trigonometry
geometry
incenter
perimeter
analytic geometry
circumcircle
geometric transformation

1. Show that $b^2 \sin 2A + a^2 \sin 2B = c^2 \sin 2B + b^2 \sin 2C = a^2 \sin 2C + c^2 \sin 2A = 4\triangle$. ($\triangle$ is the area of $\triangle ABC$.)

2. Show that $abc = aAI^2 + bAI^2 + cAI^2$. ($I$ is the incentre of $\triangle ABC$.)

[quote="Centy"]1. Show that $b^2 \sin 2A + a^2 \sin 2B = c^2 \sin 2B + b^2 \sin 2C = a^2 \sin 2C + c^2 \sin 2A = 4\triangle\ ,$
where $\triangle$ is the area of $\triangle ABC\ .$

2. Show that $abc = aAI^2 + bAI^2 + cAI^2\ ,$
where $I$ is the incentre of $\triangle ABC\ .$[/quote]=======================================================

\begin{eqnarray*}1.\blacktriangleright b^2\sin 2A+a^2\sin 2B & = & 2b\cdot b\sin A\cdot \cos A+2a\cdot a\sin B\cdot \cos B\\\ & = & 2b\cdot a\sin B\cdot \cos A+2a\cdot b\sin A\cdot \cos B\\ & = & 2ab\cdot (\sin A\cos B+\sin B\cos A)\\ & = & 2ab\sin (A+B)\\ & = & 2ab\sin C\\ & = & 4\Delta\ .\end{eqnarray*}

[b][u]Remark 1.[/u][/b]  $\sin 2x=2\sin x\cos x\ \ ;\ \ \sin (x+y)=\sin x\cos y+\sin y\cos x\ \ ;$

$x+y=\pi\Longrightarrow\sin x=\sin y\ \ ;\ \ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\ \ .$

[b][u]A geometrical interpretation.[/u][/b]

Denote the projection $D$ of the vertex $A$ on the opposite sideline $B\ .$ Then :

$AD=c\sin B=b\sin C\ ;$ $DB=c\left|\cos B\right|$ and $DC=2\left|\cos C\right|\ ;$

$\mathrm {sgn} (\sin 2A)=\mathrm {sgn}\left(90^{\circ}-A\right)\ ,$ i.e. $\sin 2A=\left|\sin 2A\right|\cdot \mathrm {sgn}(90^{\circ}-A)$ a.s.o. ;

$4\cdot [ABD]=2\cdot AD\cdot DB=c^2\left|\sin 2B\right|$ and

$4\cdot [ACD]=2\cdot AD\cdot DC=b^2\left|\sin 2C\right|\ .$ Therefore ,

$4\Delta=4[ABD]\mathrm {\ sgn}(90^{\circ}-B)+4[ACD]\mathrm{\ sgn}(90^{\circ}-C)\Longrightarrow$

$4\Delta =c^2\left|\sin 2B\right|\cdot \mathrm {sgn} (90^{\circ}-B)+b^2\left|\sin 2C\right|\cdot \mathrm {sgn}(90^{\circ}-C)\Longrightarrow$

$4\Delta =c^2\sin 2B+b^2\sin 2C\ .$

[b][u]Remark 2.[/u][/b]

$\frac{\sin 2A}{a^2(b^2+c^2-a^2)}=\frac{\sin 2B}{b^2(c^2+a^2-b^2)}=\frac{\sin 2C}{c^2(a^2+b^2-c^2)}=\frac{2\Delta }{(abc)^2}\Longrightarrow$

$b^2\sin 2C+c^2\sin 2B=4\Delta$ because

$\frac{2\Delta }{(abc)^2}\cdot \left[b^2\cdot c^2(a^2+b^2-c^2)+c^2\cdot b^2(a^2+c^2-b^2)\right]=4\Delta\ .$

=====================================================

$2.\blacktriangleright \sum a\cdot AI^2=\sum a\cdot \frac{bc(p-a)}{p}=\frac{abc}{p}\cdot \sum\ (p-a)=abc\ .$

[b][u]Remark 3.[/u][/b] $D\in BC\cap AI$ $\Longrightarrow$ $AD=\frac{2}{b+c}\sqrt{bcp(p-a)}\ ;$

$\frac{IA}{b+c}=\frac{ID}{a}=\frac{AD}{2p}$ $\Longrightarrow$ $\boxed {AI^2=\frac{bc(p-a)}{p}}\ ,$ where $2p=a+b+c\ .$

In general , for any point $X$ from the plane of the triangle $ABC$ we have

$\boxed {\ a\cdot XA^2+b\cdot XB^2+c\cdot XC^2=(a+b+c)\cdot XI^2+abc\ }\ .$

Just a clarification...is $p$ the perimeter or the semi-perimeter of the triangle?

[quote="Virgil Nicula"]

.... $\frac{IA}{b+c}=\frac{ID}{a}=\frac{AD}{2p}$ .....$(!)$

In general , for any point $X$ from the plane of the triangle $ABC$ we have

$\boxed {\ a\cdot XA^2+b\cdot XB^2+c\cdot XC^2=(a+b+c)\cdot XI^2+abc\ }\ \ \ (*)_\ .$[/quote]From the relation $(!)$ you see what is $p\ !$

The relation $(*)$ is a well-known extension of the your identity $(2)\ .$

:blush: Well now I see.

The other extension (which is where my identity came from) is that if $P$ is a point in $\triangle ABC$ with normalised areal coordinates $(\lambda, \mu, \nu)$ then for any other point $X$ in the plane \[ \lambda AX^2 + \mu BX^2 + \nu CX^2 = \lambda AP^2 + \mu BP^2 + \nu CP^2 + PX^2  \]

[b][u]Yes ![/u][/b] $\lambda\cdot PA^2+\mu\cdot PB^2+\nu\cdot PC^2=-p(P)\ ,$ where $p(P)=-\left(\mu\nu a^2+\nu\lambda b^2+\lambda\mu c^2\right)$ is the power of the point $P$
w.r.t. the circumcircle of the triangle $ABC\ .$
In general, these identities are well-known.

See in the above my reply the geometrical interpretation of the first identity.

[color=red][b][u]Lemma.[/u][/b] $bc=2Rh_a\ ;\ abc=4RS\ ;\ S=2R^2\sin A\sin B\sin C\ .$

[u]Proof.[/u] Denote the circumcircle $C=C(O,R)$ of the triangle $ABC$, the projection $D$ of the vertex $A$ on the opposite sideline $BC$ and the reflection $A'$ of the same point $A$ w.r.t. the circumcentre $O\ .$
$\widehat {ABC}\equiv \widehat {AA'C}\ ,\ AD\perp BD\ ,\ AC\perp A'C\Longrightarrow$ $\widehat {BAD}\equiv \widehat {CAO}$, i.e. the rays $[AD$ and $[AO$ are isogonals in the angle $\widehat {BAC}$ and $\triangle ABD\sim\triangle AA'C$ $\Longrightarrow$ $\frac{AB}{AA'}=\frac{DA}{CA}$ $\Longrightarrow$ $\frac{c}{2R}=\frac{h_a}{b}$ $\Longrightarrow$ $\boxed {\ bc=2Rh_a\ }$ and $ah_a=2S$ $\Longrightarrow$ $\boxed {\ abc=4RS\ }$ and $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$ $\Longrightarrow$ $4RS=2R\sin A\cdot 2R\sin B\cdot 2R\sin C$ $\Longrightarrow$ $\boxed {\ S=2R^2\sin A\sin B\sin C\ }$.[/color]

[color=blue][u]Proof of the first problem.[/u] $b^2\sin 2C+c^2\sin 2B=$ $4R^2\sin^2 B\cdot 2\sin C\cos C+4R^2\sin ^2 C\cdot 2\sin B\cos B=$
$8R^2\sin B\sin C(\sin B\cos C+\sin C\cos B)=8R^2\sin B\sin C\sin (B+C)=$
$8R^2\sin A\sin B\sin C=4\cdot 2R^2\sin A\sin B\sin C=4S\ .$[/color]


Like what stuff do you do to get ready for it? or if you don't prepare for it, like how do you do the problems (not like solutions, but your method of solving them)?

Yea, I don't think it really makes sense to really prepare for a contest where you have a month, and access to basically everything besides help from another person.  But I just usually approach it like I would other similar problems, just come up with ideas and try them to see how it works out.

what you don't do is make my mistake and try to do all five problems the day before the postmark deadline and forget to send in round four.


rotation
geometry
integration
calculus
function
calculus computations

If the curve

$ x \equal{} t \plus{} t^3$, $ y \equal{} t \minus{} \frac {1}{t^2}$, $ 1\leq t\leq 2$

is rotated about the x-axis,use your calculator to estimate the area of the resulting surface.

$ S \equal{} \int 2\pi yds \equal{} 2\pi\int (t \minus{} \frac {1}{t^2})\sqrt {(1 \plus{} 3t^2)^2 \plus{} (1 \plus{} \frac {2}{t^3})^2}dt$

How to solve this integral?

1. This is a definite integral, and you should state it with the appropriate limits of integration.

2. The likelihood that this could be expressed in closed form using elementary functions is so exceedingly remote that there is no point in wasting any effort trying. For arclengths, being expressible in closed form is very much the exception; not being so expressible is the norm. Surface areas have a few more loopholes but are nearly as bad. Besides, the instructions of the problem explicitly ask you to use a numerical method of integration (such as the numerical methods built into a number of different calculators.)


How many of you are here on aops? post here so we can know each others names.

I wasn't a student (this year!), but I did give a Prime Time Talk on the sum of the inverse squares.

I hope you all enjoyed the summer.

--Ravi (HCSSiM 1978)

This is Louis and I registered just now so I could respond.  :)

Yarr! I'm almost Michael!

EDIT: Oh hey, have the articles at portablepig been deleted? If not, can someone PM me the password? (I might be scavenging for quotes...)

Oh boy... I haven't posted here in a while...

This is Cary.

[quote="gaussianprime"]Oh boy... I haven't posted here in a while...[/quote]

Same here - this is my first post in months.

I may or may not be Dan.

This is Harrison.

Also: I lost. (Just!...kidding.)

this is andy...

wow it took me a long time to notice this. yeah, hi.


integration
real analysis
limit
topology
calculus
function
Functional Analysis

Let $C^1[0,1]$ be Banach space with the norm $||f||=\max|f(x)|+\max|f'(x)|$. Define a sequence of liner functionals $L_n(f)=2^n \int_0^1 f(x)\,\text{sign}(\sin(2^n \pi x))dx$. Is the sequence of linear functionals $(L_n)_n$ is norm-bounded?

It's norm bounded. Just integrate by parts.

$L_n(f)=\frac1{\pi}\left[ f(0)-f(1)+\int_0^1f'(x)\cos(2^n\pi x)\,dx\right],$ so

$|L_n(f)|\le\frac1{\pi}(2\|f\|_{\infty}+\|f'\|_{\infty})\le\frac3{\pi}\|f\|$

Note also that by the Riemann-Lebesgue lemma,

$\lim_{n\to\infty}L_n(f)=\frac1{\pi}(f(0)-f(1)).$

Hence in the weak-* topology of the dual space to $C^1,$ the sequence $L_n$ converges to a linear combination of point masses.

Thank you very much, Kent, for replying but now i realized that it wasn't displayed the expression "\sign", so by now i've edited it.

We still do the integration by parts. Our new antiderivative is

$2^n\int\text{ sign}(\sin(2^n\pi x))\,dx=\frac1{\pi}\arccos(\cos(2^n\pi x))-\frac12.$

I subtracted that $\frac12$ to give this piecewise-linear zigzag function an average value of zero.

We get $L_n(f)=\frac12(f(0)-f(1))-\int_0^1f'(x)(\frac1{\pi}\arccos(\cos(2^n\pi x))-\frac12)\,dx]$

So we still get norm-boundedness in the same way as in my first post, and with that average value of zero, we should still have a version of the Riemann-Lebesgue lemma.

We have $\lim_{n\to\infty}L_n(f)=\frac12(f(0)-f(1)).$


Find the number of ordered pairs (x,y) of positive integers that satisfy x :le: 2y :le: 60 and y :le: 2x :le: 60.

[hide]This is just finding all possible x,y such that x/2  :le:  y  :le:  2x for x  :le: 30 and y  :le:  30. Our answer is then  :Sigma: ([3x/2]+1) for x = 1 to 14, and then add on the solutions where x is 15 to 30, which comes out to be 23+23+22+22+21+21+...+15+15=342. This is 14 + 1 + 3 + 4 + 6 + 7 + 9 + 10 + 12.... = 14 + 4 + 10 + 16 + 22 +...+ 40 = 14 + 22*7 = 168.

