$ Let$ $ x,y,z\in{IR^{*+}}$ such that:$ x+y+z=1$
find the min of:
${ \frac{x^{2}}{x+yz}+\frac{y^{2}}{y+zx}}+\frac{z^{2}}{z+xy}$
good luck :)

$ \sum\frac{x^2}{x\plus{}yz} \ge \frac{(x\plus{}y\plus{}z)^2}{x\plus{}y\plus{}z\plus{}xy\plus{}yz\plus{}zx} \ge \frac{3}{4}$
I think this problem is old,my friend.

hello, we have $ \frac{x^2}{x\plus{}yz}\plus{}\frac{y^2}{y\plus{}zx}\plus{}\frac{z^2}{z\plus{}xy}\geq\frac{3}{4}$, the equal sign holds if $ x\equal{}y\equal{}z\equal{}\frac{1}{3}$.
Sonnhard.

Hi babyloverain!
can you plz explain why $ \sum\frac{x^2}{x\plus{}yz} \ge \frac{(x\plus{}y\plus{}z)^2}{x\plus{}y\plus{}z\plus{}xy\plus{}yz\plus{}zx}$  :?:

$ \frac{x^2}{a}\plus{} \frac{y^2}{b}\plus{} \frac{z^2}{c} \ge \frac{(x\plus{}y\plus{}z)^2}{a\plus{}b\plus{}c}$
it is cauchy-schwarz

[quote="babyloverain"]$ \frac {x^2}{a} \plus{} \frac {y^2}{b} \plus{} \frac {z^2}{c} \ge \frac {(x \plus{} y \plus{} z)^2}{a \plus{} b \plus{} c}$
it is cauchy-schwarz[/quote]

It also is a direct form of Titu's lemma. :rotfl:


quadratics
LaTeX
algebra
quadratic formula

I have this question:

Solve for x when

sqrt{17x - (sqrt{x^2 - 5})} = 7

I am told to use the quadratic formula to find x after squaring twice.

How do I find x?

NOTE: The sqrt{x^2 - 5} lies INSIDE the SAME radicand as 17x.

Clear?

you could use $\LaTeX$, it makes everyone's lives easier
square both sides, isolate the remaining square root, square both sides again, then you have a quadratic.

No biggie, just write it out again.

$\sqrt{17x-\sqrt{x^{2}-5}}= 7$


A steel sphere with a 3-inch radius is made by removing metal from the corners of a cube that has the shortest possible side lengths. How many cubic inches are in the volume of the cube?

[hide]3 inch radius, so 6inch diameter, and 6 inch side length of cube

6^3=216 cubic inches [/hide]


The sequences $ \{a_n\}, \{b_n\}, \{c_n\}$ satisfy

$ a_n\equal{}a_{n\minus{}1}\plus{}b_{n\minus{}1}, b_n\equal{}b_{n\minus{}1}\minus{}c_{n\minus{}1}, c_n\equal{}\minus{}c_{n\minus{}1}\minus{}a_{n\minus{}1}$.

Prove that all these three sequences are Fibonacci sequences from some point onwards.

$ a_n\minus{}a_{n\minus{}1} \equal{} b_{n\minus{}1}$ (Eq. A(n))
$ b_n\equal{}b_{n\minus{}1} \minus{} c_{n\minus{}1}$ (Eq. B(n))
$ c_n\equal{}\minus{}c_{n\minus{}1} \minus{} a_{n\minus{}1}$ (Eq. C(n))

A(n): $ a_n\minus{}a_{n\minus{}1} \equal{} b_{n\minus{}1}$
A(n+1): $ a_{n\plus{}1} \minus{}a_n \equal{} b_n$
A(n+1)-A(n), B(n): $ a_{n\plus{}1} \minus{}2a_n\plus{} a_{n\minus{}1} \equal{} b_n\minus{} b_{n\minus{}1}\equal{}\minus{}c_{n\minus{}1}$ (Eq. D(n))
D(n+1): $ a_{n\plus{}2} \minus{}2a_{n\plus{}1}\plus{} a_n \equal{} b_{n\plus{}1}\minus{} b_n \equal{} \minus{}c_n$
D(n+1)+D(n), C(n): $ \minus{}a_{n\plus{}1}\minus{} a_n\plus{} a_{n\minus{}1}\plus{} a_{n\plus{}2} \equal{} \minus{}c_{n\minus{}1}\minus{} c_n \equal{} a_{n\minus{}1}$
Simplify: $ \minus{}a_{n\plus{}1}\minus{} a_n\plus{} a_{n\plus{}2} \equal{} 0$ (Fibonacci)

$ \{b_n\}$ is a Fibonacci sequence, because of Eq. A(n), $ b_{n\minus{}1}\equal{}a_{n\minus{}2}$
$ \{c_n\}$ is a Fibonacci sequence, because of Eq. B(n), $ c_{n\minus{}1}\equal{}\minus{}b_{n\minus{}2}$


Is it possible to use Gauss' law to find intensity of a spherical charge distribution where the charge distribution
(i)is non-homogeneous in general...i.e. dependent on (r,theta,phi)
(ii)can be thought of as a made up of concentric spherical shells of different densities?

(i) I think -No.As the distribution varies with (r,theta,phi) the spherical symmetry is destroyed---the spherical layers should encounter the field lines symmetrically.But that is not allowed here.

(ii)the spherical layers should encounter the field lines symmetrically.So,here it is allowed.

Please confirm if I am correct

you can use gauss law yes!

in all possible cases.

but. if you dont want to waste 30 pages in a non-symmetric problem, you can use others laws for it. Like an inverse-square law

like Coulomb's law


function counting distinguishability probability

Suppose you have a list of numbers or character or any data.
The data itself are pretty much non-sense and random.
But if you have a function (actually is more like a sequence) $ f(n)$ you can make a message.
For example: $ f(n) \equal{} 2n$. You simple skip one number or character, and you form words.
This is a kinf of steganography.
But the flaw of this, is that the function used form a pattern.Does not matter how complex it is, it is a pattern.
But if you use a random key, like, the sum of the digits of $ \pi$ from the first to the $ n$ digit, it will really make ir more unbreakable? The entropy of the result will be higher and indistinguishable from a random message?
Not only for this example, but any pseudo-random determinist function?

If you use the key only once or twice then, yes, the scheme you describe won't be very susceptible to cryptanalysis.  If you use it even moderately many times, however, it becomes very easy to break.  One could look for patterns that necessarily emerge in messages written in a particular language.  For example, one might ask [i]"what is the probability that the $ n$th position is a consonant given that the $ m$th position is a vowel?"[/i]  If either positions $ m$ or $ n$ are not part of the message, then we expect the statistics at the $ n$th position to be independent of whether or not the data at the $ m$th position is a vowel.  But if they are part of the message, then the probability that the $ n$th position is a consonant should be much higher (vowels are relatively unlikely to follow vowels in English, for example).  By looking for anomalies of this sort in the set of statistics compiled over the course of many message, we can determine with high probability which locations are part of the message and which are not.

So using a scheme like this for many messages is a bad idea.  You wouldn't want to use it for just a few messages either, because in that case the problem of securely relating the key is just as difficult as relating the message itself.


geometry
3D geometry
complex numbers
combinatorics proposed
combinatorics

This has a very very nice solution.

A large regular hexagon is cut out of a triangular grid (Like in the picture) and tiled with diamonds (pairs of triangles glued together along an edge), Diamonds come in three orientations. Prove that precisely the same number of each variety must appear in the tiling.

I don't think that the right place for this problem is the Solved Section  :?
But I think something similar already appeared on the forum!!
Happy new year to everybody :)

Maybe this doesn't count as a solution, and maybe it's not very explicit, but.. :)

Dra a hexagon like the one described and a tiling. Does it remind you of anything? Maybe a $3D$ figure staring out at you from the page? :) If we interpret it as a $3D$ figure (you must draw it to see what I mean; the figure is composed from the squares obtained when we divide three faces of a cube which have a common vertex in small squares, eventually shifted), the conclusion follows from the fact that it must have $n^2$ squares on each plane ($xOy,yOz,zOx$), where $n$ is the number of segments in which a side of the large hexagon is divided.

Sprry it's so unclear it sounds like nonsense :).

You can find it here: http://www.math.u-psud.fr/%7Epansu/texte_Olivier.html

But it is in french  ;)

*Edit: The results holds for all tiling which you can recover with diamonds !*

Yes, that's what makes the problem extremely beautiful. You literally [b]see [/b]the solution after drawing the figure.

A formal proof is easy afterwards.

Another sketch of a solution: it's not hard to show that the equilateral triangles which tile the plane can be filled with complex numbers of the form $a+b\varepsilon,\ a,b\in\mathbb Z$ s.t. the orientations cover a sum of $1,\varepsilon,\varepsilon^2$ respectively (i.e each piece having the first orientation covers a sum of $1$, each piece having the second orientation covers a sum of $\varepsilon$, and each piece having the third orientation covers a sum of $\varepsilon^2$), where $\varepsilon$ is a solution of $x^2+x+1=0$.

From here, the conclusion will follow: the hexagon covers a zero sum, and if we have $a,b,c$ pieces with the three orientations respectively, the sum is $a+b\varepsilon+c\varepsilon^2$, which is $0$ iff $a=b=c$.


inequalities
limit
inequalities unsolved

For $ a,b,c>0$ with $ a\plus{}b\plus{}c\equal{}3$ prove that $ \sum{a/({b^{2}\plus{}1}})>\equal{}3/2$

[quote="mon"]For $ a,b,c > 0$ with $ a \plus{} b \plus{} c \equal{} 3$ prove that $ \sum{a/({b^{2} \plus{} 1}}) > \equal{} 3/2$[/quote]
$ \sum_{cyc}\frac{a}{b^2\plus{}1} \geq \frac{3}{2}$ $ \Leftrightarrow$ $ \sum_{cyc}(a\minus{}\frac{ab^2}{b^2\plus{}1}) \geq \frac{3}{2}$ $ \Leftrightarrow$ $ \frac{3}{2} \geq \sum_{cyc}\frac{ab^2}{b^2\plus{}1}$
by AM-GM $ \rightarrow \sum_{cyc}\frac{ab^2}{b^2\plus{}1} \leq \frac{ab\plus{}bc\plus{}ca}{2} \leq \frac{3}{2}$
because $ (a\plus{}b\plus{}c)^2 \geq 3(ab\plus{}bc\plus{}ca)$

[quote="mon"]For $ a,b,c > 0$ with $ a \plus{} b \plus{} c \equal{} 3$ prove that $ \sum{a/({b^{2} \plus{} 1}}) > \equal{} 3/2$[/quote]
My solution:
We have:
$ \sum_{cyc} \frac{a}{b^2\plus{}1}\equal{}\sum_{cyc} \frac{a^2}{ab^2\plus{}a} \geq \frac{(a\plus{}b\plus{}c)^2}{\sum_{cyc} ab^2\plus{}3}$
We need prove that $ ab^2\plus{}bc^2\plus{}ca^2 \leq 3$ or $ (a\plus{}b\plus{}c)^3 \geq 9(ab^2\plus{}bc^2\plus{}ca^2)$
We can suppose that $ c\equal{}max\{a;b;c\}$ then this inequality becomes:
<=>$ (a\plus{}b\plus{}2c)(a\minus{}b)^2\plus{}(a\minus{}c)(b\minus{}c)(4a\plus{}c\minus{}b) \geq 0$
Done!

[quote="chien than"]
We need prove that $ ab^2 \plus{} bc^2 \plus{} ca^2 \leq 3$ or $ (a \plus{} b \plus{} c)^3 \geq 9(ab^2 \plus{} bc^2 \plus{} ca^2)$
[/quote]
Which wrong. Try $ a\equal{}b\equal{}1$ and $ c\rightarrow0^\plus{}$ :wink:

[quote="Sunjee"][quote="arqady"][quote="chien than"]
We need prove that $ ab^2 \plus{} bc^2 \plus{} ca^2 \leq 3$ or $ (a \plus{} b \plus{} c)^3 \geq 9(ab^2 \plus{} bc^2 \plus{} ca^2)$
[/quote]
Which wrong. Try $ a \equal{} b \equal{} 1$ and $ c\rightarrow0^ \plus{}$ :wink:[/quote]
Dear Arqady: $ a \plus{} b \plus{} c \equal{} 3$ ![/quote]

But the inequality $ (a \plus{} b \plus{} c)^3 \geq 9(ab^2 \plus{} bc^2 \plus{} ca^2)$ is homogeneous so we don't need $ a\plus{}b\plus{}c\equal{}3$ anymore. And it's also false.

arqady: $ c\rightarrow 0^ \plus{}$ means: $ \lim_{c\rightarrow 0^ \plus{}}$ right?

[quote="FOURRIER"]arqady: $ c\rightarrow 0^ \plus{}$ means: $ \lim_c\rightarrow 0^ \plus{}$ right?[/quote]
FOURRIER, $ c\rightarrow 0^ \plus{}$ this is $ c$ approaches to $ 0$ and $ c>0.$
 I mean that $ (a \plus{} b \plus{} c)^3 \geq 9(ab^2 \plus{} bc^2 \plus{} ca^2)$ is wrong
because if $ b \equal{} c \equal{} 1$ and $ c\rightarrow0^ \plus{}$ then $ (a \plus{} b \plus{} c)^3\rightarrow8$ and $ 9(ab^2 \plus{} bc^2 \plus{} ca^2)\rightarrow9.$ :wink:

A similar inequality is $ \frac{a^2}{b^2\plus{}1}\plus{}\frac{b^2}{c^2\plus{}1}\plus{}\frac{c^2}{a^2\plus{}1}\geq\frac{3}{2}$


algebra
function
domain
logarithms
complex analysis
complex analysis unsolved

Let $ D$ be the domain obtained by deleting the ray $ \{x: x\le 0\}$ from the plane, and let $ G(z)$ be a branch of $ \log{z}$ on $ D$.  Show that $ G$ maps $ D$ onto a horizantal strip of width $ 2\pi$:
\[ \{x\plus{}iy: \minus{}\infty<x<\infty, c_0<y<c_0\plus{}2\pi\}\]
and that the mapping is one to one on $ D$.

