,.\] Note, however, that since $t$ and $s$ are related by $t=s^{3}$, we have $dt=3s^{2}\,ds$ and these two formulae actually agree.

This allows to write various integrals like $\int_\gamma\frac{d\bold{l}\times \bold{r}}{r^{3}}$ in coordinate-free form, which is very convenient. The sense of $d\bold{l}$ in this case is the same as that of $dx$ in $\int x^{2}\,dx$: a very small increment of the independent variable. Only in the case of $dx$ that independent variable is a point on the real line and the increment is a number while in the case of $d\bold{l}$, the variable is a point on the curve and its increment is a vector.

Hope, this helps a bit. The formulae above have nothing to do with the ones you wrote in your first post except they include similar symbols but, I believe, that is what was really meant.

I am so sorry, I actually meant that $d\bold{l}= 2zdz\bold{i_{x}}+2zdz\bold{i_{y}}+dz\bold{i_{z}}$ (where z could be made a seperate parameter $t$ just like you did.)  :blush:

[quote="fedja"]Note, however, that since $t$ and $s$ are related by $t=s^{3}$, we have $dt=3s^{2}\,ds$ and these two formulae actually agree.

This allows to write various integrals like $\int_\gamma\frac{d\bold{l}\times \bold{r}}{r^{3}}$ in coordinate-free form, which is very convenient. [/quote]I am not quite sure how this allows us to write various integrals in coordinate-free form? Could you please explain that to me?

[quote="fedja"]
The sense of $d\bold{l}$ in this case is the same as that of $dx$ in $\int x^{2}\,dx$: a very small increment of the independent variable. Only in the case of $dx$ that independent variable is a point on the real line and the increment is a number while in the case of $d\bold{l}$, the variable is a point on the curve and its increment is a vector.
[/quote]But does $d{l}$ has any geometric meaning in addition to this?

[quote="fedja"]
Hope, this helps a bit. The formulae above have nothing to do with the ones you wrote in your first post except they include similar symbols but, I believe, that is what was really meant.[/quote]Yes, it helped a lot. Thank you fedja.  :)


This Question for the record is from Vestavia Hills High School Mathmatics Tournament 2005: Geometry Test

What is the [b]10th[/b] pentagonal number?
[b]A)110
B)100
C)145
D)55
E)NOTA (None of the Above)
[/b]
[hide]The Awnser is C[/hide]
Also if you can post the equations for finding triagonal and pentagonal numbers or if someone could explain them to me  :blush: it would help alot!

Here are two excellent articles explaining both:
http://mathworld.wolfram.com/TriangularNumber.html

http://mathworld.wolfram.com/PentagonalNumber.html

The formula for triangular numbers is $\frac{n(n-1)}{2}+n$ where $n$ is the number of the sequence. For pentagonal numbers: $\frac{3n(n-1)}{2}+n$.

[hide="Answer to your problem"]$n=10\\ \\ \frac{3\cdot10\cdot9}{2}+10=145 \\ \boxed{C}$[/hide]

[quote="fireforce"]Vestavia Hills High School Mathmatics Tournament 2005: Geometry Test[/quote]This is geometry question? I thought simple arithmetic/counting problem...

lol i hate vestavia >< i got a 15 on that test...........  :oops: well other ppl from my school placed in geometery though  :lol: Juilet did best came in 5th i believe w/ a 38...... and our team cam in second  :lol: [hide]The forumla for pentagonal numbers is n(3n-1)/2 so the answer is c.145[/hide]

NAONAO!!!!! I went over the general formula the night before the test.  I'm pretty sure I stopped our study session and I was like everyone listen up... the formula for the N-th K-gonal number is $\frac{(k-2)n^2}{2}-\frac{(k-4)n}{2}$

Learn the derivation of this formula.  Its REALLY interesting.  The transcript on "An exploration of Figurate numbers" is available in the math jam section.

[quote="Naonao"]SOOOOOO????? u expect me 2 pay attention???? me who didn't care???????[/quote]
There are other places for private conversations: Quote from Iversonfan2005

[quote="Iversonfan2005"]If you're going to make private conversations, then pm each other, don't do it in the middle of a thread ;)[/quote]

[quote="Pakman2012"]the formula for the N-th K-gonal number is $\frac{(k-2)n^2}{2}-\frac{(k-4)n}{2}$[/quote] The one I learned is $\frac{n(n-1)(k-2)}{2}+n$, but they both simplify to the same thing.  :)


if $ a^2 \plus{}b^2 \plus{}c^2 \equal{}1$  then prove that $ 1>\equal{}ab \plus{} bc \plus{} ca\equal{}>\minus{}1/2$,$ a,b,c$ are positive reals
my solution:
by c.s inequality $ ab \plus{}bc\plus{} ca<\equal{}a^2\plus{}b^2\plus{}c^2$
and $ (a\plus{}b\plus{}c)^2>\equal{}0$ implies $ ab \plus{}bc\plus{}ca>\equal{}\minus{}1/2$

Second statement doesn't make sense for positive reals, $ ab\plus{}bc\plus{}ca$ can't be negative.

They don’t have to be positive. The result holds for all real $ a,\,b,\,c$ such that $ a^2\plus{}b^2\plus{}c^2\equal{}1.$ ;)

[quote="aadil"]...$ a,b,c$ are [b]positive[/b] reals
...[/quote]

 :wink:

I know! I’m just saying they [b]don’t have to be[/b] positive. You can drop the “positive” requirement and still do the problem. (In fact dropping the “positive” requirement would actually make the second inequality make sense.)

Of course, you are right. But I was referring to the original problem and it's wrong formulation.

[hide]$ 1\equal{}a^2\plus{}b^2\plus{}c^2 \ge ab\plus{}bc\plus{}ca \iff (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2 \ge 0$.
$ ab\plus{}bc\plus{}ca \ge \minus{}\frac{1}{2} \iff 2(ab\plus{}bc\plus{}ca)\plus{}1 \ge 0 \iff 2(ab\plus{}bc\plus{}ca)\plus{}a^2\plus{}b^2\plus{}c^2 \ge 0 \iff (a\plus{}b\plus{}c)^2 \ge 0$[/hide]

I think that is the implied solutions. But for a,b,and c being positive reals is a typo!


topology
linear algebra
linear algebra unsolved

How to construct a deformation retraction from $ GL(n)\minus{}>O(n)$?

$ GL(n)$ is homeomorphic to $ {S^{++}\times{O(n)}}$ (cf. polar decomposition).$ {S^{++}}$ is a convex, thus there exists a deformation retraction from $ {S^{++}}$ to a point.

[quote="loup blanc"]$ GL(n)$ is homeomorphic to $ {S^{ + + }\times{O(n)}}$ (cf. polar decomposition).$ {S^{ + + }}$ is a convex, thus there exists a deformation retraction from $ {S^{ + + }}$ to a point.[/quote]

What is $ {S^{ + + }}$ ?

Here $ K\equal{}\mathbb{R}$. $ S^{\plus{}\plus{}}$ denotes the $ (n,n)$ symmetric positive definite matrices.
Polar decomposition: $ \forall{A}\in{GL(n)}$ there exists an unique $ (S,O)\in{S^{\plus{}\plus{}}}\times{O(n)}$ s.t. $ A\equal{}SO$.


[quote="sapphyre571"]well if the mavs manage to win, they're gonna sweep houston[/quote]

looks like a good ol' texas showdown is out of question now  :(

on both ends of the spectrum...rockets lost in 7.

Man, I get so frustrated watching Yao play. He has trouble getting around defenders that are 10 inches shorter than him. Man why can't he play like Duncan? Then he'd be unstoppable.

It frustrates me that the Rockets never come from behind but they can lose when they ahve leads

I say we get Hakeem Olajuwon to come back  :P

I don't think you can blame his lack of speed on him. It is obviously his height which makes his nerves travel farther. Yao is also not very well conditioned, and he was even worse before he moved to America.

[quote="Art of Owna"][quote="Aznwithabrain"][quote="Art of Owna"]Alright lets see here.  yao is 7'6" and cant rebound for his life (down 2-houston with like 20 secs to go the jazz missed a shot and the ball was right in front of yao, but yaos so slow he let boozer get the rebound and two free throws, the jazz were up by 4, mcgrady made a shot, but again the rockets were too slow-like yao-and couldnt even foul the jazz until it was too late).  I mean 6 rebounds in game 6, 6 rfebounds in game 7?  when your that tall, you should get like 15 rebounds.  But hes the most unagressive player Ive ever seen and hes by far the slowest (so even if they baet the jazz they would probalby get swept by the warriors because yaos so slow and the warrios have speed), but I mean hes a great offensive player with his tournaround shots and good free throw shooting but if thats not falling, hes going to like another shawn bradley..a 7'6" bust because he has no speed, cant play defense and cant rebound.

Oh yea, Duncan is a lot differnt then yao because duncan is way more aggresive and he used to be a great swimmer (amost olympics) so hes obiviously in shape and duncan can actually block peoples shots instead of letting deron williams and carlos boozer go around him all day.[/quote]

Alot of this is true

Yao Ming has a nice 10-12 feet jumpshot.
His offensive weapons consists of hook shots, turn arounds, and slam dunks.
High field goal percentage (.524)
High free thow shooting percentage (.819)

but...

Shawn Bradley had NO offensive skills. NONE.
Have you seen a 7-6 NBA player who has great speed?
Tim Duncan. Aggressive?
He's no more aggressive than Yao Ming is

The rest of this post is dedicated to Yao Haters (not Art of Owna, he's just criticing (sp?))

[youtube]r8STO24L_NY[/youtube]
You think he's slow? Look at Dikembe.
No range? What's number 9.
Still slow? Look at number 8
Not aggressive? There's that dunk for you
Still not aggresive? How about a dunk over the Trail Blazers
Can't block? How about number 5
Don't tell me he isn't aggressive. Look at number 4. Monster dunk over Theo Ratliff
Not a team player? Look at number 3
Not a rebounder? Look at number 2 closely. He out hustles Shaq when Shaq was in his prime.
No foot work? Look at NUMBER 1!

Hope you enjoyed the video[/quote]

Im not saying hes a bad player.  hes probalby the best offensive center in the nba.  But what Im simply saying is that yao for his size, is one of the worst rebounders ever.  Look at chuck hayes, hes 6'6" but he is a lot better rebounder, especially for his size, than yao.  Im just syaing he needs to work on his defnesive skills (did you see boozer just go around him all the time) and his speed and condition.  Then he will be like hakeem olajuwan.  Remember hakeem was a great offensive player, but he was best known for his defense (only player with 2000+ steals and blocks, had a quadruple double one game), that is why Greg Oden will be number 1, for his defensive skills, so if he can develop his defensive skills sand rebounding he will be amazing and they will come out of the first round with a win.

And yes Ive seen 7 footers with great speed (and also lose in the playoffs early)-dirk nowitski :D[/quote]

Hey don't completely ignore my video :rotfl:

Greg Oden?
He'll foul out so many times in the NBA, he won't be on the court for any effective minutes.
Once he starts learning to foul less, Oden will be a superstar

A 7-6 player and a 7-0 player are COMPLETELY DIFFERENT
thats like comparing a 6-9 player and a 6-3 player
aka
comparing a front court player and a back court player

Think of a 7-6 player with great speed

Dude this sucks.  the mavs and the rockets don't make it out of the first round.  i guess i'm counting on the spurs...  of course the bucks aren't in the playoffs so i guess i have to cheer for the only other state i've ever lived in...