168 + 342 = 480 [/hide]

[hide] you can also graph this, which i find to be easier than counting and adding. its the area of the region bounded by y=2x, y=x/2, y=30, and x=30. then you have to remember to add the edges touching the horizontal and vertical lines because is it is less than or equal to. 450 + 30 = 480. [/hide]


Jensen's has to work, if someone wants solution.


system of equations
algebra
equations
IMO Shortlist

Let $ a$ and $ b$ be non-negative integers such that $ ab \geq c^2,$ where $ c$ is an integer. Prove that there is a number $ n$ and integers $ x_1, x_2, \ldots, x_n, y_1, y_2, \ldots, y_n$ such that

\[ \sum^n_{i\equal{}1} x^2_i \equal{} a, \sum^n_{i\equal{}1} y^2_i \equal{} b, \text{ and } \sum^n_{i\equal{}1} x_iy_i \equal{} c.\]

[quote=""ISL questions and solutions from 1995-2001" by A.Ajorloo"]
obviously we can assume that $ c\geq 0$ and $ a\geq b$.now we have two cases:

$ \boxed{\textrm{Case 1.}a\geq b\geq c}$
in this case let $ n \equal{} a \plus{} b \minus{} c$ and $ x_i \equal{} 1$ (for $ 1\leq i\leq a$) and $ y_i \equal{} 1$ (for $ a \plus{} 1\leq i\leq a \plus{} b \minus{} c$ or $ 1\leq i\leq c$),for the rest of $ i$'s let both $ x_i,y_i$ equal to zero.now we can easily see that $ \{x_i\},\{y_i\}$ fulfill the question's properties.

$ \boxed{\textrm{Case 2.}a\geq c > b}$
in this case we will solve the problem using induction on $ a \plus{} b$.if $ a \plus{} b \equal{} 0$ the statement is true,now assume that the problem is true for $ a \plus{} b\leq N$,we want to prove it for $ a \plus{} b \equal{} N \plus{} 1$.
let $ a' \equal{} a \plus{} b \minus{} 2c$,$ b' \equal{} b$ and $ c' \equal{} c \minus{} b$,obviously $ a' \plus{} b' < a \plus{} b$ so according to the induction's hypothesis there exist $ \{x'_i\},\{y'_i\}$ for $ (a',b',c')$,now let $ x_i \equal{} x'_i \plus{} y'_i$ and $ y_i \equal{} y'_i$,we can easily see that $ \{x_i\},\{y_i\}$ is a solution for $ (a,b,c)$.

QED[/quote]

Old problems can be elegant too. Might be the sole contender for the category: $``\text{huge gap between solution length and actual difficulty.}"$ \\
Assume $c$ to be nonnegative, and assume that $a \leq b$ unless told otherwise. We will construct the sequence $\{x_i\}$ and $\{y_i\}$ inductively/recursively (throughout this post I'd shorten the notation $\{x_i\}_{i\geq 1}$ to be only $\{x_i\}$.) \\
$\rule{25cm}{1.5pt}$
$\textbf{\text{Base case(s).}}$ $c \leq a$. \\ This is an easy sequence to construct, as we can represent $c$ as a measly sum of $1 \cdot 1 + 1 \cdot 1 + \ldots + 1 \cdot 1$ where there are $c$ quantities of $1 \cdot 1$. Expanding upon this idea, we can have $n = c + (a-c) + (b-c)$ and construct
\[ x_n= \begin{cases} 1&\text{if $ 1 \leq n \leq c$}\\ 1&\text{if $ c+1 \leq n \leq a$}\\ 0&\text{if $a+1 \leq n \leq a+b-c$} \end{cases} \quad y_n= \begin{cases}1&\text{if $1 \leq n \leq c$}\\ 0&\text{if $ c+1 \leq n \leq a$}\\ 1&\text{if $a+1 \leq n \leq a+b-c$} \end{cases} \]
$\color{magenta} \rule{25cm}{3pt}$
$\color{magenta} \textbf{\text{Recurrence.}}$ $c > a$. \\ We induct this assertion on $c$:
\[ \begin{tabular} {p{12cm}} For any $c \leq K$, whenever $ab \geq c^2$, regardless $c$'s order in relative to $a$ and $b$, then there exists two such sequences $\{x_i\}$ and $\{y_i\}$ which satisfies the problem condition. \end{tabular} \]
To construct such a sequence for $c = K+1$, we use the assertion on $a' = a, b' = b+a-2c, c' = c-a$ that there exists a sequence $\{x_i'\}$ and $\{y_i'\}$ which satisfies the problem condition (now, if $a' > b'$, reverse their positions to keep the WLOG going.) It's easy to see that $b'$ and $c'$ must be positive, therefore warranting the construction attempt. \\
From the two aforementioned sequences, we construct $\{x_i\} = \{x_i'\}$ and $\{y_i\} = \{x_i'\} + \{y_i'\}$. Again, it is easy to verify the viability of these sequences for the assertion on $(a,b,c)$. We are done. $\blacksquare$ $\blacksquare$ $\blacksquare$ \\
[hide = Motivation: Farey sequences, Vieta jumpings. This is not as easy as it looks. ]
The first three quarters of me working on this problem were to find a distinctly unique pattern that works for most values of $(a,b,c)$. As I will elaborate in the rest of the Motivation, such patterns are not easy to trace (however, in the current 2019-2020 meta where Farey sequences and the expression $a_ib_j - a_jb_i = 1$ appeared again in abundance: examples include RMM Problem 2, AMO Problem 4 and GQMO problem 3 --- I'm not so sure ... perhaps those problems subconsciously altered my initial train of thought when I tried to find patterns.) \\ Since when does a Vieta Jumping on multiple variables can appear in an A2?  \\
\[\begin{tabular}{p{12cm}}
$\rule{4cm}{0.5pt} \hspace{1cm} \bigstar \bigstar \hspace{1cm} \rule{4cm}{0.5pt}$ \\
\end{tabular} \\ \]
Before proceeding further, here is the glossary of beautiful pairs which motivated me towards the Solution --- it would be a great exercise for the reader to try to find $``\text{quantum ties}"$ along different triples --- although the last two pairs have a clear resemblance to each other.
\[
\begin{tabular}{c|c}
$(a,b,c)$&$\{x_i\}, \{y_i\}$ \\ \hline
$(2,4,2)$&$\{1,1\}, \{2,0\}$ \\ \hline
$(2,4,0)$&$\{1,1,0\}, \{0,0,2\}$ \\ \hline
$(2,3,2)$&$\{1,1,0\}, \{1,1,1\}$ \\ \hline
$(2,5,3)$&$\{1,1\}, \{2,1\}$ \\ \hline
$(5,13,8)$&$\{2,1\}, \{3,2\}$ \\ \hline
$(3,41,11)$&$\{1,1,1\}, \{4,4,3\}$ \\ \hline
$(3,41,10)$&$\{1,1,1,0,0\}, \{4,4,2,2,1\}$ \\ \hline
$(7,10,8)$&$\{2,1,1,1\}, \{2,2,1,1\}$ \\
\end{tabular}
\quad \Rightarrow \quad
\begin{tabular}{c|c}
$(a,b,c)$&$\{x_i\}, \{y_i\}$ \\ \hline
$(8,11,9)$&$8$ ones / $7$ ones, $1$ two \\ \hline
$(7,10,8)$&$7$ ones / $6$ ones, $1$ two \\ \hline
$(8,13,10)$&$\{2,2\}, \{3,2\}$ \\ \hline
$(5,6,4)$& A million ways; use zeroes! \\ \hline
$(4,10,6)$&$\{1,1,1,1\}, \{2,2,1,1\}$ \\ \hline
$(4,13,9)$&$\{1,1,1,1\}, \{2,2,2,1\}$ \\ \hline
$(9,22,14)$&$\{2,2,1\}, \{3,3,2\}$ \\ \hline
$(9,3,5)$&$\{2,2,1\}, \{1,1,1\}$ \\
\end{tabular}
\]
Here's a brief description of the tabulars above: I did not explicitly consider the cases where $ab=c^2$ (though, of course, it will add to the richness of the configurations later) because the entries of the sequences $\{x_i\}$ and $\{y_i\}$ would be boring --- it would require basically $\{x_i\}$ linearly proportionate to $\{y_i\}$, and I wanted a telling clue which works for all bounds $ab\geq c^2$, both weak and strong. \\
The first three tabular entries are just a casual observation: what would happen if I were to set $c$ equal to the smaller number between $a$ and $b$? (which, for almost the rest of the tabular entries, I'd assume to be $a$, with the exception of the last algorithmically correct one) \\
Adjusting $c$ to be zero and $b$ to be tighter yielded more interesting configurations, as when $c$ is forced to be zero then all pairs $(x_i,y_i)$ for $1 \leq i \leq n$ must contain a zero. Furthermore, as $ab$ is slided closer to $c^2$, the sequences must be at close proximity with each other: at $(a,b,c) = (2,4,2)$ the terms consist of the loosely related $\{1,1\}$ with $\{2,0\}$. Compare that with the $\{1,1,0\}$ and $\{1,1,1\}$ which must occur when $(a,b,c) = (2,3,2)$. \\
Now, aside from the weaker bounds I stretched upon in the beginning, I began to shift toward the tighter bound: what are the possible pairs when $ab = c^2+1$? It turns out that $n$ apparently must be $2$, with terms similar to Fibonacci numbers. Here the thought of Vieta Jumping has already sprouted, but here it's not immediately clear what to base it on: $a$, $b$, or $c$? \\
Up to this point, I had only considered the pairs when $b \sim a$, and I thought that $``\text{it's time to move towards the cases where} \ b = 13a!"$ In sync with the Cauchy-Schwarz equality condition, when $3 \times 41 = 123 \sim 121 = 11^2$, I put a mental note that all terms must be proportionate with each other, and when I decided to drop $c$ a bit from 11 to 10, it would seem to be a good idea when I inductively descent-ed $(a,b,c)$ to $(a,b-1,c)$. However, this only worked for loose $ab-c^2 \geq a$. When it's just close enough, I must seek for another way in. \\
\[\begin{tabular}{p{12cm}}
$\rule{4cm}{0.5pt} \hspace{1cm} \bigstar \hspace{1cm} \rule{4cm}{0.5pt}$ \\
\end{tabular} \\ \]
Back to proportionately close triples, the notice to take $(7,10,8)$ is clear enough: $7 \times 10$ is roughly equal to $8^2$. Then, I had the idea to let $c$ be a fixed variable and $a,b$ represented in terms of $c$ --- and of course, the triple $(c-1,c+2,c)$ got the pick. If this is the case, this can be done easily enough by setting $n = c-1$ and $y_1 = 2$ to propel $x_1y_1$ from being $1$ to being $2$. Interestingly, this way of thinking also brought me into the questioning of this configuration, as it's the first one to produce a quite general one which works for an infinite number of triples. \\
That questioning brought forth an interesting way to approach the problem: instead of trying so hard to find $x_i$ and $y_i$, why don't I fix the $n$ or ($\#$ terms in each sequence) first --- and can I predict with clairvoyance what $n$ will be based on the values of $a,b,c$ alone? With this, in my judgement at that time, the only clear clue as to what to consider is the $``\text{discriminant}"$ $D = ab-c^2$. \\
Surprisingly, this turned out to be a quite good predictor of $n$. Sure, when $D = 0$, $n$ could still be required to adapt with the squarefree part of $a$ and $b$, but otherwise, $n$ might as well be $1$. Moreover, for the relatively prime $(a,b,c)$ such as $(2,5,3)$ or $(8,11,9)$, $n$ is exactly $D+1$. \\
Overall, $D$ seems to be quite inaccurate when $\gcd(x_1,x_2,\ldots,x_n) \ne 1$ such as in $(a,b,c) = (8,13,10)$ --- with that said, I still tried to ask the question: what if $a$ is actually a square, but its circumstances forced it to split into smaller parts? Considering $a = 4$, I tried to form a configuration which forced $\{x_i\} = \{1,1,1,1\}$; and I succeeded. \\
\[\begin{tabular}{p{12cm}}
$\rule{4cm}{0.5pt} \hspace{1cm} \bigstar \bigstar \hspace{1cm} \rule{4cm}{0.5pt}$ \\
\end{tabular} \\ \]
From these advances alone, it would seem that no improvements are being made aside from considering the quantity $D$. After this, almost all attempts to induct $(a,b,c)$ was fruitless. The closest I've got is when $a$ is not the smallest number between $(a,b,c)$, implying $c$ to be the smallest one, I could let $(a,b,c) = (n,n+E+F,n+E)$, but any attempts of smoothing such as
\[ n(n+E+F) \geq (n+E)^2 \Rightarrow nF \geq E(E+n) \]
did not work as well as I thought it'd be. \\
Since I only have those configurations I have crafted to my control, I thought I'd reviewed it back (as what I'm doing here) and noticed that $\textbf{not only does Fibonaccis exist on tight triples, but some form of it must exist in all!}$ Take $(3,41,11)$, the loosest pair $(a,b)$ I crafted. The illusion of proportionate numbers blinded me that the $(x_i,y_i)$ is equal to $(1,4),(1,4)$ and $(1,3)$, where $4 = 3+1$. When I dissected all pairs $(x_i,y_i)$ among all configurations, it would seem that most are in the form
\[ (x_i,y_i) = (0,1), (1,1), (1,2), (2,3), (3,5), \ldots \]
and the lowest $D$'s sequences (making the most unique sequences $\{x_i\},\{y_i\}$ among all configurations) are closely approximated by the golden ratio (also see that $(7,10,8)$ has two distinct sequences, the eighth and tenth). Only using the pairs $(1,1)$ and $(1,2)$, I further assembled them into the triples $(4,10,6)$ and $(4,13,9)$ --- speaking of which, this is the first time I made a triple based on its $\{x_i\},\{y_i\}$ isntead of solving for a suitable $\{x_i\}, \{y_i\}$.
Analyzing the fourth last and third last triple, I noticed that $D$ is equal to $4$ and $3$, respectively. Now, I wondered what this could mean, and I realized that $D$ being $4$ might originate from $2 \times 2$, where there are two $(1,2)$ pairs and two $(1,1)$ pairs; and $3$ might similarly originate from $3 \times 1$. This prompted a long-forgotten manipulation of proving Cauchy-Schwarz as following:
\begin{align*} D &= ab-c^2 = (x_1^2+x_2^2+x_3^2)(y_1^2+y_2^2+y_3^2) - (x_1y_1+x_2y_2+x_3y_3)^2 \\ &= (x_1y_2-x_2y_1)^2+(x_1y_3-x_3y_1)^2+(x_2y_3-x_3y_2)^2\\ \end{align*}
and it all clicked. (And I hope that this made clear the long journey towards the apparently magical substitution in the Solution.) \\
For further exposition, I continued making the second to last triple by combining $(2,3)$ and $(1,2)$ again, and say I would like to make $D$ constant when I inductively descended from $(9,22,14)$ to something else. To that end, I would like $x_iy_j-x_jy_i$ to stay constant throughout the process, and the easiest way to fix $x_iy_j-x_jy_i$, say, to be always equal to $1$ is to change all $y_i$ to be $y_i - k\cdot x_i$, as it would ensure that
\[ x_i'y_j'-x_j'y_i' = x_i(y_j-k\cdot x_j)-x_j(y_i-k\cdot x_i) = x_iy_j-x_jy_i \]
However, if we set $k = 2$ or greater, $c$ might decrease to $c-2a$, which we didn't quite want (but still doable, since $|c| < |a|$, but I digress) --- and more dangerously, we certainly do not want $b$ to change to $b+4a-2c$, which may be boundable but a bit uglier. I stuck with $k =1$ and computed $a',b',c'$, and there I was at the Solution.
\[\begin{tabular}{p{12cm}}
$\rule{4cm}{0.5pt} \hspace{1cm} \blacksquare \hspace{1cm} \rule{4cm}{0.5pt}$ \\
\end{tabular} \\ \]
Final Comments: In theory, this problem really shouldn't be hard, and there are no big theoritical/soft tech leaps in this Solution and Motivation, but doing an old problem which combines a long-forgotten manipulation and tech might ramp up the difficulty a little. Overall, I would give this roughly 25M --- the first Section certainly contain a lot of unique configurations (and the desperation on how to induct this downwards was certainly very real), the second Section's hard tech being laughably obvious while located at a blind spot, and the final manipulation being a hard tech which may have been more well known in the past (USAMO 1999, anyone?).
[/hide]