What do you need help with here? Do you understand why it's true for the principal branch?


calculus
integration
logarithms
trigonometry
calculus computations

Just a preliminary observation: $G(1) = \int_{0}^{\frac{\pi}{3}}\ln (1-\tan^{2}x)\, dx ,$ which is undefined on the interval $(\frac{\pi}{4}, \frac{\pi}{3}).$

Perhaps the summation should begin at $n=2.$

It is $2^{n+1}$ and not $2^{n}+1$

It is a problem from IIT-JEE (Indian Institute of Technology)

no,it isn't.

if it is ,then kindly let me know the year in which it was asked.

it is probably a question manufactured by one of the coaching institutes

This is certainly not an IIT-JEE question, as pointed out by pankajsinha. What's more, the sum doesn't converge! At least, that's what Mathematica "thinks."

It is a problem given to me by my friend.He claims it is from
IIT-JEE 1982. Can somebody verify this for me?
Is there any online source which gives IIT problems of the earlier years?

Near zero, $\ln(1-\tan^{2}x)=-x^{2}+o(x^{2})$

Thus $G(n)=-\frac{\pi^{3}}{3\cdot 2^{3n+3}}+o(2^{-3n})$

The sum $\sum_{n=1}^{\infty}2^{n}G(n)$ does converge by limit comparison to $\sum 2^{-2n}.$

you can have my word for it.

you may get the iitjee questions from the tata mc graw hill's WE series or you may try to get hold of FIITJEE's book which is known as ARCHIVES but i think they distribute this book only to their students.But FIITJEE's study material(photocopied) is available in Delhi at most of the bookshops


What are ratings?
I'm not talking about FTW but just ratings.
For instance under my name and avatar and my location and all that stuff, it says rating: not rated yet.
I've been looking at ppl's posts and I've seen that nobody else says anything like that.

Can somebody explain this to me?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=146659

If you had bothered to read the topic at the top of the forum you were posting in...


Let $a,b,c$ be three positive real numbers such that $a+b+c=3$. Prove that
\[\frac{1}{7-ab}+\frac{1}{7-bc}+\frac{1}{7-ca}+\frac{1}{1+abc}\ge 1.\]

[quote="GacKiem"]Let $ a,b,c$ be three positive real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
\[ \frac {1}{7 \minus{} ab} \plus{} \frac {1}{7 \minus{} bc} \plus{} \frac {1}{7 \minus{} ca} \plus{} \frac {1}{1 \plus{} abc}\ge 1.\]
[/quote]
By the Cauchy-Schwarz Inequality, we have
$ \frac {1}{7 \minus{} ab} \plus{} \frac {1}{7 \minus{} bc} \plus{} \frac {1}{7 \minus{} ca}  \ge \frac{9}{21\minus{}ab\minus{}bc\minus{}ca}\equal{}\frac{1}{\frac{21\minus{}ab\minus{}bc\minus{}ca}{9}},$
and
$ \frac{1}{\frac{21\minus{}ab\minus{}bc\minus{}ca}{9}} \plus{}\frac{1}{1\plus{}abc} \ge \frac{4}{\frac{21\minus{}ab\minus{}bc\minus{}ca}{9}\plus{}1\plus{}abc}.$
Therefore, it suffices to prove that
$ 1\plus{}abc \plus{}\frac{21\minus{}ab\minus{}bc\minus{}ca}{9} \le 4.$
Since
$ abc \le \frac{(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)}{9}\equal{}\frac{ab\plus{}bc\plus{}ca}{3},$
we see that it is enough to check the following inequality
$ \frac{ab\plus{}bc\plus{}ca}{3}\plus{}\frac{21\minus{}ab\minus{}bc\minus{}ca}{9} \le 3,$
which is true because $ ab\plus{}bc\plus{}ca \le 3.$

[quote="GacKiem"]Let $ a,b,c$ be three positive real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
\[ \frac {1}{7 \minus{} ab} \plus{} \frac {1}{7 \minus{} bc} \plus{} \frac {1}{7 \minus{} ca} \plus{} \frac {1}{1 \plus{} abc}\ge 1.\]
[/quote]

by AM-GM inequality :

$ \frac {1}{7 \minus{} ab} \plus{} \frac {1}{7 \minus{} bc} \plus{} \frac {1}{7 \minus{} ca} \plus{} \frac {1}{3(1 \plus{} abc)} \plus{} \frac {2}{3(1 \plus{} abc)}\ge \frac {4}{\sqrt [4]{(7 \minus{} ab)(7 \minus{} bc)(7 \minus{} ca)3(1 \plus{} abc)}} \plus{} \frac {2}{3(1 \plus{} abc)}$

and again by am-gm :

$ \frac {4}{\sqrt [4]{(7 \minus{} ab)(7 \minus{} bc)(7 \minus{} ca)3(1 \plus{} abc)}} \plus{} \frac {2}{3(1 \plus{} abc)}\ge \frac {16}{24 \plus{} (3abc \minus{} ab \minus{} bc \minus{} ca)} \plus{} \frac {2}{3(1 \plus{} abc)}\ge \frac {16}{24} \plus{} \frac {2}{6} \equal{} 1$

since $ ab \plus{} bc \plus{} ca \minus{} 3abc\ge 0$ and $ abc\leq 1$

equality for $ a \equal{} b \equal{} c \equal{} 1$

:)

we have :

$ \text{LHS} \equal{} \sum_{cyc}\left(\frac {1}{7 \minus{} ab} \plus{} \frac {1}{3(1 \plus{} abc)}\right) \ge \sum_{cyc}\left(\frac {4}{10 \plus{} bc \plus{} ca}\right) \ge \left(\frac {36}{30 \plus{} 2(ab \plus{} bc \plus{} ca)}\right)\ge \frac {36}{30 \plus{} 2.3} \equal{} 1$


[quote="pvthuan"][quote="Chocolate M&M"]please help me this exercise, I can't solve it : :(
Prove that : $\frac{1}{1+x^2}+\frac{1}{1+y^2} \geq \frac{1}{1+xy}$[/quote]
That ineq is equivalent to
\[ \frac{1+y^2+1+x^2}{(1+x^2)(1+y^2)}-\frac1{1+xy}\geq0,  \]
whic simplifies to
\[ \frac{(xy-1)^2+xy(x^2+y^2)}{(1+x^2)(1+y^2)(1+xy)}\geq0.  \][/quote]

R $x$ and $y$ all positive?  :?:

[quote="bookworm_vn"]
R $x$ and $y$ all positive?  :?:[/quote]
From the proof, we see that $x,y$ are of the same sign.

[quote="pvthuan"]That ineq is equivalent to
\[ \frac {1 \plus{} y^2 \plus{} 1 \plus{} x^2}{(1 \plus{} x^2)(1 \plus{} y^2)} \minus{} \frac1{1 \plus{} xy}\geq0,
\]
whic simplifies to
\[ \frac {\color{red}(xy \minus{} 1)^2\color{black} \plus{} xy(x^2 \plus{} y^2)}{(1 \plus{} x^2)(1 \plus{} y^2)(1 \plus{} xy)}\geq0.
\]
[/quote]
That’s not quite right. It only simplifies to
\[ \frac {\color{red}1\plus{}2xy\minus{}x^2y^2\color{black} \plus{} xy(x^2 \plus{} y^2)}{(1 \plus{} x^2)(1 \plus{} y^2)(1 \plus{} xy)}\geq0.
\]:maybe:

Expanding is best solution for this problems.

[quote="Jane_Bennet"]
That’s not quite right. It only simplifies to
\[ \frac {\color{red}1 \plus{} 2xy \minus{} x^2y^2\color{black} \plus{} xy(x^2 \plus{} y^2)}{(1 \plus{} x^2)(1 \plus{} y^2)(1 \plus{} xy)}\geq0.
\]
:maybe:[/quote]
$ 1\plus{}2xy\minus{}x^2y^2\plus{}xy(x^2\plus{}y^2)\geq1\plus{}2xy\minus{}x^2y^2\plus{}xy(2xy)\equal{}1\plus{}2xy\plus{}x^2y^2\geq0$
and we just need condition $ xy\geq0$

[quote="Chocolate M&M"]please help me this exercise, I can't solve it : :(
Prove that : $ \frac {1}{1 \plus{} x^2} \plus{} \frac {1}{1 \plus{} y^2} \geq \frac {1}{1 \plus{} xy}$[/quote]
Yes.$ x\equal{}tgt.$OK!!!

The given is equivalent to
$ x^3y\plus{}y^3x\plus{}2xy\plus{}1 \ge x^2y^2$
$ xy(x^2\plus{}y^2\plus{}2)\plus{}1 \ge x^2y^2$
But $ x^2\plus{}y^2\plus{}2 \ge x^2\plus{}y^2 \ge 2xy \ge xy$.

The inequality seems to be strict. The following may be proper,
$ \frac{1}{1\plus{}x^{2}}\plus{}\frac{1}{1\plus{}y^{2}}\geq\frac{2}{1\plus{}xy}$ with equality at $ x\equal{}y$.

[quote="Shanku"]The inequality seems to be strict. The following may be proper,
$ \frac {1}{1 \plus{} x^{2}} \plus{} \frac {1}{1 \plus{} y^{2}}\geq\frac {2}{1 \plus{} xy}$ with equality at $ x \equal{} y$.[/quote]Try $ x\to 0$. This one only holds only iff $ xy\ge 1$.

[quote="hulai"]you must kown this truth:
a/b<=a-m/b-m,a,b,m>0[/quote]
1. Please use $ \text{\LaTeX}$!
2. Why do you post this here?!?
3. It's wrong and only holds if $ a\geqslant b$ and $ m \not\in [b;a]$...^^

[quote="ali666"][quote="Jane_Bennet"]
That’s not quite right. It only simplifies to
\[ \frac {\color{red}1 \plus{} 2xy \minus{} x^2y^2\color{black} \plus{} xy(x^2 \plus{} y^2)}{(1 \plus{} x^2)(1 \plus{} y^2)(1 \plus{} xy)}\geq0.\]
:maybe:[/quote]
$ 1 \plus{} 2xy \minus{} x^2y^2 \plus{} xy(x^2 \plus{} y^2)\geq1 \plus{} 2xy \minus{} x^2y^2 \plus{} xy(2xy) \equal{} 1 \plus{} 2xy \plus{} x^2y^2\geq0$
and we [color=darkred]just need condition [/color]$ xy\geq0$[/quote]
We don't need the condition ali666, as $ \ \ 1\plus{}2xy\plus{}x^{2}y^{2}\equal{}(1\plus{}xy)^{2}\geq 0$

Cheers. :lol:

But if $ xy < 0$ you can't conclude $ xy(x^2 \plus{} y^2)\ge xy(2xy)$  :!:

And for the original $ \frac{1}{1\plus{}x^{2}}\plus{}\frac{1}{1\plus{}y^{2}}\geq\frac{1}{1\plus{}xy}$, you can see that at least for $ xy\to \minus{}1^\plus{}$ it wouldn't hold.  :)

[quote="spanferkel"]But if $ xy < 0$ you can't conclude $ xy(x^2 \plus{} y^2)\ge xy(2xy)$  :!:

And for the original $ \frac {1}{1 \plus{} x^{2}} \plus{} \frac {1}{1 \plus{} y^{2}}\geq\frac {1}{1 \plus{} xy}$, you can see that at least for $ xy\to \minus{} 1^ \plus{}$ it wouldn't hold.  :)[/quote]

Why not ?
$ x^{2}\plus{}y^{2}\geq 2xy$irrespective of the sign of $ xy,$isn't it ?
Any comment as to when the equality will occur ?

[quote="Shanku"]

Why not ?
$ x^{2} \plus{} y^{2}\geq 2xy$irrespective of the sign of $ xy,$isn't it ?[/quote]
but you cannot multiply this inequality with a negative number :rotfl:
[quote="Shanku"]
Any comment as to when the equality will occur ?[/quote]OK, the inequality is strict. But try $ y\equal{}\frac 1 {\sqrt{x}}$ along with $ x\to 0.$

You mean to say $ xy$ need to be positive for this inequality to hold ?
And yes, I also see that the inequality is strict.
Thanks.

Ok, it's settled now. See here : [url]http://www.mathlinks.ro/viewtopic.php?p=1793163#1793163[/url]

Cheers.

[quote="Chocolate M&M"]please help me this exercise, I can't solve it : :(
Prove that : $\frac{1}{1+x^2}+\frac{1}{1+y^2} \geq \frac{1}{1+xy}$[/quote]
i think it is
$\frac{1}{x^2}+\frac{1}{y^2}\ge \frac{2}{1+xy}$
with $x,y\ge 1$
may be!

It's very easy. please expand and use $LaTeX$!

This inequality doesn't hold for all real number x, y.

