Nice proofs!  I was familiar with the algebra proof, but not the counting proof.  Like Rick, I prefer the counting proof.

I think this video is supposed to be before the Introducing Pascals Triangle video.

[quote=Wukong]that wasn't very necessary but ok
[img]https://cdn.artofproblemsolving.com/images/8/a/c/8ac5d58400127bc9c8871d5de788c810367fe870.gif[/img][/quote]

I could say the same for your post.

you could
also why is this in AOPS Videos

[quote=GameBot]Pascal's Identity

We prove Pascal's Identity with algebra and with a counting argument.[/quote]

how did someone use gamebot?


calculus
integration
trigonometry
derivative
logarithms
function
calculus computations

\[ \text{Problem: }\displaystyle\int\sqrt{x^2-1}dx\\
\\ \\
\text{Let }x = \sec u\\
dx = \sec u \tan u du\\ \\
\int\sqrt{x^2-1}dx\\
=\int\sqrt{\sec^2 u-1}dx\\
=\int\tan u dx\\
\text{Now using Integration by parts: }\int udv = uv - \int vdu\\
= \tan u\cdot x - \int x \sec u \tan u du\\
= \tan u \sec u - \int x dx\\
= \tan u \sec u - \frac{x^2}{2}\\
= x\sqrt{x^2-1} - \frac{x^2}{2}\]

Well, see for yourself; the derivative of what you've ended up with there is $ \sqrt {x^2 \minus{} 1} \plus{} \frac {x^2}{\sqrt {x^2 \minus{} 1}} \minus{} x$.

You've computed the derivative of $ \tan u$ incorrectly.  By the way, I don't think you need parts; you have an expression for $ dx$, so use it.

Going by the substitution the TC used;

$ \int \tan(u) \; dx \equal{} \int \tan^{2}(u)\sec(u) \; du$

let;

$ h'(x) \equal{} \sec(x)\tan(x) \;\;\;\;\;\;\; g'(x) \equal{} \sec^{2}(x)$
$ g(x) \equal{} \tan(x) \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; h(x) \equal{} \sec(x)$

Thus $ \int \tan^{2}(u) \sec(u) du \equal{} \tan(u)\sec(u) \minus{} \int \sec^{3}(u) \; du$

Using the secant reduction formula, we attain;

$ \frac {1}{2} \Big( \sec(u)\tan(u) \minus{} \ln | \sec(u) \plus{} \tan(u) | \Big)$

$ \equal{} \frac {1}{2} \Big ( x\tan(\sec^{ \minus{} 1}(x)) \minus{} \ln| x \plus{} \tan(\sec^{ \minus{} 1}(x)) | \Big) \plus{} C$

[hide]$ \int \sqrt{x^2\minus{}1}\,dx \equal{} \int \frac{x^2\minus{}1}{\sqrt{x^2\minus{}1}}\,dx \equal{} (Ax\plus{}B) \sqrt{x^2\minus{}1} \plus{} \alpha \int \frac{1}{\sqrt{x^2\minus{}1}}\,dx$.

By differentiation one gets $ \frac{x^2\minus{}1}{\sqrt{x^2\minus{}1}} \equal{} \frac{2Ax^2 \plus{} Bx \plus{} \alpha \minus{} A}{\sqrt{x^2\minus{}1}}$, which implies $ A \equal{} \frac12$, $ B \equal{} 0$, $ \alpha \equal{} \minus{} \frac12$. Therefore,

$ \int \sqrt{x^2\minus{}1}\,dx \equal{} \frac{x}{2} \sqrt{x^2\minus{}1} \minus{} \frac12 \int \frac{1}{\sqrt{x^2\minus{}1}}\,dx$

$ \equal{} \frac{x}{2} \sqrt{x^2\minus{}1} \minus{} \frac12 \cosh^{\minus{}1} x \plus{} C$.[/hide]

I will use:
(1) $ \sin^{ \minus{} 1}x \equal{} \minus{} i(\log i \plus{} \cosh^{ \minus{} 1}x)$
(2) $ \minus{} i \cos( \sin^{ \minus{} 1}x) \equal{} \sqrt {x^2 \minus{} 1}$

Integrating both sides of (2) and setting $ u \equal{} \sin^{ \minus{} 1}x$, we get:
$ I \equal{} \int \sqrt {x^2 \minus{} 1}dx \equal{} \minus{} i\int \cos( \sin^{ \minus{} 1}x)dx \equal{} \minus{} i\int \frac {\cos2u \plus{} 1}{2}dx \equal{} \frac { \minus{} i}{2}(sinucosu \plus{} u) \plus{} C_{1} \equal{} \frac { \minus{} i}{2}[x\cos(\sin^{ \minus{} 1}x) \plus{} \sin^{ \minus{} 1}] \plus{} C_{1} \equal{} \frac {x \sqrt {x^2 \minus{} 1}}{2} \minus{} \frac {\cosh^{ \minus{} 1}x}{2} \plus{} C_{2}$

hello, let $ x\equal{}\cosh(t)$ then is $ dx\equal{}\sinh(t)$ and our integral is
$ \int\sqrt{\cosh(t)^2\minus{}1}\sinh(t)dt\equal{}\pm\int sinh(t)^2dt$. Here we can use the formula $ \sinh(t)^2\equal{}\frac{\cosh(2t)\minus{}1}{2}$.
Sonnhard.

Rule of thumb involving trig substitutions:

A trigonometric substitution turns an integral with a square root into an integral with powers of trigonometric functions. Based on the even-odd parity of the exponents, trig power integrals are "easy" 3/4 of the time and "hard" 1/4 of the time. For the "hard" cases, you need reduction formulas, or double angle formulas, or some extra method. For instance an integral with sines and cosines is "easy" if either exponent is odd and "hard" if both exponents are even.

The rule of thumb: if you needed the trigonometric substitution, then you're going to wind up in a "hard" case. Expect that.

As an example: $ \int x^3\sqrt{1\minus{}x^2}\,dx.$

Doing the sine substitution gets you to an "easy" case - but you didn't really need the trig substitution. You could have let $ u\equal{}x^2$ and then $ t\equal{}1\minus{}u$ (or, in one step, $ t\equal{}1\minus{}x^2$) as a more ordinary substitution.

Two comments:

1. The rule has exceptions, such as $ \int\frac{dx}{(x^2\plus{}1)^{3/2}}.$

2. Using hyperbolic functions evades this particular rule.


modular arithmetic
number theory
prime numbers
number theory open

Sorry if I will say a stupid thing now, but I hope someone will pass this and help me  :P .

If $ p_1,p_2,...,p_n$ are prime numbers, $ a$ is a positive integer and there exist $ i_1,i_2,...,i_n$ so that

$ a^{i_1} \equiv i_1 (\mod p_1)$
...
$ a^{i_n} \equiv i_n (\mod p_n)$,

then there exist $ i$ so that $ a^i \equiv i (\mod p_1...p_n)$.

If you know when this is true or congruences like this please post.

That's not true.
The following system of congruences:
$ \begin{cases} 2^7 \equiv 7 & \pmod{11} \\
2^{10}\equiv 10 & \pmod{13}\end{cases}$
does not result in any $ 2^i\equiv i\pmod{11\cdot 13}$.


[quote="ProblemSets"]A curve in the xy-plane is defined parametrically by the equations x = t^2 + t and y = t^2 - t.

For what value of t is the tangent line to the curve horizontal?

A. t = -1
B. t = -0.5
C. t = 0
D. t = 0.5
E. t =1[/quote]
The horizontal tangent line is given when $ \frac {dy}{dx} \equal{} 0 \iff \frac {\frac {dy}{dt}}{\frac {dx}{dt}}\equal{}0$.

Thus:
$ \frac {dy}{dt} \equal{} 2t \minus{} 1$
$ \implies 2t \minus{} 1 \equal{} 0$
$ \implies t \equal{} .5$

It is important to note that if $ \frac {dx}{dt} \equal{} 0$ then $ \frac {dy}{dx}$ does not exist at this point.

$ \frac {dx}{dt} \equal{} 0$
$ \implies 2t \plus{} 1 \equal{} 0$
$ \implies t \equal{} \minus{} .5$

Thus $ \frac {dy}{dx} \equal{} 0$ at $ t \equal{} .5$.

$ \boxed{D. ~~t \equal{} .5}$


Duke
college
FTW
AwesomeMath
summer program
Mathcamp

we all know that duke kids have duketipper accounts here on aops and we know that they always come on at once and aways play by themselves. it used to be fun trying to pry out their pass but now everytime they win our computer screens freeze. i have nothing against tippers but they are killing this comp so i have decided have two groups there are the nontippers and then there are the tippers (those who play with duketippers) JOIN THIS LEAGUE
requirements: must be a non-tipper

Yeah... um, I sense immediately locking.

And the tippers could be reading this right now...

[quote="splatyango"]oh well...
ill join until this gets locked
couldnt u post the problem with the tippers on ftwq&s?[/quote]

Why are they a problem?

[quote="chenhsi"]What are duke kids and tippers?[/quote]

People taking the TIP, I think.

Duke TIP is over this year  :rotfl: And also, I wouldve recommended anti-tippers for name

i've heard of the TIP, but umm... what IS it?

[url]http://www.google.com/search?hl=en&q=TIP[/url]
Google is amazing.

It's Duke Talent Identification Program right?  like CTY, I'm pretty sure.

Yeah, it's Talent Identification Program, just like CTY. I heard it's decent.

i didnt know tippers had something going on in ftw
im in duke tip, but i don't really mention it much...
and i dont play in ftw anymore
。。。

I dont have a tipper account but I was part of Duke Tip. Most of the people with duke tip accounts are part of one of the Duke Summer Math Camps. The teachers use it for an educational exercise. Thats what I heard anyway...this post does not mean I'll join btw. I just wanted to clear things up.

crazychinesemaniac is right. The Duke TIP heavily endorses AoPS. And when they found this "goody bag" they used it.

As for this I'll join. Even though I am a Tipper, I should have gone to either Awesomemath or Mathcamp, instead of the Duke TIP math class.

ERRRRRRRRRR

I THINK I'LL JOIN

THOUGH I HAVE NOTHING AGAINST TIPPERS

AND A FRIEND I KNOW IN RL IS ONE   :rotfl:  :rotfl:


Let $ k\geq3,k \in N$, such that $ \cos{(k-1)\theta}$ and $ \cos{k\theta}$ are both rational numbers.Prove that there is $ n>k$ and $ n \in N$ such that $ \cos{(n-1)\theta}$ and $ \cos{n\theta}$ are rational numbers

Looks as if too much irrelevant details. Note that if $ \cos\theta$ is rational then $ \cos m\theta$ is rational too. Thus your problem is to find two consecutive numbers one of which is divisible by $ k-1$ and another by $ k$. The numbers $ k^{2}-1$ and $ k^{2}$ serve your purpose...


Por lo visto somos la quinta comunidad mas grande!!!

Los invito a que continuemos asi, e inviten a sus comapñeros a hacer parte del foro.

 :D  :D  :D

Felicitaciones a todos!!! Sigamos así!!!!!!!!

[color=blue]
 eso mismo digo ....
 ;)

Veo que los chicos de argentina se estan uniendo ..
espero que otros grupos más hagan lo mismo ...

Aunque recuerden que no somos solo un pais ..
eso debe definir en nosotros.. exigencias mayores de nuestra parte ..

continuemos con esta tarea..

Carlos BrAVO ;)
LIMA -PERU [/color]

[color=blue]Hablando de países… Estaba pensando en que se podría hacer una encuesta para saber cuántas personas de cada país latinoamericano participa en esta comunidad.

Esta vez escribo en un topic ya existente, pues ya pasé mucha vergüenza al enterarme de que había creado topics repetidos…  :blush:  :blush:  :blush:

Perdonen mi desatención, por favor. Era nuevo en el foro… Y no volverá a pasar.

Aunque no pueda crear una, quisiera saber qué criterios se rigen al crear una encuesta. Quisiera saber, por ejemplo, si pueden agregarse más opciones de respuesta una vez creado el programa de encuesta…[/color]


analytic geometry
combinatorics proposed
combinatorics

$A$ and $B$ are two opposite vertices of an $n \times n$ board. Within each small square of the board, the diagonal parallel to $AB$ is drawn, so that the board is divided in $2n^{2}$ equal triangles. A coin moves from $A$ to $B$ along the grid, and for every segment of the grid that it visits, a seed is put in each triangle that contains the segment as a side. The path followed by the coin is such that no segment is visited more than once, and after the coins arrives at $B$, there are exactly two seeds in each of the $2n^{2}$ triangles of the board. Determine all the values of $n$ for which such scenario is possible.

Consider a coordinate grid such that $ A\equiv(n,0)$, $ B\equiv(0,n)$, and the other two corners of the board are $ (0,0)$ and $ (n,n)$.  The conditions of the problem may be obviously restated as this: color some of the sides of the triangles such that (1) in each triangle exactly two sides are colored, and such that (2) at each vertex of the grid, an even number of colored edges meet, except for $ A$ and $ B$, where an odd number of colored edges meet.  The colored edges mark the path from $ A$ to $ B$ such that in each triangle exactly two seeds are placed.