[quote="junggi"]dont quote long posts its annoying[/quote]

Don't post off topic subjects
This is a "Go Rockets" topic
not a "Advice on Posting" topic
Private Message instead

Anyways back on subject
I hope the Rockets draft Yi JianLian
That would be awesome

[quote="junggi"][quote="Aznwithabrain"][quote="junggi"]dont quote long posts its annoying[/quote]

Don't post off topic subjects
This is a "Go Rockets" topic
not a "Advice on Posting" topic
Private Message instead[/quote]
unfortunately its easier to post one message instead of 2 pms
and also
do you like mcdonalds[/quote]

Uh...McDonalds?
This is a sports topic
Oh wait, you probably don't know a thing about sports like those guys from the ESPN News commercials

I'm going to keep posting some on topic posts
Who do you guys want the Rockets to draft?
According to http://www.nbadraft.net/admincp/profiles/morrisalmond.html
They might be drafting Morris Almond
I don't think he would be a good fit.
Who do you guys think the Rockets should draft?

[quote="junggi"][quote="Aznwithabrain"]
Uh...McDonalds?
[/quote]
well it seemed like you were on drugs[/quote]

It seems like you have no life
You just listen to drugged up metal music and stupid alternative music like Yellowcard (haha. who still listens to that other than you?)
 :rotfl:

You're completely ignorant in the areas of sports
Get off this topic and go listen to your music

More on-topic posts from me:
Who do you guys want the Rockets to sign as a free-agent the most this offseason?

The rockets will win the lottery and select Greg Oden and convert him to power forward.   :D

/jk

of course that's impossible considering they won't be in the lottery....

lol....  the rockets can't be in the lottery though since they made it to the playoffs.. :blush:

i say they should trade memphis somehow...lol

here are some interesting questions that are very on topic for you guys to consider

[quote="Aznwithabrain"]Who do you guys want the Rockets to draft?
Who do you guys think the Rockets should draft?
Who do you guys want the Rockets to sign as a free-agent the most this offseason?[/quote]

just for you guys to check that these are on-topic, here is the original post:

[quote="footballrocks41237"]Rockets lead the series against the Jazz 3-2. Tomorrow 8 ET on TNT.

GO ROCKETS[/quote]

[quote="junggi"][quote="Aznwithabrain"]bs[/quote]
haha you are funny xD[/quote]

 :rotfl:  :rotfl:
you're even more funny

You're a loser
You don't know anything about sports :rotfl:
You consider yourself male
You listen to Yellowcard :rotfl:
Do you even know who Yi JianLian is?
Do you know who Wilt Chamberlian is?
How about Bill Russel?

Get off this sports topic loser
Go listen to that heavy metal "music"

Back on topic:
The Rockets can't win the lottery but yeah if they could; they would desperately need a power forward
Chuck Hayes is good and all (I'm a Kentucky fan)
but
he would be better off as a bench player that brings energy

Edit @ Junggi:
LMFAO
 :rotfl:
Of course they're on topic you dumb piece of ----
They're about the Rockets
HAHA :rotfl:
Ignorance at it's highest level

Edit2:
[quote="noneoftheabove"]C'mon guys this argument about staying on topic is off topic.

Do you think Jeff Van Gundy is going to retire?(iono something random ABOUT the rockets)

@football: I still think a good point guard is more important right now.[/quote]

Rafer Alston is a pretty good point guard
Luther Head and Jon Lucas are great back-ups

The Rockets need more front court players
Juwan Howard and Dikembe Mutumbo are getting old

Van Gundy is not going to retire
He's still pretty young (in NBA coaching years)

[quote="Aznwithabrain"]
Of course they're on topic you dumb piece of ----
[/quote]
xD
thats what i was saying xD
why am i talking to an illiterate?
xD

edit:
me: this guy is soooooo funny
friend: dude
friend: hes not worthy of his avatar
me: LMAO

edit:
this should be my last post here, so carry on with the discussion

edit:
[quote]From: 	Aznwithabrain
To: 	junggi
Posted: 	Today, at 6:58 pm
Subject: 	Coward
You're a ignorant coward. Rotfl [/quote]
have a nice day people

[youtube]QdNqNnYWWq0[/youtube]

We'll get them next year
2008 NBA Champions Houston Rockets

@ junggi:
so you also resort to posting pm's
 :rotfl:

dude junggi, it seems like 4/20 got to you.

[quote="noneoftheabove"]Please stop this argument it is fruitless and basically spam.[/quote]

And what you just wrote isn't spam?
Don't be a hypocrite

[youtube]RY5fiHu73zk[/youtube]

Another video showing why Yao Ming WILL DOMINATE


trigonometry
calculus
complex analysis
calculus computations

Could someone show me how the following was arrived at?:

$\sin{x}=x\prod_{n=1}^{\infty}\left(1-\frac{x^{2}}{\pi^{2}n^{2}}\right)$

I feel like I've never seen this in calculus or trigonometry yet.

[quote="amcavoy"]I feel like I've never seen this in calculus or trigonometry yet.[/quote]
That's because this is essentially a complex analysis theorem; at least I have a hard time seeing how to prove it without invoking some of the tools of complex analysis.

Are you by any chance looking at Euler's method for computing $\zeta(2),\zeta(4),\dots?$

Actually, I wasn't really using it for anything; I just came upon it on the internet and was wondering where it came from (since it wasn't as obvious as other identities).  That's fine though, thanks for the info. (it looks like I'll eventually be learning it in complex analysis.)

Thanks again.


Is mathcamp good for someone who will be entering the 2010-2011 school year as a 9th grader?(I've only taken up to precalc)

Sure. There are usually quite a few rising 9th graders at camp, although that tends to be a bit younger than the average age. And you certainly don't need to have taken calculus before you start camp, a lot (most?) of the interesting classes don't require any calculus (or really much of anything). Also if you want, there will probably be some sort of "Calculus crash course" sometime in the first week or two.


Let {$ A_{k}$},{$ B_{k}$} [b]are two sequences matrices such that [/b] $ A_{k},B_{k} \in S_{n}(R)$ and
$ A_{k}B_{k} \equal{} B_{k}A_{k} \forall k \in N$ - {0} [b]and [/b]let $ a,b,c \in R$ [b]such that [/b]: $ b^{2} \minus{} 4ac < 0$
[b]Prove that :[/b]
[b]If[/b] $ \lim_{k \to \infty}(aA^{2}_{k} \plus{} bA_{k}B_{k} \plus{} cB^{2}_{k}) \equal{} 0_{n}$
[b]then[/b] $ \lim_{k \to \infty} {A_{k}} \equal{} 0_{n}$ and  $ \lim_{k \to \infty} {B_{k}} \equal{} 0_{n}$
[b]Here[/b] $ S_{n}$  [b]are set nxn symmetric real matrix[/b] .
 :(  :(

[hide]
$ 0 \leq b^2 < 4ac$, then :
Case I : If $ a > 0$ then $ c > 0$
$ lim(aA^2 \plus{} bAB \plus{} cB^2) \equal{} 0$ implies $ lim(A^2 \plus{} \frac {b}{a}AB \plus{} \frac {c}{a}B^2)\equal{}lim (A^2\plus{} \frac {b}{2a}AB \plus{} \frac {b}{2a}BA \plus{} \frac {b^2}{4a}B^2 \plus{}  (\frac {c}{a} \minus{} \frac {b^2}{4a^2})B^2 ) \equal{} 0$
$ lim(A \plus{} \frac {b}{2a}B)^2 \plus{} (\frac {c}{a} \minus{} \frac {b^2}{4a^2})B^2 \equal{} 0$
Note that $ \frac {c}{a} \minus{} \frac {b^2}{4a^2} > 0$ since :
$ 4ac > b^2$ implies $ 4ac \minus{} b^2 > 0$ implies $ \frac {c}{a} \minus{} \frac {b^2}{4a^2} > 0$.
Then :
$ lim(A \plus{} \frac {b}{2a}B)^2 \plus{} lim(\frac {c}{a} \minus{} \frac {b^2}{4a^2}B^2) \equal{} 0$
$ lim(A \plus{} \frac {b}{2a}B).(A \plus{} \frac {b}{2a}B)^T \plus{} lim (\frac {c}{a} \minus{} \frac {b^2}{4a^2})B.B^T \equal{} 0$ (Because $ A$ and $ B$ are symmetric then $ A \plus{} kB \equal{} (A \plus{} kB)^T$ and $ lB \equal{} lB^T$ for some scalar $ k$ and $ l$.
Note that for any real matrix $ A$,$ tr(A.A^T) \geq 0$ and we can prove that the equalituy occurs only if $ A \equal{} 0$, now write $ A \plus{} \frac {b}{2a}B \equal{} C$ and $ (\frac {c}{a} \minus{} \frac {b^2}{4a^2}) \equal{} h > 0$, then the expression can be re written in the form :
$ lim(C.C^T) \plus{} lim(h.B.B^T) \equal{} 0$
Note that $ tr(lim(C.C^T)) \geq 0$ and $ tr(lim(h.B.B^T))\geq 0$
$ 0 \leq tr (lim(h.B.B^T)) \leq tr(lim(h.B.B^T)) \plus{} tr(lim(C.C^T)) \equal{} 0$
then $ tr(lim(h.B.B^T)) \equal{} 0$ that implies $ B \equal{} 0$ because $ h$ is not equal to $ 0$, then :
$ 0 \leq tr (lim(C.C^T)) \leq tr (lim(C.C^T)) \plus{} tr(lim (h.B.B^T)) \equal{} 0$
then $ tr(lim(C.C^T)) \equal{} 0$ that implies $ C \equal{} 0$ then $ A \plus{} \frac {b}{2a}B \equal{} 0$ then $ A \equal{} 0$ also.
Case II : If $ a > 0$ then $ c < 0$, then $ \minus{} a > 0$ and $ \minus{} c > 0$
Then we can write as in the case I.
Sorry if I am wrong.
[/hide]

True you don't write $ Tr(\lim ({CC^T})) \equal{} 0$ then $ C \equal{} 0$ by if  $ C \equal{} diag(\frac {1}{k},\frac {1}{k},...,\frac {1}{k})$ ?
I will post my solution if you want .....  :(  (Please try prove $ Tr(\lim ({CC^T})) \equal{} 0$ then  $ \lim {C} \equal{} 0$ ?)

You have a solution pefect by LimC=O_{n} . I am stupid .... :blush:


Does there exist a sequence $ \{b_{i}\}_{i=1}^\infty$ of positive real numbers such that for each natural $ m$: \[ b_{m}+b_{2m}+b_{3m}+\dots=\frac1m\]

Can we say this?

$ \sum_{m=1}^{\infty}lim_{n\rightarrow{\infty}}(b_{m}+b_{2m}+...+b_{nm})=\sum_{m=1}^{\infty}{d(m)b_{m}}=\sum_{m=1}^{\infty}\frac{1}{m}$ (absurd)
where $ d(m)$ is the number of divisor of m

[color=orange][b]perfect_radio[/b][/color] what do you think?It's possible.

I guess, I found something interesting on this problem.
The idea is the following:
Write the sum $ b_{m}+b_{2m}+b_{3m}+...=b_{m}+(b_{2m}+b_{4m}+...)+(b_{3m}+b_{6m}+...)+...-$
$ -(b_{6m}+b_{12m}+...)-(b_{8m}+b_{16m}+...)+...=b_{m}+\frac{1}{2m}+\frac{1}{3m}+\frac{1}{5m}+...-$
$ -\frac{1}{6m}-\frac{1}{10m}-\frac{1}{14m}...=\frac{1}{m}$, from this we can see that
$ b_{m}=\frac{1}{m}(1-\sum_{i\in P}{\frac{1}{i}}+\sum_{i,j\in P, i\neq j}{\frac{1}{ij}}-\sum_{i,j,k \in P, i\neq j\neq k, i\neq k}\frac{1}{ijk}+...)$, where $ P$ is the set of all prime numbers.
Clearly the sum in the brackets is independent of $ m$ and generally of anything. There are two options, either it does not exist or it is some constant let's say $ k$. In latter case we have that
$ k(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...)=1$, which brings again to impossible result.