[quote="Chocolate M&M"]please help me this exercise, I can't solve it : :(
Prove that : $\frac{1}{1+x^2}+\frac{1}{1+y^2} \geq \frac{1}{1+xy}$[/quote]
It is Jensen's inequality! You can apply Jensen!

[quote="hoainamss"][quote="Chocolate M&M"]please help me this exercise, I can't solve it : :(
Prove that : $\frac{1}{1+x^2}+\frac{1}{1+y^2} \geq \frac{1}{1+xy}$[/quote]
Can check $x=-2, y=\frac{2}{5}$. So i think that we need to add some condition.[/quote]

Pl see here : [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=335069[/url]


geometry
3D geometry
sphere
frustum
trapezoid
Pythagorean Theorem
similar triangles

A sphere with radius $ 3$ is inscribed in a frustrum of slant height $ 10$.  What is the volume of the frustrum? (A [i]frustrum[/i] is the non-conical region formed by cutting a cone with a plane parallel to the base).

$ 1990$ MANDELBROT

Label the shorter radius $ a$ and the longer radius $ b$. Since two tangents to a sphere are equal in length, we have $ a \plus{} b \equal{} 10$. Then, using the Pythagorean Theorem, we have
\[ \left(b \minus{} a\right)^2 \plus{} 6^2 \equal{} 10^2 \implies b \minus{} a \equal{} 8.
\]
Combining the two equations, we find that $ a \equal{} 1$ and $ b \equal{} 9$.
Then if we extend the slant heights to meet at a point $ T$, we can use similar triangles to find the height of the entire cone:
\[ \frac {1}{9} \equal{} \frac {h \minus{} 6}{h} \implies h \equal{} \frac {27}{4}.
\]
Then we can find the volume of the frustum by subtracting the areas of the two cones:
\[ V \equal{} \frac {1}{3}\pi b^2h \minus{} \frac {1}{3}\pi a^2 \left(h \minus{} 6\right) \equal{} \boxed{182 \pi}.
\]

[hide]
Due to the fact that the fustrum and the sphere are perfectly symetrical around an axis we can break it down into a circle and a trapezoid, call it $ ABCD$.

The height of the trapazoid must be $ 2r \equal{} 6$
There is also a $ 6\minus{}8\minus{}10$ triangle on each side of the trapazoid. Therefore, $ \overline{CD}\equal{} 2\cdot 8\plus{}\overline{AB}$.
 [geogebra]061a7f96a26b52d597ee71e9ab254f52659832ed[/geogebra]

Now notice the two similar triangles (with the dotted lines)
$ \Rightarrow \frac{3}{6} \equal{} \frac{x}{10} \Rightarrow x\equal{}5$
Now x is half the medain line $ \Rightarrow 2x \equal{}10\equal{} \frac{a\plus{}16\plus{}a}{2} \Rightarrow a \equal{} 2$

Now we need the hight, $ h$ of the cone that makes the fustrum, finding more similar triangles
$ \frac{2}{h\minus{}6}\equal{}\frac{18}{h} \Rightarrow h \equal{} \frac{27}{4}$

So the volume of the fustrum is
$ V\equal{}\frac{1}{3}\pi 9^2 \cdot \frac{27}{4} \minus{} \frac{1}{3}\pi \left(\frac{27}{4} \minus{}6\right)\equal{}\frac{1}{3}\pi 546\equal{}182\pi$ :)
[/hide]


Let $ABC$ be a triangle with a point $P$ inside it. Let $A', B', C'$ be the intersections of $AP$ and $BC$, $BP$ and $AC$, $CP$ and $AB$. Suppose $\frac{A'B}{A'C}=x$, $\frac{B'C}{B'A}=y$, $\frac{C'A}{C'B}=z$ (the ratios are not algebraic). Denote by $S'$ be the area of $A'B'C'$. Find $\frac{S'}{S}$.

Hint:
Use this theorem: if $a$ and $b$ are sides of a triangle and $\gamma$ is an angle between them, then
an area of this triangle is equal to $\frac{1}{2}ab\sin{\gamma}$

Let
$|A'B|=kx$
$|A'C|=k$

$|B'C|=ly$
$|B'A|=l$

$|C'A|=nz$
$|C'B|=n$

If you still don't know how to do it, then give a sign, and I would try to post a full solution


Fie $ABC$ un triunghi oarecare si $I_{a}$, $I_{b}$, $I_{c}$ centrele cercurilor exinscrise triunghiului. Cu notatiile cunoscute, sa se arate ca:
\[S_{I_{a}BC}+S_{I_{b}CA}+S_{I_{c}AB}+S_{ABC}=R\cdot (a+b+c).\]

Notam cu $R_{a},R_{b},R_{c}$ razele cercurilor exinscrise.Dar $R_{a}$ este inaltimea din $I_{a}$ in triungiul $I_{a}BC$.Dar $R_{a}=\frac{S}{-a+b+c}$ de unde $S{}_{I}{}_{a}BC=\frac{Sa}{2(p-a)}$ si $R=\frac{4abc}{S}$.Adunand relatiile analoage o sa obtinem exact Heron=>Q.E.D

$S=(p-a)r_{a}=pr\Longrightarrow \boxed{\ ar_{a}=p(r_{a}-r)\ }$ $\Longrightarrow$

$\sum ar_{a}=p(4R+r-3r)=2p(2R-r)$ $\Longrightarrow$

$pr+\sum\frac{ar_{a}}{2}=2Rp$ $\Longrightarrow$ $\boxed{[ABC]+\sum [BI_{a}C]=R(a+b+c)}\ .$

[b]Observatie.[/b] $\sum\frac{a}{r_{a}}=$ $\frac{1}{r_{a}r_{b}r_{c}}\cdot\sum ar_{b}r_{c}=$ $\frac{1}{p^{2}r}\cdot\sum ap(p-a)=$ $\frac{1}{pr}\cdot\sum a(p-a)=$ $\frac{1}{pr}\cdot\left(2p^{2}-\sum a^{2}\right)=$ $\frac{1}{pr}(2r+8Rr)=$ $\frac{2(4R+r)}{p}\ge 2\sqrt 3\Longrightarrow$ $\boxed{\ \sum\frac{a}{r_{a}}\ge 2\sqrt 3\ }\ .$


algebra
polynomial
linear algebra
matrix
complex numbers
superior algebra
superior algebra solved

Let A,B be matrices from Mn(R).
Prove that if there exists U invertible in Mn(C) such that AU=UB, then there exists V inveritble in Mn(R) such that AV=VB.

Try to solve this without similarity and Jordan canonic form.

If you have a different proof i 'll take it

Let x a real number such that e^(2ix) is not a root of polynomial characteristic of -U^(-1).U*
where U* is the complex conjugate of U,  Z(x)=e^(ix)U+e^(-ix)U*
det(Z(x))=e^(-nix).detU.det(U^(-1)U*+e^(2ix)I) =/= 0
Z(x)=Z*(x),  Z(x) is reals coefficients invertible
UB=AU=> U*.B=A.U* and Z(x)B=AZ(x)

More generally:
K is infinte commutative field, L is an extension commutative field
A,B \in Mn(K) similar in Mn(L) wich means there exist P \in Mn(L)
invertible such that A=P^(-1)BP, then the matrix A,B are similar
in Mn(K) there exist Q in Mn(K) invertible, A=Q^(-1)BQ

M.Omarjee
STT


linear algebra
matrix
superior algebra
superior algebra solved

Let A \in M_3(R), symetric tA=A (where tA is the transpose of A)
Prove that tr(A^4) \geq det(A).tr(A)

I'm not sure, but I think it is enough for A to be symetric. Let a,b,c the eingenvalues of A. It is well known that they are reals. Also, Tr(A^4)=a^4+b^4+c^4>=abc(a+b+c)=det(A)*TR(A).


trigonometry
function
algebra proposed
algebra

Solve this sytem if $ x,y\in \mathbb{R}$

$ sin^2(x) \plus{} cos^2(y) \equal{} y^2$
$ cos^2(x) \plus{} sin^2(y) \equal{} x^2$

Solutions $ (\pm 1,\pm 1)$ followed from $ x^2\plus{}y^2\equal{}2$.

Adding the equations we get $ x^2\plus{}y^2\equal{}2\quad(*)$

Subtracting the first equation from the second we get

$ \cos 2x\minus{}\cos 2y\equal{}x^2\minus{}y^2\iff \sin(x\minus{}y)\sin(x\plus{}y)\equal{}\minus{}{(x\minus{}y)(x\plus{}y)\over 2}\quad(**)$

From $ (*)$ it follows $ |x|\leqslant\sqrt{2},|y|\leqslant\sqrt{2}$, hence both $ x\minus{}y$ and $ x\plus{}y$ lie within $ [\minus{}2\sqrt{2}, 2\sqrt{2}]$, which in turn lies within $ (\minus{}\pi,\pi)$. In that interval we have $ \mathrm{sgn}\,t\equal{}\mathrm{sgn}\sin t$, hence for nonzero values of $ x\plus{}y, x\minus{}y$ we can't satisfy $ (**)$, because the sides will have opposite signs. It follows that either $ x\equal{}y$ or $ x\equal{}\minus{}y$, and now it's easy to find the four solutions: $ (1,1),(1,\minus{}1),(\minus{}1,1),(\minus{}1,\minus{}1)$

[quote="Farenhajt"]In that interval we have $ \mathrm{sgn}\,t \equal{} \mathrm{sgn}\sin t$,[/quote]

[size=150]sgn=sign of [/size], right?

Signum function of a real argument is defined thus: $ \mathrm{sgn}\,t : \equal{}\begin{cases}\minus{}1&t<0\\ 0&t\equal{}0\\ \plus{}1&t>0\end{cases}$.

(For $ t\neq 0$ it can be simplified into $ \mathrm{sgn}\,t\equal{}{t\over|t|}$)

[quote="Farenhajt"]Signum function of a real argument is defined thus: $ \mathrm{sgn}\,t : \equal{} \begin{cases} \minus{} 1 & t < 0 \\
0 & t \equal{} 0 \\
\plus{} 1 & t > 0\end{cases}$.

(For $ t\neq 0$ it can be simplified into $ \mathrm{sgn}\,t \equal{} {t\over|t|}$)[/quote]

What is a real argument ? Is it real number?

sorry for asking lot of questions :blush:

"Argument" (of a function) means "any member of the set on which particular function is defined". In this case, it's the set of real numbers.


If p is n-digit prime number that has (n-1) digits '1' and one digit '7', find all p. Eg. p = 17

where did you find this question? I can't find anything, apart from the obvious fact that the numbers are divisible by $ 3$ when $ n\equiv 0 \mod 3$

[hide]
denote a number $ P(n)_i$ where $ n$ is the number of digits and $ i$ is where you place the $ 7$. e.g. $ P(4)_4 \equal{} 7111, P(5)_3 \equal{} 11711$...

The following are prime:
$ P(2)_{1,2}$
$ P(4)_{1,2}$
$ P(5)_{1,2}$
$ P(7)_{3,4,5}$
$ P(8)_{1,2,8}$
$ P(10)_{7,9}$
$ P(11)_{10}$
$ P(13)_{8}$
$ \vdots$

[/hide]

Hi;
Just from examining the problem as a repunit(n) + 6 where repunit(5) = 11111 and repunit(8) = 11111111

here is another prime repunit(23) + 6 = 11 111 111 111 111 111 111 117

and there are more;

repunit(29) + 6 is prime.
repunit(40) + 6 is prime.
repunit(131) + 6 is prime.

Can you prove that there exists infinitely many?

Hi;

Whether or not the repunits produce an infinite number of primes is I think an open problem. Your problem is precariously close to that one. I would conjecture yes there are an infinite number of primes of your form.


[quote="tranthanhnam"]let $x,y,z >0$  . find minimal value of expression :
$\frac{1}{x^2}+ \frac{1}{y^2}+ \frac{1}{z^2} + 8x^2 +108y^2 +16z^2$
          where $xy+yz+zx=1$[/quote]
If you can solve this
$xy^2+yx^2=7z^2$
$x+\frac{x^2}{z}=108-\frac{z^4}{y^4}$
$y+\frac{y^2}{z}=16-\frac{z^4}{x^4}$
your ineq will be solved.
It's very hard :D


topology
linear algebra
matrix
superior algebra
superior algebra solved

Let $k$ be an algebraically closed field. Show that the closure of the $n\times n$ matrices with entries in $k$ and rank equal to $i$ in the Zariski topology is the set of matrices with rank $\leq i$.

[quote="berti"]Is it not necessary to prove that the closed set is irreducible?[/quote]

Ooops! My bad.
I think proving that the closed set has no irreducible components outside the open set described above also works. But proving that it is itself irreducible may be easier.

[quote="amfulger"]The rank $i$ matrices are a locally closed subset of $M_{n}(k)$ given by the intersection of the open set which is the union of the fundamental open sets given by the non-vanishing of the $i$-minors and the closed set given by the zero locus of all the $i+1$-minors.
The conclusion follows.

[edit:] Because the closed set described above is the set of $\leq i$ rank matrices.[/quote]

Sorry, for misunderstanding. Is it the full proof?

If $A$ - "union of the fundamental open sets given by the non-vanishing of the $i$-minors" - our open set and
$B$ - "closed set given by the zero locus of all the $i+1$-minors" - our closed set.

Why then $cl( A \bigcap B) = B$, where $cl()$ is the closure  the Zarisky topology ?