Let us assume that the scenario is possible for some $ n$.  Since exactly two edges meet at $ (0,0)$, because of (2) they are both colored or not colored.  Then, because of (1) applied to the triangle that has $ (0,0)$ as a vertex, both these edges must be colored, while the edge joining $ (0,1)$ and $ (1,0)$ must be left uncolored.  Applying (1) again to the triangle with vertices $ (0,1)$, $ (1,0)$ and $ (1,1)$, the other two sides of this triangle must be colored.  Or, applying (2) to vertices $ (0,1)$ and $ (1,0)$, the edges joining these points to $ (0,2)$ and $ (2,0)$, respectively, must be left uncolored.  Then, applying (1) again, edges joining $ (1,1)$ to both $ (0,2)$ and $ (2,0)$ must be colored.  Considering (2) applied to vertex $ (1,1)$, either both edges joining this vertex to $ (1,2)$ and $ (2,1)$ are colored, or both are uncolored.  Because of (1) once again, both are colored, while the edge joining $ (2,1)$ and $ (1,2)$ must be uncolored.  Hence, because of (1) again, edges joining $ (2,2)$ to both $ (1,2)$ and $ (2,1)$ are colored.  Note at this point that, if $ n \equal{} 2$ we are done and the task is accomplished, or for $ n \equal{} 2$ the scenario is possible.

Let us henceforth assume that the scenario is possible for some $ n > 2$.  Then, because of (1), the edge joining $ (2,0)$ and $ (2,1)$ must be left uncolored, or because of (1) again, edges joining $ (3,0)$ to both $ (2,0)$ and $ (2,1)$ must be colored.  Then, because of (2), the edge joining $ (2,1)$ and $ (3,1)$ must be colored, or the edge joining $ (3,0)$ and $ (3,1)$ is uncolored.  Note that this makes the scenario impossible for $ n \equal{} 3$, since then there would be an even number of colored edges meeting at $ A$.  But if $ n > 3$, then because of (2), the edge joining $ (3,0)$ and $ (4,0)$ must be left uncolored.  But this makes the triangle formed by these two vertices and $ (3,1)$ violate (1).  Or the scenario is not possible either for $ n > 3$.

So the scenario is only possible for $ n \equal{} 2$, and a possible path is $ A\equiv(2,0)\rightarrow(1,1)\rightarrow(0,1)\rightarrow(0,0)\rightarrow(1,0)\rightarrow(1,1)\rightarrow(2,1)\rightarrow(2,2)\rightarrow(1,2)\rightarrow(1,1)\rightarrow(0,2)\equiv B$.


linear algebra
matrix
linear algebra unsolved

Prove that any matrix IN ANY FIELD is the product of 2 symetric matrices.

I don't know the proof for any field, but for complex matrices, it is direct consequence of another problem I solved, posted by ALekk: every complex matrix is similar to a symetric matrix. I suspect this result is true in any field, but I am not sure and definitely I don't have a proof.


I don't know enough Asy to draw this, but if we add one square to the right of the bottom one, one to its left diagonal, and one on top to fill up the whole row, we'd have it symmetrical. Thus, our answer is $ \boxed{3}$.

EDIT: Oops, misread. I didn't see it has to be symmetrical along diagonal $ \overline{BD}$. Vallon's right.

I got this problem yesterday and answered 3. It was correct.

i am not meaning to spam but ftw say s that the answer is 4

Every time I got this answer the answer was 4.

@Bugi: maybe you meant to answer 3, but you typed 4.

Problem can be changed over the span of 3 years. Errors get reported, answers get changed.

i think you guys should prove the answer by math but again i think the answer is 4


[quote="agolsme"]Mandelbrot October 2005 Round 1

6. Let $\tau$ be the golden ratio, that is, the positive solution to the equation  $x^2 = x + 1$. There is exactly one way to write $10$ as the sum of distinct, non-consecutive elements from the sequence $\{\ldots,\tau^-2,\tau^-1,\tau^0, \tau^1, \tau^2,\ldots\}$. Find the smallest exponent in this sum.[/quote]

[hide="solution"]
$-4$
$\tau^{-4}+\tau^{-2}+\tau^2+\tau^4=10$
The sequence from $\tau^{-4}$ to $\tau^{4}$ is $\{\frac{7-3\sqrt{5}}{2}, {\sqrt{5}-2}, {\frac{3-\sqrt{5}}{2}, {\frac{\sqrt{5}-1}{2}, 1, {\frac{1+\sqrt{5}}{2}, {\frac{3+\sqrt{5}}{2}, {\sqrt{5}+2}, {\frac{7+3\sqrt{5}}{2}}}}}}$
$\tau$ to the power of a negative odd number is the conjugate of the same positive odd number with the rationals cancelling out. $\tau$ to the power of a negative even number is the conjugate of the same positive even number with the irrationals ($\sqrt{5}$) cancelling out. Therefore, the exponents must be even, paired so that one is negative and one is positive and their absolute values are the same. Checking the first two pairs, we find that $3+7=10$. I hope you can all understand this. :D
[/hide]

[quote="agolsme"]how did you find that?[/quote]
Sorry. I posted a solution. I don't know if they're called conjugates, but you know: $\sqrt{5}-1$ and $\sqrt{5}+1$.


Giva two positive integer $x,y$ satisfy $x(y+1)$ and $y(x+1)$ are perfect square.Prove that exactly one of $x,y$ is a perfect square :)  :)

A solution can be found here http://www.mathlinks.ro/Forum/viewtopic.php?t=6505


inequalities
induction
inequalities proposed

I don't understand what you are writing?
Notice that:

$\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n} \ge n$

$\frac{2n\sqrt{2}}{a_1^2+a_2^2+...+a_n^2} \le 2\sqrt{2}$
 :)

Is this Inequality not enough nice to prove?
 :?

Let me rephrase his proof.
Lemma, If $x \ge p$, $p,q \ge y$ and $\xy \ge pq$, we have $x+y \ge p+q$.
Proof: Obviously we can set xy=pq, then
$x/p=y/q=k$, where $ k \ge 1$, and $x=kp$, $y=kq$, so
$x+y-p-q=(k-1)(p-y) \ge 0$.
Now, \[
\frac{1}
{{a_1 }} + ... + \frac{1}
{{a_n }} \geqslant n \geqslant 2\sqrt 2  \geqslant \frac{{2n\sqrt 2 }}
{{a_1^2  + ... + a_n^2 }}
\]
\[
\begin{gathered}
  x = \frac{1}
{{a_1 }} + ... + \frac{1}
{{a_n }} \hfill \\
  p = n \hfill \\
  q = 2\sqrt 2  \hfill \\
  y = \frac{{2n\sqrt 2 }}
{{a_1^2  + ... + a_n^2 }} \hfill \\
\end{gathered}
\]
By the lemma we immediately have the result for $n \ge 3$.
(EDIT: ah i 'm wrong... here $xy \le pq$... sorry...  :blush: )
For n=1,2 it should be easy to prove.

Wait a time?
I don't see $k=2n\sqrt{2}$ in this proof.

In case $n=3$,it's the best constant.

I think you can't  assume that $xy=pq$ in this lemma.
Beause if we decreas $p$ or $q$,the condition $x \ge p,q \ge y$ is not still satify.

I write this problem is not difficult be cause It can be proved by old method.
Try to write in sum of square.

For [tex]n\leq\ 3[/tex] it is easy.
For [tex] n\geq\ 4[/tex] i found in my notebook the following inequality:
[tex] n^2(\frac{1}{a_1}+...+\frac{1}{a_n})\geq 4(n-1)(a^2_1+...+a_n^2)+n(n-1)^2[/tex]
It's really stronger than your inequality.You can prove it by induction(i guess it was posted before,and solved by harazi)

I'm interested in zxingtan's way..
i will reprove his Lemma:Let a,b,c,d st:
[tex]a\geq\ b\geq\ c\geq\ d[/tex] and [tex]ad\geq\ bc[/tex] then    [tex]a+d\geq\ b+c[/tex]
My proof:
There exist  [tex]x,y,z\geq\ 0[/tex] such that:
[tex]c=d+x,b=c+y,a=b+z[/tex] then[tex]a=d+x+y+z,b=d+x+y,c=d+x[/tex]
Since condition, we have:
[tex]d(d+x+y+z)\geq\ (d+x)(d+x+y)[/tex]
[tex]<->d(x+y+z)\geq\ d(2x+y)+x(x+y)[/tex]
[tex]<->d(z-x)\geq\ x(x+y)\geq\ 0[/tex]
[tex]->z\geq\ x[/tex]
But:
[tex]a+d\geq\ b+c[/tex]
[tex]<->2d+x+y+z\geq\2d+2x+y[/tex]
[tex]<->z\geq\ x[/tex]
But it's true
And we are done.

[quote="hungkhtn"]
I think you can't  assume that $xy=pq$ in this lemma.
Beause if we decreas $p$ or $q$,the condition $x \ge p,q \ge y$ is not still satify.[/quote]
decrease y then x+y and xy is decreased and everything is ok.

To hungkhtn:I forget the topic but I remember the solution,I will send it to you soon

Ok.Thanks you Spdf.
For this problem,I have a nice solution.
( Use abit expanding ,we have :

${\sum{M_{i,j}(a_i-a_j)^2}} \ge 0$ )

It's not new and I will wait a solution from it!


limit
induction
calculus
calculus computations

Consider a sequence $ \{a_n\}$ defined by $ a_1 > 4, \ a_{n \plus{} 1} \equal{} \sqrt {a_n \plus{} 12}$.

(i) For $ n \equal{} 2,\ 3,\ 4,\ \cdots$, prove that $ a_n > 4$.

(ii) For $ n \equal{} 1,\ 2,\ 3,\ \cdots$, prove that $ a_{n \plus{} 1} \minus{} 4 < \frac {1}{8}(a_n \minus{} 4)$.

(iii) Find $ \lim_{n\to\infty} a_n$.

[b]i.)[/b]
$ \ a_{n \plus{} 1}>4 \Leftrightarrow \sqrt {a_n \plus{} 12}>4 \Leftrightarrow a_n \plus{} 12>16 \Leftrightarrow a_{n}>4$. Now since $ a_{1}>4$ we can make an induction and get $ a_{n}>4$ for all $ n$.

[b]ii.)[/b]
$ a_{n \plus{} 1} \minus{} 4 < \frac {1}{8}(a_n \minus{} 4)\\
\Leftrightarrow 8a_{n\plus{}1}<a_n\plus{}28\\
\Leftrightarrow 8\sqrt {a_n \plus{} 12}<a_n\plus{}28\\
\Leftrightarrow 64(a_n \plus{} 12)<a^{2}_n\plus{}56a_n\plus{}28^{2}\\
\Leftrightarrow 0<(a_n\minus{}4)^{2}$

This is true since $ a_{n}\neq 4$ for all $ n$.

[b]iii.) [/b]
Since $ a_{n}\minus{}4>0$ we have $ \frac {1}{8}(a_n \minus{} 4)<a_n\minus{}4$ and hence $ a_{n\plus{}1}\minus{}4<a_n\minus{}4$. So $ a_{n\plus{}1}<a_{n}$. Therefore we have $ 4<a_{n\plus{}1}<a_{n}$ for all $ n$. The sequence $ \left\{a_{n}\right\}$ is strictly decreasing and bounded, hence it has a limit.

Now since $ L^2\equal{}\sqrt{L\plus{}12} \Rightarrow L\equal{}4, L\equal{}\minus{}3$, the only candidates for a limit are $ 4$ and $ \minus{}3$. Since $ a_{n}>4$ for all $ n$, we must have $ \lim_{n \to \infty}a_n\equal{}4$.


Support geometry MIT college ratio calculus AMC

Reading the 'school thread' by mcrawford made me think about home school.

Having had an unexciting and non-stimulating experience in public school, I've recently thought about alternatives. While its too late for me to do something different for my self ( going to university in four months), I still think this is a VERY important issues to be resolved:

" How to best help gifted children excel, specifically regarding the viability of home school".

Home schoolers have a very odd place in our society. This bored shows that home schooling can be used to help gifted children. But growing up, I was told that home schoolers tend to have social issues. While I questioned this fact as I got older, I did meet a few of my distant relatives who home schooled their children. I felt awkward around them, once it became obvious to me that these people had home schooled their children  so that they could ensure they fallowed the same religious beliefs.
While I'm sure their intentions where good, I find it very odd to  'teach faith' to someone. To make matters odder, these kids still seemed quite bright, making the issue very grey for me.

What do you think?

(p. s. I don't want a flame war. No flames.)

[quote="rzsolt"]Reading the 'school thread' by mcrawford made me think about home school.[/quote]

That is, you started thinking about homeschooling. (Writing the term as one word helps in online searching, which is why scholars prefer that spelling.)

[quote="rzsolt"]Having had an unexciting and non-stimulating experience in public school, I've recently thought about alternatives.[/quote]

That's what got me thinking too. I had rather lousy instruction in my K-12 schooling, especially in math in some key years, and I was determined to find something better for my children. A public school assistant principal I met in ninth grade told me about John Holt's book How Children Fail, and urged me to read it. That was the beginning of deep skepticism about what schools can do for young people.

[quote="rzsolt"]"How to best help gifted children excel, specifically regarding the viability of home school".[/quote]

Homeschooling seems to be doing very well for gifted children. We enjoy a thriving gifted homeschooling support group in our community, and the learners who participate in it like one another, can join into interesting extracurricular actitivies, and are definitely advancing better in the usual school subjects than most of their contemporaries.

[quote="rzsolt"]But growing up, I was told that home schoolers tend to have social issues.[/quote]

Well, you were told that largely by people who hadn't tried it, just as you might have heard the same thing about young people who participate in math contests.

[quote="rzsolt"]While I questioned this fact as I got older, I did meet a few of my distant relatives who home schooled their children. I felt awkward around them, once it became obvious to me that these people had home schooled their children  so that they could ensure they fallowed the same religious beliefs.
While I'm sure their intentions where good, I find it very odd to  'teach faith' to someone. To make matters odder, these kids still seemed quite bright, making the issue very grey for me.[/quote]

That's a separable issue from homeschooling as such. The greatest number of young people who are explicitly taught to follow a particular religious faith are taught in classroom schools set up for that purpose. They are huge networks of Roman Catholic, Fundamentalist Baptist, Lutheran, Jewish, and Seventh-Day Adventist schools in the United States, and the aggregate number of children attending those far outnumbers the students who are homeschooled for any reason.

As to social policy, I support allowing parents to attempt to teach the deeply held beliefs that they have found trustworthy to their children. Their children are still going to grow up and meet many people with different beliefs--not to mention that most children are meeting such people all along throughout life. This social policy works very well in The Netherlands, which for more than a century has allowed broad public support through taxes for a very diverse range of schools, including schools devoting to teaching one religion or another.

As to my own practice, our homeschooling support group is religiously diverse, and our children definitely read and hear discussed ideas that originate in religions other than the one I grew up with. My children will have plenty of time to ponder their deeply held beliefs in a variety of social contexts as they grow up.

[quote="rzsolt"]What do you think?[/quote]

I think homeschooling is a good idea. It has been very helpful for a lot of AoPSers.

I am homeschooled, and I am often asked how I get socalization. It's not like homeschoolers stay at home all day; most homeschoolers I know go to places like sports, churches, and competitions. Also many areas have homeschool groups. I don't belong to one, but I've been to some of their field trips.
I am Catholic, but that's not the reason I homeschool. My brother has even taken a biology course with evolution. The main reason we started homeschooling was that the schools weren't teaching my brother very well. So my mom started homeschooling my brother and me when he was in 6th grade and I was in 2nd. He is going to MIT this fall.
Homeschooling has many benefits. One is that you can have a much more personalized curriculum, since you are not in a classroom with lots of other kids. There are also curricula put together by people which you can buy. I have read some very interesting books about history and economics that wouldn't usually be used in a public school. Also, by homeschooling, you can aviod some peer pressure found in public schools.

I agree with LadyKn1ght. I think homeschooling provides very good education, as the instructor's attention is only on one person and you set the pace for your learning. Also with an enriched curricullum one could be exposed to materials not applied in public schools. However, I think that a homeschooled student, who had none or little experience with a school institution would feel uncomfortable, when exposed to life in college, which might be a downfall. Homeschooling would be a very practical solution, if one has problems with local public schools being associated with a lot of peer pressure and violence, which is an issue that could thwart the progress of a bright student.

I, myself, have been homeschooled for all my life. I have never been to a public nor private school. And since I am finishing up my tenth year of schooling, I think it is fair to say I have a good understanding of what homeschooling is, how it works (for myself particularly), and why it is done.

Homeschooling flows from parental rights. Parents have the right to supervise their child(ren)'s education, even if they are the ones teaching. It allows parents to have the most control over a student's education. On the other hand, while parents do have some control, a public or private school definitely gives the majority of the control of a student's education to the school teachers, school system,  etc.

When I was younger, my schooling greatly constituted of being with my mother; sitting down together at a table with me, and teaching me. She would explain things such as math concepts, writing, phonics, and the like, and I would go exercises and such. While my mother was particularly gifted at teaching me as a young grade student (she majored in child education at Tuffs), as the saying goes, "you don't have to be a rocket scientist" to be able to homeschool. There are an abundant of resources available that make it feasible for parents (both mothers and fathers), to effectively teach their children.

The method I learn now is different. In fact, it is more of a supervised self-study. I am given assignments, and textbooks, which I am required to read or complete as assigned. My parents (both of them), then go over what I have done, and if explanation is needed, they help me. This is very helpful because I have learned and continuously practiced how to learn for myself. This is a valuable skill for the rest of my life. My parents have a very good knowledge of what level I am at, and how to make assignments or plan my courses.

The fact that my parents know at what level I am at, and what suites me best, is a MAJOR advantage. It allows students to go at their own pace. One reason I am on this site is because I have accelerated particularly in mathematics. And for other subjects, I have the freedom to continue at a pace that _best suits me_. Instead of a public or private school that to one degree or another has a "one size fits all" education, homeschoolers can enjoy something more of a "specialized" education. Suited for the students needs. This is not to say that public/private schools are all bad, it is just one of the things that by definition a public or private school cannot do.

I am currently in high school, and I plan to graduate as a homeschooler. Being homeschooled, I would not say that I, nor any of my siblings who are also homeschooled, have had any major disadvantages compared to public/private, etc. schooling.

My family started homeschooling (all my siblings are homeschooled), after my older brother’s first year in kindergarten. My brother was one who does not do the best in a classroom setting, and at home, my mother realized she could teach him as effectively at home. Thus, we started to homeschool. Over time, reasons for doing so have changed slightly. For one, it keeps our family together in a close way, being that we are around each other all day. Another is that homeschooling has proven to be effective with our family. Mainly, people homeschool because they find that it is more effective, for their needs, than public schools.

As to the whole thing about homeschoolers not being socialized enough, well, homeschoolers have a joke about that. At least our family does. Many homeschooling families are large ones. My family can be considered one of them. So living in close contact with six other siblings can sometimes be too much socialization!

In my area of the world ( southern Ontario ), there is a huge education system in place, but it panders to the the mediocre. I have a very hard time communicating to adults the difficulty of paying attention and learning, when 5-10 minutes of concepts are spread out over and hour and 20 minutes ( the length of a Toronto high school class ).

The other area where modern public schools are failing: Is communicating the richness of various topics. Many people write off several fields of study, or lines of work, because of one bad experience in a math class, or an uninspiring physics curriculum.

I honestly feel lucky that I caught my mistake, and gave math and computer science a second look, outside of the standard school system.

I see private individual education as a good alternative for my own children. It is obviously not a choice many people will get a chance to consider.
 One possibility is that I could skip that step, and not have any.  :D

http://www.amazon.com/gp/product/0201484021/104-3061833-0912747?v=glance&n=283155 link to the book mentioned above. Sounds fascinating, and terrifying. I think its going on my summer reading list.

[quote="José"]Well, actually, I'll never study at home.... please, do not forget the social life!!!!!!!!!!! maybe one of the most important things in life........[/quote]

I can't get work done at home very well. I generally go to the library or a coffee shop. When it comes to social life, its not to hard to make room. See my friends one or two nights a week, my girlfriend another two or three, there is almost no time left for math!

Social life is important, but it can't be forced. You either have one, or you don't. Very few people can actively 'turn it on/off'.

I think for homeschooled kids, the parents will need to take a more active role to make sure there kids get socialized. I see summer camps as a boon?

[quote="José"][quote="rzsolt"]I think for homeschooled kids, the parents will need to take a more active role to make sure there kids get socialized. [/quote]

I repeat: if I were their father, I'd not let them study at home, but well, anyone can think whatever he/she wants, that's my opinion only. nice theme anyway[/quote]

I honestly don't think i would let any chidren of mine study at home either. The best reason I can think of is that children are going to learn the hard way, that education doesn't always mean learning. In that case, they'll be much happier when I, or someone close to them actualy provideds them with a real lesson.

That and I don't know if I want kids.  :roll:

[quote="LadyKn1ght"]I am Catholic, but that's not the reason I homeschool. My brother has even taken a biology course with evolution.[/quote]  I find that phrasing really creepy.  Anyhow, Catholic Church dogma accepts evolution.

[quote="nayrb"]Being homeschooled, I would not say that I, nor any of my siblings who are also homeschooled, have had any major disadvantages compared to public/private, etc. schooling. [/quote]  But, since you've been homeschooled your entire life, how would you know?

Now, to give this post some content: I've never been homeschooled, but let me list some of the cons (as I see it) of homeschooling:
1) It's exclusive.  I am immediately skeptical of a system which is inherently limited to people of a certain income or education level.  (That said, of course, I've attended schools all my life that charge tuition or have significant entrance requirements.)
2) I feel that there are certain things that everyone really should learn, regardless of the beliefs of their parents, and I don't like the idea of opting out via homeschooling.
3) Much of homeschooling strikes me as creepy indoctrination.
4) I think public schools are incredibly important -- no matter what, they are likely to provide education to most Americans for the forseeable future.  Schools almost always benefit from involvement by committed, serious parents.  If those parents all leave the public school system, it takes away a substantial force for positive change in the existing system.

So, roughly, one count of not fitting well into my ideology, two counts of damaging students, one of damaging the public school system (albeit in a limited fashion).  Obviously, objections 2 and 3 don't apply to homeschooling generally -- a competent parent should not have problems with them.  And, for many students, homeschooling undoubtedly leaves them better off educationally than they would be under any other system.  So, all told I think I view it as a wash.  (How's that for a strong statement of opinion?)

[quote][quote]nayrb wrote:
Being homeschooled, I would not say that I, nor any of my siblings who are also homeschooled, have had any major disadvantages compared to public/private, etc. schooling.[/quote]

But, since you've been homeschooled your entire life, how would you know? [/quote]

Well, I have taken some college classes. I also have talked to people who have been to public schools. Besides, my skepticism of the public schools matchs your skepticism of homeschooling. Neither of us are very familier with the other. Therefore, you can't bee too hard on me. :)

[quote]1) It's exclusive. I am immediately skeptical of a system which is inherently limited to people of a certain income or education level. (That said, of course, I've attended schools all my life that charge tuition or have significant entrance requirements.) [/quote]

Remember, people pay for the public schools in taxes. That being said, my family spends a lot less than what the public school pays for each one of its students. And this includes activities such as private music lessons.

Also, I guess I didn't make this point well enough, it is not limited to a person of a higher education level. One of the biggest points of homeschooling is that it teaches self study. To a very large degree, I am my own teacher. I am learning by myself, but with supervision from parents. Thus, the parents need not have a very high degree of education. On top of that, there are many resources available to help those parents who do not have a good understanding of a subject. For instance, there are many books that teach young kids. But also, the teach the parent how to teach the child math, if you get my meaning. In fact, homeschooling is much of a learning process for both the parent and the child. I have seen huge catologs of homeschooling material.

Also, if you really need help, there are online courses. I have taken some for particular subjects. It may not be what you think of homeschooling, but the parents are directly supervising it.

[quote]2) I feel that there are certain things that everyone really should learn, regardless of the beliefs of their parents, and I don't like the idea of opting out via homeschooling. [/quote]

This can go the other way. Some homeschoolers may think that there are important aspects of homeschooling that public school students should have. For instance, our family is a lot stronger because we homeschool. Part of that is because we are with each other more than most other families, and need to get along.

This is also meant to say that homeschoolers may still be recieving what public schoolers are. Some of the stuff that you think "everyone" should learn.

[quote]3) Much of homeschooling strikes me as creepy indoctrination. [/quote]

My family does homeschool for what some might consider religous reasons. But think about public schools too. Why is there all the debate about evolution and inteligent design?

Also, colleges do look for homeschoolers because of their independant thinking, which flows from being homeschooled.

[quote]
4) I think public schools are incredibly important -- no matter what, they are likely to provide education to most Americans for the forseeable future. Schools almost always benefit from involvement by committed, serious parents. If those parents all leave the public school system, it takes away a substantial force for positive change in the existing system.
[/quote]

I heartily agree. Public schools are good, to a degree, and they are important. Homeschooling just another option. But then, many people who are not homeschooled still see problems in the school system.

It is good that someone who is a homeschooler can show their position.