Is there any other way to Solve this Problem? :( It seems similar to $\textbf{Cauchy-Schwarz Inequality}$
Bacause,
$$(\sum^n_{i=1} x^2_i)(\sum^n_{i=1} y^2_i) \overset{\textbf{Titu}}{\ge} (\sum^n_{i=1} x_iy_i)^{2} \Leftrightarrow ab \ge c^2$$


1.  Rearrange to x = 11.

2.  Rephrase as x^2 = 81, or x=+- 9

3.  Alternatively, x*5 = 3000 wherein x = 600


Για κάθε φυσικό αριθμό $ x$ έστω $ Q(x)$ το άθροισμα και $ P(x)$ το γινόμενο των ψηφίων του $ x$ αντίστοιχα. Να δειχτεί ότι για κάθε φυσικό αριθμό $ n$ υπάρχουν άπειροι φυσικοί αριθμοί $ x$ ώστε $ Q\left(Q(x)\right)\plus{}P\left(Q(x)\right)\plus{}Q\left(P(x)\right)\plus{}P\left(P(x)\right)\equal{}n$.

[quote="Γιώργος"]Για κάθε φυσικό αριθμό $ x$ έστω $ Q(x)$ το άθροισμα και $ P(x)$ το γινόμενο των ψηφίων του $ x$ αντίστοιχα. Να δειχτεί ότι για κάθε φυσικό αριθμό $ n$ υπάρχουν άπειροι φυσικοί αριθμοί $ x$ ώστε $ Q\left(Q(x)\right) \plus{} P\left(Q(x)\right) \plus{} Q\left(P(x)\right) \plus{} P\left(P(x)\right) \equal{} n$.[/quote]

Εστω $ u$ ένας φυσικός αριθμός με το άθροισμα των ψηφίων του ίσο με $ n$.Θεωρούμε τώρα τον αριθμό $ r\equal{}111...1$ του οποίου όλα τα ψηφία είναι ίσα με την

μονάδα και το πλήθος τους είναι $ 10u$ και έναν αριθμό $ x$ ο οποίος είναι στοιχείο του απειροσυνόλου {$ 10r,10r^{2},10r^{3},...$}. Επειδή ο $ x$ είναι

πολλαπλάσιο του $ 10$ έχουμε $ P(x)\equal{}0$ και $ Q(x)\equal{}10u$ διότι τα μη μηδενικά ψηφία του $ x$ είναι $ 10u$ μονάδες.
Επομένως έχουμε

$ Q(Q(x))\plus{}P(Q(x))\plus{}Q(P(x))\plus{}P(P(x))\equal{}Q(10u)\plus{}P(10u)\plus{}Q(0)\plus{}P(0)\equal{}Q(u)\equal{}n$

q.e.d :)


number theory
least common multiple
number theory unsolved

How to prove
$LCM(1,2,...,n) \le 3^{n}$
Or the stronger:
$\prod_{p \le n}n \le 3^{n}$

$ln(LCM(1,2,...,n)=\sum_{x=1}^{n}\Lambda (x)=\Psi (n)=n+o(n)$, were $\Lambda (x)=ln p$ if $x=p^{k}$ and else $\Lambda(x)=0$.

but we should prove for all n,but for large enough n.

Thank you,but in the link,it just shows that $lcm(1,2,...n)^{1/n}=e$,but not the result here explicit.


What is the sum of outer angles of any concave quadrilateral?360?Any proof?

for exterior angles: travel along a quadrilateral, each time you turn, you turn through the exterior angle, when you go around the whole quadrilateral, you went a full circle, that is 360 degrees

for total, just add 180n, where n is the number of verticies


By Chebyshev,

$\left(\sum x_i\right)\left(\sum x_i^3\right)\leq n\left(\sum x_i^4\right)$

This holds if all the $x_i$'s are $\geq 0$, or even arbitrary reals.

Equality holds iff $x_i=x_i^3$ for each $i$, or $x_i\in\{-1,0,1\}$.  Clearly they can only sum to $n$ if all $x_i$'s are $1$.

[quote="shobber"]Yep. Chebishev is the easiest way, but I wonder if there are any other methods?  :)[/quote]

Posted before....

See that $\sum_{i=1}^{n} x_i^3(1-x_i) = \sum_{i=1}^{n} (x_i-1)$ etc....


algebra
difference of squares
special factorizations

Find all natural $ n$ such $ n^{2}\plus{}824$ produce a total square.

factor as a difference of squares and both factors must be even for the number that is the square to be an integer  :D

[hide]n = 101
n= 205[/hide]

With $ x,n\in \mathbb{Z}$ we have

$ x^2\equal{}n^2\plus{}824\Longrightarrow (x\plus{}n)(x\minus{}n)\equal{}824$
$ \Leftrightarrow (x\pm n)\mid{824}\Rightarrow 1\le (x\pm n)\le 824$

$ (x\plus{}n)\minus{}(x\minus{}n)\equal{}2n\equal{}\{823,410,202,95\}\Longrightarrow n\equal{}\{205,101\}$.
$ (x\plus{}n)\plus{}(x\minus{}n)\equal{}2x\equal{}\{825,414,210,111\}\Longrightarrow x\equal{}\{207,105\}$

The two ordered pair soltuions $ (n,x)$ are $ (205,207)$ and $ (101,105)$.


my best advice is read the whole problem
if you try to read fast, you WILL get the prob wrong

but if you really want to read fast, just practice

Get glasses if you have bad eyesight.
That's why I got owned on this year's countdown.

i dont wear glasses.  And unfortunately didnt make cd bcause i lost tiebreaker. :mad:  :mad:  :mad:  :mad:  :mad:   i was rly slow about a month ago but ever since i started playing FTW i got so much faster. Trust me it helps alot.  And yes read the whole problem you b really mad if you get it wrong cause u didnt read one word

If the problem is long, it might be a good idea to:

1.Skim the problem, looking for important info and the actual question.
2.Read the actual question.
3.Take a look at the info.

Most of this has to be done by instinct, or you will get slowed down. I would agree that FTW helps.

instinct is complete pwnage.  Or... read a ton of books.  I did that when i was in 2nd-4th grade.  Not for math though lol.  I just liked reading =). Now it's pwnage xD

Skim the problem, find the numbers lol, look at the last few words for the units. and yep =)

Read the [i]actual[/i] question first, then read the supplament info. In doing so, you're brain automatically discards pieces of information that are worthless and have no value in actually solving the problem.

The most important thing is to find and get the question right. I practiced last years countdown questions, and also played For The Win. If the question is long, I normally misread.

The most important things are,
1. Skim for question, normally near the end.
2. Find data needed
3a) If you need paper, use it and solve completely
3b) Click buzzer, DON'T CLICK IT IF ITS A LONG PROCESS
4. Solve and shout out answer fast.

I don't know if this works, since i haven't tried it yet, other than practice with a friend. But really, you wouldn't need to think about names such as tom and activities such as tennis.

[quote="AIMEman"]Mathcrazed is right.  If you read the first sentence and it is talking about triangular numbers, you can skim down to the last sentence which might ask, "What is the 6th triangular number?"  You would hopefully get the answer much faster than others who read the whole problem.  But, you also have to read the units, or you might assume a false statement, buzz in way too early, and get it wrong.  *cough* Michael *cough*  :lol:[/quote]

i hate you so much.

but he is right about reading fast enough that you beat the other people, but just read the right stuff.

Skim read the problem, look for any numerals (obviously, since this is usually the info. you'll need to solve the problem), and read whatever the last line is CAREFULLY-it's usually the one asking the question.

You do have to be careful about reading it too quickly, though...sometimes you might misread the question. (e.g. skim read a problem too quickly, then answer for area instead of volume or something like that)

I would say that playing For The Win really helps. Yeah you should read the question first before reading any details. I definitely got better at countdown as I played it.

TWITCH.

Buzz in, then go, "Ugh..er...umm..12." as long as you don't take more than 3 seconds. You might get away with longer, depending on how lax your chapter is.

or, if the question said \[ \text{Simplify }\frac{9m^4p^5}{12m^2p^7}\]just go like *buzz in* um, 3, m squared, over, 4, p squared, or something of that sort

Just one tip for trying to buzz in really fast...
get it in your brain before you bring it out of your mouth.
One time i was playing against another kid, and the question was:
If there are three men in a lottery for seven people, what is the chance that a man will be chosen as the winner of the lottery?
The other guy buzzed in, and i was like :o because it was so easy
he got it wrong
 :rotfl:  :rotfl:  :rotfl:
Another time, in another competition
(science bowl, if you know it)
you play on teams
there was a math question:
which number is closest to the sqrt. of 217:
13   14   15  OR 16?
You should know, but the answer is
[hide]15 (you know, 225 is close to 217?)[/hide]
anyway.
So, i was about to buzz in, cause i knew the question, when my friend did.
he said 16
 :wallbash_red:
So, especially in the countdown round, where every question counts, if you have to, take your time to get the answer in your head, even if you know you know it.

That is a very good example.

There are so many variables that you have to put in here.  Here are the 2 biggest ones:

#1.  Know your opponent (if you can)
If your opponent(s) are from your school, you will probably know if they are to unsure and will check their answer before buzzing in, or if they are quick on the trigger and will buzz in really fast.  Also, if you have ever met them before, you can probably tell by their personality.

#2.  The number of people you are playing against
If you are going head-2-head, you had better make sure you know the answer before you buzz in.  Or else, you're giving your opponent plenty more time to work the problem out.  If you are playing against more people, you should probably just try to get in the first time you get the answer in your head.  But, more importantly, refer to #1.