[quote="eugene"][quote="berti"]This follows if $B$ is irreducible and $A \cap B \neq \emptyset$[/quote]

yes, and how do you prove that $B$ is irreducible?[/quote]

There is a surjective morphism $f: M_{n,i+1}(k)\times M_{i+1,n}(k)\to B$ given by $f(A,B)=AB$ for any two matrices $A\in M_{n,i+1}(k)$ and $B\in M_{i+1,n}(k)$. This is obviously a morphism of algebraic varieties. What is less obvious is that this is an onto map, but it is a largely known problem that every $n\times n$ matrix of rank (at most) $r$ is a product of a $n\times r$ and a $r\times n$ matrix.
$M_{n,i+1}\times M_{i+1,n}$ is irreducible because it is an affine space.
Let $B=\bigcup_{i=1}^{m}B_{i}$ for some irreducible closed sets $B_{i}$ and some $m\in\mathbb N$. Then $M_{n,i+1}\times M_{i+1,n}=\bigcup_{i=1}^{m}f^{-1}B_{i}$, because $f$ is surjective. Since the lefthandside is irreducible and all the sets in the righthandside are closed, $M_{n,i+1}\times M_{i+1,n}=f^{-1}B_{e}$ for some $e\in\overline{1,m}$. It follows easily that $B_{e}=B$, so $B$ is irreducible.


AMC
AIME
USA(J)MO
USAMO
MATHCOUNTS
percent
AMC 10

How do you feel about the label "nerd" or "geek" being used on yourself or others? I don't feel offended by this term and often jokingly describe myself as one or the other, but I don't know how other people feel about it. I would like to know so as not to offend others in the future.

I like the term.  I consider myself a "math geek" and I don't mind when people call me one.  It's nice to be able to make fun of yourself.

Frankly, it annoys me.

I've been stereotyped like that since last year, but this year, it's worse.
Thank god bullying isn't going on that much, but still, people constantly annoy me.

Also this guy who is considered "gifted" calls me a nerd, and i can't say the same with him. he's stupid. (all teachers who had him still calls him stupid).
and it annoys me that the world has resorted to labeling.
All because a guy wears black every day doesn't mean that you stick a store label that says, "goth"
All because someone got a perfect on AMC 10 and got a perfect on AIME doesn't mean that you can stick on a label that says, "no lifer"

my point is that people shouldn't label others.

i sometimes use the term nerd/geek jokingly, but that's about it.

I am never serious when i say, "that guy's such a geek".

I don't mind being called a geek, but nerd really annoys me.

I guess its cause I see them as different things.

[quote="davidyko"]I used to be offended by it, back in 6th grade when I was trying to be cool, but by now, in 10th, basically I've embraced being a nerd and a math and piano geek because seriously, that's what I am.[/quote]
Exact same story...except switch out the 6th with 6-7th and switch out the 10th with 9th.
Little brother attempts to call me a geek, but I pick at his B- in math  :o  :|  :huh:

One really interesting thing is, at our school if you are smart enough, and have a really quiet personality, you get labeled as genius which is pretty damn awesome :D. My friend made USAMO last year and got a 10 (8th grade), was top 8 mathcounts state (yuck but who cares), and he's really really quiet, and pretty nice, so everyone says he's a genius and extremely talented, not nerdy.

[quote="serialk11r"]One really interesting thing is, at our school if you are smart enough, and have a really quiet personality, you get labeled as genius which is pretty - awesome :D. [/quote]

Wow.  :o

I'm also not that sure about the definitions of nerd and geek among people. To me, a nerd is just a smart person while a geek is usually a person who is good at a specific skill. (Like math geek). Is this the accepted understanding?

I see a nerd a thick rimmed glasses almost cartoon-y RPG gamer who lives in a darkened room who isn't actually bright and the definition of geek follows from this by being more normal, brighter etc.

Nerd/Geek:
Used by my worst enemies or myself (as jokingly)

Genius:
Used when picking partners for projects and everyone wants to be my partner.

that's about it.. :D

Now that I think about it, I don't like my friends to use the term "genius" to describe me. I don't think I'm better than them for just studying harder for the test. I guess that's part of the reason why I don't tell others my test grades anymore.

OH. and to add to the genius thing, my friends call me that when i prove my teacher wrong. (look in the G+FF topic)

I find it offensive (only when it's meant purposely and not jokingly) because of the stereotypes attached to those names.

Sure you can like math and other "geeky" things, but people automatically assumes you are the [url=http://www.youtube.com/watch?v=-xEzGIuY7kw]uber nerd[/url], in that you like all things geeky.

Personally, I do like math, but I also like being social and hanging out with my friends. Some people never see this side of you when they call you a "nerd" and refuse to hang out with you because they think your idea of fun is doing some studying.

Definitions-

Dork- tries to act smart but fails

Nerd- stereotypical white person who wears glasses and braces, and does not interact with other people (ever seen white and nerdy?)
Nerd- person who does intellectual stuff all day and does not talk with the "popular group"

Geek- specializes in a certain field and could be considered nerd.
Geek- talks to smarter people and smart

No-lifer- devotes most of his/her time to a single passion, whether it is sports, computer games, etc.
No-lifer- devotes most of his/her time to non-social activities
No-lifer- devotes all of his/her time to activities not considered cool/popular
No-lifer- does not have a friend of the opposite gender

Plain smart- gets good grades, has high intellectual power, yet still has a social life
Genius- Ridiculously smart, yet spends little to no time on studying or learning (impossible!)

I'm in the plain smart category.

I don't mind either, although I think "nerd" to be more affectionate and "geek" to be more institutional and specific.

Nerd conjures up adorable colorful sugar clusters first, then segues into happy people who intensly enjoy what they learn. Geek has deeper strains of gamers with bad posture.

[quote="SimonM"]Don't you think that straight D students get annoyed about being stereotyped in that way?[/quote]

Very true. I have lots of friends who don't lack drive or anything--they just don't apply it to select subjects, especially mathematics. Part of the reason is the whole school system's way of categorizing and learning. So, yeah....some people aren't like that.

[quote="SimonM"]Don't you think that straight D students get annoyed about being stereotyped in that way?[/quote]
oh. oops. you get my point though. There are certain people who don't have anything to aim for and they just don't give a **** about school.

And another friend of mine gave me a different perspective. She said that I needed to learn to deal with those types of people. I understand, but I would rather work with people who actually want to succeed in life.

no offense to those undecided majors out there, but it's not like "you're undecided, you have no idea what you want to do", you just want to try a variety before actually deciding.

[quote="kimmystar94"]And another friend of mine gave me a different perspective. She said that I needed to learn to deal with those types of people. I understand, but I would rather work with people who actually want to succeed in life.
[/quote]

You have no idea how much sense that friend is talking (and I'm not sure how much of it she realised she was saying).

ok so nerd/ geek i really don't care if it's my friends that r calling me that (my REAL friends not ppls that pretend they r so they can copy my hw). i mean i'm in the "gifted" class and i have been since 3rd grade and in jr. high ppls like call me nerd and it drives me CRAZY!!  :( i  just wanna scream at them  :cursing: U WANNA GET A GOOD GRADE THEN WORK HARDER!!! I WORKS!!! NOW STOP CALLING PPLS THAT GET GOOD GRADES NERD AND STOP COPYING OFF THEM!!! GO ACTUALLY WORK HARDER!! THEN UR GRADES MIGHT GO UP!!!!!  :yup: yeah this one girl in my science class is like always trying to cheat off of me and it's annoying cuz when we do partner work she just doesn't try and then copies my work cuz i don't want a bad grade!! ok i'm ranting too much i could go on for hours about this...

i mean i really don't care about nerd if they're not just trying to get answers. so anyone that wants to call me nerd/geek on this website go ahead

[quote="SimonM"][quote="kimmystar94"]i hate the "stereotypical asian" thing.
....
The others, (straight D-students), just do things spur-of-the-moment. They make wrong decisions, etc.
[/quote]

Don't you think that straight D students get annoyed about being stereotyped in that way?[/quote]

Each party has stereotypes against the other.  Straight-A students see straight-D students as stupid and lower on the social chain, while straight-D students see straight-A students as nerds who have no lives because they take 5 or 6 AP classes.

I remember doing this sort of activity on the first day of my US History class.  Its purpose was to show that history can be analyzed from two different points of view.  In fact, we use Howard Zinn's [url=http://en.wikipedia.org/wiki/A_People%27s_History_of_the_United_States][i]A People's History of the United States[/i][/url] as a second source (the first is our textbook).  We analyzed our stereotypes on blacks vs. whites, Indians vs. cowboys, and low-level students vs. AP students.  It was amazing how biased we all were toward these different groups.

Meh, I dont really think people that study too much because they want a good grade or are advanced at something because they really like it and are really good at it are considered nerds/geeks...

i consider people that do something they dont like but still do it becuase "its good for college" nerds...like these kids that join Science Olympiad not because they like it, but because they think it will help them in college applications...or volunteer on their weekeds because it looks good to colleges, not because they actually want to give back to the community...

also like people that pay for classes or take classes that are unnecssary.  If you are really good at math and want to take advaced classes normally not offered to your peers then go ahead, if you truly or good at it or like it...

but like theres a couple of people in my grade that are taking pre-calc and learning pre-calc outside of school with a tudor.  why would you be wasting your time doing that (everything is going to be covered in school)  and its not like they are stuggling in pre-calc or something.  They probalby can get like a 95 with out a tudor, but they always strive for a 100 so they pay and waste a whole bunch of money and time just trying to impress everyone and get 100s...actually those people are more idiots than nerds...and its also not like they use that math knowledge for good use.  They say they dont compete in any math competitions because it sounds nerdy to them...hmm yea...

no one really calls me a nerd but thats mostly because i dont get the top (95-100 percent) grades for my classes, and also if i do get a really good grade, I dont go around bragging to other people...

Nerd is a good title/mostly used in a good sense in my school.
I've never heard geek used.

I am a jock, play most sports, but still get awesome grades in math, im doing 4 grades ahead of my class in math. competed in AMC 10, Mathcounts, but people call me a jock

What is your definition of jock then? I consider a jock someone who is just totally sports-obsessed, and your case doesn't sound like a jock.

Maybe people in your school only see you playing sports and assume you're a jock.

It's kinda stupid. I am not that nerdy, but I am labled as one.

I know people who are really nerdy that no one ever labels.

I really don't care about being called a nerd because Im used to it.

I've been called many things — “nerd” is among the nicest ones. :) No, I don't have pimples or glasses, though I [i]am[/i] considerably awkward and socially inept.

I am neither a nerd nor a geek

it is pretty awesome

if someone referred to me as such I would correct them

I think this falls under the rule againist reviving a topic no one has posted in over a month


calculus
integration
calculus computations

Integrate $ \int\frac {dx}{ x^5 \plus{} 1 }$ .

Actually i find that its really difficult for me to calculate  the integrals of the above form when the [b]power of x is odd and greater than 5 .[/b]

Is there a short way ??

One method is to factor $ x^5 \plus{} 1$ into a partial fraction decomposition integrate terms. This gives
[code](sqrt(5) - 3)*log(2*x^2 + (sqrt(5) - 1)*x + 2)/(10 - 10*sqrt(5)) - (sqrt(5) + 3)*log(2*x^2 + (-sqrt(5) - 1)*x + 2)/(10*sqrt(5) + 10) + (sqrt(5) + 1)*arctan((4*x + sqrt(5) - 1)/sqrt(2*sqrt(5) + 10))/(sqrt(5)*sqrt(2*sqrt(5) + 10)) + (sqrt(5) - 1)*arctan((4*x - sqrt(5) - 1)/sqrt(10 - 2*sqrt(5)))/(sqrt(5)*sqrt(10 - 2*sqrt(5))) + log(x + 1)/5+C
[/code]

Thats what i said factoring is very big problem , because as the power of x increases say to 9 or 11 its really difficult . that why i asked if there is a quicker method.

It is a standard approach for this:
Polynom $ X^{2n+1} + 1$ has roots $ x_k = cos (\alpha_k) + i sin(\alpha_k), \alpha_k = \frac{2k\pi + \pi}{2n+1}, k = 0,1,...,2n$
These roots can be re-written: $ -1, x_k, \bar{x_k}, k = 0,1,...,n-1$ [exercise]
So: $ X^{2n+1} + 1 = (X + 1)\prod_{k=0}^{n-1}{(X-x_k)(X-\bar{x_k})}$
and from here we should find real constants $ c, a_k, b_k$ such that:
$ \frac{1}{X^{2n+1} + 1} = \frac{c}{X+1} + \sum_{k=0}^{n-1}\frac{a_kX + b_k}{(X-x_k)(X-\bar{x_k})}$ [1]

The trick here is to multiply by $ X+1$, for example, then compute the left side for $ X = -1$ after simplifying the $ X+1$ term. In the right side we get our $ c$ constant. You may also apply l'Hospital rule for left side and take the limit when $ x -> -1$. We get $ c = \frac{1}{2n+1}$

The same goes for the other constants $ a_k, b_k$: multiply by $ X - x_k$, then compute the both sides for $ X = x_k$: the right side will equal $ \frac{a_kx_k + b_k}{x_k - \bar{x_k}}$. For the left side we use the following more general result:

If polynom P has m roots [real or complex], all distinct, $ a_1,...a_m$, and if $ Q(X) = \frac{X - a_k}{P(X)}$, then $ Q(a_k) = \frac{1}{P'(a_k)}$
[just write $ P(X) = \prod_{i=0}^{m-1}(X - a_i)$, $ \frac{P'(X)}{P(X)} = \sum_{i=0}^{m-1}\frac{1}{X - a_i}$, then multiply by $ (X - a_k)$]

After these we get: $ \frac{1}{(2n+1){x_k}^{2n}} = \frac{a_kx_k + b_k}{x_k - \bar{x_k}}$

But $ x_k^{2n + 1} = -1$, $ x_k - \bar{x_k} = 2isin(\alpha_k)$, equal real and imaginary parts and find: $ a_k = \frac{-2cos(\alpha_k)}{2n+1}, b_k = \frac{2}{2n+1}$

[1] will be then:

${ \frac{1}{X^{2n+1} + 1} = \frac{1}{2n+1}\frac{1}{X+1} + \sum_{k=0}^{n-1}\frac{2}{2n+1}\frac{-Xcos(\alpha_k) + 1}{X^2 - 2cos(\alpha_k)X + 1}}$

You may split then the main sum termen in 2 fractions that will integrate using $ ln$ and $ arctan$

Please double check again all these and try to directly compute it for $ n = 1$, just to be sure


This is kind of a pointless question, but what do people mean when they say "Edit:_____________________"

They didn't really edit the post, so what does it mean?? :huh:

Usually it means that they, indeed, edited the original post some time after they posted it. Sometimes it is just an abbreviation for "afterthought".