Yes. Exclusiom-Inclusion Principle kills it. :)

Sure, the solution of post 5 is correct?
As there has been working with infinitely many infinite series, it seems there will be trouble with the order of executing that sum.
Also post 3 seems like not concluding anything as $d(m)$ isn't bounded.

[quote]Determine whether there exists a sequence $a_1,a_2,\ldots$ of nonnegative reals such that
\[ a_{n}+a_{2n}+a_{3n}+\cdots=\frac1n\]
for every positive integer $n$. [/quote]
No. Suppose there did. Let $S_k=\{ n\in \mathbb{N}:p_1\mid n, \text{ or } p_2\mid n, \ldots, \text{ or } p_k\mid n\}$ where we order the primes $\{p_1<p_2<\cdots\}$. The idea is to ``interlace'' the sequences for $n=p_1,\ldots,p_k$. Using CRT and Inclusion-Exclusion:
\[ \sum_{n\in S_k} a_n = \sum_{\substack{T\subset S_k \\ T\not=\emptyset}} (-1)^{|T|+1}\prod_{i\in T}\frac{1}{p_i} = 1-\prod_{i=1}^k\left(1-\frac{1}{p_i}\right) \]
by expansion. Using the estimate $\log(1-\varepsilon)\le -0.01\varepsilon$ for $\varepsilon\in (0,1)$, we have
\[ \log \prod_{i=1}^k\left(1-\frac{1}{p_i}\right) = \sum_{i=1}^k \log\left(1-\frac{1}{p_i}\right) \le \sum_{i=1}^k -0.01\cdot \frac{1}{p_i}=-0.01P_k\]
where $P_k:=\sum_{i=1}^k 1/p_i$. Therefore, since $\log$ is an increasing function, we have
\begin{align*}
 \prod_{i=1}^k\left(1-\frac{1}{p_i}\right) \le e^{-0.01P_k} &\implies \sum_{n\in S_k}a_n \ge 1-\frac{1}{\left(e^{0.01} \right)^{P_k}}.
\end{align*}
Hence
\begin{align*}
 1-a_1=\sum_{i\ge 2}a_i \ge \sum_{n\in S_k} a_n \ge  1-\frac{1}{\left(e^{0.01} \right)^{P_k}}
 \implies a_1 \le \frac{1}{\left(e^{0.01} \right)^{P_k}}.
\end{align*}
But it is well-known that $P_k$ grows arbitrarily large varying $k$, so this forces $a_1=0$. Now, fix an $\ell$. Let $b_n=\ell a_{\ell n}$ for each $n$. Then $(b_n)$ is also a valid sequence, since
\[ b_n+b_{2n}+\cdots = \ell a_{\ell n}+\ell a_{2\ell n} + \cdots = \ell \cdot \frac{1}{\ell n} = \frac{1}{n}.\]
Therefore, $b_1=0$, so $\ell a_\ell=0$, i.e. $a_\ell=0$. Therefore, the entire sequence is $0$, contradiction.

[hide=Remarks]
The main idea was the CRT interlacing, which follows easily from experimentation, and works well with PIE. It is intuitive to see why we should have $a_1=0$ from here (this is the only $a_i$ not in any $S_k$), but formalizing it takes some work but it straightforward. Finally, the finish requires almost no work, by choosing an appropriate sequence which must also have first term $0$.
[/hide]

The answer is no. Assume for contradiction such a sequence exists, and let \(P_n\) denote the first \(n\) primes. By the Principle of Inclusion-Exclusion, \[\frac1n-a_n=\lim_{k\to\infty}\sum_{T\subset P_k}\frac{(-1)^{|T|}}{n\prod_{p\in T}p}=\frac1n\lim_{k\to\infty}\sum_{T\subset P_k}\frac{(-1)^{|T|}}{\prod_{p\in T}p}=\frac{1-a_1}n.\] Solving, \(a_n=\tfrac1na_1\) for all \(n\), so \[1=\sum_{k=1}^\infty a_k=\sum_{k=1}^\infty\frac{a_1}k=a_1\sum_{k=1}^\infty\frac1k.\] Since \(\sum_{k=1}^\infty a_k\) diverges, \(a_1=0\), which means that \(a_n=0\) for all \(n\). This yields the desired contradiction.

i think the idea of this problem related to Euler formula in Zeta Function

Indeed.

Note that $a_1 + a_2 + \dots = 1$ and that $a_n \le \frac{1}{n}$. However, we also have that
\[
    \sum_{k}^{N} \mu(k) (a_k + a_{2k} + \dots) = \sum_{k}^N \frac{\mu(k)}{k}
\]
eventually contains all of $a_1$ through $a_N$ and thus has a limit of $1$. This contradicts the fact that $\lim_{s \to 1^+} \frac{1}{\zeta(s)} = \lim_{s \to 1^+} \sum_k \frac{\mu(k)}{k}$ approaches $0$ which implies that this has a limit of $0$.


analytic geometry
geometry
area of a triangle
combinatorics unsolved
combinatorics

In the coordinate plane, color the lattice points which have both coordinates even black and all other lattice points white. Let $ P$ be a polygon with black points as vertices. Prove that any white point on or inside $ P$ lies halfway between two black points, both of which lie on or inside $ P$.

I think I got it  :)

First, use any triangulation of $ P$ to decompose $ P$ into triangles with black vertices and no other black vertex inside or on the boundary of any of these triangles. Thus, we only have to prove the desired result in the case where $ P\equal{}ABC$ is such a triangle.

Let $ A',B',C'$ be the respective midpoints of $ BC,CA,AB$. Note that $ A',B',C'$ are integer white points.

Any other white point in $ P$ must be inside the triangle $ A'B'C'$ : If $ M$ is any white point inside or on the boundary of $ AB'C'$, distinct from $ B'$ and $ C'$, then the symetric of $ A$ with respect to $ M$ is a black point inside or on the boundary of $ P$ which is not one of its vertices. A contradiction. In a same manner, there is no white point inside or on the boundary of $ BC'A'$ or $ CA'B'$.

Now, there are three types of white points, according to the respective parities of their coordinates, namely (e,o),(o,e),(o,o), where e means even and o means odd. Since there is no integer point on the boundary of $ A'B'C'$, these three points must have pairwise distinct types.

Now, assume that there exists a white point, say $ D$, inside $ A'B'C'$.
From the pigeon-hole principle, $ D$ must have the same type than one among $ A',B',C'$, say $ A'$. Then, the midpoint of $ A'D$, say $ E$, is an integer white point.
Then, consider the triangle $ B'C'E$, with $ D$ as an interior point. As above, among these four white points, at least two must have the same type, so that their respective midpoint is an integer white point. It is easy to verify that, in any case, we may strictly reduce (with respect to inclusion) triangle $ B'C'E$ to another non degenerated triangle $ T'$ whose vertices are white points, and $ D$ is inside or on the boundary of $ T$.
Using this procedure, we may construct and infinite sequence of such non-degenerated triangles, which is strictly decreasing for the area. But the area of a triangle has the form $ \frac a 2$ where $ a$ is a positive integer (use Pick's formula). Thus, we have found an infinite decreasing sequence of positive integers. A contradiction.
(Note : another way to look at it is that we found an infinite number of integer points inside $ T$, which is a bounded figure...).

Thus, there is no other white point inside or on the boundary of $ T$ than $ A',B',C'$. But, these white points are clearly midpoints of two black points, and we are done.

Pierre.

As a consequence, and from Pick's formula, we may deduce that any triangle with three black vertices and no other black point inside it or on its boundary must have area equals to $ 2$.

Pierre.

Your statement about area 2 follows just from Pick's theorem, right?  We don't need the problem posted here.

By Pick's theorem, every triangle with lattice points as vertices and no interior or other boundary lattice points has area $ \frac {1}{2}$.  By scaling the $ x$ and $ y$ coordinates by a factor of 2, it follows that every triangle with black points as vertices and no interior or other boundary black points has area $ 2$.

Maybe I misunderstood your last comment.

Well, in fact, you have to prove that if there is no black point in the interior, there is no white point too...

Pierre.

But when we scale the x and y coordinates down by a factor of 2, the white points are no longer lattice points, and so are irrelevant to Pick's theorem.

Sorry if I am writing nonsense.

No problem Ravi B.
The consequence I mentioned is the first way I thought to prove the desired result (after the triangulation). I thought about the scaling idea, but the fact that white points were no more integer points to be considered for Pick's formula did not enter my mind... :blush:

So, I used the above way.
Clearly, the Pick's formula is shorter to solve the problem.

Pierre.


group theory
abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

well-known: if $ L/K$ is a field extension, and $ E,E'$ are intermediate fields such that $ E/K,E'/K$ are finite galois, then also $ E*E'/K$ is finite galois and in the case of $ E \cap E' \equal{} K$ there is a canonical isomorphism $ Gal(E*E'/K) \to Gal(E/K) \times Gal(E'/K)$.

what about a generalization? let's omit $ E \cap E' \equal{} K$, but assume that $ E \cap E' / K$ is galois. then there is a canonical monomorphism

$ Gal(E*E'/K) \to Gal(E/K) \times_{Gal(E \cap E'/K)} Gal(E'/K)$

is this an isomorphism?

it's true for cyclotomic fields:

this comes down to: given positive integers $ n,m$, then $ (\mathbb{Z}/lcm(n,m))^* \to (\mathbb{Z}/n)^* \times_{(\mathbb{Z}/gcd(n,m))^*} (\mathbb{Z}/m)^*$ is an isomorphism. this can be derived from

[b]lemma.[/b] given a ring $ R$ with two ideals $ I,J$, we have $ R/(I \cap J) \cong R/I \times_{R/(I\plus{}J)} R/J$.
[hide="Proof"]There is a canonical map $ R/(I \cap J) \to R/I \times_{R/(I\plus{}J)} R/J$, which is clearly injective. to show surjectivity, choose $ ([r],[s])$ in the target ring, i.e. we have $ r\minus{}s \in I\plus{}J$, say  $ r\minus{}s \equal{} i\plus{}j$. then $ [s\plus{}j] \in R/(I \cap J)$ is a preimage. [/hide]
(I wonder why I can't find this lemma in any introduction to algebra, it seems to be only well-known for $ I\plus{}J\equal{}R$?)

I think it's true :-)

I use the following well-known facts:

(1) $ Gal(E'/K)/Gal(E'/E \cap E') \cong Gal(E \cap E'/K)$
(2) $ Gal(E' / E \cap E') \cong Gal(E*E' / E)$

now we can count the elements in the fiber product:

$ \# Gal(E/K) \times_{Gal(E \cap E'/K)} Gal(E'/K)$
$ \equal{}\sum_{\sigma \in Gal(E/K)} \#\{\tau \in Gal(E'/K) : \tau|_{E \cap E'} \equal{} \sigma|_{E \cap E'} \}$
$ \stackrel{(1)}{\equal{}} \sum_{\sigma \in Gal(E/K)} \# Gal(E'/E \cap E')$
$ \stackrel{(2)}{\equal{}} |E/K| |E*E'/E|\equal{}|E*E'/K| \equal{} \# Gal(E*E'/K)$

thus, the injective map $ Gal(E*E'/K) \to Gal(E/K) \times_{Gal(E \cap E'/K)} Gal(E'/K)$ is an isomorphism!

there is a common(!) generalization of the chinese remainder theorem and the galois group of a compositum. namely, if $ H,K$ are normal subgroups of a group $ G$, then $ G/(H \cap K) \cong G/H \times_{G/(H \cdot K)} G/K$. similarily for submodules of a $ R$-module or ideals of a $ R$-algebra. if we put $ G \equal{} Gal(E E' / K) , H \equal{} Gal(E E' / E), K \equal{} Gal(E E'/E')$, we get the result for the galois groups.