If you want a sure win... Practice. It's not about your opponent's skills, it's about your skills.  If you mess up, who's fault is it? If you're fast enough, is anyone going to beat you? So pretty much: get faster.  I like contradicting myself =).

My chapter had a system where about 12 people compete at once, and 3 advance from each group to the next round, but a person can answer more than once for each question as long as they don't answer twice consecutively. My strategy was to buzz every other answer unless I got it right. It worked for a couple questions because a LOT of the people buzzed fast and got it wrong.  :ninja:

But in a head to head competition, I'd probably take more time doing the problem, and if my opponent screws up, I'd take nearly all of the time to get the answer and check it. I wouldn't want to miss an opportunity for a free point.

Use FTW.  Before I could barely finish the sprint round, but now I can finish it within 30 min after about a month of FTW. HOWEVER, this is not great for accuracy.  BTW you do not need 2 months for it to be effective.  This is also great for typing skills.


Bonne Année de la part d'un roumain, français et (nouveau) européen ! :wink: Pourvu que 2007 amène plus de MathLinkers s'exprimant dans la langue de Molière...  :blush:


quadratics
trigonometry
function
absolute value
complex numbers

[quote="tim1234133"][hide="If x is allowed to be complex"]
$x+\frac{1}{x}=-1$
$x^{2}+1=-x$
$x^{2}+x+1=0$
The (complex) roots of this equation are $\omega$ and $\omega^{*}$, the third roots of unity. (Multiply this quadratic by $(x-1)$, and you get $x^{3}-1=0$).
Hence $x^{999}=x^{3}=1$
and so $x^{999}+\frac{1}{x^{999}}=2$.[/hide]

There are, of course, no real x that satisfy the condition, since $x$ and $\frac{1}{x}$ are both the same sign and one of them has an absolute value of at least 1[/quote]
How can you just multiply by (x-1)?
 :maybe:
thats like saying:
$(x-2)=0$
$(x-2)(x+3)=0$
So,$x=2,-3$
You've changed the problem completely...or havent you...
Instead of $x-1$ you could have $x+2$

And Draconis:
$x^{4}=x$
 lead to $x=0$ and $x=1$ and some other solution
So:
$0+\frac{1}{0}$
is also the answer?

riddler, those solutions are completely right. They say

$x^{2}+x+1 = 0\ \Rightarrow\ x^{3}= 1$, i.e. if $a$ is a root of $a+\frac{1}{a}=-1$, then $a^{3}=1$.

You say that they say

$x^{2}+x+1 = 0\ \Longleftrightarrow\ x^{3}= 1$, i.e. the solutions of $x+\frac{1}{x}=-1$ are the solutions of $x^{3}= 1$.

I hope you understand.

[quote="Kurt Gödel"]riddler, those solutions are completely right. They say

$x^{2}+x+1 = 0\ \Rightarrow\ x^{3}= 1$, i.e. if $a$ is a root of $a+\frac{1}{a}=-1$, then $a^{3}=1$.
[/quote]
How does
$x^{2}+x+1=0$
imply $x^{3}=1$?
Why?

[quote="neelnal"][quote="riddler"][quote="Kurt Gödel"]$x^{3}-1 = (x-1)(x^{2}+x+1)$, so $x^{2}+x+1 = 0$ implies that $x^{3}-1 = 0$...I really fail to see what's so mysterious about that.  :)[/quote]
the fact that you multiply by $(x-1)$
Are you allowed to do that?[/quote]

Riddler, he multiplied x-1 on both sides. It just so happened that the RHS is 0, hence the lack of change on the RHS.[/quote]
Look I know what happened to the R.H.S
But that changes the problem completely, doesn't it?
It's like saying changing $x+2=0$
Multiplying by $x-1$ gives
$x=1,-2$
????
By multiplication your changing the problem completely aren't you?

[quote="riddler"][/quote][quote="neelnal"][quote="riddler"][quote="Kurt Gödel"]$x^{3}-1 = (x-1)(x^{2}+x+1)$, so $x^{2}+x+1 = 0$ implies that $x^{3}-1 = 0$...I really fail to see what's so mysterious about that.  :)[/quote]
the fact that you multiply by $(x-1)$
Are you allowed to do that?[/quote]

Riddler, he multiplied x-1 on both sides. It just so happened that the RHS is 0, hence the lack of change on the RHS.[/quote][quote="riddler"]
Look I know what happened to the R.H.S
But that changes the problem completely, doesn't it?
It's like saying changing $x+2=0$
Multiplying by $x-1$ gives
$x=1,-2$
????
By multiplication your changing the problem completely aren't you?[/quote]

It is a theorem/concept that when you have these two certain complex numbers, you can do stuff, yea... here:

[quote]The (complex) roots of this equation are  and , the third roots of unity. (Multiply this quadratic by , and you get ). [/quote]

Or you can also just stick to draconis' solution, which is better I think.

@riddler: No, your not changing the problem. The only thing we say is that if $x$ satisfies $x+2 = 0$, then $x$ also satisfies $(x+2)(x-1) = 0$. Come on, this really isn't hard to understand...  :huh: I think you're just confused. The only reason why that would be confusing, is that you could think that we're expressing an equivalence...but we're not  :roll:

I think maybe I understand where I was wrong...
$x^{3}=1$
is different from $x=1$
Thanks

If n=0 then no matter what you multiply by n, you still get 0.

[quote="riddler"]Let x be a real number with $x+(1/x) =-1$. What is the value of $x^{999}+\frac{1}{x^{999}}$?[/quote]

[hide="proof"]
I will not use euler's identity at all to make ppl happy \[cis(\theta )=\cos \theta+i\sin \theta\] using algebra and cosine sum, sine sum formulas we obtain: \[cis (a+b) = cis (a) cis (b) \ \ (*)\] So this random little $cis$ function is a nice guy, seems like an exponential function.

First notice that if we plug $b=-a$ then we have $cis(0)=1=cis (a) cis(-a), \implies cis (a)=\frac{1}{cis (-a)}$ \[x=cis(k), x+\frac{1}{x}=-1=cis(k)+cis(-k)=2\cos (k) \ \ \text{(direct algebra)}\] Which lets us know $k=\frac{2\pi}{3}$ works
Now notice $cis(nk)=cis(k)^{n}$ as we have shown above, at least for integer $n$.
Therefore, $cis(nk)+cis(-nk)=2\cos(nk)=x^{n}+\frac{1}{x^{n}}$
for $n=999$ we find that $2\cos(999k)$ is the answer to the problem, and it equals $2\cos(2*999\pi)=2\cos(0)=\boxed{2}$
[/hide]

Riddler, here it is:

If you have $x+2=0$ and you multiply it by $x-1$, you do get $x^{2}+x-2=0$. But then you must make a proviso: "The solution of $x+2=0$ is the solution of $x^{2}+x-2=0$ which is [i]different from $1$[/i]." (Because you really don't want to be multiplying equations by zero.)

The same thing is in the given problem:

The equation $x^{2}+x+1=0$ is equivalent to $x^{3}-1=0$ with the condition that $x-1\neq 0$.

remember that x is not a variable...it is fixed between two values, it might be more clear to say this:

let $r$ be a root of $x^{2}+x+1$, then

$r^{3}=1$

nice trick! now solve for any integer power without any complex algebra.

Either way you look at it, this problem you are giving is the exact same as the one we justed solved (999 has nothing to do with the problem except for its remainder mod 3)

We could alsodo it like this, which is a different way of expressing the same concept shown earlier:

$x+\frac{1}{x}=-1$

$(x+\frac{1}{x})^{2}= 1$
$x^{2}+2+\frac{1}{x^{2}}= 1$
$x^{2}+\frac{1}{x^{2}}=-1$

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

Continuing with the same idea, we see

$x^{4}+\frac{1}{x^{4}}=-1$

At which point we see that the pattern will be:

$x^{n}+\frac{1}{x^{n}}=-1$ when n is an even integer $\ge 2$
$x^{n}+\frac{1}{x^{n}}=2$ when n is an odd integer $\ge 3$

Thus $x^{999}+\frac{1}{x^{999}}= 2$


Last night I was on the forum and every other topic I clicked on would come up "This page can not be displayed". Then after a lot of hitting back and clicking a different topic the home page wouldn't come up. All the other Websites I went to(google, msn, yahoo, etc.) worked fine and now the forum is working fine for me is there anything I can do to prevent this from happening again?

Hmm. . . I'd suggest cleaning out your cookies, but I'm not really sure what might cause that sort of behavior.  Perhaps our server had a little trouble last night.


integration
calculus
function
geometry
rectangle
advanced fields
advanced fields unsolved

Hi! I'm looking for a nice solution to

$ \int_{-\infty}^\infty \frac{e^{i\xi x}}{\cosh{\frac{\xi}{2}}}\frac{\mathrm{d}\xi}{2\pi}=\frac{1}{\cosh(\pi x)}$

Can you please help me?

In order to calculate that integral, I'd calculate the integral of that same function around the border of a rectangle with vertices $ -R,R,R+2\pi i,-R+2\pi i$ by residues. The vertical pieces of the integral go to zero as $ R\to\infty$, and the horizontal pieces differ only by a real constant multiplier.

[Edit: My rectangle wasn't tall enough]

Note: I am typing this really fast and am so tired. It is VERY possible that I made a mistake. But it is the idea that is more important.

Jmerry gave you the basic idea. You cant use Jordan's Lemma nor treat this as a semi-infinite circle because it has [i]infinitely[/i] many poles. So that is why Jmerry gave his idea. By considering the function $ f(z) = \frac{e^{iaz}}{\cosh \frac{z}{2}}$ there are no poles and by Cauchy's theorem the integral is zero.

The contour can be evaluated as:
$ \int_{-R}^{R}\frac{e^{iax}}{\cosh \frac{x}{2}}dx-\int_{-R}^{R}\frac{e^{ia(x+\pi i)}}{\cosh \frac{x+\pi i}{2}}dx+\int_{0}^{\pi}\frac{e^{ia(R+iy)}}{\cosh \frac{R+iy}{2}}idy-\int_{0}^{\pi}\frac{e^{ia(-R+iy)}}{\cosh \frac{R+iy}{2}}i dy$

The first two integrals are what you want, in fact they are basically the same.
The second two integrals hopefully go to zero as $ R\to \infty$.

We will show that,
$ \int_{0}^{\pi}\frac{e^{ia(R+iy)}}{\cosh \frac{R+iy}{2}}idy \to 0$

First note that $ \left| \cosh \left( \frac{R}{2}+i \frac{y}{2}\right)\right| \geq \left| \cosh \frac{R}{2}\right|$

So, $ \left| \frac{e^{ia(R+iy)}}{\cosh \frac{R+iy}{2}}i \right| \leq \left| \frac{e^{-iay}}{\cosh \frac{R}{2}}\right| \leq \left| \frac{1}{\cosh \frac{R}{2}}\right| \leq | \mbox{sech}\frac{R}{2}|\to 0$ as $ R\to \infty$.

So,
$ \left| \int_{0}^{\pi}\frac{e^{ia(R+iy)}}{\cosh \frac{R+iy}{2}}idy \right| \to 0$

Similarly we can show the 4th integral chose to zero.
The rest is boring  :sleeping: algebra which I'll surly mess up on at my state.


group theory
abstract algebra
algebra unsolved
algebra

Hello, just a stupid question ...
I have an exercise which says:
Prove that subgroups of $\mathbb{R}$ are monogene or dense.

Thanks

I don't know what "monogene" means, but I'm assuming it must mean something like "discrete" or $\mathbb Z$-like :).

If there's no smallest positive number in your group then there are positive numbers in the group arbitrarily close to $0$, because there are numbers in the group which are arbitrarily close to each other; then, it's clearly dense: for any $0<a<b$ (it's the same for negative numbers) take some $0<\alpha<b-a$ in the group, and there will be some $n\in\mathbb N$ with $n\alpha\in(a,b)$. If, on the other hand, there's a smallest positive number $\alpha$, the group is easily seen to be equal to $\{n\alpha\ |\ n\in\mathbb Z\}$.

Ok I understand
But now, I don't see why after this question it's asked to find groups that are dense and finite, I think there is none groups like that ...

of course (finite + dense) is impossible.
(and 'monogene" is a french word to say that the group is generated by only one element...)

It's strange to have this question, it's an exercise of algebra for the first year in ENS Lyon, statement is:
a) Montrer que les sous groupes de R sont soit monogènes soit denses
b) Donner un exemple de sous groupe de R dense de type fini
c) Tous les sous groupes de R sont-ils de type fini ?

Questions b and c looks trivial and I don't think it's normal.

In english:
a) Show that subgroups of R are discrete or dense
b) Give an example of subgroup of R dense of finite type
c) Are every subgroups of R of finite type ?

Maybe "of finite type" means finitely generated?