How many five-digit numbers are divisible by 3 and also contain 6 as one of their digits?

i) 6 is the first digit:

(1+x+..+x^9)^4

ii) 6 is not the first digit
4*(x+x^2+..+x^9)(1+x+...+x^9)^3

do a 3rd roots of unity filter to find the coefficients of the x^{3k} terms [k an integer]

Is your method called generating functions? or what is it called?

In the context of the book, I don't think a generating functions bash is necessary for this problem. At first glance I thought it would be $5*3000$, for the $3000$ 4-digit numbers divisible by $3$ and $5$ places to put the $6$, but there's some overcounting that's not too easy to resolve.

its not like the "generating function bash" takes that long
1/3( 10^4+1+1)+1/3(4(9*10^3)+0+0)=15334

At first I thought that balls and urns would work:

You have some permutation of the digits 6, a, b, c, d.

$0\leq a+b+c+d\leq 36$, and $a+b+c+d$ must be a multiple of 3.

So $a+b+c+d=0$, $a+b+c+d=3$, etc.

But I think that still overcounts in the end...


complex analysis
complex analysis unsolved

Write f(x) = 2 sin(2x) - 3i cos(2x) as a linear combination of exponentials.

The answer is given

-2.5 i e^(2ix) - 0.5 i e^(-2ix)

I don't understand this. Help?


conics
parabola
Putnam
ellipse
hyperbola
induction
advanced fields

Here's a cute one (although not difficult):

The parabole of equation $y=x^2$ rolls without any sliding on the parabole $y=-x^2$. What's the locus of the focus of the first parabola?

I think it's from an old Putnam contest, but I can't remember which one.

Hmm... can this be generalized to other conics? just wondering. It seems that a conic rolling on a congruent conic always makes its focus(es) describe a circle or a straight line.

circle on circle: focus = center describes a circle.
ellipse on ellipse: focuses describe circles around the fixed ellipse's focuses.
hyperbola on hyperbola: focuses describe circular arcs???, i'm not sure, but obviously follows by incomplete induction and complete intuition :lol:

There's a nice lemma stating that the foot of the perpendicular to a tangent to a parabola from the focus always lies on the tangent at the vertex (see [url=http://mathworld.wolfram.com/Parabola.html]MathWorld[/url]).

Call $F$ the focus of the parabola $y=-x^2$. Then, the focus $G$ of the other parabola is always the symmetric of $F$ respect to the line passing through the common point of the two parabolas and tangent to both parabolas. So, from the lemma above follows that the locus of $G$ is the directrix of the $y=-x^2$ parabola, namely the line $y=\frac 1 4$.

Indeed! The locus of the symmetric of the focus wrt a variable tangent is the directrix of the parabola.

For the ellipse, the locus of the symmetric of the focus $F_1$ wrt a variable tangent is a circle centered at $F_2$, of radius $2a$ (usual notation; $a$ is the largest radius of the ellipse). I'm not sure about th hyperbola right now, but it is, of course, something similar (a circle centered at the other focus).

hyperbola works as well:

let EF be focuses, P on hyperbola. then E', the image of E by mirroring wrt the tangent to the hyperbola in P, lies on PF (since the tangent bisects EFP). E'F = PF - PE' = PF - PE = const. = VV' (V and V' are the hyperbola's vertices) by definition of the hyperbola. thus the locus of E' is a circle around F with radius VV'.


t² + 2t + lnt = 0        I know it doesn't exactly qualify for this topic but it is from Calc class and I can't seem to elliminate the natural log to solve for t. Help?

There's no way to find the exact solution (there is exactly one solution, which is between zero and 1).

Perhaps you should be looking for an approximate solution?

well, the whole problem is for me to find the value of the unit tangent vector at the point (1,1,1) when I am given r(t) = <t²,2t, lnt > so I don't think I can use just an approximation. Thus my problem. :lol:

That question doesn't make sense, and you definitely shouldn't be setting $x+y+z=0$. $(1,1,1)$ isn't on the curve at all; are you sure you've got the question right?

Yeah, I talked to the teacher last night and she said that the problem was incorrect! She wrote it wrong. The point should have been (0,0,1) not (1,1,1)! Now there's no problem at all :D !! So thanks so much for trying to help on an unsolvable problem :oops: Don't ya just love typos?
Thanks again!


parameterization
inequalities
inequalities proposed

Let  $ x + y + z = 3,x,y,z\ge{0},e\ge{1},k\in{N}(k\ge{1}),$ prove that

  $ (x^{k + 1} + e^{k + 1})(y^{k + 1} + e^{k + 1})(z^{k + 1} + e^{k + 1})\geq{(x^{k} + e^{k})(y^{k} + e^{k})(z^{k} + e^{k}).}$

If you feel difficult, please just to prove the inequality for some specific values of parameters,which is holds.

[quote="Materazzi"][quote="fjwxcsl"]Let  $ x + y + z = 3,x,y,z\ge{0},e\ge{1},k\in{N}(k\ge{1}),$ prove that

  $ (x^{k + 1} + e^{k + 1})(y^{k + 1} + e^{k + 1})(z^{k + 1} + e^{k + 1})\geq{(x^{k} + e^{k})(x^{k} + e^{k})(x^{k} + e^{k}).}$

If you feel difficult, please just to prove the inequality for some specific values of parameters,which is holds.[/quote]

I think it must be :

 $ (x^{k + 1} + e^{k + 1})(y^{k + 1} + e^{k + 1})(z^{k + 1} + e^{k + 1})\geq{(x^{k} + e^{k})(y^{k} + e^{k})(z^{k} + e^{k}).}$

???[/quote]

Thank you! Has edited.


trigonometry
geometry
circumcircle
IMO Shortlist
cyclic quadrilateral

Let $ ABCD$ be a convex quadrilateral. The diagonals $ AC$ and $ BD$ intersect at $ K$. Show that $ ABCD$ is cyclic if and only if $ AK \sin A \plus{} CK \sin C \equal{} BK \sin B \plus{} DK \sin D$.

Let r be the radius of the circle. Obviously SinA=SinC, SinB=SinD, so the problem is reduced to AC SinB = BD Sin A. But AC=2rSinB, BD=2rSinA, so ACSinB= BDSinA=2rSinASinB

Yes, of course, but the converse is (in my opinion) substantially more difficult.

Thats's why it's sometimes easier to work backwards from what you are trying to prove.  Sort of like starting a maze from the end.

After all that, I got it out myself. If $A+C > 180$, then $AC \sin A > BD \sin B$ and also $AC \sin A > BD \sin B$, so $AC \sin A = \frac{AC.BK \sin A}{BD} + \frac{AC.DK \sin A}{BD} > BK \sin B + DK \sin D$, and doing a similar thing with $C$ we easily get that $AK \sin A+CK \sin C > BK \sin B + DK \sin D$.

I think your assertion that if $A+C>180$, then $AC \sin A<BD \sin B$ could be wrong.
Suppose $A$,$B$,$D$ are on circle $O$ with radius $r$, while $C$ is inside the circle, then $A+C>180$. The segment $BC$ is extended and crosses the circle at $E$. Now we have \[AC \sin A=\frac {AC \cdot BD}{2r}\],\[BD\sin B=\frac {AE \cdot BD}{2r}\].
Therefore if $AC>AE$(which is absolutely possible), then $AC \sin A>BD \sin B$. The conclusion turns to be in the opposite.

[quote="Fiachra"]ABCD is a convex quadrilateral. The diagonals AC and BD intersect at K. Show that ABCD is cyclic if and only if $AK \sin A + CK \sin C=BK \sin B + DK \sin D$.

Obviously the "only if" part is trivial.[/quote]

Actually, the IMO Shortlist 1997 problem 23 asked to prove that if AK sin A + CK sin C = BK sin B + DK sin D, then the quadrilateral ABCD is cyclic. This is a really diabolic problem, since it is quite easy to find a solution that works for the "usual" case (in most solutions, this is the case when the triangles ABC, BCD, CDA and DAB all are acute-angled, i. e. when each diagonal of the quadrilateral ABCD and each side form an acute angle), but not all of these solutions properly work in the general case of an arbitrary convex quadrilateral ABCD. Anyway, you can look at http://www.mathlinks.ro/Forum/viewtopic.php?t=15584 post #2 file IMO_1997_shortlist.pdf to find the proposed solution of the problem (scroll down to page 52), and you can also look at http://www.kalva.demon.co.uk/short/soln/sh9723.html (that's practically the same solution). Both times, you will find solutions working for arbitrary convex quadrilaterals ABCD.

Proving the converse, i. e. proving that if ABCD is a cyclic quadrilateral, then AK sin A + CK sin C = BK sin B + DK sin D, is really a piece of cake, and it was done by Zengmao in post #2 of this thread. It could also be done as follows:

Let X, Y, Z, W be the orthogonal projections of the point K on the sides AB, BC, CD, DA of the quadrilateral ABCD. Then, since < AWK = 90° and < AXK = 90°, the points W and X lie on the circle with diameter AK. In other words, the circumcircle of triangle AWX has diameter AK. Thus, by the extended law of sines (which states that a sidelength of a triangle equals the diameter of the circumcircle multiplied with the sine of the opposite angle), applied to the triangle AWX, we have $WX=AK\cdot\sin\measuredangle WAX$. In other words, $WX=AK\cdot\sin A$. Similarly, $XY=BK\cdot\sin B$, $YZ=CK\cdot\sin C$ and $ZW=DK\cdot\sin D$. Hence, instead of proving the equation AK sin A + CK sin C = BK sin B + DK sin D, it will suffice to establish the equation WX + YZ = XY + ZW.

Since the points W and X lie on the circle with diameter AK, the chordal angle theorem yields < KWX = < KAX. In other words, < KWX = < CAB. Similarly, < KWZ = < BDC. But since the quadrilateral ABCD is cyclic, the chordal angle theorem yields < CAB = < BDC. Thus, < KWX = < KWZ. It follows that the point K lies on the angle bisector of the angle ZWX. Similarly, the point K lies on the angle bisectors of the angles WXY, XYZ and YZW. Thus, the angle bisectors of the angles WXY, XYZ, YZW and ZWX of the quadrilateral XYZW concur at one point (namely, at the point K). Hence, the quadrilateral XYZW has an incircle (with center K); thus, WX + YZ = XY + ZW. And this completes the proof.

I'm actually posting this proof because it gives a subtle hint towards the proof of the rather difficult part of the problem (AK sin A + CK sin C = BK sin B + DK sin D $\Rightarrow$ the quadrilateral ABCD is cyclic)...

  Darij

I have found that the problem 2149 of CRUX MATHEMATICORUM it is asked prove that a proposition equivalent to problem of Shortlist.
How I can To attach files? :)

For my own ease in labeling, let $K=X$. Drop perpendiculars from $X$ to $AB, BC, CD, DA$ with feet $K, L, M, N$ respectively. Then the midpoint of $AX$ is the center of the circle that passes through points $A, K, X, N$. Also, $KN$ subtends an angle $2\angle A$ at the center, so $KN=2r\sin\frac{2A}{2}=XA\sin A$ where $r$ is the radius of the circle. From the given condition in the question we have $KN+LM=MN+KL (**)$. But we also have \begin{align*}KN&=XK\cos\angle XKN+XN\cos\angle XNK\\&=XK\cos\angle XAN+XN\cos\angle XAK (*)\end{align*}

Now, let $\angle XAD=\alpha_1$, $\angle XAB=\alpha_2$, $\angle XBA=\beta_1$, $\angle XBC=\beta_2$, $\angle XCB=\gamma_1$, $\angle XCD=\gamma_2$, $\angle XDC=\delta_1$, $\angle XDA=\delta_2$. Then $(*)$ becomes $KN=XK\cos\alpha_1+XN\cos\alpha_2$. Similarly, $KL=XL\cos\beta_1+XK\cos\beta_2$, $LM=XM\cos\gamma_1+XL\cos\gamma_2$, $MN+XN\cos\delta_1+XM\cos\delta_2$. So $(**)$ becomes \[XK(\cos\alpha_1-\cos\beta_2)+XL(\cos\gamma_2-\cos\beta_1)+XM(\cos\gamma_1-\cos\delta_2)+XN(\cos\alpha_2-\cos\delta_1)=0 (***).\]

WLOG, assume $\alpha_1\le\beta_2$ i.e. $A$ lies outside the circumcircle of $\triangle BCD$, which means $\alpha_2\le\delta_1$. But we have $\alpha_1+\delta_2=\beta_2+\gamma_1$, so $\gamma_1\le\delta_2$. Similarly $\alpha_2+\beta_2=\gamma_2+\delta_1$, so $\gamma_2\le\beta_1$. The cosine function decreases over $[0^{\circ}, 180^{\circ}]$, so it follows that all the terms within parenthesis in $(***)$ are nonnegative. Hence they must all be zero, and all pairs of angles are equal. It follows that $ABCD$ is cyclic, as desired. $\Box$

ABCD cyclic
sinA=sinC   sinB=sinD  we need prove that
ACsinA=BDsinB
AC=2RsinB  BD=2RsinA proved.


topology
real analysis
real analysis unsolved

$X$, $Y$, and $Z$ are topological spaces. If $X \times Y$ is homeomorphic to $X \times Z$, does it follow that $Y$ is homeomorphic to $Z$?