Let $ ABC$ be an acute triangle, points $ D$ and $ E$ lie on segments $ BC$ and $ AD$ respectively, such that:

$ \frac{AE}{ED} \equal{}  \frac{CD}{DB}$

A perpendicular drawn from $ D$ meets segment $ BE$ at $ F$. Suppose $ F$ belongs to segment $ BE$ and quadrilateral $ AFDC$ is cyclic. Prove that $ E$ lies on one of $ ABC$´s altitudes.

[quote="livetolove212"]Please check your problem again![/quote]

why should i check it??

It seems its not clear for some people..

Theres a line drawn from D, which is perpendicular to sement BE, the point of intersection of perpendicular with BE is F.

You can see this figure below. It's not true!

Are you sure you constructed D and E such that:

$ \frac{AE}{ED}\equal{}\frac{CD}{DB}$?

I actually proved E was not on any of ABC´s altitudes, but I still think thats wrong...

Yes.I constructed $ \frac{AE}{ED}\equal{}\frac{CD}{DB}\equal{}2$.


real analysis
geometry
analytic geometry
function
combinatorial geometry
real analysis unsolved

Let K be a compact of $ R^2$ and $ \lambda(K)$ its area (its Lebegue measure).
I suppose that forall $ a,b,c \in K$ we have
\[ \min( \parallel{}a \minus{} b\parallel{},\parallel{}a \minus{} c\parallel{},\parallel{}b \minus{} c\parallel{}) < 1\]
where $ \parallel{}a\parallel{} \equal{} \max(|x_a|,|y_a|)$. Is it true that $ \lambda(K)\leq 2$  ?

Thanks

One should start by proving, or recalling, that the area of a set of diameter $ d$ is at most $ \pi d^2/4$. (Pass to the convex hull and compute in polar coordinates centered on the boundary.)

Let $ d$ be the diameter of the given set $ K$; also, A and B are points that realize the diameter. The part of $ K$ that lies outside of the unIt disk centered at A has diameter at most 1, hence area at most $ \pi/4$. Same goes for the part outside the unit disk centered at B. If $ d>2$, this gives the upper estimate of $ \pi/2$ for the area of $ K$. Otherwise we have the estimate $ \pi/2\plus{}f(d)$, where $ f(d)$ is the area of the intersection of the above-mentioned disks. Clearly, $ f$ is a decreasing function, which can be computed explicitly.

So we have two upper estimates: $ \pi d^2/4$, which is increasing with $ d$, and $ \pi/2\plus{}f(d)$, which is decreasing. At $ d\equal{}\pi/2$ both are least than 2, which yields the result.

It seems that the sharp estimate is $ \pi/2$...

Thanks.

But I think there's a confusion betwenn the norm $ \parallel{}a\parallel{} \equal{} \max(|x_a|,|y_a|)$ and $ \parallel{}a\parallel{}_2 \equal{} \sqrt {x_a^2\plus{}y_a^2}$.
In fact, the maximal value of a compact of diametre $ d$ is $ d^2$, think about a square of side $ d$.

I did not read the questions completely. :blush: Would a similar argument work for your norm?

Very cute problem! :) Where did it come from?

Consider any family of parallel vertical lines spaced by $ 1$ (like the family of vertical lines through the integers on the $ x$-axis, only shifted by some number $ t\in[0,1)$ to the right). It is easy to observe that the total length of the intersection of $ K$ with this family is at most $ 2$ ($ K$ cannot intersect 3 or more lines, if it intersects 2, then the diameter of each intersection is at most $ 1$, and if it intersects just 1, then the cross-section can be covered by 2 intervals of length $ 1$). Integrating over $ t$, we get the desired conclusion.

Of course, the estimate is sharp: the union of any two disjoint coordinate squares with side length $ a < 1$ is an example of such set of area $ 2a^2$.

[color=red]P.S.[/color] Another, even shorter, approach is to notice that the intersection of $ K$ with any lattice with step $ 1$ contains at most 2 points...

So the problem is easier (and nicer) for infinity-norm, but I still wonder if $ \pi/2$ (union of two disks of diameter 1) is extremal for Euclidean norm. If it's not, then the answer is probably complicated and not as interesting.

In general, one can require than between any $ n$ points, at least $ k$ distances are at most 1. Then the question would be: as $ n\to\infty$, how should $ k$ behave to maintain uniform control of area? Here the choice of norm does not matter.

The only source of the problem is me. I imagined this and I thank both of you for your ideas.

what do you think of this generalisation :

let K be a compact of $ R^2$ and forall $ x_1,...,x_n \in K$ there are $ i\neq j$ such that $ \parallel{}x_i \minus{} x_j\parallel{}_\infty < 1$.
Is it true that $ \lambda(K)\leq n \minus{} 1$.

I hope it's true, but I don't see.

Yes, by fedja's remark (P.S. above).
Consider the map of the plane onto the unit square that takes fractional part of each coordinate.
When restricted to the given set, this map is at most (n-1) -to- 1.

[quote="mlok"] I still wonder if $ \pi/2$ (union of two disks of diameter 1) is extremal for Euclidean norm. [/quote]
I'm not sure about "2 out of 3" case but for the "2 out of $ n$" case the sharp estimate has to be close to $ \frac{\sqrt 3}{2}(n\minus{}1)$, which is the lattice bound.

The problem for Euclidean distance was posed by Croft and Falconer in 1981, [url=http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=2082788]link[/url].
Their Theorem 1 gives the $ \pi/2$ bound that I wanted. Then they formulate the general problem $ P^d(n,m)$: maximize the volume of a $ d$-dimensional compact set $ S$ subject to the constraint: among the $ \binom{n}{2}$ distances between any $ n$ points in $ S$, at least $ m$ of them are not greater than $ 1$. But $ P^d(n,m)$ turns out to be the same problem as $ P^d(n\minus{}1,m')$ where $ m'$ is the integer part of $ \frac{(m\minus{}1)(n\minus{}1)}{n\plus{}1}$. So it suffices to consider $ P^d(n)\equal{}P^d(n,1)$. Finally, they study $ P^2(n)$ for large $ n$ and find what [b]fedja[/b] observed above.


analytic geometry
number theory
greatest common divisor
linear algebra
matrix
vector

Let $A = (0, 0, 0)$ in 3D space. Define the [i]weight[/i] of a point as the sum of the absolute values of the coordinates. Call a point a [i]primitive lattice point[/i] if all of its coordinates are integers whose gcd is 1. Let square $ABCD$ be an [i]unbalanced primitive integer square[/i] if it has integer side length and also, $B$ and $D$ are primitive lattice points with different weights. Prove that there are infinitely many unbalanced primitive integer squares such that the planes containing the squares are not parallel to each other.

Consider the square $(0,0,0), (a,b,0), (a,b,c), (0,0,c)$ where (a,b,c) is a primitive pythagorean triple.

Since c is an integer, it has integer sidelengths  $\checkmark$
B, D have different weights $(a+b \neq a+b+c)$ and gcd(a,b,c) = gcd(a,b,0) = 1  $\checkmark$
The squares are not parallel to each other since $b/a$ is different for each different pythag. triple $\checkmark$
There are infinitely many primitive pythagorean triples $\checkmark$

edit:  oops it is wrong lol

is $D$ $(a,b,c)$ or $(0,0,c)$ in your definition of the square?

The last condition of the problem is absolutely weird, is it? All the planes are supposed to contain $A$, how can they be parallel?

To singular: I don't think $\gcd (0,0,c)=1$...

OK, I think this should work. Here's a sketch:

Consider the orthogonal matrix A =

[ 2 -2  1 ]
[ 2  1 -2 ]
[ 1  2  2 ]

Note that det(A) = 27, so for any prime p besides 3, there are no non-trivial solutions to Ax = 0 in mod p. Setting $t_{1}= [2 \ 2 \ 1]^\mathrm{T}, t_{2}= [-2 \ 1 \ 2]^\mathrm{T}$, and working in mod 3, we have that

$At_{1}\equiv [1 \ 1 \ 2]^\mathrm{T}\equiv t_{2}$
$At_{2}\equiv [2 \ 2 \ 1]^\mathrm{T}\equiv t_{1}$.

Thus, if we consider anything of the form $A^{k}t_{1}$ or $A^{k}t_{2}$, it will be a primitive point. Furthermore, since $|Ax| = 3|x|$ for any vector x, and $t_{1}\cdot t_{2}= 0$, we know that we can take $B = A^{k}t_{1}, D = A^{k}t_{2}$ to make a primitive integer square. As k varies, we can show using an eigenvectors argument that none of these squares will share the same plane. Finally, it shouldn't be hard to show that B and D do not always have the same weight.