It's very classical, the one who gets that at ENS Lyon is too lucky ;)


inequalities
search
inequalities unsolved

ab+bc+ca=1 and a, b, c>0.    show the minimun value of

1/(1-a) + 1/(1-b) + 1/(1-c)

thanks.

[quote="siuhochung"]\[
\sum {\frac{1}
{{1 - a}}}  \geqslant \frac{3}
{{1 - \frac{{a + b + c}}
{3}}} \geqslant \frac{3}
{{1 - \sqrt {\frac{{ab + bc + ca}}
{3}} }} = \frac{{3\sqrt 3 }}
{{\sqrt 3  - 1}}
\]
By Jensen's inequality.[/quote]

could you tell me more details? And what is so called "Jensen's inequality?"

I think you can search the forum about Jensen's inequality..


[quote="Drunken_Math"]For what real numbers $n$ is this property true:

$\frac{n}{2}>n$  :D[/quote]

[hide]  All negative numbers apply. [/hide]

Ha Ha Ha!!!!!! :rotfl:  :rotfl:  :rotfl:
Is this a question even for middle level??? (No offence plz)

Actually, elementary students in my area have no concepts of negative numbers and inequalities.

same thing in my area. I learned about negatives and inequalities in 5th grade.  But everyone else in my grade learned it in 6th grade.


Could somebody tell me where I can find the proofs of the sos theorem ?

Thanks

http://www.comp.nus.edu.sg/~anhcuong/products.html
the method is relatively unknown, so it is likely you'll have to prove a few facts to use it.

[quote="nickthegreek"]Could somebody tell me what exactly is this sos theorem???

(also,the link doesn't work...)[/quote]
Why the link does not work?
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80127

thank you very much arqady!!!

there could actually be a problem in my computer,but still not able to see that link.However, the new one you gave me tells me pretty much all i wanted to know...


For every problem, VN's contestants either all solved or all couldn't solve. Is there anything strange? Many VNese have wondered and eventually there are some bad rumors about it.

Any insider's information?

[quote="corrupted"]For every problem, VN's contestants either all solved or all couldn't solve. Is there anything strange?[/quote]

Yes, there is. Someone seems to have hacked into the IMO random scores generator.

  darij

I don't think Viêt-Nam needs corruption. See results of last years, and you will see Viêt-Nam has generaly good results  :)

Problem set of IMO 2007 can be splited up to EASY (I mean VERY EASY, Pro. 1, 2, 4, 5) and HARD (I mean VERY HARD, Pro. 3, 6). In such easy problems, VIEese contestants score round at 7pt. Thus, it is not difficult to see that VIEese contestants will score round at 28pt at all. I think that this is normal and there is no unfair trick here.

I think the home court advantage has a little to do with it. The students are at easy when they are at home; thus they do better on the exam. Also, VN is known to be one of the top countries in the IMO, so there is no surprise here.

Please don't make silly accusation and cloud up a great result from VN.

Just a thought

[quote="darij grinberg"]Yes, there is. Someone seems to have hacked into the IMO random scores generator.

  darij[/quote] :rotfl:  :w00t:

Hi guys,

I personally don't know anything. That's why I asked you. If there's no corruption then rumors are just rumors and the VN's contestants surely deserve their honors.


[b]Να αποδείξετε ότι σε κάθε τρίγωνο $\bigtriangleup ABC$ ισχύει:[/b]

$\frac{1}{cos\frac{B-C}{2}}$ $+$ $\frac{1}{cos\frac{A-B}{2}}$ $+$ $\frac{1}{cos\frac{A-C}{2}}$ $\leq$ $\frac{\sqrt{3}(a^{2}+b^{2}+c^{2})}{4E}$             [b] όπου Ε το εμβαδόν του τριγώνου.[/b]

Δεν ημουν στα μαθηματα της ΕΜΕ,αρα μπορω να απαντησω...
Ειναι αρκετα καλη...
Δινω ενα hint...
[hide]$\frac{1}{cos\frac{B-C}{2}}+\frac{1}{cos\frac{A-B}{2}}+\frac{1}{cos\frac{A-C}{2}}$=δα/υα+δβ/υβ+δγ/υγ οπου δα,δβ,δγ ειναι οι διχοτομοι και υα,uβ,uγ τα υψη..[/hide]

 :D  :D  :D

TELOS..... :mad:  :wink:  :P

Geia sou re Ralli!!!

Ανισότητα (ευκολάκι τρελό)
Παιδιά επανέρχομαι δρυμίτερος με μια αρκετά εύκολη ανισότητα που είναι δική μου επινόηση ελπίζω αυτό να μην σας τρέψει υπονοούμενα για την (μειωμένη) μαθηματική μου ικανότητα (αστειεύομαι) λοιπόν χωρίς αργοπορία
Να δείξετε ότι
(χ^2+ψ^2+ω^2)(1/χ+ψ  +1/ψ+ω +1/ω+χ )>= 3/2 (χ+ψ+ω)
Καλή διασκέδαση
χαιρετώ τους φίλτατους Νίκο Ράπανο κι Ράλλη Καραμιχάλη κι όλους τους υπόλοιπους

Λοιπον να η λύση μου στου Αμβρόσιου.(παρότι δεν καταλαβαίνω που κολλάει αυτή με το τόπικ)  :maybe:
Έχουμε από Cauchy
$(x^{2}+y^{2}+z^{2})(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x})\geq (\frac{z}{\sqrt{x+y}}+\frac{y}{\sqrt{x+z}}+\frac{x}{\sqrt{y+z}})^{2}$

Λόγω ομοιογένειας θέτουμε $x+y+z=1$ και άρα αρκεί να δείξουμε οτι
 $(\frac{z}{\sqrt{1-z}}+\frac{y}{\sqrt{1-y}}+\frac{x}{\sqrt{1-x}})^{2}\geq\frac{3}{2}$

Θεωρούμε τη συνάρτηση $f(x)=\frac{x}{\sqrt{1-x}}$
Είναι κυρτή άρα από Jensen
$\frac{z}{\sqrt{1-z}}+\frac{y}{\sqrt{1-y}}+\frac{x}{\sqrt{1-x}}\geq 3\frac{\frac{x+y+z}{3}}{\sqrt{1-\frac{x+y+z}{3}}}=\sqrt\frac{3}{2}$

άρα $(\frac{z}{\sqrt{1-z}}+\frac{y}{\sqrt{1-y}}+\frac{x}{\sqrt{1-x}})^{2}\geq\frac{3}{2}$
 το ζητούμενο

Χρησιμοποιούμε τις ανισότητες :

$\ 2(x+y+z) \left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right) \geq 9$

$\ 3(x^{2}+y^{2}+z^{2}) \geq (x+y+z)^{2}$ .

η λύση του ramanujan είναι σωστή αλλά του σιλουανού δεν την καταλαβαίνω λόγω άγνοιας σσυναρτησεων κι ανισοτητων πχ jensen αλλά δεν πειράζει
χαιρετώ μετα΄τιμής

[quote="Nick Rapanos"]Φυσικά,όμως η λύση του Σιλουανού είναι ολόσωστη!![/quote]

Ευχαριστώ  :)  .
Πάντως λύθηκε μια άσκηση  σε λάθος τοπικ ενώ για την αρχική του τόπικ έχει δωθεί μόνο hint  :P
Επειδή από  την εκφώνηση (εγώ αποκλείομαι  :lol:) ας  το κανει κάποιος άλλος......


SFFT search AoPSwiki MATHCOUNTS function algebra polynomial

[quote="jli"]for number three it depend how many of each coin there are[/quote]

Um, isn't the point to find all the combinations of coins (doesn't matter each's amount) that will make the desired amount?

Other than that,

1. What do you mean?

2. Easy. (What is Simon's favorite factoring trick).

3.  Can be solved faily quickly, but it's 11:30 and I need to go to sleep.

[quote="asianhottie8293"]Hey, have no time, so on same page, sry.

1. How do u quickly calculate partitions?

2. 1/m +1/n= 2/15

m>n and not multiple of n. Find m+n

3.Suppose a coin purse contains 30 coins which are nickel, dime, or quarter. How many different combinations of coins value 1 dollar?[/quote]
1. Well I don't know? Work with them a lot. There is no "easy way" if I think I understand you.

[hide="3."]
the least amount is 30 nickels=1.50
$0$ ways to get 1 dollar
[/hide]
[hide="1."]
Unless you are really good at math, the formula for partitions is too complicated
[/hide]

Is number 3-[b]29[/b]?

I probalby did it wrong-considering I had these weird equations with cases..

vamathletes, you dont have to use all of the coins

For 1, it is usually not too bad to find all the partitions. I'll use an example-7. First, let yourself use every #-7-only 1. Then, don't let yourself use the highest #-6,1-only 1. And so on...

EDIT:when u can use only 1-n, you must use at least one of n.

For #1, I would suggest AoPSWiki-ing it.

Usually, for MATHCOUNTS problems, we would just find all of them.  For more advanced partitions problems, generating functions would be required (but thats over MATHCOUNTS) so can we just discuss problems #2 and #3?

alright...this is probably wrong..but number 3
[hide]5a+10b+25c=100, so a+2b+5c=20 where a,b,c are nonnegative
We break it into two cases-when c is even or when c is odd-so

even

a+2b+10c=20, that means a is even so 2x+2b+10c=20,
x+b+5c=10, c can either be 0,1, or 2.

when c is 0, x+b=10, so there are 11 solutions
when c is 1, x+b=5, so there are 6 solutions
when c is 2, x+b=0, so there is 1 solutions

odd

a+2b+10k+5=20, so a+2b+10k=15, so a must be odd so
2w+1+2b+10k=15, w+b+5k=7, k can be 0 or 1
when k is 0, w+b=7, there are 8 solutions
when k is 1, w+b=2, so there are 3 solutions

all the solutions contain three values in which their sum is less than 30, so [b]11+6+1+8+3=29?[/b][/hide]

there is no closed form for the partitions. for the other formula that is complicated and stuff, i used wikipedia

you have to use generating functions...

and for 3, do you have to use all 30 coins?

[quote="asianhottie8293"]2. 1/m +1/n= 2/15

m>n and not multiple of n. Find m+n[/quote]
You're often saying you're in a hurry.  Please slow down a little bit, and carefully consider the responses.  I think you'll get more out of this Forum.

Anyway, here's a solution to #2, using SFFT:

[hide]\begin{eqnarray}\frac{1}{m}+\frac{1}{n}&=& \frac{2}{15}\\ 15n+15m &=& 2mn \\ 0 &=& mn-\frac{15n}{2}-\frac{15m}{2}\\ 0 &=& (m-\frac{15}{2})(n-\frac{15}{2})-\frac{225}{4}\\ 225 &=& (2m-15)(2n-15) \end{eqnarray}

[size=75]SFFT amounts to a specific way of factoring the given terms, usually including the addition or subtraction of an additional term.  There's a big "Search" button, above, which you can use to find several extensive descriptions of SFFT.  You can also find a description of SFFT in the AoPSWiki.  Now back to the solution...[/size]

So, we find each pair of factors, $pq=225$.  Let $p=2m-15$ be one of those factors, and $q=2n-15$ be the other.  Solve for $m$ and $n$, and see if you can find a pair that satisfies the other constraints of the problem.

${\begin{array}{ccccl}p=2m-15 & q=2n-15 & m=\frac{p+15}{2}& n=\frac{q+15}{2}& \text{remarks}\\ \hline 225 & 1 & 120 & 8 & \text{120 is a multiple of 8}\\ 75 & 3 & 45 & 9 & \text{45 is a multiple of 9}\\ 45 & 5 & 30 & 10 & \text{30 is a multiple of 10}\\ 25 & 9 & 20 & 12 & \text{Looks good}\\ & & & & \text{No others, where }m>n>0}\end{table}$

So $m+n=\boxed{32}$[/hide]

[quote="BFG"][quote="rcv"]here's a solution to #2, using SFFT:

[hide]\begin{eqnarray}\frac{1}{m}+\frac{1}{n}&=& \frac{2}{15}\\ 15n+15m &=& 2mn \\ 0 &=& mn-\frac{15n}{2}-\frac{15m}{2}\\ 0 &=& (m-\frac{15}{2})(n-\frac{15}{2})-\frac{225}{4}\\ 225 &=& (2m-15)(2n-15) \end{eqnarray}

[size=75]SFFT amounts to a specific way of factoring the given terms, usually including the addition or subtraction of an additional term.  There's a big "Search" button, above, which you can use to find several extensive descriptions of SFFT.  You can also find a description of SFFT in the AoPSWiki.  Now back to the solution...[/size]

So, we find each pair of factors, $pq=225$.  Let $p=2m-15$ be one of those factors, and $q=2n-15$ be the other.  Solve for $m$ and $n$, and see if you can find a pair that satisfies the other constraints of the problem.

${\begin{array}{ccccl}p=2m-15 & q=2n-15 & m=\frac{p+15}{2}& n=\frac{q+15}{2}& \text{remarks}\\ \hline 225 & 1 & 120 & 8 & \text{120 is a multiple of 8}\\ 75 & 3 & 45 & 9 & \text{45 is a multiple of 9}\\ 45 & 5 & 30 & 10 & \text{30 is a multiple of 10}\\ 25 & 9 & 20 & 12 & \text{Looks good}\\ & & & & \text{No others, where }m>n>0}\end{table}$

So $m+n=\boxed{32}$[/hide][/quote]

You are way too complicated. Here is my easy way:
let m = 3 and n = - 5. You get the answer rightway.[/quote]
you sir, are a genius

[quote="asianhottie8293"]Hey, have no time, so on same page, sry.

1. How do u quickly calculate partitions?

2. 1/m +1/n= 2/15

m>n and not multiple of n. Find m+n

3.Suppose a coin purse contains 30 coins which are nickel, dime, or quarter. How many different combinations of coins value 1 dollar?[/quote]

[hide]1. You can't, just list it.

2. Simon's favorite factoring trick

3. The best you can do is set up the two equations, eliminate a variable, then list all ordered pairs of the equation you are left with and then you can go from there.[/hide]

[quote="binomial_4eva"]i think there is a formula for finding the partition of n
p(n)=something
i will find that something very soon.[/quote]
I don't think there is.
No mathematician has come up with a closed form of it in hundreds of years.

What about
http://en.wikipedia.org/wiki/Partition.
Click on the number theory.

[quote="indianamath"][quote="Arvind_sn"][quote="frostmourne"]what are partitions?[/quote]

Partitions are the ways that you can add different positive integers up to make that number. for example, the partitions for 1 are:

1

For 2:
2
1+1

For 3:
1+1+1
2+1
3

And so on. however, do not think that the number of partitions for a number n is n. it is actually something more complex.

Here are a few values of the partition function (the number of partitions a given number n has)
    * p(1) = 1
    * p(2) = 2
    * p(3) = 3
    * p(4) = 5
    * p(5) = 7
    * p(6) = 11
    * p(7) = 15
    * p(8) = 22
    * p(9) = 30
    * p(10) = 42
    * p(100) = 190,569,292
    * p(1000) = 24,061,467,864,032,622,473,692,149,727,991 ≈ 2.4 × 1031

Asianhottie, you're best off posting your question on one of the high school math forums. :wink:[/quote]

hmm how did you get those numbers so fast? what's the formula?[/quote]
there isn't one.
You can look up large numbers partitions on the internet I'm sure.
http://home.att.net/~numericana/data/partition.htm
this one has p(x) up to x=4096
http://www.btinternet.com/~se16/js/partitions.htm
this one has a java applet

[quote="jhredsox"]can someone explain partitians on here?

jorian[/quote]
partitions are the number of ways that they can divide up a number where order doesn't matter.
Partitions of 1: 1
2: 1+1, 2
3: 1+1+1, 2+1, 3
4: 1+1+1+1, 2+1+1, 2+2, 3+1, 4
etc

[quote="indianamath"]hmm how did you get those numbers so fast? what's the formula?[/quote]
dude  man care to read others' posts before you?

i think the formula has something to do with combinatorics, permutation and combination. but the formula in this page http://en.wikipedia.org/wiki/Partition_%28number_theory%29
doesnt have any combinatorics init. its wierd

i also think that formula for partition has something to do with formula for factorial.

[quote="13375P34K43V312"][quote="asianhottie8293"]wowwwwww how did u find p(1000) I feel so low level and stupid here, and i'm almost in high school. i don't even know what mod is XD[/quote]

um he either used a computer or just looked it up.
there is no formula that will be useful to you at this point, it's not something like a polynomial where you just plug in numbers, it's pretty complicated, and even stating the specific way to compute the number of partitions of a given integer is far above mathcounts.[/quote]

yea i just looked it up. but i guess you could use the generating function. (of course it would take forever and ever and ever and...)

But Mathcounts does ask partitions, doesn't it?

yea it does...

[size=75][color=red]<snipped by moderator rcv.  This material is too advanced for MATHCOUNTS.>[/color][/size]

EDIT: If you don't get that, either look it up or disregard it. The best way to do the partition problem on mathcounts is to count it out, because:

1. They don't usually give big numbers for partitions.

2. It's not as hard as you think it is.

3. [size=75][color=red]<snip>[/color][/size] you're better off reading AoPS and stuff


Καλησπέρα παιδιά, μπορείται να με βοηθήσετε με αυτήν την άσκηση?

Να δειχθεί ότι η σειρά $ \sum_{n\equal{}0}^{\infty} \frac{x}{1\plus{}n^2x}$
α) συγκλίνει ομοιόμορφα, όταν χ>=0 σε μια αύξουσα συνάρτηση f(x) ,

β) Nα βρεθούν τα  $ f(oo)$ και $ f(0\plus{})$
γ) και να δειχθεί ότι $ f'(x)\equal{}\sum_{n\equal{}0}^{\infty}( \frac{1}{1\plus{}n^2x})^2$
     όταν χ>=α>0

Νοστιμότατη :)

Νομίζω ότι η παρακάτω λύση δουλεύει, εκτός και αν έχω χάσει κάπου κάτι προφανές

[hide]
Για να μη γράφουμε πολλά, έστω η ακολουθία των (προφανώς συνεχών) συναρτήσεων $ f_n(x) \equal{} \sum_{k\equal{}0}^n \frac{1}{k^2\plus{}\frac{1}{x}}$ για $ x>0$ και $ f_n(0)\equal{}0$.

Από σύγκριση με την $ \sum_{k\equal{}0}^\infty \frac{1}{k^2}$, είναι εύκολο να δειχθεί ότι η παραπάνω σειρά συγκλίνει (κατά σημείο) για κάθε $ x$, και έστω $ f(x) \equal{} \lim f_n(x)$.

Τώρα, έστω $ x_1 < x_2$ οπότε, προφανώς, $ f_n(x_1) < f_n(x_2)$ και, συνεπώς, $ f(x_1)\leq f(x_2)$ περνώντας στο όριο $ n\to\infty$. Άρα η $ f$ είναι αύξουσα.

Για την ομοίομορφη σύγκλιση, έστω $ \epsilon>0$, οπότε πρέπει να βρεθεί $ N$ τέτοιο ώστε αν $ n\geq N$ να έχουμε $ \sup_{x\geq0}|f_n(x) \minus{} f(x)|<\epsilon$. Ωστόσο:
\[ \sup_{x\geq0}|f_n(x) \minus{} f(x)| \equal{} \sup_{x>0}\sum_{k\equal{}n}^\infty \frac{1}{\frac{1}{x}\plus{}k^2} \leq\sup_{x>0}\sum_{k\equal{}n}^\infty \frac{1}{k^2}\]
και το παραπάνω $ \sup$ μπορεί προφανώς να φραχθεί από τη σύγκλιση της σειράς $ \sum\frac{1}{k^2}$.

Συνεπώς, έχουμε ομοιόμορφη σύγκλιση $ f_n\to f$ και τα υπόλοιπα ερωτήματα προκύπτουν σχετικά εύκολα. Δηλαδή, $ f(0) \equal{} 0$, $ \lim_{x\to\infty} f(x) \equal{} \frac{\pi^2}{6}$ (ή όσο είναι το άθροισμα της σειράς, ποτέ δεν τα θυμάμαι αυτά τα πράγματα ;) ) και η παράγωγος προκύπτει απο παραγώγιση κάτα όρους για $ x>0$.[/hide]

Hope this helps. Cheerio,

Durandal 1707

σε ευχαριστώ πολύ Durandal!
νομίζω πως αυτή είναι η λύση, με βοήθησες πάρα πολύ,
ευχαριστώ και πάλι καληνύχτα!

den xriazete theoritika na kanoume ta pragmata poio diskola apo oti einai...To na dikseis omiomorfa einai poli apli efarmogi tou kritiriou tou Wiestrass....

[quote="Durandal"]Hehe, profanos Sorrowman, dikio exeis... aplos pote den thymamai ta diafora kritiria syglisis kai protimo na exo mia heiropiasti apodeiksi brosta mou gia na min yparxei periptosi na kano lathos epeidi den thymamai sosta to tade i to deina kritirio ;)

Cheerio,

Durandal 1707[/quote]

E kala to sigkekrimeno kritirio omws einai san na leme to evagkelio , sto na dixneis omiomorfi sigklisi seirwn!

Hehe, mia kai to ethikses, to ti theorimata/kritiria/limmata/whatnot theorei kapoios os eyaggelia einai mallon prosopiko zitima, no? ;) :)

Se prosopiko epipedo loipon, teino na hrisimopoio kai na epikaloumai mono theorimata pou den exoun endelos profani kai syndomi apodeiksi. Diladi, an i apodeiksi enos theorimatos de mou prosferei mia kainouria mi profani texniki kai einai kai misi grammi, eh, teino na to ksexnao (ah, ayti i alzheimer ;) ).

Ayto sigoura einai endelos prosopiko, epeidi protimo na vlepo to periexomeno tis apodeiksis brosta mou para to onoma tou kritiriou. Etsi, kritiria opos to Weierstrass ta theoro perissotero san hrisima ergaleia kodikopoiisis kai formalismou, kai ta psaxno gia na ta epikalestho mono otan exo poly makroskeli epixeirimata. Kata t'alla, prospatho na ta apofeygo kai, logo athlias mnimis, ta ksexnao, ayto einai pou prospathousa na po proigoumenos ;)

My tuppence anyway... Stin en logo askisi, proteinoume akrivos ton idio tropo me diaforetiko onoma, opote let's move on :) Thanks a lot pou mou thymises pos to lene to kritirio, imoun sigouros oti kati tetoio tha eprepe na yparxei kodikopoiimeno, alla to eixa ksexasei endelos!

Cheerio,

Durandal 1707

PS: Apologies pou ayto bike sto forum kai oxi se PM, alla to software mou kanei nera kai de mou to estelne :(


Well next month I have a serious test coming up open to all 8 and 9th grade residents of New York City. This ONE test determines admission to the top schools, referred to as the "Specialized high schools". I currently attend the Bronx High school of Science (http://en.wikipedia.org/wiki/Bronx_High_School_of_Science), and am a freshmen. I personally DON'T want to be there. I am planning on taking the test this up coming November 8. At first I was eager to take the test. But looking at the statistics to the number of applications and offers is ... Of 5200 9th grade applications only a paltry 233 received offers. 10 of which were for the school I want. My goal is Stuyvesant high school. I know how to approach ALL the questions on the test, but my bloody nervous ego completely ruins it with my nervous-ego.

As the test date nears, I'm getting more and more nervous. I feel that my chances are too low, and I may not be good enough. Its really been bugging me lately. Even to the point where I can't sleep. I despise waking up at 4, to go to a school I DON'T want to be in.

I really need some advise, what to do, should I even take the test, if so how can I do help myself do the best I possibly can. This is my final chance to get into the school of my choice, and I REALLY don't want to lose this opprutunity...

Any advise, pep-talk, anything will be greatly appreciated, I will also post my score in mid February when I get my results.

In conclusion, I really want to take the test as a confident individual, and not whimping over what will or may happen.

My advice would be to take the test or else you would be wondering for the rest of your life if you would have made it. The important thing is that you try your hardest. Don't let your parents, your friends, or even yourself say that you are terrible. Because it's the effort that counts. You should only beat yourself up if you didn't put 110% effort.

Stress is primarily created through the constant thoughts of a particular subject. Take your mind off this test during the weekdays, and only study during the weekends. Don't even think about it during school, or after school. Hang out with friends and talk about anything but school or play some video games (Call of Duty 4 or Halo 3 always helps me take my mind off things  :P ).

I'd suggest meditation or running or both. Not only are both great for stress relief, they also are great for increasing mental capacity (and hence doing better on tests). Meditation techniques in particular can be applied during the test (obviously running can't) to keep you focused and not nervous.

Thanks to the above posters, I'll take both ideas into consideration.  :)

well, think of it this way

what is the worst thing that can ever happen?

you die

not so bad now is it?

and why would you want to go to this school anyways?

because you want to get into a good college.

why would you wanna get into a good college?

so you can have a good job with high pays

why would you want high pays?

so you can be stuck with a high paying busy boring job wiht no time to relax and enjoy life

and remember that watever happens, you die anyways

so chill

Don't worry too much about failure.  As Thomas Edison said, you have to fail your way to success.

Character is determined by what you do on your third and fourth tries.

I'd get some sleep, and keep this in mind: Bronx Science is a great school, in many respects on par with Stuyvesant (and in some respects I'd say better).  Whether or not you get into Stuy, you're already obviously an excellent student, and you're better off concentrating your efforts on getting the most out of what Bronx Science has to offer.  If you genuinely don't like it there (and not merely because you feel it's second best) then I would wager that you will almost certainly not like it at Stuy either -- the atmosphere is very similar.

People will eventually know you for how hard you work and what you have accomplished. This is almost completely independent of what school you go to. You can go to a crappy high school (I'm not familiar with the NYC system, but it sounds like you aren't in that bad of a situation), then you can go to a crappy college, and even that won't really be that much of a set back. I did exactly that, but worked hard through both and eventually ended up at a really good grad school.

Don't get me wrong. Going to good school's makes the process easier, but it is never the end of the line if you don't get in. It is what you do there. So don't worry too much about it. Always try your best, and what happens happens. Not much you can do about it once it happens.

[quote="HilbertThm90"]it sounds like you aren't in that bad of a situation[/quote]  No, indeed: Bronx Science is one of the top handful of New York City high schools.  (Although for reasons that are unclear to me, its math team hasn't been so great in recent years -- I figure largely a product of poor advertising and recruiting in school.)  See [url]http://en.wikipedia.org/wiki/Bronx_High_School_of_Science[/url] if you're curious.  They have a rather impressive list of alumni, including 7 Nobel laureates in physics.

Out of curiosity, why do you dislike Bronx Science so much?


algebra
polynomial
function
induction
system of equations
linear algebra

Let $P_{n,k}$ denote the set of all $n$-variable polynomial functions that are homogeneous of degree $k$. Prove that the set $\{ f(x_{1},\ldots,x_{n}) : f(x_{1},\ldots,x_{n})=x_{1}^{i_{1}}x_{2}^{i_{2}}\ldots x_{n}^{i_{n}}$ where $i_{1},\ldots,i_{n}$ are integers satisfying $0 \leq i_{j}\leq k$ for all $j$, and $i_{1}+\ldots+i_{n}=k \}$ forms a basis for $P_{n,k}$.

It's in the correct forum. It just hasn't been answered yet. Sometimes that happens.

Still no reply. Can we find more people that know how to approach this problem in the abstract alg. forum?

It's not true. If the polynomial functions are over $\mathbb F_{2}$ (you didn't mention anything about the field), then, for example, \[x^{2}y+x y^{2}\equiv 0 .\]

[quote="perfect_radio"]It's not true. If the polynomial functions are over $\mathbb F_{2}$ (you didn't mention anything about the field), then, for example,
\[x^{2}y+x y^{2}\equiv 0 . \]
[/quote]

sorry, i do not understand what you mean. i have
only been through a little bit of abstract algebra (not field theory yet).

the functions are from $R^{n}$ to $R$.

for example, with $n=2$ and $k=3$ we know that

$x^{3},x^{2}y,xy^{2},y^{3}$ form a basis for $P_{2,3}$.

I was thinking about performing computations modulo $2$.

Anyway, for the real case: Suppose that a linear combination of polynomials in $P_{n,k}$ gives the null function. Write this in the form $\sum_{i=0}^{k}x_{1}^{i}Q_{i}\left( x_{2}, \ldots, x_{n}\right) \equiv 0$, where $Q_{i}\in \mathbb R \left[ x_{2}, \ldots, x_{n}\right]$ are homogeneous of degree $n-i$.

Fix $x_{2}, \ldots, x_{n}$. We are dealing with a single variable real polynomial polynomial in $x_{1}$, that is identically zero. A well-known theorem tells us that all of its coefficients must be zero!!! Hence, $Q_{i}\equiv 0, \, \forall i \in \overline{0,k}$. The conclusion then follows by induction over $n$, the number of variables.

Concrete example: If $a x^{3}+b x^{2}y+c x^{2}z+d y^{2}z+e y^{2}x+f z^{2}x+g z^{2}y+h xyz+i y^{3}+j z^{3}\equiv 0$, then $a x^{3}+x^{2}(by+cz)+x(e y^{2}+f z^{2})+(d y^{2}z+g z^{2}y+i y^{3}+j z^{3}) \equiv 0$.

PS: Using the same method we can prove that the monomials from $\mathbb R \left[ x_{1}, x_{2}, \ldots, x_{n}\right]$ are a basis of the space composed of the corresponding polynomial functions. Homogeneity was never the issue.

ok. i follow what you are saying. but then how are we sure that $Q_{i}\equiv 0$ for all $i$ implies that the original coefficients are $0$. Doesn't that transform the original linear combination into a new nonlinear system of equations?

now, about your final comment. so do you mean that we can generalize to the case where we deal with a basis like $\{x,x^{2},xy,z^{3},wxyz \}$ for some set of polynomial functions, where each term in the basis is not necesarily the same degree, and the number of variables is arbitrary?

[quote="perfect_radio"][quote="drunner2007"]ok. i follow what you are saying. but then how are we sure that $Q_{i}\equiv 0$ for all $i$ implies that the original coefficients are $0$. Doesn't that transform the original linear combination into a new nonlinear system of equations?[/quote]
Remember that we are doing induction by the number of variables. The $Q_{i}$'s are polynomials of $n-1$ variables. As I said, $Q_{i}\equiv 0$, so, by the induction hypothesis, all the coefficients of the different monomials in $Q_{i}$ are equal to $0$. When we expand that sum from $i=1$ to $k$ there's no way to get a non-zero coefficient.

As for the other question, yes.[/quote]

oh, ok. that's a nice argument.

so for example you would conclude that $by+cz =0$ for all $y$ and $z$. then we fix $z$ and conclude that $b=0$. then we get $c=0$.

after that do the same thing with the other $Q_{i}$'s ?


Prove that $\forall a,b,c > 0$ and $a,b,c \in \mathbb{R}$ \[\sum_{cyc}{\frac{a^{5}}{a(a+3b+3c)+2bc}}\geq \frac{a^{2}b+b^{2}c+c^{2}a}{9}\]

According to Cauchy in Engel form: \[\sum_{\text{cyc}}\frac{a^{5}}{a(a+3b+3c)+2bc}\geq \frac{\left(a^{3}+b^{3}+c^{3}\right)^{2}}{\sum_{\text{cyc}}\left(a^{2}(a+3b+3c)+2abc\right)}= \frac{\left(a^{3}+b^{3}+c^{3}\right)^{2}}{(a+b+c)^{3}}.\] But $\frac{\left(a^{3}+b^{3}+c^{3}\right)^{2}}{(a+b+c)^{3}}\geq \frac{a^{3}+b^{3}+c^{3}}{9}$ because of Power Mean and $a^{3}+b^{3}+c^{3}\geq a^{2}b+b^{2}c+c^{2}a$ because of Rearrangement. The result follows.

Edit: I knew I saw it [url=http://www.q-e-d.be/index.php?p=forum&topic=3809]somewhere[/url] before.  :P

Exactly what I had in mind  :lol:  Nice work!


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what do you guys think about it. i dont think its the games fault at all.  its up to the person to do something.

Well what you like to play is somewhat related to what you prefer to do.... not to say that video games are the conspirators.

I do agree its the person's own decisions. And the parents suing video games (i know I didn't spell sueing right ... suing... sueing...su...) should really be complaining to themselves about why they didn't train their kids against this kinda stuff.h

I personally think there's a lot more violence in TV, it's just covered up. I mean just look at the Powerpuff Girls or Teen Titans or Justice League, they all seem to be about "Justice" and "Being nice" but instead they're just beating up people who they label "evil"...we live in such a moronic society...

On another note in Bowling for Columbine some people blamed violent video games for corrupting the kids...

I love violent games, you name it I'll play it (and eat while I'm at it :lol:) As long as you can seperate out the real world from the virtual game world you'll be fine, but as you know some people can't do that and as a result... :(

right then. how do you beat the last level of GTA 2? for the first level, i just drove around bumping into cars, and that was pretty much it (well no, i did jobs). for the second level, i got a *8 multiplier, then busted out the tank. i don't know how to beat the third level. right now i've got a *5 multiplier.

i never have played gta2, but on another note. you can get gta3 and gta vice city for 30 bucks at best buy! woot. im gonna have to get that, if im allowed. but hopefully my parents can see the light that i can tell the diff, which i can

I thought that your parents were less strict than mine... but guess not.

so your parents being more lenient is locking your computer and not telling you the password?

well.. they let me buy those kinda stuff... well... maybe one's less strict in one area... and the other is less strict in the other area...

There's nothing wrong with it.  If you can't handle the violence don't play it.  You can't blame street violence on a video game.  It's up to the player to find out the difference between a video game and REAL LIFE.  Such a difficult concept.

Suing.

And I agree with Dark_Shadow - people have a choice.

True...people have a choice, but they don't always see that...

...how can you not be aware that you have a choice. i mean duh. you can play the game or you can not play the game. wow.

sure...you can play or not play the game, but how you react to the game is your choice. If a person grows up on violent videogames, they become "immuned" to the violence, I mean, they see it whenever they play their videogames, and since no one actually gets hurt, they don't see the harm. Those kids are more likely to be immuned to violence in the outside world as well. They might take a gun to school/work/or wherever because when they use guns w/ the videogames, no one gets hurt, so why should someone get hurt in real life? Believe it or not, it's true. Many people take guns somewhere to make a point and then use them because no one gets hurt when they're at home. Yes, many people [i]do[/i] believe this stuff...I don't but have met people who do and it's rather scary. They have no concept of violence, it's all normal for them. They support war and underage weapon holders etc. They just don't see the harm...they [i]can't[/i] see the difference between real life and a video game...

Well most families are closed towards sex...for example in my family we've never discussed sex and in school we only discuss sex in 6th science and family/consumer science (though the second is not an "official" place to discuss it but we still learn a lot like what an "aphrodisiac" is)

Mmm this might evolve into another topic; sex in TVs (I doubt there are video games with sex; they say "adult content" quite a lot but mainly they have girls dancing in skimpy clothes, I doubt there's even any nudity in it) and movies (which also rarely has sex but rather nudity which is now considered offensive :?:)
What I mean is violence and sex is completely different topic and they aren't often in video games but rather TV or movie or songs. Therefore it might be better to make a new topic rather than continue on here since it might be labeled "off-topic" It's still along the same line though.

Yup I did what you said.

NPR = national public radio.  It doesn't have (much) advertising, so it doesn't have to try to get ratings.  As a result, it's much higher quality and much less sensationalistic.  The NPR news station in the Boston area is WBUR, broadcasting from Boston University.  I don't remember the place on the band it has, but I wake up to it every morning.  Their news is excellent (they get some of it from the BBC, which gives you an outside-America point of view, as well as occasionally telling you international cricket scores :) ).

Not sure which bit you are referring too with the second comment (feel free to go back and edit instead of starting anew), but I've only been a college student for a few months, and anything I can say about this as a college student I could have said as a high school student.  My viewpoint (and my knowledge on these matters) have not changed at all.

Lara Croft?  (Not exactly sex, I know.  I'm not really being serious.  There are probably video games with 'adult content,' but I don't know of any.)

WBUR currently broadcasts at 90.9 FM Boston, 1240 AM West Yarmouth, 91.5 WSDH Sandwich, and 90.3 WCCT Harwich.  (From their website, http://www.wbur.org)

Being Christian, I'll admit that in the past we've killed, been killed, you name it.  But it still doesn't make violence okay.  The reason the crusaders killed was to get people to Christ, which if you think about it, doesn't make sense.  "You won't be saved, we'll kill you"?  And sorry, sex doesn't happen before marriage in a Christian's case, or at least not usually.  It's not supposed to.

I have to agree with raccoon...we Christians have killed and been killed, and like raccoon said, that doesn't make it ok...

and also like raccoon said, Christians practice abstinence, I know I do...as do many many others...sex is something that should only happen between a husband and wife, and definately should not occur before marriage...that is what many many Chrustians, including myself, believe...

But you also believe that violence isn't something that should happen at all, with the possible exception of 'just wars.'  So -- why is there violence on TV but no sex?  Either both or none, I say.

Since this thread still hasn't been split I'll keep talking:

Okey dokes. My mom usually listen to WBG for news, is that a good source of news?

Well Lara Croft is sex[b]y[/b] but then most other video games have sexy character too. Besides what's sexy and what's not depends on the person.

Um even if you're not a Christian you're still way too young...

I don't know about WBG -- it certainly [i]might[/i] be a good news station.  It probably isn't any worse than any alternatives.  I don't know.

NPR is nationally syndicated -- that is, much of the same programming ends up being played in NPR stations in New York and Boston -- so I knew before coming here that NPR was good.

I think that violence should not be in vidio games.  Violence is something that should be dicoraged and putting it in games does not dicorage it.

Behold, the philosophical guru has returned....
Violence in video games is not necessarily evil in any way.  If you analyze the entire history of the human race, you will see humanity engaging in wars for supreme dominance and dictatorship over all.  Such games as "God" games promote the thought that humans are dominant and godlike, which to a certain extent may be true.  However, when one takes this into view, one must then see humanity's merciless butchering of animals.  Now, this also suggests that humanity is a very violent and bloodthirsty species, bent on destruction and greed to further achieve their own twisted corrupted goals.
Therefore, one can safely conclude from these findings : Humans love a good war.  Humanity loves to fight.  This is why the banning of violent video games would be next to impossible, as violence transcends into one of the more primitive pleasures, and thus one of the most satisfactory.  The same can be said of sex : humanity loves lust, and each individual has some desire to show what they can do, for in such a twisted and corrupted society, humanity has only one choice: to show it is supremely dominant.  Therefore when taking this also into account, one can see that violence and sex have the same ultimate goals, and thus must fall along the same perspective.  Such mindless games as GTA3 (gotta love it) are one of the epitomes, where you can easily run around and terrorize a city with ongodly weapons (oooh man machine rocket launcher, how i love thee.. let me count the ways...gameshark + infinite ammo + happy trigger finger...)  Of course, this then falls back to the category of greed.  In the drive of humans to prove themselves, they must have some way to show their lordly powers.  And this is the interesting part...whether by instinct or by years of conditionaing, all humans are cowards.  The reason why violent games sell so much is that human beings realize they cannot beharmed or killed.  If they die,theres always a save and reset button.  Violent games, though they do indeed affect society, do so in a very minimal way by the fact that humans do not want to get hurt, and to be on the run, and to be cornered.  Thus one can see that only the extremists actually go out onto the streets with twin katanas and mercilessly slaughter people to sate their violent appetite.  Such people are extremely rare, and usually have some kind of mental trauma (such as thinking too philosophically...now whered i put those katanas?...)
When taking all these factors into account, violent and sexed up video games are not necessarily evil or bad.  The points about TV shows also having violence which children grow up watching are well founded.  In this corrupt society, one cannot merely put a classification of "evil" and "good" on certain aspects.  To do so shows their own weakness in succumbing to society.  Even in popular shows like looney tunes, this mysterious and shady company known only as Acme is a premier supplier of terrorist weapons, such as explosives, anvils, and bananas.  It also shows man hunting animals, yet animals always win.  In a twisted, sadistic way, this is meant to show that man is not all powerful.  Therefore what you get from this show is mixed.  Other shows portray superheroes beating up bad guys for they have something morally unjust in society.  And then the MPAA comes in (their history is incredibly interesting....all cartoon characters springing up from adult flicks?...) and their biased way of looking at things easily puts an imbalance into society.
And while im on this particular thread, let us look at japanese animation, or anime.  It has been extremely biased and shunned, for one does not see the beauty of miyazaki, or the sadness of some of true anime.  DRAGONBALL Z IS NOT ANIME AND WILL NEVER EVER BE ANIME.  THE SAME APPLIES TO YUGIOH AND POKEMON.  Though this is technically anime, it is not anime as defined in Japan.  Likewise, anime is also seen as steamy erotic hentai, with japanese perverts running around tearing off their clothes.  Americans shun this type of anime also.  Anime is neither of these.  Many people do not even watch anime, for it is aimed at kids.  All Americans automatically think of cartoons as kids shows, when there is some very deep and thought-provoking anime out there.  Likewise, hormonally induced teenagers watch anime hoping to catch a sight of bare skin.  As such, Americans have shunned anime without truly knowing its roots.  The reason why anime and manga is so explicit in its content lies deep within japanese roots and culture.  Nudity in Japan is not regarded as some obscene misdeed.  There are public baths in Japan, and the history of Japan is rich and colorful.  Anime is just another facet of japanese culture.  If you get so biased, go watch some Trigun or some Serial Experiments Lane.  If you want to see anime's beauty, go watch Grave of the Fireflies, a world war 2 epic, or Princess Mononoke, one of the greatest animes of all time.  (watch these in japanese with english subs, as english dubbed anime has been rectally violated).
And so from all of these conclusions put together, violence is something humanity is immersed in at all levels.  Violence in video games does not usually turn normal people into raving mad lunatics who rant about anime.  It is merely to obtain profits since it sells so well.
I rest my case.  Ill bet anyone that NO ONE is going to read all the way through this.  And before anyone argues, [size=150]IT IS NOT SPAM.[/size]  Heh...good luck.

I read through all that [size=59]spam[/size] and I agree to most except too things. One, violence does affect you subliminary whether it's for the good or the bad like having dream about killing/raping/etc. people. (which may not be a bad thing to some people, remember I'm not being judgemental here) Two, you are sadly misinformed about the definition of "anime" First of all, "anime" is short for (and thereby having the same definition as) "animation" which can go anywhere from Digimon to Toy Story (fyi Pixar is short for "Pixar Animation Studios") to Cinderella 2. Animation can be 3-D and does not necessarily have to be Japanese. Also anime is very little part of Japanese culture considering the entire shlong. Secondly "average" teenagers (which may or may not include those from your area) are not really attracted to 2-D people/cartoon characters and people who do are considered weird. Besides there are tons of alternatives then to go on an endless search for nudity in cartoons (a.k.a. Google image search). Third, the frase "hentai" means "perverted" in Japanese fyi so  usually it is an adjective describing a person not an object. Therefore hentai does not originally mean anime at all, it might now in America but not originally. Fifth, most animes are meant for kids: Miyazaki films and other more artistic creations (which tend to be unpopular because of its artistic feature) are more classified as artwork or a film rather than anime even though technically it is still an anime. I'd rather divide anime into "artwork" and "others" or "artwork" "anime" and "porn" rather than how you classify them. Finally you make anime look like a good thing but subjectively it has some good but many bad; currently many middle-aged male are now attracted to cartoon characters which I can not even describe: they are vile creatures who stay in a stinky air-conditioned apartment all day doing terrible things. However it is impossible to know this as an outsider so I don't blame you for not knowing but just to let you know.

Final thought: Welcome back o great one.

heh..i thank you for your welcome, and of course i know i erred somewhat.  Arigato Gozaimasu.  However i was not talking about the standard definition of anime (which i know) but rather anime that is more widely accepted as anime.  Films such as Miyazaki in fact have incredible approval and admiration and are indeed classified as artwork, but also as anime.  As for japanese culture, i am quite well aware of that.  I have studied the japanese, and the various anime relations in said culture.  As for hentai, it is generally accepted as anime porn, however i did indeed know it means "perverted".  I know a bit of japanese myself :-P  In fact, it is an interesting fact that much of hentai is actually made by the women, not by the men.  Of anime in general, i prefer to focus on the good aspects of it, rather than point out the negative.  After all, im using it in my argument :)  however im well aware that anime is a mix of both good and bad, and is not perfect.  I also agree with the teenagers..again i was exaggerating.  However anime is based off of japanese culture.  Read that paragraph again.  I did not say it was a type of japanese culture, merely that it is influenced by japanese culture.  Sumimasen.
Of course, my views are a little extreme, and i can see the various flaws.  However, overall i was stressing the point of violence.  I do indeed agree with you on all that you said.  Just know that I know of japanese culture and japanese anime :)  speaking of which, do you watch any?  And not DBZ (AKU, SOKU, ZAN!)

Heh sorry for the flaws :)

P.S.  Ive been hearing differing opinions, but...are you a boy or a girl?

:lol: "different opinions" :lol: hahahahaha...I choose not to answer, just to keep things a bit more interesting.

Oooh good to know there's one more person here who speaks Japanese. (though that only makes three)

Yeah I guess, still channel like Cartoon Network and Kids WB! clearly portraits anime as a cartoon, besides not a lot of people see these films, just look at how much Spirited Away made.

It is interesting, especially when you see an anime with the male being more superior to the female though that is another part of Japanese culture (while Americans go shouting Women's Rights Japanese consider women to be better if they are shy and obedient, kinda like people in Sierra Leone)

Whoops, when I see a post that long my eyes tend to wade a bit...then it gets refocused after seeing words like "Japanese" :lol:

Let's see, I watched Pokemon, Zoids (little), Detective Conan, Crayon Shinchan, Doraemon, and other ones I saw very little (like Galaxy Train 999 and Sailor Moon; I actually watched the last episode of Dragonball GT) back when I was in Japan, I saw the entire first and second season of Digimon here in the U.S. not to mention that I still watch Yu-Gi-Oh! here. (I'm 95% possitive Teen Titans is American though there's Japanese voice actors/actresses and Japanese in the opening sequence)

You know speaking of violence there's a lot of that in Tom and Jerry and no one said anything about that. (and yet people ban books like Harry Potter and Captain Underpants)

Just something I failed to say in my last post:
Reading Ten Commandments.
Noticing two that connect to this thread.
5)  Thou shalt not kill.
6)Thou shalt not commit adultery.
TRANSLATIONS
5) Don't kill people.
6)No sex unless you're married to your partner.

Oh yeah, a random fact: "adultery" originated from Latin (fyi which was popularly used in Rome, also the capital of the Roman Catholics) which was on the Top 100 words on the SAT. Think about it...


geometry
3D geometry
pyramid
inequalities
LaTeX
inequalities proposed

$SO$ is height of pyramid $SABC$, where $\measuredangle ASB=\measuredangle ASC=\measuredangle BSC=90^{\circ}.$
Let $SO=h,$ $AO=a,$ $BO=b,$ $CO=c.$
Prove that $h\leq\frac{\sqrt[3]{abc}}{\sqrt2}.$

Hello there,
After some computations (if hopfully I don't have any mistakes) using that
$h=\frac{xyz}{\sqrt{x^2y^2+y^2z^2+z^2x^2}}$
the inequality turns to
$8(xyz)^2 \leq (x^2+y^2)(y^2+z^2)(z^2+x^2)$
Where $x=SA, y=SB, z=SC$. Then by using $x^2+y^2 \geq 2xy$ this last inequality is proven. :)

קרינה , פרטי , בבקשה !                          1 :)

[quote="kar"]
Please tell me if there is something wrong in it.[/quote]
All correct. :)


Find all finite polynomials whose coefficients are all $+1$ and  $-1$ and whose roots are all real.

It is sufficient to find all polynomials with the leading coefficient is $1$.
the modulus of free coefficient is $1$, $n\leq 3$ (why?), the rest is simple.


[url=http://en.wikipedia.org/wiki/Tryptamine]Tryptamine[/url]+ $ \ce{CH3COCl}$ ----->A
1.A+ $ \ce{P2O5}$ ----->B
2.B + $ \ce{Pd, C}$ ------>C

As it stands, this problem seems too vague. Aren't there any more data?

nothing except that last step involves heating :)

Is the last Last step some aromatisation?

can u please post ur probable answer :huh:

the problem is i cant draw the diagrams  :(
Let me explain what happens?
1st step only the NH2 gets acylated
2nd step it becomes NC?  :maybe:
3rd step cyclisation?  :maybe:

2ns step cyclisation
3rd step aromatisation in the last ring  :)

im not used to proposing da like you  :P   :rotfl:   :D

Wait ill edit this and write the mechanism after sometime if i get it

then get used to it soon......
[hide="sorry"]  :rotfl:  :rotfl:  :rotfl:  i forgot u r a baby :P
[/hide]

anyways the answer to this is no matter to me ....the mechanism proposed is extraordnary :D

dei vetti naye! if this wasnt the chem forum..........!!!!!  :mad:
And why have you asked the question? Bcos you feel that the mechanism given in the book is absurd or wrong?  :maybe:   :D

yes it's as magical as the one using the 'cyclising agent $ \ce{HF}$

Maybe its like this?
1st step direct acylation of  $ \ce{NH2}$ only
2nd step the 5 membered ring is formed where the H $ \beta$ to the N of the ring eliminates itself with an O and a H from $ \ce{NHCOCH3}$
3rd step Aromatisation of the new ring formed with C2H4 eliminated?  :maybe:   :)

be more exact that's not teh escat proposed mecahnism

[quote="madness"]the first step is right isnt it?
the 2nd step i propose is the elimination of the H $ \beta$ to the N of the ring to form a 5 membered ring witha commom edge with the 2nd ring
3rd step is maybe something related to surface chem by adsorption of $ \ce{C2H4}$ on Pd and thereby its elimination forming the thirs aromatic ring  :)[/quote]
then ur answer itself is wrong....there is no reduction in carbon number in the final product

no man i meant the C's of $ \ce{NHCOCH3}$. Dont they escape?  :maybe:

And 3rd step may also be aromatisation due to adsorption of H on Pd  :)

consider the enol form hoh is eliminated and it cyclizes

H from where? $ \beta$ to N of the initial ring ?  :maybe:

but pardesi says i have some product wrong?  :maybe:
Then what happens to the remaining C's of $ \ce{NHCOCH3}$ is my question  :maybe:

ch3 will be an exocyclic methyl group thats all

Oh ok  :D

The 3rd part is right no? Aromatisation of the new ring only?  :maybe:

well, yes. thats waht i think it is. will have to wait for mythili to confirm :D

vetti dogs...yes there is adsorption of H not C2H4as rohit stated initially..
varun ur meachanism is rite