(The topology on the product spaces is taken to be the usual product topology.)

Certainly not. Take $X$ to be any infinite set with the discrete topology, and $Y=X$, while $Z$ is a one-point set with the discrete topology. In fact, if you work only with the discrete topology on our sets, this question turns into: "if the sets $X\times Y, X\times Z$ have the same cardinality, is it true that $Y$ and $Z$ also have the same cardinality?"

Yes. This is my solution also. Now for the more interesting question (to which I don't know the answer) -- what are sufficient conditions for the claim to be true? It is certainly true if $X=\mathbb{R}^k$, $Y=\mathbb{R}^\ell$, $Z=\mathbb{R}^m$, for any nonnegative $k,\ell,m$, and the difficulty of finding counterexamples with the "usual" spaces suggests that something like this might be true with some additional assumptions. What do you think?


modular arithmetic
number theory
prime numbers
number theory unsolved

1.let  $ n \in N$ , $ n>6$ and $ n\plus{}1$ , $ n\minus{}1$ are prime numbers.
prove that $ 720 / n^2 (n^2\plus{}16)$
2.let  $ \forall n \in N$  such that  $ 720 / n^2 (n^2\plus{}16)$ . is it true that  $ n>6$ and $ n\plus{}1$ , $ n\minus{}1$ are prime numbers?

[b]#1[/b]

Any prime number greater than 6 is of the form $ 6k\pm1$ for some $ k\in\mathbb{Z}$. Hence, if $ n\plus{}1$ and $ n\minus{}1$ are both primes, $ n$ must be divisible by 6. So $ n^2$ is divisible by 36.

If $ n$ is not divisible by 5, then, since $ n$ is even, $ n^2\equiv\pm1\pmod{5}$. Suppose $ n^2\equiv1\pmod{5}$. Then 5 divides $ n^2\minus{}1\equal{}(n\plus{}1)(n\minus{}1)$ $ \Rightarrow$ 5 divides $ n\plus{}1$ or $ n\minus{}1$. Since $ n>6$, this would mean that either $ n\plus{}1$ or $ n\minus{}1$ is not prime, a contradiction. Hence, if $ n$ is not divisible by 5, we must have $ n^2\equiv\minus{}1\pmod{5}$, i.e. $ n^2\plus{}16$ must be divisible by 5. Thus $ n^2(n^2\plus{}16)$ is always divisible by 5.

$ n^2\plus{}16$ is also divisible by 4. Thus $ n^2(n^2\plus{}16)$ is divisible by $ 36\times5\times4\equal{}720$.

For the second part, a counterexample is $ n\equal{}90$.


55 numbers are chosen from the integers 1,2,3...................100.

Prove that there are two of the numbers which differ by 12.

[hide]

I think we can improve it to $53$...

We can look at the sets

$\{1,2,\ldots,24\}$, $\{25,26,\ldots,48\}$, $\{49,50,\ldots,72\}$, $\{73,74,\ldots,96\}$, and $\{97,98,99,100\}$.

At most $4$ elements can be in the last set (obvious), so there are at least $49$ elements distributed among the other four sets. By Pigeonhole, we have at least $13$ in one of the sets. Now pair up elements that differ by $12$ in the set with $13$. We have

$\{a, a+12\}, \{a+1, a+13\}, \ldots, \{a+11, a+23\}$.

There are $12$ pairs and $13$ elements so by Pigeonhole again two elements are in a pair. Hence they differ by $12$, as desired.[/hide]


calculus
real analysis
real analysis theorems

I studied real analysis in undergraduate and was never good at it. I never got a pass in this topic and it really bothers me. As a result, i would like to study on my own and give it another try. any suggestion would be helpful, thanks! and recommand me a good textbook please.

[quote="fedja"]I consider "Calculus" by Michael Spivak a fairly good one for the beginners. At least, among all the textbooks I tried for the beginners' rigorous analysis course, this one seems to be the most liked by the American students.  Or, maybe, you prefer something in French? Then ask Moubinool. He will, probably, be able to give a good advice.[/quote]
Thank you very much! i will give it a good read :)

I think Real  Analysis by Carothers is a good one.


Five parcels of land are given. In each step, we divide one parcel into three or four smaller ones. Assume that, after several steps, the number of obtained parcels equals four times the number of steps made. How many steps were performed?

[hide="Solution"]
x-number of dividing into 3 parts
y-number of dividing into 4 parts
$ 5\plus{}2x\plus{}3y\equal{}4x\plus{}4y$
$ 5\equal{}2x\plus{}y$
$ x\equal{}2$, $ y\equal{}1$- 3 steps
$ x\equal{}1$, $ y\equal{}3$- 4 steps
$ x\equal{}0$, $ y\equal{}5$- 5 steps
[/hide]


Ring Theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose $ k$ is any ring and consider $ R \equal{} k<x,y>$ the free k-ring on two generators.
Show that for each $ n > 0, R$ contains a subring that is isomorphic to the free $ k$-ring on $ n$ generators.

Consider the subring generated by $ x, xy, xy^2, ... xy^{n \minus{} 1}$.  (This is inspired by a standard example in commutative algebra and it just so happens to work in the noncommutative case as well.)

Yea I had thought of that because they're not commutative, so they may be considered as "new" indeterminates.
I just don't know how to really write that out.

The subring in question is the free $ k$-module on words in $ x$ and $ y$ starting with $ x$ such that there are at most $ n \minus{} 1$ consecutive $ y$s between $ x$s.  Any such word has a unique factorization into ordered products of the generators and this defines the isomorphism you want.  (In particular, this argument immediately implies that there are no relations among the generators.)


Solve the equation

sqrt(4+sqrt(4 -sqrt(4 +sqrt(4 - x))))  = x

sqrt   mean   (   )^1\2

It is equavalent to
(1) $f(f(x))=x$,
were $f(x)=\sqrt{4+\sqt{4-x}}$, because f(x) decrease function (1) equavalent to $f(x)=x$ or $(x^{2}-4)^{2}+x=4$.

can you complet your soluation rust

\[\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}= x\]

I am sorry I cannot edit my post I write f(x)=\sqrt{4+\sqt{4-x}}, must be
$f(x)=\sqrt{4+\sqrt{4-x}}$.


geometry
trapezoid
circumcircle
geometry unsolved

Figure, $ AB < AC$, $ H$ is the orthocenter of △$ ABC$, $ A$ is the midpoint of $ HH'$ . $ AB,AC$ intersect $ CH',BH'$ at $ D,E$ respectively .

Prove that： ∠$ BAC \equal{} 60 < \equal{} > BD \equal{} CE$ .

[hide]Or You can consider locus[/hide]

The below proof was provided to me by my son, but he's too busy to type it by himself.

On $ CA$ produced we take $ F| AF \equal{} AC$, on $ BA$ produced we take $ G| AG \equal{} AB$. Obviously, $ BCGF$ is an isosceles trapezoid, with $ BG \equal{} CF$.
We shall get $ EF \equal{} DG$ iff $ \frac {EF}{CE} \equal{} \frac {DG}{BD}$ which, in turn is true iff $ \frac {S_{BFH'}}{S_{BCH'}} \equal{} \frac {S_{CGH'}}{S_{BCH'}}$, the last equality being true iff $ S_{BFH'} \equal{} S_{CGH'}\ ( \ * \ )$.
Seing that $ BF$ and $ CG$ are parallel to the bisector of the $ \angle BAC$, let's draw a parallel to this lie through $ H'$ and call $ P$ its intersection with $ FG$.
The relation (*) is true iff $ \frac {FP}{PG} \equal{} \frac {AC}{AB}\ ( \ 1 \ )$, because $ \triangle BAF \sim \triangle CAG$, but relation (1) shows that, $ H'$ is the symmetrical of the middle of the arc $ FG$ about $ FG$, i.e. $ H'$ is the circumcenter of $ \triangle AFG$, which happens iff $ \angle BAC \equal{} 60^\circ$.

The reverse of the problem, iff $ \angle BAC \equal{} 60^\circ$, then $ H'$ is the circumcenter of the $ \triangle AFG$, etc.

[b]Note: Here $ S_{XYZ}$ stands for the area of $ \triangle XYZ$.[/b]

Best regards,
sunken rock

Thanks for your proof .
Is there anybody want to give different way? Nex I'll show how I design this problem .

[b]Lemma (well-known):[/b] $P$ is a point inside the parallelogram $ABCD$ and $DP,BP$ cut $AB,AD$ at $X,Y,$ respectively. Then $DX=BY$ $\Longleftrightarrow$ $PC$ bisects $\angle BPD.$ $\blacksquare$

Back to the problem. Let $M$ and $L$ be the midpoints of $\overline{BC}$ and the arc $BC$ of the circumcircle $(O) \equiv \odot(ABC).$ If $T$ is the point forming the parallelogram $H'CTB,$ then from the previous lemma, we get $BD=CE$ $\Longleftrightarrow$ $AT$ bisects $\angle BAC$ $\Longleftrightarrow$ $A,T,L$ are collinear $\Longleftrightarrow$ $\tfrac{AH'}{ML}=\tfrac{TH'}{TM}=2$ $\Longleftrightarrow$ $ML=\tfrac{1}{2}AH'=\tfrac{1}{2}AH=OM$ $\Longleftrightarrow$ $L$ is reflection of $O$ on $BC$ $\Longleftrightarrow$ $2\angle BAC=180^{\circ}-\angle BAC$ $\Longleftrightarrow$ $\angle BAC=60^{\circ}.$


inequalities
LaTeX
inequalities proposed

Prove that for any real numbers a, b, c such that[b] $a^2+b^2+c^2=9$[/b],
the following inequality is true:  [b]3.min {a, b, c} $\le$ 1 + abc[/b]      (*).

[b]Remark.[/b] Sir Dang Tran Nam proposed the inequality (Vietnam, 2002):
" [u]For any real numbers a, b,c such that   $a^2+b^2+c^2=9$[/u]
 [u]the inequality 2 (a + b + c )$\le$ 10 + abc is true[/u] ".
[i]Solution. [/i] From C.B.S. and (*) results 2(a+b+c)=(2b+2c-a)+3a$\le$
sqrt[9($a^2+b^2+c^2$)]+3a=9+3a. For any a$\le$b$\le$c results
 2(a+b+c)$\le$9+3a$\le$9+(1+abc)=10+abc, i.e. (a+b+c)$\le$10+abc.

First of all please post in the inequalities area. Secondly, I see that you know your way around $\LaTeX$ so please use it to write full inequalities and equations in it, not just the $\leq $ sign!

PS I have posted this problem before on the forum right here: http://www.mathlinks.ro/Forum/viewtopic.php?t=34669 thus this topic is closed.


trigonometry
number theory
relatively prime
Irrational numbers

[quote="Peter"]An integer-sided triangle has angles $ p\theta$ and $ q\theta$, where $ p$ and $ q$ are relatively prime integers. Prove that $ \cos\theta$ is irrational.[/quote]

 The reference for this problems is "[b]CRUX(No. 1. Vol. 24, 1998), Problem 2305, Richard I. Hess[/b]".

Added. :) Now all we need is a solution.

What about equilateral triangle?
It can be $p=q=1, \theta = \frac{\pi}{3}$
It is integer-sided, but $\cos \frac{\pi}{3} = \frac{1}{2} \in \mathbb{Q}$, Then this can disprove the proposition. :(

maybe is:

"p and q are distinct primes"

instead of:

"where p and q are relatively prime integers"

Unless I'm mistaken, I have a proof that $\cos \theta$ is [b]rational[/b].

Let $S = \{n \in \mathbb{N} \ | \ \cos n\theta \in \mathbb{Q}  \}$. Using that the angles of the triangle are $p\theta, \ q\theta, \ \pi - (p+q) \theta$, and the cosine rule, we have that $p,q,p+q \in S$.
Now, $\cos |(x-y)| \theta = 2 \cos x\theta \cdot \cos y\theta  - \cos (x+y)\theta$, means that $x,y,x+y \in S \implies |x-y| \in S$. But this means that if $x,y,x+y \in S$, then gcd$(x,y) \in S$. But, then since $p,q$ are coprime, so $1 \in S$.


function
inequalities
analytic geometry
graphing lines
slope
topology
triangle inequality

Here's an exersise I was given in class. The problem is that our teacher is not sure if it is true and I'm having trouble proving one little detail. I also have not been able to find a counterexample if there is one.