Here's another solution, more elementary and more motivatable:
First off, if $ B$ is $ (x_{1}, x_{2}, x_{3})$ and $ D$ is $ (y_{1}, y_{2}, y_{3})$, $ AB\equal{}AD$ and equal to some integer is equivalant to the following: $ x_{1}^2 \plus{} x_{2}^2 \plus{} x_{3}^2 \equal{} y_{1}^2 \plus{} y_{2}^2  \plus{} y_{3}^2\equal{}t^2$ (throughout this proof all variables are integers). The fact that $ ABCD$ is a square is equivalant to $ BA$ and $ DA$ are perpendicular, i.e. $ BA^2 \plus{} DA^2 \equal{} BD^2$, which upon expansion is equivalant to: $ x_{1}y_{1} \plus{} x_{2}y_{2} \plus{} x_{3}y_{3} \equal{} 0$ (*). Also, we require $ (x_{1}, x_{2}, x_{3}) \equal{} (y_{1}, y_{2}, y_{3}) \equal{} 1$, $ |x_{1}| \plus{} |x_{2}| \plus{} |x_{3}| \neq |y_{1}| \plus{} |y_{2}| \plus{} |y_{3}|$, and the planes through $ ABCD$ are distinct.
To cheaply kill this problem, just note that the following fulfills all criteria: $ x_{1} \equal{} y_{1} \equal{} 4(d^2\plus{}1)$, $ x_{2}\equal{}\minus{}(d^2\plus{}16d\minus{}1)$, $ x_{3} \equal{} \minus{}(8d^2\minus{}2d\minus{}8)$, $ y_{2} \equal{} 8d^2\plus{}2d\minus{}8$, $ y_{3} \equal{} d^2\minus{}16d\minus{}1$, for some $ d$ even. Here's how we can find such a solution set:
To simplify things, let $ x_{1}\equal{}y_{1}\equal{}a$. For convenience switch the parities of $ x_{2}, x_{3}$, so the condition (*) becomes $ a^2 \equal{} x_{2}y_{2} \plus{} x_{3}y_{3}$.
We require $ x_{2}^2 \plus{} x_{3}^2 \equal{} y_{2}^2 \plus{} y_{3}^2$, $ (x_{2} \minus{} y_{2})(x_{2}\plus{}y_{2}) \equal{} (y_{3} \minus{} x_{3})(y_{3} \plus{} x_{3})$. To satisfy this, let $ x_{2} \minus{} y_{2} \equal{} 2tu, x_{2}\plus{}y_{2} \equal{} 2vw, y_{3} \minus{} x_{3} \equal{} 2tv, y_{3}\plus{}x_{3} \equal{} 2uw$, so $ x_{2} \equal{} tu\plus{}vw, y_{2} \equal{} vw\minus{}tu, x_{3} \equal{} uw\minus{}tv, y_{3} \equal{} tv\plus{}uw$.
We now have $ a^2 \equal{} x_{2}y_{2} \plus{} x_{3}y_{3} \equal{} (tu\plus{}vw)(vw\minus{}tu)\plus{}(tv\plus{}uw)(uw\minus{}tv) \equal{} (w^2\minus{}t^2)(u^2\plus{}v^2)$. We also require $ t^2 \equal{} a^2 \plus{} x_{2}^2 \plus{} x_{3}^2 \equal{} (w^2\minus{}t^2)(u^2\plus{}v^2) \plus{} (tu\plus{}vw)^2 \plus{} (uw\minus{}tv)^2 \equal{} (w^2\minus{}t^2)(u^2\plus{}v^2)\plus{}(t^2\plus{}w^2)(u^2\plus{}v^2) \equal{} 2w^2(u^2\plus{}v^2)$.
For this to occur, we then need $ u^2\plus{}v^2 \equal{} 2l^2$, so then $ t\equal{}2wl$. Since we also require $ a^2 \equal{} (w^2\minus{}t^2)(u^2\plus{}v^2) \equal{} 2l^2(w^2\minus{}t^2)$, we also require $ w^2\minus{}t^2 \equal{} 2k^2$, and $ a\equal{}2kl$.
If $ w^2\minus{}t^2 \equal{} (w\minus{}t)(w\plus{}t) \equal{} 2k^2$, let $ w\minus{}t \equal{} 2f^2, w\plus{}t \equal{} 4g^2$, so $ w\equal{}f^2 \plus{} 2g^2$ and $ t\equal{}2g^2\minus{}f^2$.
For $ u^2\plus{}v^2 \equal{} 2l^2$, we require $ (v\minus{}l)(v\plus{}l) \equal{} (l\minus{}u)(l\plus{}u)$, so let $ v\minus{}l \equal{} 2cd, v\plus{}l \equal{} 2ef, l\minus{}u \equal{} 2ce, l\plus{}u \equal{} 2df$, so $ l \equal{} ef\minus{}cd \equal{} ce\plus{}df$, so $ f(e\minus{}d) \equal{} c(d\plus{}e)$, so let $ c \equal{} e\minus{}d, f\equal{}e\plus{}d$. Substituting back in, gives us $ l\equal{}d^2\plus{}e^2, u\equal{}e^2\plus{}2ed\minus{}d^2, v\equal{}d^2\plus{}2ed\minus{}e^2$.
Now we have constructed satisfactory numbers $ x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, y_{3}$ satisfying the main conditions. It suffices to check that they satisfy the other conditions. So substituting everything back in:
$ k\equal{}2fg, w\equal{}f^2\plus{}2g^2, t\equal{}2g^2\minus{}f^2, u\equal{}e^2\plus{}2ed\minus{}d^2, v\equal{}d^2\plus{}2ed\minus{}e^2, l\equal{}d^2\plus{}e^2, a\equal{}2kl$
We now have:
$ x_{1} \equal{} y_{1} \equal{} 4fg(d^2\plus{}e^2)$
$ x_{2} \equal{} tu\plus{}vw \equal{} (2g^2\minus{}f^2)(e^2\plus{}2ed\minus{}d^2) \plus{} (d^2\plus{}2ed\minus{}e^2)(f^2\plus{}2g^2)$
$ x_{3} \equal{} uw\minus{}tv \equal{} (f^2\plus{}2g^2)(e^2\plus{}2ed\minus{}d^2) \minus{} (d^2\plus{}2ed\minus{}e^2)(2g^2\minus{}f^2)$
$ y_{2} \equal{} vw\minus{}tu \equal{} (f^2\plus{}2g^2)(d^2\plus{}2ed\minus{}e^2) \minus{} (e^2\plus{}2ed\minus{}d^2)(2g^2\minus{}f^2)$
$ y_{3} \equal{} tv\plus{}uw \equal{} (2g^2\minus{}f^2)(d^2\plus{}2ed\minus{}e^2) \plus{} (e^2 \plus{} 2ed \minus{} d^2)(f^2 \plus{} 2g^2)$.
To obtain the parametrisation shown above, simply plug in $ f\equal{}1, g\equal{}2, e\equal{}1$, and reverse the signs of $ x_{2}$ and $ x_{3}$ back.

Set $ B \equal{} (1, 2n, 2n^2)$ and $ D \equal{} (2n, 2n^2\minus{}1, \minus{}2n)$, and C the 4th vertex of the square. It can be verified directly that the distances $ AB, AD$ equal $ 2n^2 \plus{}1$. $ AB, AD$ are perpendicular since the dot product of the position vectors is 0.

[quote="silversheep"]Set $ B \equal{} (1, 2n, 2n^2)$ and $ D \equal{} (2n, 2n^2\minus{}1, \minus{}2n)$, and C the 4th vertex of the square.[/quote]
Motivation: We parameterize solutions $(u,v,w)$ to $xu+yv+zw=0$ as $(ty-sz,rz-tx,sx-ry)$ for rational $r,s,t$.

For simplicity, take $r=0$: then we need $x^2+y^2+z^2 = (ty-sz)^2+t^2x^2+s^2x^2\implies x^2 = \frac{y^2+z^2-(ty-sz)^2}{s^2+t^2-1}$. But any $x,y,z,s,t$ satisfying this lead to $x^2+y^2+z^2 = \frac{(s^2+t^2)(y^2+z^2)-(ty-sz)^2}{s^2+t^2-1} = \frac{(sy+tz)^2}{s^2+t^2-1}$, so in general it's enough to take $s^2+t^2-1$ a perfect square. The simplest is just $s=t=1$; then we want $x^2=2yz$, so we take $(x,y,z)=(2\alpha\beta,2\alpha^2,\beta^2)$ and $(u,v,w)=(2\alpha^2-\beta^2,-2\alpha\beta,2\alpha\beta)$ (with $\gcd(2\alpha,\beta)=1$). Then $\beta=1$ gives your solutions.

(The cross product of these two vectors is $(2\alpha^2+\beta^2)\langle{2\alpha\beta,-\beta^2,-2\alpha^2}\rangle$.)


Triangle ABC is acute -angled.

Prove that if  (CosA)^2.CosB.SinB = (CosB)^2.CosC.SinC = (CosC)^2.CosA.SinA, then the triangle is equilateral.

I hope someone solves this problem .Five questions on trigonometry that I have submitted on this forum have gone unanswered. This is disappointing.

[quote="Michael Niland"]Triangle ABC is acute -angled.

Prove that if  (CosA)^2.CosB.SinB = (CosB)^2.CosC.SinC = (CosC)^2.CosA.SinA, then the triangle is equilateral.[/quote]

$\cos^2 A \cos B \sin B = \cos^2 B \cos C \sin C = \cos^2 C \cos A \sin A$

$\Rightarrow \frac{\cos^2 A \cos B \sin B}{\cos A \cos B \cos C} = \frac{\cos^2 B \cos C \sin C}{\cos A \cos B \cos C} = \frac{\cos^2 C \cos A \sin A}{\cos A \cos B \cos C}$

***We can divide by $\cos A \cos B \cos C$ because the cosine of none of $A,B$ or $C$ can be $0$ in the defined interval -- $(0,90)$ of an acute-angled triangle.

$\Rightarrow \frac{\cos A \sin B}{\cos C} = \frac{\cos B \sin C}{\cos A} = \frac{\cos C \sin A}{\cos B}$

Now, for arbitrary $x,y,z$ such that $x + y + z = \pi$, we have:

$\frac{\cos x \sin y}{\cos z} = \frac{\cos x \sin y}{\cos (\pi - (x+y))}$

$\rightarrow \cos(\pi - k) = \cos \pi \cos (-k) - \sin \pi \sin (-k)$

$= (-1)(\cos k) - (0)(-\sin k) = -\cos k$

$\Rightarrow \frac{\cos x \sin y}{\cos (\pi - (x+y)} = \frac{\cos x \sin y}{-\cos (x+y)} = \frac{\cos x \sin y}{\sin x \sin y - \cos x \cos y}$

$=  \frac{1}{\frac{\sin x \sin y}{\cos x \sin y} - \frac{\cos x \cos y}{\cos x \sin y}} =  \frac{1}{\frac{\sin x}{\cos x} - \frac{\cos y}{\sin y}} = \frac{1}{\tan x - \cot y}$

***Once again, bringing the cosines and sines to the denominator to make tangent and cotangent is fine because we can't divide by 0 due to the fact that $\triangle ABC$ is acute-angles as stated.

$\Rightarrow \frac{\cos A \sin B}{\cos C} = \frac{1}{\tan A - \cot B}$

etc.

$\Rightarrow \frac{1}{\tan A - \cot B} = \frac{1}{\tan B - \cot C} = \frac{1}{\tan C - \cot A}$

$\Rightarrow \tan A - \cot B = \tan B - \cot C = \tan C - \cot A$

Say: $\tan A = a, \tan B = b$ and $\tan C = c$.
We get: $a - \frac{1}{b} = b - \frac{1}{c} = c - \frac{1}{a}$

$a - \frac{1}{b} = b - \frac{1}{c} \Rightarrow \frac{1}{c} = b + \frac{1}{b} - a$

$a - \frac{1}{b} = c - \frac{1}{a} \Rightarrow c = a + \frac{1}{a} - \frac{1}{b}$

$\Rightarrow 1 = (c)\left(\frac{1}{c}\right) = \left(a + \frac{1}{a} - \frac{1}{b}\right)\left(b + \frac{1}{b} - a\right)$

$= ab + \frac{b}{a} - 1 + \frac{a}{b} + \frac{1}{ab} - \frac{1}{b^2} - a^2 - 1 + \frac{a}{b}$

$\Rightarrow ab + \frac{b}{a} + \frac{2a}{b} + \frac{1}{ab} - \frac{1}{b^2} - a^2 = 3$

$\Rightarrow 3ab^2 = ab^2\left(ab + \frac{b}{a} + \frac{2a}{b} + \frac{1}{ab} - \frac{1}{b^2} - a^2\right)$

$= a^2 b^3 + b^3 + 2a^2b + b - a - a^3b^2$

$\Rightarrow a^2 b^3 - a^3b^2 + b - a + b^3 - a^2b = 3ab^2 - 3a^2b$

$\Rightarrow a^2b^2(b - a) + (b - a) + b(b^2 - a^2) = 3ab(b - a)$

$\Rightarrow (b - a)(a^2b^2 + 1 + b(b + a) - 3ab) = 0$

$\rightarrow a^2b^2 + 1 + b(b + a) - 3ab = a^2b^2 + 1 + b^2 + ab - 3ab$

$= a^2b^2 -2ab + 1 + b^2 = (ab - 1)^2 + (b)^2$

So, in order for $(b - a)(a^2b^2 + 1 + b(b + a) - 3ab) = 0$ to be true, one of the factors must be zero. If $b-a = 0$ we know $b = a$.

If the other factor, $a^2b^2 + 1 + b(b + a) - 3ab = (ab - 1)^2 + (b)^2$, is equal to $0$, we know $(ab - 1)$ and $b$ must also be equal to $0$ because any number squared is positive or zero, so the sum of 2 squares is the sum of 2 non-negatives. In order for that sum to be 0, both must be 0, because a positive would need a negative to offset it and there is no negative value produced when reals are squared. So, as we said, $ab - 1 = 0$ and $b = 0$, but $b = 0 \Rightarrow ab - 1 = a(0) - 1 = -1$, so $ab - 1 \neq 0$ and the factor $a^2b^2 + 1 + b(b + a) - 3ab$ can never be $0$, thus the other factor must ALWAYS be $0$ which means $b$ ALWAYS equals $a$.

From our equalities with tangents, we found $c = a + \frac{1}{a} - \frac{1}{b}$, but since we now know that $a = b$, we get $c = a + \frac{1}{a} - \frac{1}{a} = a$ which completes $a = b = c$. So, back-substituting, we have $\tan A = \tan B = \tan C$.

Since the problem states $\triangle ABC$ is an acute-angled triangle, $0 < A,B,C < 90$. Tangent is positive and one-to-one on this interval, which tells us if $\tan x = \tan y$ on said interval, then $x = y$. From this and $\tan A = \tan B = \tan C$ we get $A = B = C$ which is an equilateral triangle as desired.

Could anyone else offer a maybe easier or more insightful proof. This question bothered me when posted and still does. I don't really like my method. There has to be an easier way.

If a-1/b =b-1/c =c-1/a we wish to prove a=b=c

a-1/b=b-1/c leads to (ab-1)/b =(bc-1)/c which gives

 abc =(b^2)c -b +c. Similarly by symmetry we have

 abc =(c^2)a -c +a  and

 abc=(a^2)b -a + b.

Adding these three equations we get

(a^2)b + (b^2)c +(c^2)a  =3abc. *

  But the arithmetic-geometric mean inequality states  that (a^2)b +(b^2)c +(c^2)a >= 3abc with equality iff a=b=c

Equation * is precisely the case of equality in this inequality so we conclude a=b=c

Thanks Daniel for solving this question


Can anyone help me about canada universities??
What is the best university in mathematics in canada?
any information about this topic may heplful. please send me .
Thank you.

Saman Gharib

The best two are the University of Toronto and the University of Waterloo.  I would give Toronto the edge for pure math (analysis, algebra, differential equations, differential geometry) and Waterloo the (big) edge for combinatorics and computational mathematics (and for CS, for that matter).  Waterloo has slightly better co-ops.

Moving more east, McGill is a good school, although I confess I don't know a lot about it.  To the west, Simon Fraser University and the University of British Columbia are good schools for mathematics.

[quote="samangharib"]Why anybody just answer this question completely?[/quote]

Because nobody likes you for your too clever question :lol:

[quote="samangharib"]Why anybody just answer this question completely?[/quote]

Post #2 is a fairly complete answer to the original question. Perhaps other people familiar with Canadian universities are still to notice this thread and express their opinions.

For an applicant from Iran, which countries are the easiest countries outside of Iran to which to apply for university studies? How do visa policies and the level of expense in different countries affect the choices of Iranian applicants?

Any new suggestion about canada univ?
thank you

Saman Gharib


anybody here interested in films, shor films and film making?
anybody can recommend me good directors to search for good films..
my favorite directors are: Ingman Berman, Tarkovsky, Sakurov, Ceilan, Sally Potter and many more..
lately i've seen some great short films in film festivals but dunno how to get them...

i am also interested in photography too.. so anybody that likes it just tellme to chat about it

[quote="manuel"]anybody here interested in films, shor films and film making?
anybody can recommend me good directors to search for good films.[/quote]

Definitely look for films by Akira Kurosawa. His most famous film is Seven Samurai, but my favorite is High and Low.

I didn't see Francois Truffaut on your list. He is quite a famous director.

I took a course in film-making in high school in the 1970s, and made a short 8mm film with my best friend.

I saw Ikiru by Kurosawa, it was good, and yes Truffaut is great too. I'll look for high and low.
I just saw to excelent films, the first one is Russian Ark. it is great it was shut in a single shut without cuts! and the other one is YES directed by sally potter, excellent film too.
annd of course I also like Luis Buñuel, and Fellini.. classics..

any other good film you recommend me to see?

i think Alfred hichkad (if i type wrong excuse) is one of the best.
but i think you should see $A$ $Walk$ $to$ $remember$,in it very Romantic,but it doesn't have famous director or ... ?
see it.

The spelling is Alfred Hitchcock, and, yes, he was a very famous director with a good eye for cinematography.

I like the cinematography in Orson Welles's movie Citizen Kane, which is definitely one of the ten best movies of all time.

first thank you.i am agree with you i think Alfred Hitchcock is one of the best and his movies are great.but unfortunately i didn't see any movie of Orson Welles,but if i can i will see. :D

Jensen you seem like a real movie fanatic

However more and more I am turning my back to hollywood

Try these :
The idiots  : extremely graphic for overal rudeness and "extreme sensuality", i only found it good because of its painful harshness
Festen, also a dogma film : really hard....

Cure (kurosawa) : japanese weird horror movie

Uninvited : south korean horror movie, really thought provoking

Tesis : Spanish horror movie about snuff movies : it is really okay what concerns the violence, even I could stand it, fear is induced in a much more chilling way

Le fabuleux destin d'amelie poulain : nice french movie about a young woman trying to make other people happy : just so original

Existenz : Canadian movie about virtual reality games : really weird

Lola Rennt : already famous german rollercoaster movie, the whole movie is a really nice puzzle

Take care of my cat : south korean movie, just nice

Hero : Chinese sword fighting movie, but with an interesting storytelling technique, not so easy, thought provoking

Do you also like certain tv series?  I like the shield, 24, simpsons , most of the american series that have  required some thought by the creators (although the simpsons has been producing some worthless stuff over the years now...)

see the phantom of the opera; i do not remeber the director but it is just perfect!


In one of the adjoining figures a square of side 2 is dissected into four pieces so that $E$ and $F$ are the midpoints of opposite sides and $AG$ is perpendicular to $BF$. These four pieces can then be reassembled into a rectangle as shown in the second figure. The ratio of height to base, $XY/YZ$, in this rectangle is

$\text{(A)} \ 4 \qquad \text{(B)} \ 1+2 \sqrt 3 \qquad \text{(C)} \ 2 \sqrt 5 \qquad \text{(D)} \ (8 + 4 \sqrt 3)/3 \qquad \text{(E)} \ 5$

[hide="Answer"]$EC=1$ and $DC=2$, so $DE=\sqrt{5}$. We can see that the height of the rectangle must be $2\sqrt{5}$. The width is length $AG$. $\triangle ABF$ has $b\times h=1\cdot 2$, and this also equals $\sqrt{5}\cdot AG\Rightarrow AG=\frac{2}{\sqrt{5}}$. Therefore, we have $\frac{2\sqrt{5}}{\frac{2}{\sqrt{5}}}=5\Rightarrow \boxed{E}$.[/hide]


Let a,b,c,d be real numbers such that (1+a^2)(1+b^2)(1+c^2)(1+d^2)=16.Prove that:
                                        -3<=ab+bc+cd+da+ac+bd-abcd<=5

Could help perhaps:

Since $(1+a^2)$ and the others are all positive, we can apply AM-GM:
\[ \frac{4+a^2+b^2+c^2+d^2}{4} \geq \sqrt[4]{(1+a^2)(1+b^2)(1+c^2)(1+d^2)} \]
So we get \[ a^2+b^2+c^2+d^2 \geq 4 \]

So we have $-2 \leq a,b,c,d \leq 2$

$(a+b+c+d)^2 \geq 0$, so we get $ab+ac+bc+ad+bd+cd \geq -4$

Sorry if it's useless

Jan

can you post the detail solution ,Jan.I don't see the right ine.Anyone has the nice solution ????Plese help me

It's a problem of Tutu Andresscu and Gabriel Dospoinesscu ( Harazi). I had post in this forum the short and nice solution by C-S inequality. Try to seach.


from what I know, I think changing it back to base 10, calculate, then change back to base 6 will be the easiest.

If you were dividing 243 base 10 by 9, what would you do?

Treat 24 as 2*10+4 = 24 base 10 (duh.).  Divide and get 2r6.  Then treat 63 (6 from the remainder above) as 6*10+3 = 63 base 10.  Divide and get 7.
The answer is 27.

So do the same with this one, except replace 243 by 345, 9 by 5, and 10 by 6.

[quote="mihail911"]im doing intro to NT and right now im having trouble with division of integers in bases other than the decimal.

e.g. $ 345_6 \div 5_6$

can somebody explain this process to me?[/quote]

[code]
 __43_R2__
5|345.000
  -32
   ---
     25
    -23
    ---
       2
[/code]

So $ 345_6/5_6 \equal{} 43_6 \text{ R}2 \equal{} 43_6\frac {2_6}{5_6}\equal{} 43.\overline{2}_6$


Let ABC be a triangle in which BE and CF are the angle bisectors of the interior angles at B and C. X is arbitrary point on EF. By d(Y,ZT) we denote the distance from point Y to line ZT. Prove that d(X,BC)=d(X,CA)+d(X,AB)

see http://www.mathlinks.ro/Forum/viewtopic.php?t=78547 .

  darij


induction
number theory
number theory unsolved

I don't know if this is Number Theory, but...

there is a sequence defined as following.,
$ i)a_1\equal{}1$
$ ii)$when $ n$ is a natural number, $ a_{2n\plus{}1}\equal{}a_{2n}\plus{}1$, $ a_{2n}\equal{}3a_n$

how many numbers in this sequence is smaller than 2007?

[quote="bjh7790"]I don't know if this is Number Theory, but...

there is a sequence defined as following.,
$ i)a_1 \equal{} 1$
$ ii)$when $ n$ is a natural number, $ a_{2n \plus{} 1} \equal{} a_{2n} \plus{} 1$, $ a_{2n} \equal{} 3a_n$

how many numbers in this sequence is smaller than 2007?[/quote]

It's rather easy to see that if $ n \equal{} \sum 2^k$ (all $ k$ differents), then $ a_n \equal{} \sum 3^k$

So $ a_n$ is the number whose representation in base 3 is the binary representation of n.

Example : $ n \equal{} 13 \equal{} 1101_2$ then $ a_n \equal{} 1101_3 \equal{} 37$

$ 2007 \equal{} 2202100_3$

So the greatest $ a_n$ below $ 2007$ is $ 1111111_3 \equal{} a_{127}$

So, there are 127 $ a_i$ below $ 2007$

[quote="dgreenb801"][quote="pco"]
It's rather easy to see that if $ n \equal{} \sum 2^k$ (all $ k$ differents), then $ a_n \equal{} \sum 3^k$
[/quote]
Can you explain this?[/quote]

Sure : it's easy to show thru induction on the number of elements in the sum:

It's true with one element : $ a_{2^k}\equal{}3^ka_1\equal{}3^k$

Now consider that it's true for any sequence up to $ p$ elements, that's to say : If $ n\equal{}\sum_{i\equal{}1}^{p}2^{k_i}$ with $ k_i<k_{i\plus{}1}$ then $ a_n\equal{}\sum_{i\equal{}1}^{p}3^{k_i}$

Let then $ n\equal{}\sum_{i\equal{}1}^{p\plus{}1}2^{k_i}$ with $ k_i<k_{i\plus{}1}$. Then $ n\equal{}2^{k_1}(1\plus{}\sum_{i\equal{}2}^{p\plus{}1}2^{k_i\minus{}k_1})$ with $ k_i\minus{}k_1<k_{i\plus{}1}\minus{}k_1$.

We have $ n\equal{}2^{a_1}(1\plus{}t)$ with $ t$ even (since $ k_2>k_1$) and so : $ a_n\equal{}3^{k_1}a_{1\plus{}t}\equal{}3^{k_1}(1\plus{}a_t)$

But, since $ t\equal{}\sum_{i\equal{}2}^{p\plus{}1}2^{k_i\minus{}k_1}$ with $ k_i\minus{}k_1<k_{i\plus{}1}\minus{}k_1$, we have $ a_t\equal{}\sum_{i\equal{}2}^{p\plus{}1}3^{k_i\minus{}k_1}$ (induction hypothesis)

And so $ a_n\equal{}3^{k_1}(1\plus{}a_t)$ $ \equal{}3^{k_1}(1\plus{}\sum_{i\equal{}2}^{p\plus{}1}3^{k_i\minus{}k_1})$ $ \equal{}\sum_{i\equal{}1}^{p\plus{}1}3^{k_i}$

Hence the result.


geometry
ratio
LaTeX
trigonometry
geometric transformation
reflection
geometry unsolved

problem see: http://web.mit.edu/hmmt/www/datafiles/problems/2008/problems-geometry-2008.pdf
No.10

There is a solution in that website, but I can not understand it from the beginning, such as what is "harmonic" or "pole". Any geometry masters in this site can help me with that?

[quote="greentreeroad"]problem see: [url]http://web.mit.edu/hmmt/www/datafiles/problems/2008/problems-geometry-2008.pdf[/url], No.10[/quote]
Let $ ABC$ be a triangle with $ BC \equal{} 2007, CA \equal{} 2006, AB \equal{} 2008.$ Let $ \omega$ be an excircle of $ ABC$ that touches the line segment $ BC$ at $ D,$ and touches the extensions of lines $ AC$ and $ AB$ at $ E$ and $ F,$ respectively (so that $ C$ lies on segment $ AE$ and $ B$ on segment $ AF$). Let $ O$ be the center of $ \omega.$ Let $ l$ be the line through $ O$ perpendiucular to $ AD.$ Let $ l$ meet line $ EF$ at $ G.$ Compute the length $ DG.$

[quote="greentreeroad"]There is a solution in that website, but I can not understand it from the beginning, such as what is "harmonic" or "pole". Any geometry masters in this site can help me with that?[/quote]

If a line $ l$ intersects a circle $ \omega$ at $ B, C,$ the intersection $ A$ of tangents to $ \omega$ at $ B, C$ is the pole of $ l$ with respect to the circle $ \omega.$ Conversely, $ l$ is the polar of the point $ A.$

If the line $ l$ passes through the center $ O$ of $ \omega,$ the tangents at $ B, C$ are parallel and the pole of $ l$ is a point at infinity, identical with the direction perpendicular to the line $ l$.

If the line $ l$ touches $ \omega$ at $ A,$ the tangency point is the pole of $ l$ and the tangent $ l$ is the polar of $ A \in \omega.$

If the line $ l$ does not intersect $ \omega,$ drop a perpendicular from $ O$ to $ l,$ intersecting $ l$ at $ M.$ Tangents from $ M$ touch $ \omega$ at $ B, C$ and the midpoint $ A$ of $ BC$ is the pole of $ l.$ Conversely, $ l$ is the polar of $ A.$
________________________________

Basic theorems about poles and polars:

1. If lines $ l_1, l_2, l_3$ are concurrent at $ P$, theirs poles $ A_1, A_2, A_3$ are collinear. $ P$ is the pole of the line $ A_1A_2A_3.$ Conversely, if the points $ A_1, A_2, A_3$ are collinear, their polars $ l_1, l_2, l_3$ are concurrent at a point $ P.$ $ A_1A_2A_3$ is polar of the point $ P.$

2. If lines $ l_1, l_2$ are parallel, their poles $ A_1, A_2$ are collinear with the center $ O$ of the circle $ \omega.$ Conversely, if the points $ A_1, A_2$ are collinear with the center $ O$ of $ \omega,$ their polars $ l_1, l_2$ are parallel.

3. If $ A_1, A_2$ are poles of $ l_1, l_2$ (same as if $ l_1, l_2$ are polars of $ A_1, A_2$), the angles $ \angle A_1OA_2 \equal{} \widehat{(l_1,l_2)}$ are equal modulo $ 180^\circ.$
_________________________________

In the problem, point $ A$ is the pole of the line $ EF$ with respect to the circle $ \omega,$ point $ D$ is the pole of the line $ BC,$ and the pole of the line $ l$ through $ O$ and perpendicular to $ AD$ is a point $ L$ at infinity, identical with direction perpendicular to $ l,$ parallel to $ AD.$ The poles $ D, A, L$ are collinear, therefore the lines $ BC, EF, l$ are concurrent at $ G.$ By Menelaus theorem for the $ \triangle ABC$ cut by the transversal $ EFG,$

$ \frac {\overline{EC}}{\overline{EA}} \cdot \frac {\overline{FA}}{\overline{FB}} \cdot \frac {\overline{GB}}{\overline{GC}} \equal{} 1$

$ \frac {\overline{GB}}{\overline{GC}} \equal{} \frac {\overline{EA}}{\overline{EC}} \cdot \frac {\overline{FB}}{\overline{FA}} \equal{} \frac {s}{s \minus{} b} \cdot \frac {s \minus{} c}{s} \equal{} \frac {s \minus{} c}{s \minus{} b} \equal{} \minus{} \frac {\overline{DB}}{\overline{DC}}$

The cross-ratio $ G(G, D, B, C) \equal{} \frac {\overline{GB}}{\overline{GC}} \cdot \frac {\overline{DC}}{\overline{DB}} \equal{} \minus{} 1$ for directed line segments is called harmonic, when equal to $ \minus{} 1.$ Since the segment lengths $ \overline{BD} \equal{} \minus{} \overline{DB} \equal{} s \minus{} c,$ $ \overline{DC} \equal{} s \minus{} b,$ $ \overline{BC} \equal{} a$ are known, it allows to calculate the segment length $ \overline{GD}.$

$ \frac {\overline{GB}}{\overline{GB} \plus{} a} \equal{} \frac {s \minus{} c}{s \minus{} b}\ \Longrightarrow\ \overline{GB} \equal{} \frac {a(s \minus{} c)}{b \plus{} c}.$

$ \overline{GD} \equal{} \overline{GB} \plus{} \overline{BD} \equal{} \frac {a(s \minus{} c)}{b \plus{} c} \plus{} s \minus{} c \equal{} \frac {2s(s \minus{} c)}{b \plus{} c}.$
__________________________________

As for your past questions about LATEX: You do not have to download anything. Just type equations, in LATEX code, sandwiched between dollar signs. For example,

$ \$$\frac{a^2}{b^2} = \frac{\sin A}{\sin B}$ \$$ becomes $ \frac {a^2}{b^2} \equal{} \frac {\sin A}{\sin B}$

$ \$$A_1 \in (O)$ \$$ becomes $ A_1 \in (O)$

$ \$$X = AB \cap CD$ \$$ becomes $ X \equal{} AB \cap CD$

$ \$$\angle ABC = \widehat{ABC} \neq 90^\circ$ \$$ becomes $ \angle ABC \equal{} \widehat{ABC} \neq 90^\circ$

$ \$$\triangle ABC \sim \triangle A'B'C' \cong \triangle A''B''C''$ \$$ becomes $ \triangle ABC \sim \triangle A'B'C' \cong \triangle A''B''C''$

See the LATEX help for more codes.

btw, how can we type blank space in Latex code without repeatedly using dollar sign. (I just type blank spaces between but they all disappears) :maybe:

like $ 2008 Chinese Math Olympiad$

$ \$$2008\ Chinese\ Math\ Olympiad$ \$$ = $ 2008\ Chinese\ Math\ Olympiad$ (LATEX code for blank is backslash-blank)
$ \$$\text{2008 Chinese Math Olympiad}$ \$$ = $ \text{2008 Chinese Math Olympiad}$

Also, try the Test Forum.

Some more questions, thanks :lol:

1.how to prove the first basic theorem for pole and polar (the collinear and concurrent one)?
2.I still didn't quite get the definition for "harmonic", what is cross ration, and what is the order?
3.I've heard that Brianchon's theorem about convex hexagon can be proved using pole and polar or radius axis. How does that proceed?

1. There are many different proofs. For example, let $ l$ be a polar of $ A$ with respect to a circle $ (O, r)$ and $ A'$ the feet of a perpendicular from $ O$ to $ l.$ Then $ A'$ is the image of $ A$ in an inversion with center $ O$ and power $ r^2$ (inversion in the circle $ (O)$). Therefore, the image of $ l$ in this inversion is a circle $ (L)$ with diameter $ OA.$ Let $ l_1, l_2, l_3$ be polars of collinear points $ A_1, A_2, A_3.$ The images of $ l_1, l_2, l_3$ are circles $ (L_1), (L_2), (L_3)$ with diameters $ OA_1, OA_2, OA_3.$ They all pass through $ O$ and their centers $ L_1, L_2, L_3$ are also collinear $ \Longrightarrow$ the circles $ (L_1), (L_2), (L_3)$ intersect at $ O$ and at the reflection $ P$ of $ O$ in the line $ L_1L_2L_3$ $ \Longrightarrow$ the lines $ l_1, l_2, l_3$ intersect at the image $ P'$ of $ P.$ Obviously, $ P \in A_1A_2A_3$ and $ P'$ is the pole of $ A_1A_2A_3.$

2. If $ A, B, C, D$ are any four points on a line, the cross ratio of directed line segments $ G(A, B, C, D) \equal{} \frac {\frac {\overline{AC}}{\overline{AD}}}{\frac {\overline{BC}}{\overline{BD}}} \equal{} \frac {\overline{AC}}{\overline{AD}} \cdot \frac {\overline{BD}}{\overline{BC}}.$ This is different from the cross ratio $ (A, C, B, D) \equal{} \frac {\frac {\overline{AB}}{\overline{AD}}}{\frac {\overline{CB}}{\overline{CD}}} \equal{} \frac {\overline{AB}}{\overline{AD}} \cdot \frac {\overline{CD}}{\overline{CB}},$ etc. The cross ratio $ (A, B, C, D)$ is harmonic, if equal to $ \minus{} 1.$ In addition to $ \frac {|AC|}{|AD|} \equal{} \frac {|BC|}{|BD|} \equal{} 1,$ this means that either $ B$ is between $ C, D$ and $ A$ is not, or $ A$ is between $ C, D$ and $ B$ is not, otherwise the cross ration would not be negative.

3. Again, there are different proofs of Brianchon theorem. For example, given a hexagon $ ABCDEF$ tangential to a circle $ (O)$ at $ P \in AB, Q \in BC, R \in CD, S \in DE, T \in EF, U \in FA,$ then $ UP, PQ, QR, RS, ST, TU$ are polars of $ A, B, C, D, E, F$ $ \Longrightarrow$ $ AD, BE, CF$ are polars of the intersections $ X \equiv UP \cap RS, Y \equiv PQ \cap ST, Z \equiv QR \cap TU.$ The points $ X, Y, Z$ are collinear by Pascal theorem $ \Longrightarrow$ their polars $ AD, BE, CF$ are concurrent at a point $ W,$ the pole of the line $ XYZ.$


calculus
integration
function
calculus computations

This was the first thing that came to mind.  I suppose there is a shorter solution that does not appeal to beta functions?

[hide]For positive integers $ k$, let

$ N(k) \equal{} \int_{x\equal{}0}^1 x^k(1\minus{}x)^k \, dx$, $ D(k) \equal{} \int_{x\equal{}0}^1 x^k(1\minus{}x^{2k\plus{}2})^k \, dx$.

With the substitution $ t \equal{} x^{k\plus{}1}$, $ dt \equal{} (k\plus{}1)x^k \, dx$, the second integral becomes

$ D(k) \equal{} \frac{1}{k\plus{}1} \int_{t\equal{}0}^1 (1\minus{}t^2)^k \, dt \equal{} \frac{1}{k\plus{}1} \int_{t\equal{}0}^1 (1\minus{}t)^k (1\plus{}t)^k \, dt$.

Note that both of these integrals have the form

$ S_a(k,m) \equal{} \int_{x\equal{}0}^1 (x\plus{}a)^{k\minus{}m} (1\minus{}x)^{k\plus{}m} \, dx$,

where $ a$ is a constant.  Integration by parts with the choice

$ u \equal{} (x\plus{}a)^{k\minus{}m}$, $ du \equal{} (k\minus{}m)(x\plus{}a)^{k\minus{}m\minus{}1} \, dx$, $ dv \equal{} (1\minus{}x)^{k\plus{}m} \, dx$, $ v \equal{} \minus{}\frac{(1\minus{}x)^{k\plus{}m\plus{}1}}{k\plus{}m\plus{}1}$

yields the recursion relation

$ S_a(k,m) \equal{} \left[\minus{}\frac{(x\plus{}a)^{k\minus{}m} (1\minus{}x)^{k\plus{}m\plus{}1}}{k\plus{}m\plus{}1}\right]_{x\equal{}0}^1 \plus{} \frac{k\minus{}m}{k\plus{}m\plus{}1} S_a(k,m\plus{}1)$

$ \equal{}\frac{a^{k\minus{}m}}{k\plus{}m\plus{}1} \plus{} \frac{k\minus{}m}{k\plus{}m\plus{}1} S_a(k,m\plus{}1)$.

Since we are only interested in the case where $ a \equal{} 0$ or $ a \equal{} 1$, we can write

$ S_0(k,0) \equal{} \frac{(k!)^2}{(2k)!}S_0(k,k)$,

$ S_1(k,0) \equal{} \frac{1}{k\plus{}1} \plus{} \frac{k}{(k\plus{}1)(k\plus{}2)} \plus{} \cdots \plus{} \frac{(k!)^2}{(2k)!}S_1(k,k)$.

But $ S_a(k,k) \equal{} \int_{x\equal{}0}^1 (1\minus{}x)^{2k} \, dx \equal{} \frac{1}{1\plus{}2k}$,

so it follows that

$ S_0(k,0) \equal{} N(k) \equal{} \frac{(k!)^2}{(2k\plus{}1)(2k)!} \equal{} \frac{(k!)^2}{(2k\plus{}1)!}$,

$ S_1(k,0) \equal{} (k\plus{}1)D(k) \equal{} \sum_{j\equal{}0}^k \frac{(k!)^2}{(k\minus{}j)!(k\plus{}j\plus{}1)!} \equal{} \frac{(k!)^2}{(2k\plus{}1)!} \sum_{j\equal{}0}^k \binom{2k\plus{}1}{j}$

$ \equal{} \frac{(k!)^2}{(2k\plus{}1)!} \cdot \frac{1}{2} \sum_{j\equal{}0}^{2k\plus{}1} \binom{2k\plus{}1}{j} \equal{} 2^{2k} N(k)$.

Therefore,

$ 2^{2k\plus{}1} \frac{N(k)}{D(k)} \equal{} 2(k\plus{}1)$.

The given expression corresponds to the case $ k \equal{} 1004$, hence it evaluates to $ 2010$.[/hide]

I have a better solution.

[hide]$ D(k) \equal{} \frac {1}{k \plus{} 1} \int_{x \equal{} 0}^1 (1 \minus{} x^2)^k \, dx$, from my first post.

From here, we write

$ \equal{} \frac {1}{2(k \plus{} 1)} \int_{x \equal{} \minus{} 1}^1 (1 \minus{} x^2)^k \, dx \equal{} \frac {2^{2k \minus{} 1}}{(k \plus{} 1)} \int_{x \equal{} \minus{} 1}^1 \left(\frac {1 \minus{} x^2}{4}\right)^{\!\!k} \, dx$

$ \equal{} \frac {2^{2k \minus{} 1}}{k \plus{} 1} \int_{x \equal{} \minus{} 1}^1 \left(\frac {1 \minus{} x}{2}\right)^{\!\!k} \left(\frac {1 \plus{} x}{2}\right)^{\!\!k} \, dx$.

With the substitution $ u \equal{} (1 \plus{} x)/2$, $ x \equal{} 2u \minus{} 1$, $ dx \equal{} 2 \, du$, we obtain

$ D(k) \equal{} \frac {2^{2k}}{k \plus{} 1} \int_{u \equal{} 0}^1 u^k (1 \minus{} u)^k \, du \equal{} \frac {2^{2k}}{k \plus{} 1} N(k)$.

This immediately gives us

$ 2^{2k \plus{} 1} \frac {N(k)}{D(k)} \equal{} 2(k \plus{} 1)$.[/hide]

Yes, [color=blue]the answer[/color] is :

A Happy New Year!

Good wishes for the year [color=blue]2010[/color].


Suppose a1, a2, ..., a10 are given integers.
Prove that there exist m1, m2, ..., m10 in {-1, 0, 1} such that
:Sigma:1:le:k:le:10mkak
is divisible by 1001.

I particularly like this one.

[quote="Arne"]
:Sigma:1:le:k:le:10mkak
[/quote]

For clarification, you mean

m1a1+m2a2+m3a3+m4a4+m5a5+m6a6+m7a7+m8a8+m9a9+m10a10 Correct?

Yes I believe so...I'm just used to seeing stuff on top and below the sigma and it didnt register at first...

How about [hide]mi = 0 for all i?[/hide]

I forgot one condition :

m1, m2, ..., m10 are not all zero.

Sorry :mrgreen:

[color=cyan]Figures.  You're sure it's divisible by 1001?  1001 seems too big, off hand.[/color]

I can assure you it's correct. It's an IMO longlist question.

It's not too difficult in fact, but very elegant. That's why I like it that much. But "Pigeonhole 4" is even more nice.

what is the difference between IMO shortlist and Longlist? and why do they have two lists?

Each country is free to propose problems, max. 6 problems.

All the proposed problems together (> 100) form the longlist.

Then the jury selects about 30 problems.

These form the shortlist.

That does not mean longlist questions are not nice or hard !
Sometimes, a problem does not make the shortlist since it can give an unfair advantage to well - trained students or because it might be familiar to some students or so.

Only before 1992, longlist were published. Now they are not published anymore since some people want to use them in their national competitions.

If each m_i is restricted to {0,1}, then there are 2^10 = 1024 choices for the vector m.  Thus, by the pigeonhole principle, there exist two different vectors (say m' and m'') that lead to the same remainder mod 1001.  Now subtract m' from m'' to get the desired vector m.

how do you know that won't give all m's equal to 0?

Because, by Pigeonhole, m' and m'' are two different vectors.

...and confusing...how did he prove that they must be divisible by 1001?...just by the fact that they have the same mod 1001?

There are 1001 residue classes and 1023 sums for mi in {0,1}.
Hence, by the pigeonhole principle, two sums must be equal modulo 1001.
So there are sums a, b such that a = b (mod 1001).
This means a - b = 0 (mod 1001). So 1001 divides a - b.
Now it suffices to notice that a - b is again of the form
:Sigma:mkak
but this time all mi are in {-1, 0,1}.

[color=cyan]Right, that's 3^10 total, not 10^3 . . . ooops.[/color]

ahh...tricky little problem we have there...


geometry
rectangle
calculus
integration
geometry unsolved

a. ) Prove that a rectangle with integral side length can be extended to a parallelepiped with integral side length and integral space diagonal if the value of the rectangle area is an even number.

b.) Prove a.) with the additional property that the rectangle's diagonals are of integral lengths as well.

i think you mean box by parellelepiped. part a seems extremely easy.
(a) this is equivalant to : if ab is even then there exists an integer c s.t. a^2+b^2+c^2=d^2 for some d, i.e. a^2+b^2=(d-c)*(d+c) which is obviously doable unless a^2+b^2 = 2 (mod 4), i.e. a and b are both odd which contradicts the evenness of ab.
(b) do you mean rectangle or parellelepiped? if its just that the rectangle's diagonals are integer lengths then it doesnt really change anything. if you mean the box's outer diagonals are integer lengths it is equivalant to: for any a,b, s.t. ab is even, there exists a c s.t. a^2+b^2+c^2, a^2+c^2 and b^2+c^2 are all squares. is this what you mean?

Part b.) is as follows:

Prove that a rectangle with integral side length and integral diagonal lengths can be extended to a parallelepiped with integral side length and integral space diagonal if the value of the rectangle area is an even number.

I think it is a very hard problem.  :)


Let $g: \mathbb{R}\rightarrow \mathbb{R}$ be lipschitz and $f: \mathbb{R}\rightarrow \mathbb{R}$ continuous.show that the system $x'=g(x),y'=f(x)y$ has at most one solution on any interval,for a given initial value.


number theory
Diophantine equation
number theory solved

Show that the diophantine equation

$\frac{1}{{x_{1}}}$ + $\frac{1}{{x_{2}}}$ + ... + $\frac{1}{{x_{n}}}$ + $\frac{1}{{x_{1}}{x_{2}}...{x_{n}} }$=1

has at least one solution for every positive integers n.

Suppose we have a solution $x_{1},\ldots,x_{n}$ for some $n$. Define $x_{n+1}=x_{1}x_{2}\ldots x_{n} +1$. Then,
\[ \frac{1}{x_{n+1}} +\frac{1}{x_{1}x_{2}\ldots x_{n}x_{n+1}} = \frac{1}{x_{1}x_{2}\ldots x_{n}}  \]

 So that we have a solution for $n+1$ variables. Example - solutions for $n=1,n=2,n=3$ are $(1), (2,2), (2,3,6)$ respectively.


If a+b+c=1, a,b,c>0, prove that

$ \frac{ab\plus{} \sqrt{a^3 c} \plus{}\sqrt{b^3c}}{a\plus{}b} \plus{}\frac{bc\plus{} \sqrt{b^3 a} \plus{}\sqrt{c^3a}}{b\plus{}c} \plus{}\frac{ca\plus{} \sqrt{a^3 b} \plus{}\sqrt{c^3b}}{c\plus{}a}  \leq \frac{3}{2}$

By AM-GM ,
$ \sqrt{a^3c} \le \frac{a^2\plus{}ac}{2}$
and,
$ \sqrt{b^3c} \le \frac{b^2\plus{}bc}{2}$,
therefore -
$ \sum_{cyclic}\frac{ab\plus{}\sqrt{a^3c}\plus{}\sqrt{b^3c}}{a\plus{}b} \le \sum_{cyclic}\frac{2ab\plus{}a^2\plus{}b^2\plus{}c(a\plus{}b)}{2(a\plus{}b)} \equal{} \sum_{cyclic}\frac{(a\plus{}b)(a\plus{}b\plus{}c)}{2(a\plus{}b)} \equal{} \sum_{cyclic}\frac{a\plus{}b\plus{}c}{2} \equal{} \frac{3}{2}$
equality for $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$
$ Q.E.D$


He looks like his avatar  :rotfl:
just kidding  :P  what's all those books?

:rotfl:  :rotfl:

...so, why was he wearing a dress?

:rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:
why is he wearing a dress anyways?
 :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:

*sigh*

Why did you have to post that?

We where doing a play, and I was an alien who dresses like a girl, and that's me rehearsing.

Please don't think of me as a crossdresser or anything like that.  :(

And aznmonkey1992, please don't do this again. I personally don't want my picture out on the internet and especially my name.

Can someone please delete this topic?  :(

HAHA!!!

aznmonkey goes to the same school as you?

[quote="Drunken_Math"]
Can someone please delete this topic?  :([/quote]

No, but I've taken the image down and will lock the thread.

And this is a warning not to post the image of anyone else without their permission, ESPECIALLY another community member. I'll contact the admin about banning if this becomes a constant problem, which I hope it isn't.


[i]Cyberchase[/i] is a 30-min TV show designed to teach the kids about math. I've watched the first several episodes, the math problems in it were very easy. For most of you it's certainly a waste of time.

I hate to admit it but......I LOVE Cyberchase! lol. I know it's very basic and easy, but many of the principles used on that show are on MC problems. It's cool to watch an episode then go to a similar problem the next day.  :)

i've barely ever admitted this, but i like cyberchase too. it is easy and seems to be a waste of time, but it's actually fun to watch though.

Cyberchase is one of the few shows I watch as well my sister watches it too though unfortunatly and I don't know how many of you caught the episode where they did line graphing but now my sisters correcting my homework since we are doing that in algebra...go figure...
I get teased a lot for watching cyberchase by kids at school who call it a baby show but I don't care its not like they don't tease me about liking math in the first place!

I've watched one or two episodes, but my 9-year-old brother watches it all the time, and he hadn't shown much interest in math before.  It is basic, but does anyone here remember Square One?  It was a bit more involved but not too hard (although it was too hard for me, I was only 7 or so when it was cancelled), and it also got kids interested.

I've seen it before, but I stopped watching it.  It's a little too basic for me, and I have too many other things to do.

Btw, that story about your sister is really funny BHorseMath.

happy that we have so many Cychase fans!