[quote="JBL"]Now, to give this post some content: I've never been homeschooled, but let me list some of the cons (as I see it) of homeschooling:
1) It's exclusive.  I am immediately skeptical of a system which is inherently limited to people of a certain income or education level.  (That said, of course, I've attended schools all my life that charge tuition or have significant entrance requirements.)
2) I feel that there are certain things that everyone really should learn, regardless of the beliefs of their parents, and I don't like the idea of opting out via homeschooling.
3) Much of homeschooling strikes me as creepy indoctrination.
4) I think public schools are incredibly important -- no matter what, they are likely to provide education to most Americans for the forseeable future.  Schools almost always benefit from involvement by committed, serious parents.  If those parents all leave the public school system, it takes away a substantial force for positive change in the existing system.
[/quote]
I have homeschooled for 3 years, and have gone to both public and private schools before I started homeschooling.
Homeschooling [b]might [/b]be limited to people in a certain range of income levels, but it is certainly not limited to people of a certain education. I am sure than many people homeschool because the pace of the school system is too fast for them, or there might be a completely different reason. (I know someone who has to homeschool because she had a severe concussion)
Many people might avoid having their children learn certain things by sending them to a private school.
What do you mean by "creepy indoctrination"?
If all those who are homeschooling were to stop, I think that it would be most likely that most of them would go to private schools rather than public.

nayrb, that was a good effort, but you aren't consoling me.  I'll hide my [hide="Explanation of why not."]
1.  Taxes are irrelevant: My objection was the exclusivity, not the notion that someone must pay (or forgo pay) for education.  Everyone has access to the public school system, regardless of how much they pay in taxes.  Families in which neither parent (if there are two parents) makes enough money to support the family cannot homeschool.  I don't know anything about your family, but I imagine that one of your parents is not working full-time: many families, probably most, cannot afford that.

[quote="nayrb"][quote]2) I feel that there are certain things that everyone really should learn, regardless of the beliefs of their parents, and I don't like the idea of opting out via homeschooling. [/quote]
This can go the other way. Some homeschoolers may think that there are important aspects of homeschooling that public school students should have. For instance, our family is a lot stronger because we homeschool. Part of that is because we are with each other more than most other families, and need to get along.
This is also meant to say that homeschoolers may still be recieving what public schoolers are. Some of the stuff that you think "everyone" should learn.[/quote]
That is not sufficiently analogous that I would use the phrase "going the other way."  I believe there are certain things everyone in a liberal democratic state should know in order to be a "good citizen."  Schooling, in other words, should be not just for the benefit of the individual but for the benefit of society.  If I say, "Some homeschoolers are being deprived of important information in their schooling," and you say, "But they are really tight with their families," you haven't done anything to assuage my worries.
In your second-to-last sentence, it is the "may" that disturbs me.  Certainly, there is nothing in the notion of homeschooling that [i]prevents[/i] students from being well-educated, in this very specific sense.  My complaint here is that there seems to be very little to [i]ensure[/i] it.

[quote="nayrb"][quote]3) Much of homeschooling strikes me as creepy indoctrination. [/quote]
My family does homeschool for what some might consider religous reasons. But think about public schools too. Why is there all the debate about evolution and inteligent design?
Also, colleges do look for homeschoolers because of their independant thinking, which flows from being homeschooled.[/quote]Your second and third sentences are a non-sequitur.  Your fourth is unsupported on two counts.

[quote="nayrb"]Homeschooling just another option. But then, many people who are not homeschooled still see problems in the school system.[/quote]Yes, homeschooling is another option, but that doesn't address my concern.  Again, this strikes me as something of a non-sequitur as a response to a very specific complaint on my part.[/hide]

Now, to daine and nayrb:
You both claim that the parents of homeschoolers don't require any particular level of education.  I think, frankly, that you are flat-out wrong about this.  Parents of homeschoolers may not need higher degrees, but there are essential skills that it is important that homeschooling parents be able to impart.  The most obvious of these is a basic appreciation of how to learn.  Students in schools have parents, teachers and peers guiding them in the process of learning, but homeschoolers have only their parents.  If their parents aren't equipped to give a model in acquiring knowledge, where will the student pick that up from?  The fact that there are resources available doesn't matter if the parent doesn't know where to get those resources, or which ones to use.

Finally, to daine:
I agree that people probably use a "private school dodge" frequently.  I think that is a serious problem with private schooling, just as I think it is a problem with homeschooling.
Creepy indoctrination -- I was hoping to avoid further definition in order to not draw the emnity of those who don't think it's either creepy or indoctrinating.  Suffice it to say that I strongly dislike the idea of religion as part of a formal education.
And as for your last point, I think that is very likely true -- but, again, I am not thrilled with the notion of private schooling, either.

JBL,

Two points I do not think I made clear enough. I didn't develop them well enough. Excuse me for that. Again, both of us must understand that we are coming from two different sides of the debate, and don't have much of the same experience of the other side. Oh, and public schools also “refer” similarly to private schools.

(1) You say that homeschoolers are lacking in "schooling" that makes them a good "citizen." Or, at least suggesting that is the case. I suppose I was hinting (not stating clearly) that (1) you did specifically mention what schooling homeschoolers lacked, and (2) because you did not give an example, you cannot say that homeschoolers are not receiving that particular need for homeschooling, (3) even if you did state what public schools had that homeschoolers lack (in the context you were talking about), you would not have the same authority I have to know, since you have never been homeschooled, nor (I am guessing) had much contact with homeschoolers.

My second part to this point was, again, not stated clearly. I said that while you may think that homeschoolers are lacking in some critical area, homeschoolers may think that people going to public schools are lacking in some critical area. For example, a person who is homeschooled may think that a person who goes to a public school does not have the same quality of a family, which is important. I am not taking this stand, but it is a supposition. My statement, "On the other hand." fits better here. Or what about the custom pace at which homeschoolers can go through their studies? Some people may think that public are critically lacking because of that.

(2) My second point was about your concern about "creepy indoctrination." My point was, and again, merely hinted at, was that the teaching of evolution at public schools can be considered such a thing. Because many have decided not to teach ID, can't that be considered indoctrination? It is not letting the other side of the discussion to be heard.

Again, homeschoolers may have concerns that students are being indoctrinated at public schools, or exposed to unwholesome things (for instance, issues about school prayer, pledge of allegiance, bad language), and thus decide to homeschool them. This is a parents right, though.

The second part wasn't too well founded. "Also, colleges do look for homeschoolers because of their independent thinking, which flows from being homeschooled." I believe, though, that  [url=http://www.hslda.org]HSLDA[/url] has done some studies on such things, though.

I understand your concern about the potential that some people take their kids out of school and do nothing with them. These people are called un-schoolers. I haven’t come across any, but I have heard of them. I do not approve of them. However, people who homeschool generally have a good reason for doing so. There are few homeschoolers who are irresponsible with their children. Also, [url=http://www.hslda.org/docs/nche/000010/200410250.asp]HSLDA[url] has some studies that show that most homeschoolers are good performers.

Also, I hear that there can be failure at public schools too. For instance, I have heard of kids who could not read until high school. I have also heard statistics that in cities, one third of highschoolers drop out. I do not the ratio of failure with homeschoolers and with public schools, but I am not guessing it is any worse, considering such statistics.

But let me put it this way, those who are going to be irresponsible will be so in public schools, or in homeschooling.

Finally, you have been making the feasibility of homeschooling look harder than it is. That is one point I want to dispel. Yes, parents do need some degree of knowledge to homeschool. But I really think that not a very high degree is needed. There are countless resources available to help parents learn if they need to. I have said this before, homeschooling has a lot of learning for the parents as well. Financially as well, homeschooling is the least expensive form of schooling I know. I do know of several single parents who have, or are homeschooling.

I will encourage you to poke around at http://www.hslda.org. I have referenced it several times. It deals with the legal issues of homeschooling. Its full title Home School Legal Defense Association.

Just half now -- more later :)

[quote="nayrb"](1) You say that homeschoolers are lacking in "schooling" that makes them a good "citizen." Or, at least suggesting that is the case. I suppose I was hinting (not stating clearly) that (1) you did specifically mention what schooling homeschoolers lacked, and (2) because you did not give an example, you cannot say that homeschoolers are not receiving that particular need for homeschooling, (3) even if you did state what public schools had that homeschoolers lack (in the context you were talking about), you would not have the same authority I have to know, since you have never been homeschooled, nor (I am guessing) had much contact with homeschoolers.[/quote]Did I say that?  I understand why you've interpreted it that way, but it's certainly not what I said.  I said, "I feel that there are certain things that everyone really should learn, regardless of the beliefs of their parents, and I don't like the idea of opting out via homeschooling."  I [i]didn't[/i] say, "I think that homeschoolers are all lacking important pieces of their education."  So I think your entire first paragraph is defending something which wasn't under attack.  My concern here is for the potential of homeschooling to be used to avoid particular important pieces of information about the world.  Do you see the difference?

[quote="nayrb"]My second part to this point was, again, not stated clearly. I said that while you may think that homeschoolers are lacking in some critical area, homeschoolers may think that people going to public schools are lacking in some critical area. For example, a person who is homeschooled may think that a person who goes to a public school does not have the same quality of a family, which is important. I am not taking this stand, but it is a supposition. My statement, "On the other hand." fits better here. Or what about the custom pace at which homeschoolers can go through their studies? Some people may think that public are critically lacking because of that.[/quote]Here, I think I didn't express myself clearly.  I think there is a [i]body of knowledge[/i] that every adult citizen of this country should possess, because it is important to society.  Whether or not we agree on the specifics of that body of knowledge, I think we can agree that there are things which homeschooling parents would be negligent not to teach (like, say, how to read).  All of society is the interested party here -- how does society ensure that homeschoolers are being taught these fundamental pieces of knowledge?

Ok, point taken. You are mainly concerned that homeschoolers have less liability to society, or for teaching important pieces of knowledge, because of the very nature of it.

I think that to a degree you can't ensure this with homeschooling. Although, it is a bit hard to understand specifically what you mean without enough examples. Frankly, I haven't seen any problem with my, or any of my friends homeschooling education. I could understand it better if you clarify your argument more specifically.

Wow, we have certainly been at it.  :)

I have been to a private school once, and all it gave me was disease, unhappiness, and boredom. The school wouldn't raise the difficulty of the subjects, so I was learning my times table at home, but I came home with 1+2=3. My parents knew that I was gifted.

They pulled me out of there and stuck me in the Learn at Home curriculum, but it was still too easy. I was learning eighth grade math when they pulled me out of there.

They finally found the Robinson curriculum, and it was independent study. I got to choose my mathematics and my writing, but the reading my dad was in charge of. I learned Algebra 1 and 2, pre-calculus, language skills and reading arts. I learn a lot faster, and I am still content with the curriculum.

I get to go to school competitions, like spelling bees, AMC 8 and 12, AIME, NYSML, ARML, Mathcounts, and I'm thinking about going to MOEMS. These contests are also social.

I also get to spend more time with my family. My dad and I watch movies, my mom and I solve puzzles, I play with my little brother and I enjoy my dog's company.

I also get 33% of the illnesses I used to get when I was in private school. I got earaches, allergies and colds by the truckload.

I actually never learned about religion, but I think that I follow my father's religion, Catholic. I was never baptized, but dedicated. I am actually confused with my religion.

I think that homeschooling is a very good idea. You get to meet new people and help other people. You also get to have new opportunities.


Duke college ARML AMC AIME USA(J)MO USAMO

Well, DMM 07 is only a little over a month away, so I decided to make a discussion topic :)

How is everyone doing in terms of getting a team, etc?  We haven't done any practice yet, but plan to soon.

What day is DMM07?  What day is PuMAC07?  I keep on forgetting if there is a conflict.

I plan to come in second to Haitao again. :D

erm: PuMAC is 11/10, Duke is 11/17. BTW Arnav demands that his problem be solved.

dude yeah wtf guys go do math

NC PERFORMANCE IS PROVING TO BE EXTREMELY DISAPPOINTING DO YOU GUYS WANT ME TO BE AN ARML COACH OR NOT SHEESH

haha.. yeah
east's team is practically dead.. i just hope we have enough ppl to compete
i'll be happy if we did

[quote="MysticTerminator"]dude yeah wtf guys go do math[/quote]

Sheesh, I've been working on it... some.

We have a team - probably two, as usual.

[quote="MysticTerminator"]NC PERFORMANCE IS PROVING TO BE EXTREMELY DISAPPOINTING[/quote]

How can it be disappointing if we have not yet competed in any contests?

Anyway, NC won't be terrible at ARML this year, but we probably won't be as good as last year/the year before.

It'd be pretty interesting with you as a coach, but I would not object.

dude you guys all failed at reading the post which was [i]immediately above mine[/i]

also east? who is on the team

yea ncssm is sending a couple of teams so I'll see you guys there  :)

I think Mr. Benton is planning on sending a North Buncombe High (my highschool) team.  Of course we'll get killed, but it should be fun.

Yeah... my school (Cary Academy...) is sending 2(?) teams. Sadly, everyone but me and one other kid got at most 2 points last year on individual :(

Enloe is sending a couple of teams.  Me and Vivek will carry our first team to victory, or so we hope.  The other team might not do quite as well.

I'm interested in East's team, because Jeremy will be on my team...

And why is that?  You don't refer to East Chapel Hill's team as 'our team', so I presume that you're not on their team.  But Jeremy is still a senior at East Chapel Hill, right?

[quote="Tenoreoz"]hm does anyone remember what number 4 was?[/quote]

Individual?  Or team?

If it was individual, then it was:

There are two people, A and B.  The probability that A wins a game is 70%.  They are playing ping-pong and the current score is 9-9.  The game ends when either one player reaches 11 points and the other is 2 or more behind, or when one player gets to be 2 points ahead of the other.  Find the probability that player B wins.

From the results posted by b-flat, Ihave determined the following:

Question difficulty (# of correct answers out of 139 ( % of correct answers))

1: 93 (66.9%)
2: 86 (61.9%)
3: 57 (41.0%)
4: 25 (18.0%)
5: 39 (28.15)
6: 15 (10.8%)
7: 1 (0.7%)
8: 0 (0.0%)

Score distribution:
8: 0
7: 1
6: 4
5: 13
4: 14
3: 20
2: 36
1: 22
0: 29

The average score was about 2.2.

dude yo where are these problems posted these look pretty tasty

I will add them to my slate of problems that I might want to do and never will

[quote="Karth"]Georgia >> NC now[/quote]

individual schools in NC<Georgia<<NC

[quote="MysticTerminator"]hmm let's see who has more MOPpers

MOPpers I know from North Carolina

jeremy

uh ok I can't remember any more right now

MOPpers I know from Georgia

hmmm

hmmmm

give me a sec I might think of one

hmmmmmmm

the guy who wears a suit

no wait he was in red

hmmmmmmmmm

ok that was kind of silly

let us go ahead and include red MOPpers

berman
damien

drobny

hmm[/quote]

We still won without any Moppers (apart from Harrison, who is in fact, a blue Mopper). However, I probably agree with jb05, though I don't understand how NC < Georgia with individual schools. >> meant to majorize.

[quote]Karth wrote:
Georgia >> NC now

individual schools in NC<Georgia<<NC[/quote]

well it depends... are the NC people still wearing pink shirts?

Well, [i]some[/i] of us wore them to Duke... including Mr. #1, Mr. #4, and Mr. loser = me.

[quote="Karth"]Georgia >> NC now[/quote]
This is hubris.  I am pretty sure that Georgia does not majorize North Carolina, and I disavow this claim.

Anyhow, usually $ \succ$ means "majorizes", whereas $ \gg$ usually means "is much greater than (i.e., this is a terrible bound)".

uh...majorization does not make any sense in this context

you realize one talks about sequences majorizing one another, yes?

I suspect that $ \{\text{NC's ARML Team}\} \succ \{\text{Georgia's ARML Team}\}$ though... :P

assuming the sums of the sets are equal, which may or may not be true

The sum of the sets is clearly not equal.

Oh by the way GUYS I WILL NOT BE GOING TO ARML BECAUSE APPARENTLY I LIED AND ACTUALLY LIVE IN VIRGINIA AND GO TO THOMAS JEFFERSON.  so see ya at ARML on the TJ team.  k?

[hide=" :D "]the joke behind this is that apparently my name changed to "John" temporarily, and then my school magically became TJ on the certificate.[/hide]

click the smiley

oh dude teki [color=red](mod edit)[/color] did you see that guy on the atlanta math club A team that looked kinda like you? i was like huh...didn't he just go up already?

No I did not.  I don't usually go around saying "Hey, man, you look a lot like me."  Also, please delete my name from your post.  You can replace it with 'teki' if you want - I just don't like people using my real name.


vector
function
Functional Analysis
advanced fields
advanced fields unsolved

$H$ a Hilbert. Prove that the only closed two-sided ideal of $L(H)$ is the ideal of compact operators in $L(H)$.

I knew of this result for separable Hilbert spaces (I saw it as a problem in Halmos's "A Hilbert Space Problem Book", where it was stated explicitly that the result doesn't sound this simple in general). Is it true in the most general setting? :?

sorry, I only know it is true for separable Hilbert.  :blush:

I forgot about this problem :).

A non-zero ideal $\frak a$ always contains an operator of rank $1$: if the range of $A\in\frak a$ contains a non-zero vector $e$, then $PA$ is an operator of rank $1$, where $P$ is the projection onto the subspace spanned by $e$. It follows from here that all operators of rank $1$ belong to $\frak a$: suppose, for instance, that the operator of rank $1\ T\in\frak a$ is defined by $Tf=(f,g)e$, and let $g',e'$ be arbitrary non-zero vectors. If $P,Q$ are unitary operators which send $e$ to $e'$ and $g$ to $g'$ respectively, then $T'=PTQ^*\in\frak a$ is defined by $T'f=(f,g')e'$.

Now let $T$ be an operator of finite rank, and let $e_1,\ldots,e_n$ be an orthonormal basis for its range. Then, if $P_i\in\frak a$ is the orthogonal projection onto the subspace spanned by $e_i$, we have $T=\sum_{i=1}^nP_iT\in\frak a$. This means that $\frak a$ contains all operators of finite rank. Finally, since every compact operator is the uniform limit of a sequence of operators of finite rank, $\frak a$ contains the ideal $\frak q$ of compact operators.

Now suppose it also contains a non-compact operator $A$. The Spectral Theorem can easily be avoided, but I like using it, and I think it makes things more transparent here :). The operator $A^*A$ will also be non-compact ($A$ is compact iff $A^*A$ is), and clearly, it's also Hermitian. This means that we can represent our Hilbert space as $L^2(X,\mu)$ for a finite measure $\mu$ such that $A^*A$ is a multiplication operator $M_\phi$ induced by $\phi\in L^\infty(X,\mu)$. Since $A^*A$, which is not a limit of operators of finite rank, is the limit of $M_{\phi\chi_n}$, where $\chi_n$ is the characteristic function of the set $S_n$ on which $|\phi|\ge\frac 1n$, there is an $n$ such that the range of $\phi\chi_n$ is infinite dimensional. Take $\tilde\phi$ to be $\frac 1\phi$ on $S_n$, and $0$ on $X\setminus S_n$. Then $\tilde\phi\in L^\infty$, and $M_{\tilde\phi}M_\phi=M_{\chi_n}\in\frak a$, i.e. $\frak a$ contains a projection $P=M_{\chi_n}$ onto an infinite dimensional subspace $\mathcal K$ of $\mathcal H$.

There are now two possibilities: either $\mathcal K^\perp$ is finite-dimensional, in which case we add to $P$ the projection onto $\mathcal K^\perp$ (which we know belongs to $\frak a$) to get the identity (i.e. $\frak a=\mathcal H$), or $\mathcal K^\perp$ is infinite dimensional, in which case there is a unitary operator $U$ mapping $\mathcal K$ onto $\mathcal K^\perp$ and vice-versa (because both $\mathcal K$ and its orthogonal complement are separable Hilbert spaces). Then $UPU^*$ will be the orthogonal projection onto $\mathcal K^\perp$, and $P+UPU^*=1$. We thus get $\frak a=\mathcal H$ again.

[quote="grobber"] Is it true in the most general setting? :?[/quote]
No. Another "obvious" closed two-sided ideal is that of operators with separable range.


calculus integration logarithms limit trigonometry function derivative

Calculate the following integrals:

1. $\int_{0}^\infty\frac{\ln^{2}x}{x^{2}+1}dx=$

and

2. Let $a$ be a real number such that $a^{2}-4\leq0$.  Calculate

$\int_{0}^\infty\frac{\ln^{2}x}{x^{2}+ax+1}dx=?$

Didi

Here is an almost complete solution to [b]problem (1)[/b]:

$I = \int_{0}^\infty\frac{\ln^{2}x}{x^{2}+1}dx = \int_{0}^{1}\frac{\ln^{2}x}{x^{2}+1}dx+\int_{1}^\infty \frac{\ln^{2}x}{x^{2}+1}dx$

We make the substitution $x=\frac1{y}$ in the second integral above. After some simplification, we obtain

$I = 2\int_{0}^{1}\frac{\ln^{2}x}{x^{2}+1}dx$

$= 2\int_{0}^{1}\ln^{2}x \sum_{k=0}^{\infty}(-1)^{k}x^{2k}dx = 2\sum_{k=0}^{\infty}(-1)^{k}\int_{0}^{1}(\ln^{2}x) x^{2k}dx = 2\sum_{k=0}^{\infty}(-1)^{k}I_{k}$ ... $(*)$,
where $I_{k}= \int_{0}^{1}(\ln^{2}x) x^{2k}dx .$

Now, $I_{k}= \frac1{2k+1}\int_{0}^{1}\ln^{2}x\ d(x^{2k+1}) =\frac1{2k+1}\left[ (\ln^{2}x) x^{2k+1}\Big|_{0}^{1}-\int_{0}^{1}x^{2k+1}\frac{2\ln x}{x}\ dx \right]$

$= \frac1{2k+1}\left[ 0-\lim_{x\rightarrow 0}(\ln^{2}x)x^{2k+1}-2 \int_{0}^{1}(\ln x)x^{2k}\ dx \right]$

$=-\frac{2}{(2k+1)^{2}}\int_{0}^{1}\ln x\ d(x^{2k+1})$

$=-\frac{2}{(2k+1)^{2}}\left[ (\ln x)x^{2k+1}\Big|_{0}^{1}-\int_{0}^{1}x^{2k+1}\cdot \frac1{x}\ dx \right]$

$=-\frac{2}{(2k+1)^{2}}\left[ 0-\lim_{x\rightarrow 0}(\ln x)x^{2k+1}-\int_{0}^{1}x^{2k}\ dx \right]$

$= \frac{2}{(2k+1)^{3}}$

So, the above result and $(*) \Rightarrow I = 4 \sum_{k=0}^{\infty}(-1)^{k}\frac{1}{(2k+1)^{3}}.$

Now, the above series, of course, converges (absolutely), and its value is $\frac{\pi^{3}}{8}$ (courtesy Mathematica)  but I don't know how to compute it. Anyone?

Maybe, a simple way of doing the problem could be by putting $x=\tan \theta.$ I will see if it can be done that way later.

FieryHydra,
That is my solution too. You do not have to calculate the sum since it is known that the value of that series is $\frac{\pi^{3}}{32}$.
However, I think that there is a more elegant solution of the first integral by using your substitution $x=\tan^{2}t$.
I will make use of the following well-known formulae involving Gamma function:
$\int_{0}^{\frac{\pi}{2}}\sin^{2p-1}x\cos^{2q-1}x dx =\frac{\Gamma(p)\Gamma(q)}{2\Gamma(p+q)}$,  for $Re p>0$, $Re q >0$,

and

$\Gamma(p)\Gamma(1-p)=\frac{\pi}{\sin p\pi}$
Now we have that
$\int_{0}^{\infty}\frac{\ln^{2}x}{1+x^{2}}dx=\int_{0}^{\frac{\pi}{2}}\ln^{2}\tan t dt$
Then,
$\int_{0}^{\frac{\pi}{2}}\tan^{\alpha}tdt=\int_{0}^{\frac{\pi}{2}}\sin^{\alpha}t \cos^{-\alpha}t dt = \frac{1}{2}\Gamma(\frac{1+\alpha}{2})\Gamma(\frac{1-\alpha}{2})=\frac{\pi}{2 \sin \frac{1+\alpha}{2}\pi}$
Take the second derivative of the above equality:
$\int_{0}^{\frac{\pi}{2}}\tan^{\alpha}t \ln^{2}\tan t dt = \frac{\pi^{3}}{8}\frac{\cos^{2}\frac{1+\alpha}{2}\pi}{\sin^{3}\frac{1+\alpha}{2}\pi}$
If $\alpha =0$, then
$\int_{0}^{\frac{\pi}{2}}\ln^{2}\tan t dt = \frac{\pi^{3}}{8}$.
Didi si Diana

Thanks, didilica.

One of these days, I must learn all about the Gamma function! Anyone good problem book on that one?


inequalities
geometry
inradius
circumcircle
inequalities unsolved

Dear all, I would like to see the proof of the inequality

$ 5 r^{2}-16 R r+s^{2}\geqslant 0.$

(with, as usual, $ r =$inradius, $ R=$circumradius, $ s=$semiperimeter)

Rigby and Bottema both cite this inequality from

J. Steinig. Inequalities concerning the inradius and circumradius of a
triangle. Elem. Math. 18 (1963), 127-131

Has anybody access to this source?

Best regards,

Francisco J. García Capitán.
http://garciacapitan.auna.com

denote by $ A$ its area.
Then $ A=sr=\frac{abc}{4R}$ where $ a,b,c$ are the sidelenghts of the triangle. So what we want to prove is \[ 5\frac{A^{2}}{s^{2}}+s^{2}\geq 4 \frac{abc}{A}\frac{A}{s}\] or equivalently \[ 5A^{2}+s^{4}\geq 4 abc s\]. Now use the identity $ A^{2}=s(s-a)(s-b)(s-c)$ simplify by $ s$ and make Ravi-substitution $ a=x+y, b=x+z, c=y+z$, $ x,y,z \geq 0$ such that $ s=x+y+z$. We want to prove that \[ 5xyz+(x+y+z)^{3}\geq 4(x+y)(y+z)(z+x)\] which after expanding is just Schur

Thank you, Crux.

I arrived to this inequality from this equivalent one

$ 2(a+b+c)(a^{2}+b^{2}+c^{2}) \geqslant 3 (a^{3}+b^{3}+c^{3}+3abc).$

Possibly it can be proven directly and we have another  proof of Steinig inequality.

Indeed this one is also true :) It can be proved by Schur... Actually it is completely equivalent to the one I posted beofre (after Ravi-substitution)


[quote="Mitzah"]Find all natural numbers $ x$ so that $ x^{4}\minus{}16x^{2}\plus{}100$ is a natural prime.[/quote]

Here are some hints:
$ f(x)\equal{}x^{4}\minus{}16x^{2}\plus{}100\equal{}(x^{2}\minus{}6x\plus{}10)(x^{2}\plus{}6x\plus{}10)$ so if $ f(x)\equal{}p$ then one of these two polynomials must be equal to $ 1$ and the other equal to $ p$ (here we do not need to worry about plus of minus since the polynomials are always positive)

[hide]$ x^{4}\minus{}16x^{2}\plus{}100 \equal{} (x^{2}\minus{}6x\plus{}10)(x^{2}\plus{}6x\plus{}10)$, and since of them must be one, we have. $ x^{2}\minus{}6x\plus{}10 \equal{} 1\Rightarrow{}x^{2}\minus{}6x\plus{}9 \equal{} 0$. Thus $ x \equal{}\boxed{3}$. We can check and $ 3^{4}\minus{}16\cdot3^{2}\equal{} 100 \equal{} 81\minus{}144\plus{}100 \equal{} 37$, sp we are correct.[/hide]


AMC
AIME
AMC 12
AMC 12 B
USA(J)MO
USAMO
percent

This is where you'll be able to vote and share the scores of 2006 AMC-12 B.

137

22 right 1 wrong 2 omit

(if anyone has a solution for 25, i'd really appreciate the help :D)

SO DARN CLOSE TO MAKING THE AIME!!! :mad:   i KNOW I SHOULDN'T WORRY CAUSE I'M ONLY IN 5TH/6TH GRADE BUT @*#%!!!  2 MORE PROBLEMS!!!  PLEASE REPLY TO MY OTHER POST :wallbash:  :wallbash_red:

126.5...EXACTLY the same as 12A.  (made a stupid mistake on the floor log one...)

[quote="surge"]Does anyone have a well-organized list of the answers (for the AMC12B? I want to know what score I recieved.[/quote]Yes, the AMC web site has the answers posted.

[url]http://www.unl.edu/amc[/url]

[quote="surge"]Does anyone have a well-organized list of the answers (for the AMC12B? I want to know what score I recieved.[/quote]

The UNL AMC site will have the official answers up soon.

However, by the end of tonight, AoPS forumers will most likely have posted all of the problems from the 12B, and complete with explained solutions :D

Edit: Actually the answers ARE up, you just have to dig a little :D :
http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2006-12a/06AMC12AB-answers.html

124  :(  :(  :(
19 right, 2 missed, 4 blank

Got #4 and #21 wrong because of adding errors.
And I spent 10 minutes on the increasing number one, because I couldn't figure out if increasing meant that in abc, a>b>c, or a<b<c. And I thought thaat 7 choose 2 =7*8/2=28.

There I go.... making excuses for my stupidity again. I don't know what to say anymore.

Well, other than my chances of qualifying for USAMO are shot. I'm stupid.

I got either 103 or 109... what I circled in the test booklet was 109, but I think I may have bubbled wrong :( .

130 for me here, I answered 1-20 all correctly and decided to try 22, but I missed it.

I'm pretty sure I'm in the 130-140 range. I didn't check my scores exactly, but I could have made one stupid mistake and still have 130. (I answered 21 and left 4 blank.)

I got 143. I omitted 23 and 24 and got the rest right.

[quote="1234567890"]
And I spent 10 minutes on the increasing number one, because I couldn't figure out if increasing meant that in abc, a>b>c, or a<b<c.
[/quote]

I did the same thing, so I kept getting 70.  I spent about 10 minutes on that one as well :(.  I eventually did get it though.

136, I was playing it conservative and decided to check over my work for the last 10-15 minutes. I'm happy, since I don't think I couldve done better under this strategy and given my skill level.

I heard Arnav Tripathy got a 150 on the A too, right?

[quote="Pakman2012"]I think this poll could've been a little more specific...

Yeah, thats what I've heard too, Arnav got a perfect on the A.[/quote]
I really don't think it needs to be considering that the title says, in big bold letters "AMC-12 [b]B[/b] Score Poll"...  ;)

[quote="Pakman2012"]I meant in terms of scores in the ranges.

I kinda wanted to see how I ranked out.  There was a pretty big difference this year between a 130 and a 137.  I guess I can wait a little while longer haha.[/quote]
Ah, okay, sorry.  Yeah, I agree it would have been nice to see a more precise breakdown, but it also would have made the poll itself painfully long - this poll at least gives a rough estimate of the distribution (and you can look at what people self-report for more info).

118.5 (130 on 10A)
Is that good for a freshman? >_>
EDIT: I actually got a 112.5, must have misbubbled something.

Yeah, I think 118.5 is really good for a freshman :)

I got 122, I think I did ok, considering the fact that I'm in 9th grade...

Congrats to those who passed on to the AIME.  And if you didn't, there is almost assuredly next year. Work hard, and you can make 130+ too.

[quote="57_factorial"]I only got 102 :(  I'm in grade 11 too... is that horrible? At least I made AIME and I'm school champion :)

12 right, 1 error, 12 blank.[/quote]It's not horrible, just making AIME is a great accomplishment.  On this forum the level of ability tends to be [b]extremely [/b]high compared to the general high school population (for example, on the SAT score polls the average is 1400-1500), so don't get discourages if you aren't tops here.

Yeah, the people on this forum do not represent a typical cross-section of the general student population. As eryaman noted, simply making AIME is a great accomplishment, so congrats!

[quote="57_factorial"]I only got 102 :(  I'm in grade 11 too... is that horrible? At least I made AIME and I'm school champion :)

12 right, 1 error, 12 blank.[/quote]

It's a great accomplishment to make AIME at all. However, it's great that you've registered on here, because I can guarantee that you'll learn a lot if you're determined. While it may not help you much for this year, this site will give you a big improvement for next. ;)

114.5, but according to what i circled i got 120.5, which is worse than what i got last year
17 right, 1 miss bubbled, 2 wrong 5 omitted
Did 1-13, 15-20 and 22

132. It's ridiculous: I solved the last 3 correctly but made stupid mistakes on 19,20, and 22. :wallbash:
What do you think the cutoff index for USAMO will be?


calculus
integration
function
complex numbers
complex analysis
complex analysis unsolved

Let $ f$ be an analytic function on a simple closed curve $ C$.  Show that $ Re\left(\int_C\bar{f}(z)f'(z)dz\right)\equal{}0$

no one?! :maybe:
this is a nice problem!

Let $ f(z)=f(x,y)=u(x,y)+iv(x,y)$ then $ f'(z)=u_x+iv_x$ and $ \overline{f}=u-iv$ so that
\begin{align*}
\overline{f}f'&=(u-iv)(u_x+iv_x) \\
&=(uu_x+vv_x)+i(uv_x-vu_x) \\
&=U+iV
\end{align*}
So that:
\begin{align*}
\oint (U+iV)dz&=\oint (U+iV)(dx+idy) \\
&=\oint (Udx-Vdy)+i\oint (Udy+Vdx)
\end{align*}

Now apply Green's Theorem to the real part:

\[ \oint (Udx-Vdy)=\mathop\iint\limits_{A}\bigg[-\frac{\partial V}{\partial x}-\frac{\partial U}{\partial y}\bigg] dx dy\]

Now, the partial of $ V$ wrt x is ok, but the partial of $ U$ wrt y would lead to mixed partials.  So I'll use the Cauchy Riemann equations to express $ U$ in terms of partials wrt $ y$:

\begin{align*}
U&=uu_x+vv_x \\
&=uv_y-vu_y
\end{align*}
Then:
\[ \text{Re}\oint \overline{f}(z)f'(z)dz=\mathop\iint\limits_{A} \left\{-\frac{\partial}{\partial x}\bigg(uv_x-vu_x\bigg)-\frac{\partial}{\partial y}\bigg(uv_y-vu_y\bigg)\right\}dxdy\]

That partial expression is zero.


Does anyone know why authors of math books tend to have "solutions to selected exercises" or "answers, hints, and solutions" instead of plain "solutions?" For me, a math book is piece of junk if it doesn't have solutions. You might as well throw it in the trash because you can't learn anything from it if they don't have solutions! What is the point of leaving out solutions to exercises-- that is the biggest part of learning a subject: working through problems.

I know.  All my math books have answers to only select problems.  They don't even have all the answers!  There for I only work the ones with answers and miss out on the good problems that don't have answers!

Because most math books are used in classes where the exercises are given for homework. They wouldn't work if people could just copy answers to their homework out of the book. College textbooks in particular get very stingy about giving answers to exercises.

Some books come with separate solution manuals though (usually available only to teachers). Otherwise, you just have to find someone who knows enough to tell you whether you got the questions right or not.

my maths book has the answers to all the odd problems, and all the practice tests in it...it comes in very handy and helps u see if ur doing the prob right or not

They usually get graduate students to write solutions or at least answers for the backs of textbooks. My differential geometry textbook has full solutions to every problem, but I don't think it's an especially good book.

[quote="tokenadult"][quote="mathnovice"]Does anyone know why authors of math books tend to have "solutions to selected exercises" or "answers, hints, and solutions" instead of plain "solutions?"[/quote]

The higher the level of the math book, the harder it is to prepare the solutions, and the smaller the market there is for the book in the first place. So preparing solutions just isn't done past a certain level.[/quote]

Yes, I've noticed that, and I absolutely hate it. For me, if there are examples in the text, then I don't really need solutions, only answers. When you don't have the answer, there is no way for you to know whether you are right or wrong. At least when there is an answer and you get it wrong, you can ask someone else.


logarithms
inequalities
function
limit
inequalities proposed

If $ a,b $ are positive numbers, let us denote $L(a,b)=\displaystyle  \frac{b-a}{\ln{b}-\ln{a}}\; ,\;  A(a,b)=\frac{a+b}{2}\; . $ [b] For $ 0<a<b $ prove that[/b]
\[\frac{(b-a)^2}{2b(b+a)(b+2a)}\le
\frac{1}{L(a,b)}-\frac{1}{A(a,b)}\le \frac{(b-a)^2}{2a(a+b)(a+2b)}\;
.
\]

hello,the left- handside of the inequality is equivalent to
$0 \le ln\left(\frac{b}{a}\right)-\frac{(a+5b)(b-a)}{2b(b+2a)}$. By setting $x=\frac{b}{a}$ we get: $0 \le ln(x)-\frac{5x^2-4x-1}{2(x^2+2x)}$. By defining the function
$f(x)=ln(x)-\frac{5x^2-4x-1}{2(x^2+2x)}$ we get desired result from $\lim_{x\to 1+0}f(x)=0$ and $f'(x)=\frac{(x-1)^2}{x^2(x+2)^2}>0$ for $x>1$. Analogously can the right-handside of the given inequality be proven.
Sonnhard.


algebra
linear equation
system of equations

Some fun easy problems (from Chinese 8th grade math text  :)  ):

1. Knowing $ a \plus{} b \plus{} c \equal{} 0$, find the value of $ a(\frac {1}{b} \plus{} \frac {1}{c}) \plus{} b(\frac {1}{a} \plus{} \frac {1}{c}) \plus{} c(\frac {1}{a} \plus{} \frac {1}{b})$.

2. If $ a^2 \minus{} 4a \plus{} 1 \equal{} 4$, find the value of $ \frac {a^3 \minus{} a^2 \plus{} a \minus{} 1}{a^2 \minus{} 1}$.

3. Find $ x$ for $ \sum_{i \equal{} 1}^{1994}\frac {1}{(x \plus{} i)(x \plus{} i \plus{} 1)} \equal{} \frac {2x \plus{} 3987}{3x \plus{} 5985}$.

4. Reduce $ \frac {\dfrac{x \plus{} 1}{x \minus{} 1} \minus{} \dfrac{x \minus{} 1}{x \plus{} 1}}{\dfrac{x \plus{} 1}{x \minus{} 1} \plus{} \dfrac{x \minus{} 1}{x \plus{} 1}}$.

5. Factor $ x(x \plus{} 1) \plus{} y(y \plus{} 1) \plus{} 2xy$.

6. Factor $ a^2 \plus{} b^2 \minus{} c^2 \minus{} 2ab$.

7. Factor $ x^2 \minus{} 6x(a \plus{} b) \plus{} 9(a \plus{} b)^2$.

8. Knowing $ \frac {1}{x} \plus{} \frac {1}{y} \equal{} \frac {1}{x \plus{} y}$, find the value of $ \frac {y}{x} \plus{} \frac {x}{y}$.

Have fun!!! (Chinese math is so fun  :lol: )

EDIT: I made a huge messup on #2. It's fixed now. I'm worry for wasting all you guys' time and  :(  :oops: .
EDIT 2: I couldn't find the original problem for #5, which leads me to assume I typoed again and again is sorry for all my stupid errors and such.

did the easy ones first  :lol:

[hide="6"]
(a-b-c)(a-b+c)
[/hide]

[hide="7"]
(x-3a-3b)^2
[/hide]

[hide="8"]
-1
[/hide]

1. $ a(\frac {1}{b} \plus{} \frac {1}{c}) \plus{} b(\frac {1}{a} \plus{} \frac {1}{c}) \plus{} c(\frac {1}{a} \plus{} \frac {1}{b}) \equal{} \frac {1}{a}(b \plus{} c) \plus{} \frac {1}{b}(a \plus{} c) \plus{} \frac {1}{c}(a \plus{} b)$

$ \frac {1}{a}(b \plus{} c) \plus{} \frac {1}{b}(a \plus{} c) \plus{} \frac {1}{c}(a \plus{} b) \plus{} 3 \minus{} 3 \equal{}$

$ \frac {1}{a}(b \plus{} c) \plus{} \frac {1}{a}\times a \plus{} \frac {1}{b}(a \plus{} c) \plus{} \frac {1}{b}\times b \plus{} \frac {1}{c}(a \plus{} b) \plus{} \frac {1}{c}\times c \minus{} 3 \equal{}$

$ \frac {1}{a}(a \plus{} b \plus{} c) \plus{} \frac {1}{b}(a \plus{} b \plus{} c) \plus{} \frac {1}{c}(a \plus{} b \plus{} c) \minus{} 3 \equal{} (a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \minus{} 3 \equal{} \minus{} 3$

answer : -3

3. $ \sum_{i\equal{}1}^{1994}\frac{1}{(x\plus{}i)(x\plus{}i\plus{}1)} \implies \frac{1}{x\plus{}1}\minus{}\frac{1}{x\plus{}1995}\equal{}\frac{2x\plus{}3987}{3x\plus{}5985}$

$ (x\plus{}1)(x\plus{}1995)(2x\plus{}3987)\equal{}1994\times3(x\plus{}1995)$

$ x\equal{}\minus{}1995, \frac{1}{2}$

4. $ \frac {2x}{x^2 \plus{} 1}$

Will edit later.

[hide=" 3 "]On #3, I telescoped it, then got totally screwed over, large cubic equation  :( ...I need a nicer way.[/hide]

[hide="2"]On #2,I know we shouldn't solve for a, but I get to $ \frac{4\minus{}4a}{a\plus{}1}$
I brute-forced it, and got $ 4\sqrt{5}\minus{}8$[/hide]

[hide="5"]On #5, should it be y(y+1)?[/hide]

#1, you can expand combine the ones like $ a^2b\plus{}ab^2$ factor then get the answer

help?

Nope, zapi, not right.

On #3, well, you have to make everything neater first.

On #2, you cant brute it, especially because solving for x wont get you the right answer.

For #5 DOH youre right  :oops:
Sorry  :(
Fantasy did #1 correctly, BTW.

#1
$ a\left( \frac{1}{b}\plus{} \frac{1}{c} \right) \plus{} b\left( \frac{1}{a}\plus{} \frac{1}{c} \right) \plus{} c \left( \frac{1}{a}\plus{} \frac{1}{b} \right) \\ \\ \equal{} \minus{}\left(\frac{a^2}{bc}\plus{}\frac{b^2}{ac}\plus{}\frac{c^2}{ab}\right) \\ \\ \equal{} \minus{} \left(\frac{a^3\plus{}b^3\plus{}c^3}{3abc}\right) \\ \\ \equal{} \minus{} \left(\frac{(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)\plus{}3abc}{abc}\right) \\ \\ \mbox{Since} \ a\plus{}b\plus{}c\equal{}0 \ \mbox{we have} \\ \\ \equal{} \minus{}3$

[hide="Nice solution #3"]$ S\equal{}\sum_{i \equal{} 1}^{1994}\frac {1}{(x \plus{} i)(x \plus{} i \plus{} 1)} \equal{} \frac {2x \plus{} 3987}{3x \plus{} 5985}$
$ \equal{}\frac{1}{(x\plus{}1)(x\plus{}2)}\minus{}\frac{1}{(x\plus{}2)(x\plus{}3)}\ldots\ldots\frac{1}{(x\plus{}1994)(x\plus{}1995)}$
$ \equal{}\frac{1}{x\plus{}1}\minus{}\frac{1}{x\plus{}2}\plus{}\frac{1}{x\plus{}2}\minus{}\frac{1}{x\plus{}3}\plus{}\frac{1}{x\plus{}3}\ldots\ldots\frac{1}{x\plus{}1994}\minus{}\frac{1}{x\plus{}1995}$

The alternate terms get canceled.

$ S\equal{}\frac{1}{x\plus{}1}\minus{}\frac{1}{x\plus{}1995}\equal{}\frac{2x\plus{}3987}{3(x\plus{}1995)}$
So,

$ \frac{1994}{x\plus{}1}\equal{}\frac{2x\plus{}3987}{3}\ \ \ \ \ [x\ne\minus{}1995]$
$ \Rightarrow 2x^2\plus{}3990x\minus{}x\minus{}1995\equal{}0$
$ \Rightarrow 2x(x\plus{}1995)\minus{}1(x\plus{}1995)\equal{}0$

Or,$ x\equal{}\frac{1}{2}$[/hide]

[quote="zapi2007"]
On #2,I know we shouldn't solve for a, but I get to $ \frac {4 \minus{} 4a}{a \plus{} 1}$
I brute-forced it, and got $ 4\sqrt {5} \minus{} 8$
help?[/quote]

This is not correct because $ a^2 \plus{} 4a \plus{} 1 \equal{} 4\iff a \equal{} \minus{} 2\pm\sqrt {7}$, but the result $ \frac {a^3 \minus{} a^2 \plus{} a \minus{} 1}{a^2 \minus{} 1} \equal{} \frac {4 \minus{} 4a}{a \plus{} 1}$ is safe.

I think, there is a mistake in the constraint. If we take $ a^2\minus{} 4a\plus{}1\equal{}4$ then $ \frac {a^3 \minus{} a^2 \plus{} a \minus{} 1}{a^2 \minus{} 1}\equal{}4$.

[quote="FantasyLover"][quote="Euclid_great"][hide="Nice solution #3"]$ S \equal{} \sum_{i \equal{} 1}^{1994}\frac {1}{(x \plus{} i)(x \plus{} i \plus{} 1)} \equal{} \frac {2x \plus{} 3987}{3x \plus{} 5985}$
$ \equal{} \frac {1}{(x \plus{} 1)(x \plus{} 2)} \minus{} \frac {1}{(x \plus{} 2)(x \plus{} 3)}\ldots\ldots\frac {1}{(x \plus{} 1994)(x \plus{} 1995)}$
$ \equal{} \frac {1}{x \plus{} 1} \minus{} \frac {1}{x \plus{} 2} \plus{} \frac {1}{x \plus{} 2} \minus{} \frac {1}{x \plus{} 3} \plus{} \frac {1}{x \plus{} 3}\ldots\ldots\frac {1}{x \plus{} 1994} \minus{} \frac {1}{x \plus{} 1995}$

The alternate terms get canceled.

$ S \equal{} \frac {1}{x \plus{} 1} \minus{} \frac {1}{x \plus{} 1995} \equal{} \frac {2x \plus{} 3987}{3(x \plus{} 1995)}$
So,

$ \frac {1994}{x \plus{} 1} \equal{} \frac {2x \plus{} 3987}{3}\ \ \ \ \ [x\ne \minus{} 1995]$
$ \Rightarrow 2x^2 \plus{} 3990x \minus{} x \minus{} 1995 \equal{} 0$
$ \Rightarrow 2x(x \plus{} 1995) \minus{} 1(x \plus{} 1995) \equal{} 0$

Or,$ x \equal{} \frac {1}{2}$[/hide][/quote]

eh thats exactly what i did...[/quote]

Sorry oveerlooked your post. :blush:  :(

[quote="fishythefish"]What's the problem with the constraints, m@ri@n? We're supposed to find the value of the second equation, so 4 would be a solution (if it's correct), and not a problem.[/quote]

The problem is with the value $ \sqrt {5}$ which can not occurs in the final value of the original term. The value $ 4\sqrt {5} \minus{} 8$ is false.

I think it is not possible to find the value of the original term with the original constraint without the solution of the equation $ a^2\plus{}4a\plus{}1\equal{}4$.

What do you mean by the original term?

Besides, if this problem is like the others of its kind, then the key to the solution is manipulating the first equation to get the second one.

[quote="M@ri@n"][quote="zapi2007"]
On #2,I know we shouldn't solve for a, but I get to $ \frac {4 \minus{} 4a}{a \plus{} 1}$
I brute-forced it, and got $ 4\sqrt {5} \minus{} 8$
help?[/quote]

This is not correct because $ a^2 \plus{} 4a \plus{} 1 \equal{} 4\iff a \equal{} \minus{} 2\pm\sqrt {7}$, but the result $ \frac {a^3 \minus{} a^2 \plus{} a \minus{} 1}{a^2 \minus{} 1} \equal{} \frac {4 \minus{} 4a}{a \plus{} 1}$ is safe.

I think, there is a mistake in the constraint. If we take $ a^2 \minus{} 4a \plus{} 1 \equal{} 4$ then $ \frac {a^3 \minus{} a^2 \plus{} a \minus{} 1}{a^2 \minus{} 1} \equal{} 4$.[/quote]If it was supposed to be fully based on a clever trick, $ a^2 \minus{} 4a \plus{} 1 \equal{} 4$ should have been given.
Otherwise, you dead end at $ \frac {4 \minus{} 4a}{a \plus{} 1}$, and can only substitute $ a \equal{} \minus{} 2\pm\sqrt {7}$ to get $ \frac {\minus{} 8\pm4\sqrt {7}}{3}$.
The linear equation $ \frac {4 \minus{} 4a}{a \plus{} 1}\equal{}4\sqrt {5} \minus{} 8$ has $ a\equal{}\frac {\sqrt {5}\minus{}1}{2}$ as a solution, which does not satisfy $ a^2 \plus{} 4a \plus{} 1 \equal{} 4$.

Okay, I'm thoroughly confused now. What's the problem with the problem, where did (4-4a)/(a+1) come from, and where did someone come up with $ 4\sqrt{5}\minus{}8$?

[hide="Solution to Number 2"]
We have $ \frac{a^3\minus{}a^2\plus{}a\minus{}1}{a^2\minus{}1}$. We group the numerator like this: $ \frac{a^3\plus{}a\minus{}a^2\minus{}1}{a^2\minus{}1}$. So now we can factor.
$ \frac{a^3\plus{}a\minus{}a^2\minus{}1}{a^2\minus{}1}\equal{}$

$ \frac{(a\minus{}1)(a^2\plus{}1)}{a^2\minus{}1}$.

We also factor the denominator.

$ \frac{(a\minus{}1)(a^2\plus{}1)}{(a\plus{}1)(a\minus{}1)}$

Simplify.

$ \frac{a^2\plus{}1}{a\plus{}1}$.

Now we take $ a^2\minus{}4a\plus{}1\equal{}4$, and turn it into $ a^2\plus{}1\equal{}4a\plus{}4$.
And substitute that into our original equation.

$ \frac{a^2\plus{}1}{a\plus{}1}\equal{}\frac{4a\plus{}4}{a\plus{}1}$

The top side is obviously 4 times the bottom side, so our final answer is $ \boxed{4}$.  :) [/hide]

ugh, you made a typo..

i got to $ \frac{a^2\plus{}1}{a\plus{}1}$ but i couldnt solve it unless i know the actual value of a...

Oh no, I did. Sorry people  :oops:  :blush: !

[hide]
[quote="KMST"][quote="M@ri@n"][quote="zapi2007"]
On #2,I know we shouldn't solve for a, but I get to $ \frac {4 \minus{} 4a}{a \plus{} 1}$
I brute-forced it, and got $ 4\sqrt {5} \minus{} 8$
help?[/quote]

This is not correct because $ a^2 \plus{} 4a \plus{} 1 \equal{} 4\iff a \equal{} \minus{} 2\pm\sqrt {7}$, but the result $ \frac {a^3 \minus{} a^2 \plus{} a \minus{} 1}{a^2 \minus{} 1} \equal{} \frac {4 \minus{} 4a}{a \plus{} 1}$ is safe.

I think, there is a mistake in the constraint. If we take $ a^2 \minus{} 4a \plus{} 1 \equal{} 4$ then $ \frac {a^3 \minus{} a^2 \plus{} a \minus{} 1}{a^2 \minus{} 1} \equal{} 4$.[/quote]If it was supposed to be fully based on a clever trick, $ a^2 \minus{} 4a \plus{} 1 \equal{} 4$ should have been given.
Otherwise, you dead end at $ \frac {4 \minus{} 4a}{a \plus{} 1}$, and can only substitute $ a \equal{} \minus{} 2\pm\sqrt {7}$ to get $ \frac { \minus{} 8\pm4\sqrt {7}}{3}$.
The linear equation $ \frac {4 \minus{} 4a}{a \plus{} 1} \equal{} 4\sqrt {5} \minus{} 8$ has $ a \equal{} \frac {\sqrt {5} \minus{} 1}{2}$ as a solution, which does not satisfy $ a^2 \plus{} 4a \plus{} 1 \equal{} 4$.[/quote]
[/hide]

Exactly!

Problem 5: Factor $ x(x + 1) + y(y - 1) + 2xy$.

The factorization (if exists) has the form
$ x(x + 1) + y(y - 1) + 2xy = (ax + by + c)(dx + ey + f)$.
This is equivalent to
$ x(x + 1) + y(y - 1) + 2xy = acx^2 + (ad + bc)xy + bdy^2 + (af + ec)x + (bf + ed)y + ef$.

[b]Case 1:[/b] $ \boxed{e = 0}$. We shall find the numbers $ a,b,c,d,f\in\mathbb{R}$ such that
\begin{align} bf & = 1, \\
af & = 1, \\
bd & = 1, \\
ad + bc & = 2, \\
ac & = 1. \end{align}
But (1) is equivalent with $ abf = - a$ and (2) with $ abf = b$. Hence $ - a = b\iff b - a = - 2a$. Similarly, from (2) and (5) we obtain $ c = f$ and (1) and (3) imply $ f = - d$. The equation (4) can be rewritten as
$ ad + bc = 2\iff (b - a)c = 2\iff - ac = 1\iff ac = - 1$
which contradicts the fact that $ ac = 1$ in (5). Hence the system of equations (1)-(5) has no solution.

[b]Case 2:[/b] $ \boxed{f = 0}$. See [b]Case 1[/b].

A slight (solvable) modification of Problem 5:
$ x(x + 1) + y(y + 1) + 2xy = (x + y + 1)(x + y)$. :)

[quote="Textangle"]There's actually a... simpler way to do #1[/quote]

I didn't see this solution, perhaps this is it?

[hide="Number 1"]
$ a\left( \frac{1}{b}\plus{} \frac{1}{c} \right) \plus{} b\left( \frac{1}{a}\plus{} \frac{1}{c} \right) \plus{} c \left( \frac{1}{a}\plus{} \frac{1}{b} \right) \\
\\
\equal{} a\left(  \frac{1}{a} \plus{} \frac{1}{b}\plus{} \frac{1}{c} \right) \minus{} 1 \plus{} b\left( \frac{1}{a}\plus{} \frac{1}{b} \plus{} \frac{1}{c} \right) \minus{} 1 \plus{} c \left( \frac{1}{a}\plus{} \frac{1}{b} \plus{} \frac{1}{c} \right) \minus{} 1 \\
\\
\equal{} (a \plus{} b \plus{} c) \left( \frac{1}{a}\plus{} \frac{1}{b} \plus{} \frac{1}{c} \right) \minus{} 3 \\
\\
\equal{} \minus{}3$
[/hide]

[quote="FantasyLover"]um simpler way=

$ a(\frac {1}{b} \plus{} \frac {1}{c}) \plus{} b(\frac {1}{a} \plus{} \frac {1}{c}) \plus{} c(\frac {1}{a} \plus{} \frac {1}{b}) \equal{} a(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \minus{} 1 \plus{} b(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \minus{} 1 \plus{} c(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \minus{} 1 \equal{} \minus{} 3$?[/quote]

...exactly what i did.

EDIT : whoa its exactly same

Because it got changed to $ a^2\minus{}4a\plus{}1\equal{}4$, move the 4a over to the other side, plug it in, then cancel the $ a\plus{}1$ to get 4.


[quote="pencilkarthik"]Find the number of non-negative integral solutions in (x,y) for

$x^{3}+8x^{2}-6x+8 = y^{3}$[/quote]
$(x+1)^{3}<y^{3}<(x+3)^{3}.$ Hence, $y=x+2.$
$(0,2)$ and $(9,11).$


calculus
integration
algebra
function
domain
abstract algebra
superior algebra

$ R$ is a commutative, infinite, unitary ring. For every non-trivial ideal $ I$ in $ R$, the quotient ring $ R/I$ is finite. Prove that $ R$ must be an integral domain.
[Example of such a ring: $ R = \mathbb{Z}$ ].

[hide]for any $ a,b$, the canonical $ R/(ab) \rightarrow R/(a) \times R/(b)$ is an injection. (Note that the given condition implies that $ (a) \cap (b) \equal{} (ab)$, thus the map is injective.

If $ ab \equal{} 0$ but $ a,b \neq 0$, then left hand side is infinite, right hand side is finite.[/hide]

I don's see $ (a)\cap (b) \equal{} (ab)$. How do you get that?

Anyway if $ x\not \equal{} 0$, then $ R/Rx$ is finite by assumption.
Also the kernel $ I$ of $ R\to Rx,y\mapsto yx$ is an ideal with $ R/I\cong Rx$ as $ R$-modules.
So for $ I\not \equal{} 0$ we have $ |R| \equal{} |Rx|\cdot |R/Rx| \equal{} |R/I|\cdot |R/Rx|$ finite. Contradiction!
So $ I \equal{} 0$ and $ x$ is not a zero-divisor.

[quote="olorin"]I don's see $ (a)\cap (b) \equal{} (ab)$. How do you get that?
[/quote]

I'm sorry, I actually needed $ (a)\plus{}(b) \equal{} R$ to conclude.

[quote="olorin"]
Also the kernel $ I$ of $ R\to Rx,y\mapsto yx$ is an ideal with $ R/I\cong Rx$ as $ R$-modules.
[/quote]
Sorry, but I don't understand why $ R/I\cong Rx$ - I don't think the mapping $ y \mapsto yx$ is a homomorphism unless $ x$ is idempotent, so the First Isomorphism Theorem isn't applicable - how else can we conclude that $ R/I\cong Rx$?.

$ R\to Rx, y\mapsto yx$ is certainly an epi of abelian groups with kernel $ I$.
So $ R/I\cong Rx$ as abelian groups. (or as $ R$-modules, not as rings)
So $ |R/I|\equal{}|Rx|$.


this is sumthing really stupid

but go 2 the fun factory, and go 2 TOP COUNTRY of AoPS

india is failing :(

i need ur help!

 :rotfl:

hey but i think i already won one round for India...those guys won't stop taking pts off india :mad:

i wud love to end it today

its this game to see what is the top country of aops (as the name suggests)

i cant win it by myself...

but i see that readnow and rituraj have contributed...

[quote="ernie"]
but i see that readnow and rituraj have contributed...[/quote]

readnow = ritu! isnt that obvious there?  :huh:

oh my bad

 :rotfl:

remember... im new to this forum :P

[quote="Da Vinci"]
readnow = ritu! isnt that obvious there?  :huh:[/quote]
how can i be online on 2 id's at a time?? :rotfl:

get on one

open a new window

log on as a different user :P

well anyways, india just lost 10 points that ritu and read gained for us

but we're tied for first :thumbup:

with 19 points :noo:

ritu $ \neq$ read

oh btw, we're like 10th... JOIN THE REBELLION!

Thanks @ernie and ritu for keeping india in the competition!
i posted once,but a member 1=2 instead deducted two points from india!

so you all carry on!
 :)

he did that as his turn

and akashnil and I got india up to 43 points!!!

Congrats for India!

India is owning with 54 points...

Mwahahahahaha... India pwned with 100 points...

well ,i wud say we shud leave US this time
i think they finished at bottom(almost) in all 3 earlier games

india supposedly the winner with 112 points .
thanks to Reon, rt_08, ernie, gaurav and all others who supported INDIA.

[quote="ernie"]Too bad, the US finished 15th, and two newcomers are PWNING the game...[/quote]

what do u mean by PWNING the game? :huh:

PS:Has any country ever won with such a huge score?
If not we made history! And shall do so again!

OMG u guys are so awesome, posting some 60 pages or something in just 2 1/2 hours and making india win. :rotfl:  :rotfl:  :rotfl:

The highest score before this was $ 96$. (first game)

And as for [url=http://en.wikipedia.org/wiki/Pwn]pwning[/url]...

At last the final score.......
India won with 132 pts !!!!

[quote="apratimdefermat"]1 [b][color=orange]Indyeah[/color][/b]! - 132  :first:

2 Ireland
3 Brazil
4 Australia
5 Canada
6 Russia
7 Mexico
8 Spain
9 France
10 Poland
11 China
12 United Kingdom
12 Finland
14 Japan
15 U.S.
16 Belgium[/quote]

beating india is not a childs play :P  :D

(omly Aussie cud do that :( )

[quote="rituraj007"]beating india is not a childs play :
(only Aussie cud do that [/quote]

Too bad they finished 4th :rotfl:

India has crushed :bruce: them.   :stretcher:

Beating INDIA is impossible coz we stand united!

[quote="ernie"]And as for [url=http://en.wikipedia.org/wiki/Pwn]pwning[/url]...[/quote]
OMG they have an article for [i]this[/i] :o  :P  :rotfl:

i was  talking abt the prev game

and well u can urself rite an article on wikipedia--so dont expect limits

[quote="ernie"]In order to give us a break, I removed India from the list.[/quote]
Did the right thing. You ppl need a break. I'll surely join the next game for India.

hm... one of us is adding india back...?

its reon
the boy has a point,and is a little emotional on this topic
cool reon :)  :)  :)

lol everyone got so fed up with us, that they changed top country to top planets  :D

chak de india! (is that necessary, I don't know much :oops: )


A Weston galvanometer has a resistance of 120 ohms; an 0.05-ohm shunt converts it into an ammetet. If the ammeter reads 1.2 amperes, how much current is in the galvanometer?


Given an orientated graph with no cicles, each node has a cost associated to it. If there is an edge from node A to node B,  pointing from A to B, we say that node B depends on A. all nodes are initially in set X. We need to get all the nodes in set Z, through set Y. at one step, we can bring one node from X into Y,  or from Y into Z. a node A can get from X to Y if all the nodes that A depends on are already in Y, or A depends on no nodes. the same node A can get from Y to Z if all the nodes that depend on A are in Y.
we define the norm of Y as the sum of the costs of all the nodes in Y.
The problem is this:
determine the order in wich nodes pass from X into Y and from Y into Z so that the maximum norm of Y is minimum.
this problem came up while building a program. to me, the cost of the nodes is memory.
please give it some thought. thanks.

Well the problem description is a bit ambiguous to me. What’s ur goal? To minimize the |Y| of any time?

Consider, A depends on B and B depends on C.

Then C moves to Y. Current Cost 1. Then B moves to Y. Current cost 2. Then C moves to Z. current cost 1. then A moves to Y. current cost 2. then B moves to Z. current cost 1. then A moves to Z. current cost 0. That means in this case ur output is 2. Am i right?

Let me know that..

But one thing: I think this problem can be solved with the idea of DAG (directed acyclic graph) :D

Mahbub

my goal is not to minimize |Y| at all times, but to minimize the maximum of |Y|. so for me, it's better if |Y| stays around say 20 and below 25 all the time, rather than if it stays around 10,  but gets as high as 30; and consider more complicated graphs. for example
A depends on A1 and A2, B depends on B1 and B2, C depends on A and B. costs are : A1 2,  A2 2, A 1, B1 1,  B2 1, B 3, C2. this is still a simple example. but we could add a node  D depending on A1, B and C and that would complicate things. i need an algorithm able to solve this for any acyclical graph.

[quote="rareshb"]costs are : A1 2,  A2 2, A 1, B1 1,  B2 1, B 3, C2.[/quote]

HOW?? In this case ur answer is: 3 right?

i will be away for next 2 days. but i am pretty confident it is use of DAG. Ok I will try!

no, in this case the answer is 6:
A1 X->Y         |Y| = 2
A2 X->Y         |Y| = 4
A   X->Y         |Y| = 5
A1 Y->Z         |Y| = 3
A2 Y->Z         |Y| = 1
B1 X->Y         |Y| = 2
B2 X->Y         |Y| = 3
B   X->Y         |Y| = 6    <------ max
B1 Y->Z         |Y| = 5
B2 Y->Z         |Y| = 4
C  X->Y          |Y| = 6   <------- max
A  Y->Z          |Y| = 5
B  Y->Z          |Y| = 2
C  Y->Z          |Y| = 0

if you take b1, b2, b first then max cost is 8


Does anyone know where this problem is from?

Let [tex]a_1, a_2, ..., a_n (n > 3)[/tex] be real numbers.
Suppose:

[tex]a_1 + ... +a_n \geq n[/tex]

and

[tex]a_1^2 + ... a_n^2 \geq n^2.[/tex]

Prove that [tex]Max\{a_1, ..., a_n\} \geq 2.[/tex]

Thanks.

[quote="fuzzylogic"][quote="Ravi B"]It's from the 1999 USAMO, Problem 4.  It was discussed on the following link:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10074[/url][/quote]
[quote="Ravi B"]Here's yet another method: [/quote]

Your solution is very nice![/quote]
Thanks!  It took me many pages of false starts to come up with, so I'm glad you liked it.

This is my favorite problem! Thanks for that great solution Ravi.