Let $ (X,d)$ be a metric space, and let $ M_1$ and $ M_2$ be compact, disjoint subsets of $ M$. Let $ A$ and $ B$ be closed subsets of $ M_1$ and $ M_2$. Let $ \phi: A\to B$ be a homeomorphism. Let $ M \equal{} M_1\cup M_2$ and build the equivalence relation $ a\equiv b$ if $ a \equal{} b$ or $ \phi (a) \equal{} b$ or $ \phi (b) \equal{} a$. Prove that $ M/\equiv$ endowed with the distance $ \tilde d$ is a metric space, where $ \tilde d$ is a distance function given by:
\[ \tilde d([x], [y]) \equal{} \inf\{d(p_1,q_1) \plus{} d(p_2,q_2) \plus{} \dots d(p_n,q_n)\}\]
(where the infimum is taken among all the tuples $ (p_1,\dots,p_n)$ and $ (q_1,\dots, q_n)$ such that $ [p_1] \equal{} [x]$, $ [q_n] \equal{} [y]$ and $ [q_k] \equal{} [p_{k \plus{} 1}]$ for $ k\in \{1,\dots, n\}$)

OK, let's go over it.

1) Non-negativity of $ \widetilde d$ is obvious.

2) The triangle inequality is rather obvious too (just take the union of 2 chains).

3) Now (and I assume that is the little detail you are talking about) we want to show that $ \widetilde d([a],[b])\equal{}0$ implies $ a\equiv b$ or to build a counterexample to this. It turns out that the counterexample exists. Just take $ X$ ot be the real line, $ M_1\equal{}[0,1]$ and $ M_2\equal{}[2,3]$. We will construct a strictly increasing continuous function from $ M_1$ to $ M_2$ such that the length totally collapses on $ M_1$ and $ M_2$. To this end, just start with the function $ f(x)\equal{}x\plus{}2$. Now, suppose that we have a function that is piecewise linear and strictly increasing. Take any linear interval in its graph and replace it like this:
[asy]size(100);
D((0,0)--(1,1));
D((0,0)--(0.1,0.4)--(0.9,0.6)--(1,1),red);[/asy]
if the slope is at least $ 1$ and like this:
[asy]size(100);
D((0,0)--(1,1));
D((0,0)--(0.4,0.1)--(0.6,0.9)--(1,1),red);[/asy]
otherwise.
Such replacement will decrease the factor-distance between the endpoints twice if the replacement Z-curves are chosen appropriately and the corresponding sequence of functions will have a limit that will be continuous, strictly increasing, and totally colapsing the distance on $ M_1$ and $ M_2$.


I'm using C++ and whatever I input, it flashes on the screen for a fraction of a second and then it disappears.  The problem should not be my source code because I'v tried really basic codes, as well as codes I've copied off the internet that should certainly work.  I probably did something stupid so please help!

What 's the complier you are using ? If you use DevC++ and want to view your output longer ,you should insert the command"system("pause")" near the end of your sourcecode (just above the last "}" ) :D .

otherwise, initialize any variable and at the end cin>>whatever variable you chose, that way you can tell it when to close.

Old-school used getch() for keeping the window open - interesting enough, that practice was suppressed/discouraged by TAs and professors during my programming courses (and they did have a point :wink:).


group theory
abstract algebra
function
superior algebra
superior algebra unsolved

1.  a) List all the subgroups of [b]Z[/b][size=75]12[/size]/H where H={0,6}

I know that {0,2,4,6,8,10} and {0,3,6,9} are ones but are there others?

    b) List all the subgroups of [b]Z[/b][size=75]20[/size]/K where K={0,4,8,12,16}

2.  Let G be an abelian group
    a) Show that K={a an element of G such that |a|<= 2} is a subgroup of G
    b) Show that H={x^2 such that x is an element of G} is a subgroup of G
    c)  Prove that G/K is isomorphic to H (by defining a surjective homomorphism from G to H with Kernel K)

For 1 (a), you're missing the two obvious subgroups, but those are the only other ones: $ \mathbb{Z}_{12}/H$ is essentially $ \mathbb{Z}_6$, as there's a natural isomorphism between the two (namely, reduction modulo 6). (b) is similar.

For 2, if you can prove (c) (that is, find a homomorphism from $ G$ to itself with kernel $ K$ and image $ H$), you get (a) and (b) automatically. Now $ H$ is explicitly given as the image of a [i]function[/i] from $ G$ to itself...


What percent of 63 is 0.1? Express your answer as a decimal percent to the nearest tenth of a percent.

[hide] (.1/63)*100 is what we are looking for.  This goes to 10/63, which is about .15%.  This rounds to .2%.  [/hide]

[quote="The QuattoMaster 6000"][hide] (.1/63)*100 is what we are looking for.  This goes to 10/63, which is about .15%.  This rounds to $.2\%$.  [/hide][/quote]
Oh, I was doing $(63*.1)/100=x$ and I got 6.3%, which was wrong.

BTW, to get a % in $\LaTeX$, use the \% $\%$

[quote="Adam4444"][hide]$\frac{0.1}{63}* 100 \approx 0,015873016$

$0,015873016 \approx 0.02\%$[/hide][/quote]
I think you must have multiplied or divided wrong. The answer is $.2\%$.

[quote="Quevvy"][quote="Adam4444"][hide]$\frac{0.1}{63}* 100 \approx 0,015873016$

$0,015873016 \approx 0.02\%$[/hide][/quote]
I think you must have multiplied or divided wrong.[/quote]Yes I did... :blush:
But I edited my post now to the correct answer. :wink:


Euler
trigonometry
geometry proposed
geometry

I have come up with a nice problem.
Suppose the Euler line of $\triangle ABC$ meets the edge of $\triangle ABC$ at $D,E$. WLOG, let $\angle C=\max\{\angle A,\angle B,\angle C\}$, show that
(1) $\angle C<90^{\circ}$, then we have $8DE^2>AB^2+BC^2+CA^2$;
(2) $\angle C=90^{\circ}$, then we have $8DE^2=AB^2+BC^2+CA^2$;
(3) $\angle C>90^{\circ}$, then we have $8DE^2<AB^2+BC^2+CA^2$.

$D\in AB,\ E\in AC,\  \frac {DB}{DA}=x,\ \frac {EC}{EA}=y\Longrightarrow x+y=1;\ for\ A\le90^{\circ}$ $, \ x\sin 2B+y\sin 2C=\sin 2A\Longrightarrow$

$calc.\ x,y\Longrightarrow calc.\ AD,AE\Longrightarrow calc.\ DE,\ a.s.o.$

Dear levi,could you show me the source?Thanks! :)

Sir Darij can know the source of these used relations. I offer you the following lemma (you can prove easily):

$Line\ d,\ points\ P\not\in d,\ A,B,C\in d,\ X\in PA,\ Y\in PB,\ Z\in PC,\ Z\in XY\Longrightarrow $ \[\overline {\underline {\left| \frac {\overline {XA}}{\overline {XP}}\cdot \overline {BC}+\frac {\overline {YB}}{\overline {YP}}\cdot \overline {CA}+\frac {\overline {ZC}}{\overline {ZP}}\cdot \overline {AB}=0.\right| }}\]


induction
modular arithmetic
combinatorics unsolved
combinatorics

Some pairs of towns are connected by a road. At least 3 roads leave each town. Show that there is a cycle containing a number of towns which is not a multiple of 3.

This looks very familiar, and the following proof does too :). I think it's around here somewhere.

We consider only simple cycles, i.e with no repeated vertices. Let $C=(x_1,x_2,\ldots,x_{3k},x_1)$ be a $3k$-cycle (of course, there must be a cycle, since our graph $G$ cannot be a forest). Now look at the graph $\widetilde G$ obtained by collapsing $x_i$ into a single vertex $x$. $\widetilde G$ still has all vertices of degree $\ge 3$, so by induction, it has a cycle $\widetilde C$ whose size is not a multiple of $3$. If $\widetilde C$ does not contain $x$, we're done: it's the image, after the collapsing, of a cycle in $G$ with the same size. Suppose now that it does contain $x$. Then, since $x_i,x_j$ cannot be connected by a path outside $C$ if $i\not\equiv 0\pmod 3$, the two edges of $\widetilde C$ which contain $x$ are adjacent to $x_i,x_j$ with $i\equiv j\pmod 3$ in $G$, so in $G$ we can obtain a cycle with the same size as $\widetilde C$ modulo $3$ by considering all sides in $\widetilde C$, plus $0\pmod 3$ edges connecting $x_i,x_j$ in $C$.

I hope it's correct.


Compute the number of positive integers $a$ for which there exists an integer $b$, $0 \le b \le 2006$, such that both of the polynomials $x^{2}+ax+b$ and $x^{2}+ax+b+1$ have integer roots.

[hide]a^2-4b=p^2 and a^2-4b-4=q^2 are perfect squares...but this is tricky because they are 4 apart...so we have to do some 1337 factoring and some case work... (p,q)=(2,0), so a^2=4b+4, then we are like omg 88^2<4*2006+4<89^2

so there are 88 solutions[/hide]

[quote="mna851"]Yes, but I think you messed up a little at the end.  $\ a^{2}=4b+4$ so $\ a=2\sqrt{b+1}$ so $\ a \in{2,4,6, \ldots , 86, 88}$ since $\ 44^{2}< 2007 < 45^{2}$.  Therefore there are only $\ \boxed{44}$ such values of $\ a$.  By the way, I think this problem originally appeared at ARML.[/quote]

Hey that's what we guessed on the test... lol.

My original guess was $2$, but we kept finding solutions for even $a$, and then we figured it out  :rotfl:

haha...eastlake contest rules, cause i wrote all the questions...ha.

(sorry for the off topic note. :P)

the correct answer is indeed 44


[b]Thanks PCO.
and once again Thanks to hsbhatt sir(for unique solution)[/b]

[quote="hsbhatt"]We have $ \frac {f(x) \plus{} 1}{f(x) \minus{} 1} \equal{} x$ and hence $ \frac {f^{2009}(x) \plus{} 1}{f^{2009}(x) \minus{} 1} \equal{} \frac {f^{2008}(x) \plus{} 1}{f^{2008}(x) \minus{} 1} \equal{} ... \equal{} \frac {f(x) \plus{} 1}{f(x) \minus{} 1} \equal{} x$[/quote]

I'm affraid there are some mistakes here :

1) $ \frac {f(x) \plus{} 1}{f(x) \minus{} 1} \equal{} \minus{}\frac 1x$ and not $ x$

2) $ \frac {f^{2009}(x) \plus{} 1}{f^{2009}(x) \minus{} 1} \equal{} \frac {f^{2008}(x) \plus{} 1}{f^{2008}(x) \minus{} 1}$ is wrong (and I dont understand how you could find it) :

$ \frac {f^{2p}(x) \plus{} 1}{f^{2p}(x) \minus{} 1} \equal{}\frac {x\plus{}1}{x\minus{}1}$

$ \frac {f^{2p\plus{}1}(x) \plus{} 1}{f^{2p\plus{}1}(x) \minus{} 1} \equal{}\minus{}\frac 1x$

[quote="jagdish"]Thanks for correction PCO.
actually here I am going  for Method not for solution.[/quote]

Huh ? And writing $ \frac {f^{2009}(x) \plus{} 1}{f^{2009}(x) \minus{} 1} \equal{} \frac {f^{2008}(x) \plus{} 1}{f^{2008}(x) \minus{} 1}$ is a wrong solution but a good method ?

yeah i muffed that, dont remember what i was thinking  :blush:

To redeem myself:

$ f(x) \equal{} \frac{1\minus{}x}{1\plus{}x} \Rightarrow  \frac{1\minus{}f(x)}{1\plus{}f(x)} \equal{}x$

Hence $ f^{2009} x \equal{} \frac{1\minus{}f^{2008}x}{1 \plus{} f^{2008} x} \equal{} f^{2007}x \equal{}...\equal{} f(x) \equal{} \frac{1\minus{}x}{1\plus{}x}$

Apologies again to Patrick and Jagdish


algebra
difference of squares
special factorizations

Rationalize the denominator of $ \frac{\sqrt{x}}{3\sqrt{x\minus{}\sqrt{y}}}$

Interesting! I've never seen a problem like that before.

[hide="My Solution"]
We still use our old difference of squares method throughout.

$ \frac{\sqrt{x}}{3\sqrt{x \minus{} \sqrt{y}}} \cdot \frac{\sqrt{x \plus{} \sqrt{y}}}{\sqrt{x \plus{} \sqrt{y}}} \equal{} \frac{\sqrt{x(x \plus{} \sqrt{y})}}{3\sqrt{x^2 \minus{} y}} \equal{} \frac{\sqrt{(x^2 \plus{} x\sqrt{y})(x^2 \minus{} y)}}{3x^2 \minus{} 3y} \equal{} \boxed{\frac{x^4 \plus{} x^3\sqrt{y} \minus{} x^2y \minus{} xy\sqrt{y}}{3x^2 \minus{} 3y}}$
[/hide]

why is it not $ \sqrt{x^2\plus{}\sqrt{xy}}$ over the denominator in the second step?

Remember that $ x$ and $ (x \plus{} \sqrt{y})$ are both under square roots themselves so their product would be $ x(x \plus{} \sqrt{y}) \equal{} x^2 \plus{} x\sqrt{y}$ under one square root.


algebra
polynomial
complex analysis
complex analysis unsolved

Hello everyone I need help with this exercises, if you can help me i really appreacite it:

"Let $ f(z)\equal{}z^2\minus{}2.$ Prove that the Julia set of $ f$ is the closed interval on the real axis between $ \minus{}2$ and $ 2.$ "

thanks

Adriana

The idea is that $ f$ is conjugate to $ g(z)\equal{}z^2$ by something like $ h(z)\equal{}z\plus{}1/z$.

Ok, thanks, I got it, It's easy to show that $ Julia(z^2)\equal{}S^1$ and under $ h$ its image is $ [\minus{}2,2]$.   However I was looking in the book where I took the exercises and it says (As hint) that actually $ Julia(z^2\minus{}2)\equal{}closure(Per(z^2\minus{}2))$.  I could prove that $ Per(z)$ is contained in the real axis, but I cannot conclude the equality.

If it's bigger than 2, it goes to $ \infty$. One step gets you to the positive side, and from there each step takes you higher.


Find all $f: Z\rightarrow R$ such that $f(1)=\frac{5}{2}$ and that $f(x)f(y)=f(x+y)+f(x-y)$, for all $x,y\in Z$

[hide="hint strong"]Let $x=y=0$ , then $f(0)=0\;\; \text{or}\;\; 2$.
Let $y=1$, we will have a relation of $f(x+1),f(x),f(x-1)$.
Now, use induction! :)[/hide]


Two decks of playing cards are shuffled and placed side by side. The top cards of both decks are flipped one by one. What is the probability that there will be at least one matching pair?

[hide]Suppose there are $k$ pairs of matching cards.  There are $\binom{52}{k}$ ways to choose the pairs, and $(52-k)!$ ways to match up the rest of the cards.  By PIE, the number of possibilities in which there is at least one pair of matching cards is $\sum_{k=1}^{52}(-1)^{k+1}\binom{52}{k}(52-k)!=\sum_{k=1}^{52}(-1)^{k+1}\frac{52!}{k!}=52!\sum_{k=1}^{52}(-1)^{k+1}\frac{1}{k!}$.

There are $52!$ total ways to match up the cards, so the probability is $\sum_{k=1}^{52}(-1)^{k+1}\frac{1}{k!}$.[/hide]


where can i get my hands on good e-novels?
also tell me if there are gud books(not just ebooks) worth reading that i mite've missed. now dat all exams are over, im pretty jobless.
and ofc- no books on philosophy. :D

Disclaimer: I'm not encouraging you to do anything illegal.
Okay, if you like fanfiction there are loads of fanfiction sites out there which are fantastic - fictionalley, fanfiction.net, mugglenet to name a few. Many of the HP fanfictions that I've read (H/Hr ones especially) are awesome, sometimes even better than the original books! Some fanfiction is really awful too, so explore at your own risk!
 You can search live journal and wordpress for authors whom you may like. Many authors choose to put their work online and some of them are very good.
 Rapidshare hosts several illegal copies of novels either scanned or (rarely) typed out. You can take your pick of them :).
 But despite the internet - I read mostly printed versions. In my opinion, the joy of actually sitting down and reading a book is unparalleled.
Just my two cents.
Some of my favourite books are : To kill a mockingbird, Rebecca, Forward the Foundation(and all the other foundation books by Asimov), PG Wodehouse's novels on Blandings and Jeeves and Wooster, 2001: A Space Odyssey and 2010: Odyssey 2(I haven't read the other two books in the series but they're supposed to be very good), Murder of Roger Ackroyd (and most other Hercule Poirot books by Agatha Christie) and so many more. I think there was a post in the Round Table forum a couple of years ago about favourite books. You can look that up.
Ashwath

[quote="VARUNARASU"]and ofc- no books on philosophy. :D[/quote]
you know, our similarities in taste is growing everyday, mate. i HATE philosophy too  :mad:
i read the alchemist one day (no offence sanjith  :P) or rather TRIED reading it, and i nearly died!

i mean wtf, a guy turns himself to the wind?! even asimov couldn't have thought of that!  :rotfl:  :rotfl:

even i've tried reading some of paulo coelho's stuff. 'eleven minutes' was the name of the book i think. utter crap!!!!
and oh yes, i tried another book by robin sharma the monk who sold his ferrari due to recommendations by friends. another crappy book. whcih complete idiot wud be thick enuf to give away his ferrari? :o . and wisely since then i've never turned towards philosophy. :lol:

ROFL... Fanfiction = portkey :D

But as for good novels, there are loads.... Michael Crichton's books, for instance. Amazing.... And then there is Kahlil Gibran, for some serious reading. Try 'The Broken Wings' - its not philosophical, but its not your usual masala book either.

[quote="VARUNARASU"]i've read michael crichton's books. some of them are good, but i dint like one book of his-"state of fear". too much of crap without hard evidence. :cleaning: but timeline was good. :)
and as for kahlil gibran, never heard of him/her. will try to get that.[/quote]

Michael Crichton!! I am not reading him again.
After I suffered with his "Next"
It wasn really spun well , lot of misconnected stories

Now coming to good Novels
Have you read  Jeffrey Archer ? He's  my favorite.
Start from his first and the best- "Not a penny more Not a penny less"
"Shall we tell the president" is a gripping piece.
"Kane and Abel" is a masterpiece.
"The eleventh commandment" is also superb
I've read all of  his books except the 2 short story collections and his latest A prisoner of Birth :P

And read all Dan Brown titles if u havent already read dem
all of dem are good and worth their money. Dont forget the commotion that Da Vinci code created.

Robert Ludlum is another good bet. Some of his novels are just too good.
The Osterman weekend was a nice one. Tristan betrayal was also good.
Few of his books like Lazarus Vendetta and Moscow vector are laid with the same flavor.
Just two days bac I read this book called The Arctic incident. Nice one.

Now if you like action packed novels :roll:
ud like to go for Mathew Reilly and Tom Clancy and also Clive cussler.
Ice Station and Scarecrow by Mathew Reily are simply the best.
And I hate Robin cook.

As for the classic mystery tales Agatha Christie and Sherlock Holmes anyday

And as for the source:
Download dis software calld Microsoft Reader. :arrow:
And get Limewire. Lot of novels are available in Limewire in the Reader format.
Audiobooks are also an option but I had enough with the Scarlatti Inheritance.

Disclaimer: Im  rt now jobless
so the list may have grown too long and some u may find too old..

[quote="VARUNARASU"]i've read jeffrey archer, dan brown, robert ludlum, and robin cook.
i havent read any of tom clancy's books but i've played games based on his books(rainbow six). and tx for informing abt that micrososft reader.[/quote]

Rainbow Six :ninja:
I never quite got the feel of the game.  :(
The first few time I played I ended up killing my own team :rotfl:
but that was years ago

E-novels: http://www.gutenberg.org is good, I got a few from there... Project Gutenberg is quite good I guess..
I also remember way way back when I was on Orkut there was some community which had about a thousand links to e-book sites... sorry I'm being so vague though  :D
I just LOVE the Artemis Fowl collection and the Supernaturalist by the same dude...the Bartemius Trilogy is also really good...
I just got my hands on a 95 year old version of the Phantom of the Opera, waiting to read it!! Turn of the Screw is another book no-one seems to have heard of, which is quite good...but ofc Arthur Clarke's Space Odyssey Series and Asimov's stuff is THE BEST...Fantastic Voyage I and II are really good as well...
Mystery-No one beats Agatha Christie...but I love the courtroom drama in Perry Mason, by Earl Stanley Gardner...that's also a nice series to read...
DOES ANYONE READ DARREN SHAN???
I found it amazzingly bone-chilling and quite gory...good if you like that sorta stuff  :D
Oh yea..I had downloaded quite a few Wodehouse e-books sometime back...the not-so-famous ones, Psmith, Mulliner, the less well known short stories and early publications, they're quite good(And much refreshing compared to all the other books of his which I almost know by heart  :lol: ) and I have them in ZIP files...can always send it to you...

Do any of you have/know about the Sony Reader? I am trying to convince my parents to buy one(Amazon Kindle is pretty awful from what I've seen) but as usual there's no guarantee that there's proper after-sales in India...anyone know a good shop selling these here?

Michael Crichton's NEXT was a disappointment, but Timeline and Congo were master pieces, not to mention The Jurassic Park Series. Words that brought out the descriptions much better than the movies actually.

As for Jeffery Archer, he is a fantastic writer.. Prodigal Daughter is still one of my favorite books :)

And then there is also Sons of Fortune. And books by Sidney Sheldon. There is always a female protagonist in them though, but his novels are all pretty interesting reads over all. (But never read his autobiographical - The other side of me. Its irritating)

You could read Feynman's books. Trust me, its a lot better than fiction. "Surely you're joking, Mr.Feynman" for instance, its a brilliant book!!!! :)

Kahlil Gibran is an awesome writer. Beauty in words

:spam: how did ya double post? :o  i thought consecutive posts werent allowed within ten minutes.

i dint like 'sons of fortune' but i liked some of his other books not a 'penny less not a penny more', and 'kane and abel'. 'the prodigal daughter' was ok. dan brown is ok, but once you read a couple of his books, you can know what to expect from the rest. the same old -gud guy meets gud girl,falls in love and the bad guy turns out to be one of their friends.
john grisham, neone?
some of his books are realy nice. 'the partner' or 'the summons' for eg.
then there's alistair maclean-
i love his style of writing. i've only read a coupla his books and i've started growing on him already. 'fear is the key' and 'south by java head' are really good.
and sidney sheldon is ok. used to enjoy his books when i was younger but dont enjoy them now. and yes his autobiography was terrible. it is so  g o d d a m n boring. :sleeping:  but not all autobiographies are bad. you guys shud read'its not about the bike' by lance armstrong. thats a really nice book.

surely you are joking Mr. Feynman is awesome! simply awesome!
i have read a few Jeffrey Archer books, and also Fist of God by Frederick Forsyth. its like dan brown novels, a fictional story around a thing and characters which actually exist!
i heard someone mention bartimaeus trilogy! it is THE ultimate trilogy ever. the protagonist Bartimaeus has an incredible sense of humour!!
contrary to many here, i liked The Alchemist, but trashed The Monk Who Sold His Ferrari.

i know quite a lotta english purists exist here, but still asking, ru going to read Chetan Bhagat's next one, due to be released on May 8?

im not an english purist but im not gonna read chetan bhagats book.
does neone read erich segal. im sure shruthi must have read 'doctors' by him.
also does neone read classics? ofc the ultimate classic is 'gone with the wind' and thats one of my fav books. or ayn rand?
and ofc im sure every one over here has read the HP series ten times over.
btw, what is the bartimaeus trilogy. i've never even heard of that. :(

Well, I have no idea how I double posted. Really :D

And as for the Bartimaeus Trilogy, one of my friend read it and had all rave reviews for it. I havent read it though. I havent read LoTR yet... (*ducks stones*)

John Grisham is a great writer, I cant believe I forgot to mention him!!!! As for Chetan Bhagat, well, he is alright at the best. I didnt really like any of his books. And Robin Sharma, had a good idea for 'The Monk who sold his Ferrari" and it just, i dont know, backfired on him.

Atlas Shrugged is a great book too.

I have so many books that I want to mention here, but its just not striking me now. Will think about it, and post later :)

u can get a CD of ebooks delivered to ur doorstep free of cost.....go and add a request in projectgutenberg.com :D

i have already got two of their CDs......for time pass..  :P  :rotfl:

but most of the books are classic literature books  :P

i cant blive they give cds free of cost. they wud be broke in less than a month.

yes even i coulnt believe it...........but to my surprise i received the cds in less than a month :roll:  :)

why? they give free dvds also

esnips.com has a good collection of illegal copyrighted material

How come i forgot to post in this topic? :o I recently rediscovered a book that i had (and still love) loved more than most of the books i have read and it is shocking that i forgot that i had read the book and that i even had it :rotfl: Well its an excellent book by R.M. Ballantyne and its called The Coral Island. I read the abridged version by Puffin Classics and i dont think that the original one will be nice as it'll involve too much nature description and all so i think its better to read the abridged version :coolspeak:

lol. thats for kids da.i read that book when i was in fifth. forgotten most of it now. dunno whether i read the abridged or unabridged version. liked it then but surely cant read it now.

no its about 10th level ok! Not a kids story surely! The maturity shown by the people in the story is really excellent and to comprehend the finer aspects of the story isnt possible of a kid! :huh:

You guys done with ur exams right.....too bad Delhi guys lyk me.....still hav ONE MORE EXAM TO GO...........DCE! :mad:  :o

However, I also recommend Hitchhiker's Guide To The Galaxy by Douglas Noel Adams!!!!!!Its absolutely hilarious :rotfl:  ! I half way through book 1 coz of my entrance on 31st may

hmm, i expect a few stones to be thrown at me for saying this.. but anyway, i read the first 70 odd pages of Hitchhikers Guide, and found it extremely lame.  :huh:  It was not even interesting, let alone funny  :|
That's why i chucked the book after that  :P

All depends on one's tastes :P I saw one fellow in another forum say that he hates HP :rotfl: If he had dared to say it out in the open crowd, imagine his pitiable situation :rotfl:

[quote="madness"]How come i forgot to post in this topic? :o I recently rediscovered a book that i had (and still love) loved more than most of the books i have read and it is shocking that i forgot that i had read the book and that i even had it :rotfl: Well its an excellent book by R.M. Ballantyne and its called [b]The Coral Island[/b]. I read the abridged version by Puffin Classics and i dont think that the original one will be nice as it'll involve too much nature description and all so i think its better to read the abridged version :coolspeak:[/quote]

i had read that book when i was in seventh std!!  :D  :lol: