inequalities
algebra
polynomial
function
4-variable inequality

[quote="Kenny_cz"]Thanks a lot :).

BTW Reading in Chinese is tough, but I will find my way  :lol:[/quote]

can you tell me your E-mail address?

[quote="frankvista"]Let $ a \plus{} b \plus{} c \plus{} d \equal{} 0,$ Prove that

$ 7\left(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2\right)^2 \minus{} 12\left(a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4\right)\ge 0.$[/quote][quote="hedeng123"]$ 7\left(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2\right)^2 \minus{} 12\left(a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4\right)$

$ \equal{} 7\left[a^2 \plus{} b^2 \plus{} c^2 \plus{} ( \minus{} a \minus{} b \minus{} c)^2\right]^2 \minus{} 12\left[a^4 \plus{} b^4 \plus{} c^4 \plus{} ( \minus{} a \minus{} b \minus{} c)^4\right]$[/quote]$ \equal{} \frac {2}{3}\sum{\left[2a(a \plus{} b \plus{} c) \minus{} b^2 \minus{} c^2 \minus{} 4bc\right]^2}\geq0.$

[quote="Ji Chen"] $ \frac {2}{3}\sum{\left[2a(a \plus{} b \plus{} c) \minus{} b^2 \minus{} c^2 \minus{} 4bc\right]^2}\geq0.$[/quote]

Is this result of pure ingenuity or also application of some method? This reminds me of sums of squares which prove Vasc's ineq3 and these are (to me at least) very difficult to find. Even now, when I kinda remember what they look like  :P

What about Vasc's ineq $ (a^2\plus{}b^2\plus{}c^2)^2 \geqslant 3(ab\plus{}bc\plus{}ca)$ How can we get it's equivalent to $ \Sigma_{cyc}(a^2\minus{}2ab\plus{}bc\minus{}c^2\plus{}ca) \geqslant 0$ ?

[quote="Keehlzver"]What about Vasc's ineq $ (a^2 \plus{} b^2 \plus{} c^2)^2 \geqslant 3(ab \plus{} bc \plus{} ca)$ How can we get it's equivalent to $ \Sigma_{cyc}(a^2 \minus{} 2ab \plus{} bc \minus{} c^2 \plus{} ca) \geqslant 0$ ?[/quote]

yeah...

\[ 2(a^2 \plus{} b^2 \plus{} c^2)^2 \minus{} 6(a^3b \plus{} b^3c \plus{} c^3a) \equal{} \sum_{cyclic}{(a^2\minus{}2ab\plus{}bc\minus{}c^2\plus{}ca)^2}\]

is true, how to get this... :o

[quote="frankvista"]Let $ a \plus{} b \plus{} c \plus{} d \equal{} 0,$ Prove that

$ 7\left(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2\right)^2 \geq 12\left(a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4\right).$[/quote]sujun1993's Identity :

$ 7\left(\sum{a^2}\right)^2 \minus{} 12\sum{a^4} \plus{}\left(\sum{a}\right)\left[5\sum{a^3} \minus{} 5\sum{a^2(b \plus{} c \plus{} d) \plus{} 6\sum{bcd}}\right]$

$ \equal{} 2\left[(a\minus{}b)^2(c\minus{}d)^2\plus{}(a\minus{}c)^2(b\minus{}d)^2\plus{}(a\minus{}d)^2(b\minus{}c)^2\right]\geq0.$

See also : http://forum.cnool.net/topic_show.jsp?id=8218751&thesisid=494

Seriously, I still don't know how to find methods about completing squares of hard ineq  :( I know that it's about homogeneous function and look for the equality... but the above of Ji Chen's or Vasc's ineq is too hard for me. (The PDF file is very good method but I can't read it clearly in Chinese version, anyway, I will try it! :mad: )

Can anyone teach me about Vasc's ineq?
Keehlzver@hotmail.com

[quote="frankvista"]Suppose we know that
$ a \plus{} b \plus{} c \plus{} d \equal{} 0$,
Let
$ f \equal{} 7(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)^2 \minus{} 12(a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4)$.

a).Prove that $ f \ge 0$
b).Try to write $ f$ as the sum of squares.[/quote]
$ \equal{} 7(a^2 \plus{} b^2 \plus{} c^2 \plus{} ( \minus{} a \minus{} b \minus{} c)^2)^2 \minus{} 12(a^4 \plus{} b^4 \plus{} c^4 \plus{} ( \minus{} a \minus{} b \minus{} c)^4)$
$ \equal{} \frac {(2\sum{a^3} \minus{} \sum{a^2(b \plus{} c) } \minus{} 12abc)^2 \plus{} 3(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{\sum(a^2 \minus{} ab)}$

[quote="hedeng123"][quote="frankvista"]Suppose we know that
$ a \plus{} b \plus{} c \plus{} d \equal{} 0$,
Let
$ f \equal{} 7(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)^2 \minus{} 12(a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4)$.

a).Prove that $ f \ge 0$
b).Try to write $ f$ as the sum of squares.[/quote]

The inequality holds!

[size=150]while[/size] $ a,b,c\geq 0$and  $ a,b,c\le 0$

$ d \equal{} \minus{} a \minus{} b \minus{} c$

$ f \equal{} 7(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)^2 \minus{} 12(a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4)$

$ \equal{} 7(a^2 \plus{} b^2 \plus{} c^2 \plus{} ( \minus{} a \minus{} b \minus{} c)^2)^2 \minus{} 12(a^4 \plus{} b^4 \plus{} c^4 \plus{} ( \minus{} a \minus{} b \minus{} c)^4)\ge0$

$ \Longleftrightarrow 4\sum a^2(a \minus{} b)(a \minus{} c) \plus{} 12\sum a(b \plus{} c)(a \minus{} b)(a \minus{} c) \plus{} 36\sum bc(a \minus{} b)(a \minus{} c)\geq 0$

obvious true.[/quote]

$ 7(a^2 \plus{} b^2 \plus{} c^2 \plus{} ( \minus{} a \minus{} b \minus{} c)^2)^2 \minus{} 12(a^4 \plus{} b^4 \plus{} c^4 \plus{} ( \minus{} a \minus{} b \minus{} c)^4)$

$ \equal{}(2a^2\plus{}12ab\plus{}2b^2\plus{}16c^2)(a\minus{}b)^2\plus{}(2b^2\plus{}12bc\plus{}2c^2\plus{}16a^2)(b\minus{}c)^2$ $ \plus{}(2c^2\plus{}12ac\plus{}2a^2\plus{}16b^2)(c\minus{}a)^2\geq 0$

or $ \equal{}(2a^2\minus{}4ab\plus{}16ac\plus{}2b^2\plus{}16bc)(a\minus{}b)^2\plus{}(2b^2\minus{}4bc\plus{}16ab\plus{}2c^2\plus{}16ac)(b\minus{}c)^2$ $ \plus{}(2c^2\minus{}4ac\plus{}16bc\plus{}2a^2\plus{}16ab)(c\minus{}a)^2\geq 0$

or $ \equal{}(2a^2\plus{}12ac\plus{}2b^2\plus{}12bc\plus{}4c^2)(a\minus{}b)^2\plus{}(2b^2\plus{}12ab\plus{}2c^2\plus{}12ac\plus{}4a^2)(b\minus{}c)^2$ $ \plus{}(2c^2\plus{}12bc\plus{}2a^2\plus{}12ab\plus{}4b^2)(c\minus{}a)^2\geq 0$

 :wink:

[quote="daigiaga1994"][quote="hedeng123"][quote="Kenny_cz"]Is there some theory about writing homogenous polynomial inequalities in this say "Schur" form? I know there is an algorithm which transforms expressions in SOS form, but I couldnt find anything like that for this form.[/quote]

Yes ,This is called the Schur decomposition

Look at this,This is a relatively simple form of Schur decomposition,

In China in the form of the general has done a lot of work is Shengli Chen and Fangjian Huang.[/quote]
hi hedeng123,
Can you translate your file into English, please?
I can't understand Chinese :([/quote]

I'm sorry my English is poor,Translation is very difficult for me,I am so sorry.

 :lol:

[quote="daigiaga1994"][quote="hedeng123"]

I'm sorry my English is poor,Translation is very difficult for me,I am so sorry.

 :lol:[/quote]
It isn't really lucky, but by the way, I want to thank you. I hope that on certain day, I can understand your file :wink:[/quote]

That document is talking about [size=150]Schur decomposition [/size],Any of a ternary homogeneous symmetric polynomial,Can always be decomposed into a linear combination of several schur-base,These schur-based non-negative,We just need to verify that the coefficients to be non-negative,Document describes how to decomposition of a polynomial degree of 3-6.

In fact, we can use computers and the undetermined coefficient method decomposition of a high degree of polynomial.

I am sorry,my english is very poor.

 :blush:

[size=150]A paper about Schur Partition (Schur Decomposition) in English was published in the Proceedings of ISSAC 2005. You can also download from

http://www.irgoc.org/viewtopic.php?f=10&t=378&sid=695e51bc0f6a621cf0d24d12bca5e2da[/size]

@hedeng123: Yes, I understand you, dear friend  :lol:

@Guest: Thank you very much. It's very helpful  :)


linear algebra
matrix
linear algebra unsolved

Let $A$ be an $n\times n$ matrix, and define $B=e^A$ as usual. Find the determinant of $B$.

It's a classical result : det (e^A)=e^(tr(A)).
Try trigonalising.


calculus
integration
trigonometry
real analysis
real analysis unsolved

Determine:
1. $ \int\frac{xe^{\arctan x}}{\left(1\plus{}x^2\right)^{\frac{3}{2}}}dx$
2. $ \int {\sqrt {\frac{{1 \minus{} \sqrt x }}{{1 \plus{} \sqrt x }}} } dx$
3. $ \int {\sqrt {\frac{x}{{1 \minus{} x\sqrt x }}} } dx$

Hi :)

for 1)

$ Let$   $ t\equal{}arctan(x)$

than $ \int \frac{xe^{arctan(x)}}{(1\plus{}x^2)\sqrt{1\plus{}x^2}}dx \equal{} \int sin(t)e^t dt \equal{} \frac{sin(t)\minus{}cos(t)}{2}e^t \plus{} cte$
...
Ps i calculed that quickly than i'm not sure..
________________________________________
lahoucine

hello, for 2) you can set ${ t=\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}}$ then you will get the following integral
$ 8\int\frac{t^2(t^2-1)}{(t^2+1)^3}\,dt$
Sonnhard.

hello for 3) setting $ t\equal{}1\minus{}x^{\frac{3}{2}}$ you will get the integral $ \frac{2}{3}\int\frac{1}{\sqrt{t}}\,dt$
Sonnhard.

For 3
 Let $ y = x^{\frac{3}{2}}$ then $ \sqrt{x}$.dx =$ \frac{2}{3}$.dy and the integral becomes

$ \int\frac{1}{\sqrt{1-y})$.dy =$ -2.(1-y)^{\frac{1}{2}}$ +C etc

Thanks for your solutions. Please help me to determine these integrals.

4. $ \int\frac{dx}{1\plus{}a^x}$
5. $ \int\frac{x\plus{}4}{\sqrt{x^2\plus{}2x\minus{}3}}dx$
6. $ \int\frac{dx}{\sqrt 2\plus{}\sin x\minus{}\cos x}$
7. $ \int\cos^6xdx$
8. $ \int\frac{dx}{(x\plus{}1)\sqrt{x^2\plus{}2x\plus{}2}}$

Hi :) :

you're welcome

for 1)

$ Let$  $ a\in \mathbb{R}_ \plus{} ^* \minus{} {1}$

$ \int \frac {dx}{1 \plus{} a^x} \equal{} \frac { \minus{} 1}{ln(a)} \int \frac { \minus{} ln(a)a^{ \minus{} x}}{a^{ \minus{} x} \plus{} 1}dx \equal{} \minus{} \frac {ln|a^{ \minus{} x} \plus{} 1|}{ln(a)}\plus{}cte$

to be continued ...
_________________________________
lahoucine

5.$ \int\frac{x\plus{}4}{\sqrt{x^{2}\plus{}2x\minus{}3}}dx\equal{}\frac{1}{4}\int\frac{5(x\plus{}3)\minus{}(x\minus{}1)}{\sqrt{(x\minus{}1)(x\plus{}3)}}dx$

6.$ u\equal{}\tan\frac{x}{2}$

7.$ \cos^{2}x\equal{}\frac{1\plus{}\cos 2x}{2}$

8.$ x\plus{}1\equal{}\tan u$

[quote="Timestopper_STG"]5.$ \int\frac {x \plus{} 4}{\sqrt {x^{2} \plus{} 2x \minus{} 3}}dx \equal{} \frac {1}{4}\int\frac {5(x \plus{} 3) \minus{} (x \minus{} 1)}{\sqrt {(x \minus{} 1)(x \plus{} 3)}}dx$
...[/quote]

Hi Timestopper :

i'm not agree with you because we can't simplified by $ \frac {A(x) \minus{} B(x)}{\sqrt {A(x)B(x)}} \equal{} \sqrt{B(x)}\minus{}\sqrt{A(x)}$ that's is not correct because in our exemple we have$ Df \equal{} \mathbb{R} \minus{} \{[ \minus{} 3,1]\}$ than we can't separ $ A(x) \ and \ B(x)$

$ \int \frac {x \plus{} 4}{\sqrt {x^2 \plus{} 2x \minus{} 3}}dx \equal{} \int \frac {2x \plus{} 2}{2 \sqrt {x^2 \plus{} 2x \minus{} 3}}dx \plus{} 3 \int \frac {dx}{\sqrt {(x \minus{} 1)(x \plus{} 3)}}$

$ \equal{} \sqrt {x^2 \plus{} 2x \minus{} 3} \plus{} 3 \int \frac {dt}{\sqrt {t^2 \minus{} 1}}$

with in 2nd integral we put $ x \equal{} 2t \minus{} 1$

than the second integral are very easy to calcul it ....

hello, to 4) we have $ \int\frac{1}{1\plus{}a^x}\,dx\equal{}\int1\,dx\minus{}\int\frac{a^x}{1\plus{}a^x}\,dx\equal{}x\minus{}\frac{\ln(1\plus{}a^x)}{\ln(a)}\plus{}C$
Sonnhard.

Find value of this integral: $ \int\limits_0^1 {\frac{{dx}}{{(1 \plus{} x^n )\sqrt[n]{{1 \plus{} x^n }}}}}$, với $ n$, with $ n$ is natural number. ;)

[quote="Dr Sonnhard Graubner"]hello, to 4) we have $ \int\frac {1}{1 \plus{} a^x}\,dx \equal{} \int1\,dx \minus{} \int\frac {a^x}{1 \plus{} a^x}\,dx \equal{} x \minus{} \frac {\ln(1 \plus{} a^x)}{\ln(a)} \plus{} C$
Sonnhard.[/quote]

that's the same result of me :)

hello, for 5) setting $ \sqrt{x^2\plus{}2x\minus{}3}\equal{}t\plus{}x$ we get $ x\equal{}\frac{1}{2}\frac{3\minus{}t^2}{1\minus{}t}$ further we have $ dx\equal{}\minus{}\frac{(\minus{}3\plus{}t)(1\plus{}t)}{2(t\minus{}1)^2}\,dt$, after an ugly calculation we have
$ \int \frac{x\plus{}4}{\sqrt{x^2\plus{}2x\minus{}3}}\,dx\equal{}\int\frac{11\minus{}8t\plus{}t^2}{8(t\minus{}1)^4}\,dt$.
Sonnhard.

hello, for 6) setting $ z\equal{}\tan(\frac{x}{2})$ and $ \sin(x)\equal{}\frac{2z}{1\plus{}z^2}$, $ \cos(x)\equal{}\frac{1\minus{}z^2}{1\plus{}z^2}$ so we get the following integral $ \int\frac{2}{\minus{}1\plus{}\sqrt{2}\plus{}2z\plus{}z^2(1\plus{}\sqrt{2})}\,dz$
Sonnhard.

hello, for 7) use the following formula $ \cos(x)^6\equal{}\frac{1}{32}(10\plus{}15\cos(2x)\plus{}6\cos(4x)\plus{}\cos(6x))$.
Sonnhard.

hello, for 8) use the substitution $ \sqrt{x^2\plus{}2x\plus{}2}\equal{}t\plus{}x$
Sonnhard.


The average age of the $ 6$ people in Room A is $ 40$.  The average age of the $ 4$ people in Room B is $ 25$.  If the two groups were combined, what is the average age of all the people?

$ \textbf{(A)}\ 32.5\qquad
\textbf{(B)}\ 33\qquad
\textbf{(C)}\ 33.5\qquad
\textbf{(D)}\ 34\qquad
\textbf{(E)}\ 35$

[hide]

The total sum of the ages in Room A is $ 6 \times 40 \equal{} 240$, and the total sum of the ages in Room B is $ 4 \times 25 \equal{} 100$.  Total sum of all the ages is $ 240 \plus{} 100 \equal{} 340$, and with $ 10$ people, the total average is $ \boxed{\textbf{(D)}\ 34}$[/hide]


[color=darkred][b][u]Show the inequality holds for a,b,c positive real numbers such that abc=1.[/u][/b][/color]

$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{b+a}}>1$

[i]This one is also the problem was idealised in the marathon.[/i]

Davron

Davron, very easy to prove that

$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{b+a}}\geq\frac{3}{\sqrt2}$ ;)

[i][u]ahhh[/u][/i] [i]Yes it is i forgot that statement.[/i] [color=darkred]Thank you Arqady !!![/color]

[i]Anyway my proof is:[/i]    $\frac{a}{\sqrt{b+c}}>\frac{a}{\sqrt{a+b+c}}$ Then summing the another three terms we get the desired value.

[i][color=blue]My proof is coming from Darij's one for the CMO's problem as i remember[/color].[/i]
[u]Also my teamate Maruf has a goos proof for this inequality. Maybe he will write it for us.[/u]

Davron Latipov

Maybe you should not bother posting problems that are this easy in the first place.

Yes you are right they are really easy but they give strategy for us. Anyway they are really interesting.

Davron

Yes you are right they are really easy but they give strategy for us. Anyway they are really interesting.
Please try this one also by me.http://www.mathlinks.ro/Forum/viewtopic.php?t=82911

Davron Latipov

you're right Davron  :)     they give strategy for us

I don't really care what you think, actually; if you want to post problems this easy, please do it in the Pre-Olympiad section.

And certainly don't link me to posts in the algebra section here; if I wanted to do algebra problems I would be looking in the algebra section, not the inequality section.


calculus
integration
trigonometry
calculus computations

can anyone help me find this integral:
\[ \int \sqrt{r^2-2pr\cos(t)+p^2}dt \]

Hello. Begin with the substitution, u = arccos(t), and manipulate the resulting elliptic integral to express it in terms of the three basic elliptic integrals.

Is this actually a definite integral? In particular, is it the integral "all the way around"? By that I would mean

$\int_0^{2\pi}\sqrt{r^2-2p r\cos t+p^2}\,dt$

If that is the case, there are some complex variable or Fourier series techniques that could be brought to bear.

well actually this expresses the length $\int \sqrt{(dx)^2+(dy)^2}$ of the general cycloid:

$x(t)= rt-p\sin t$
$y(t)= r-p\cos t$

i dont know if that helps any ...

ps: i wrote a paper about that and now only could find the length if $r=p$ ...

Hello again. For a scenario as general as yours, it is probably most convenient to leave the integral in definite-integral form.

If you do make that the integral from $0$ to $2\pi$ - in other words, the arclength on one complete arch of the curve - the integral, under a suitable change of variables, turns out to exactly the same as the integral for the arclength of an ellipse. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=55166]this recent topic[/url] for more on that subject.


Έστω $ p\geq 3$ πρώτος και $ a_i$, $ 1\leq i \leq p \minus{} 2$ ακολουθία φυσικών αριθμών για την οποία ισχύει οτι το $ p$ δεν διαρει ουτε το $ a_k$ ουτε το $ a_k^k \minus{} 1$ για κανενα $ k$ οπου, $ 1\leq k \leq p \minus{} 2$.

νδο το γινομενο καποιων ορων της ακολουθιας ειναι ισουπολοιπο με $ n(modp)$ για καθε $ n$, οπου $ n$ φυσικος με $ 2\leq n \leq p \minus{} 1$
 :wink:

Δινω hint
[hide]Κατασκευαστε επαγωγικα μια ακολουθια $ b_n$, $ p\minus{}1$ ορων διαφορων $ \mod p$ οπου καθε ορος θα ειναι ειτε ισος με $ 1$ ειτε ισος με το γινομενο καποιων ορων της ακολουθιας $ a_n$[/hide]

Σε hide γιατι καποιοι ισως να την προσπαθουνε ακομα.

[hide]Θεωρω τα συνολα $ \{a_1, a_2\}, \{a_1, a_1a_2\}$, τοτε ενα απο αυτα θα αποτελειτε απο 2 διαφορετικα μεταξυ τους στοιχεια και διαφορα του $ 1 (modp)$, πραγματι αν τα $ a_1, a_2$ ειναι διαφορετικα $ (modp)$ τοτε επηδη ολα τα $ a_i$ ειναι $ \neq 1 (modp)$ το πρωτο συνολο εχει την παραπανω ιδιοτητα, αν παλι $ a_1 \equal{} a_2 (modp)$ τοτε απο την υποθεση ειναι $ a_1a_2 \equal{} {a_2}^2 \neq 1 (modp)$ και επισεις $ a_2 \neq 1 (modp) \Rightarrow a_1a_2 \neq a_1 (modp)$ οποτε το 2ο συνολο εχει την παραπανω ιδιοτητα.

Θεωρω τωρα οτι, χρησημοποιοντας μονο τους πρωτους $ k$ ορους της ακολουθιας $ a_i$ σε εκθετες < 2, εχω κατασκευασει ενα συνολο $ \{b_1, b_2, ... b_k\}, 2 \leq k \leq p \minus{} 3$ του οποιου τα στοιχεια ειναι διαφορετικα ανα δυο και διαφορα του $ 1 (modp)$. Θα κατασκευασω τωρα ενα συνολο με $ k \plus{} 1$ στοιχεια διαφορετικα ανα δυο και διαφορα του $ 1 (modp)$, χρησημοποιοντας μονο τους πρωτους $ k \plus{} 1$ ορους της ακολουθιας $ a_i$ σε εκθετες < 2 ως εξης:
Θεωρω το συνολο $ \{1, b_1, b_2, ... b_k\}$ το οποιο περιεχει $ k \plus{} 1$ διαφορετικα ανα δυο στοιχεια $ (modp)$, και πολλαπλασιαζοντας καθε στοιχειο αυτου του συνολου με $ a_{k \plus{} 1}$ λαμβανω το συνολο $ \{a_{k \plus{} 1}, a_{k \plus{} 1}b_1, ... a_{k \plus{} 1}b_k\}$ το οποιο επισεις περιεχει $ k \plus{} 1$ διαφορετικα ανα δυο στοιχεια $ (modp)$, αν τα δυο αυτα συνολα ταυτιζονταν τοτε τα γινομενα των στοιχειων τους θα ταυτιζονταν επισεις και θα ειχαμε $ {a_{k \plus{} 1}}^{k \plus{} 1} \equal{} 1 (modp)$ το οποιο ειναι ατοπο, αρα στο δευτερο συνολο υπαρχει στοιχειο $ m$ που δεν ανηκει στο πρωτο και επομενως ειναι διαφορετικο απο $ 1, b_1, b_2, ... b_k (modp)$, ετσι το συνολο $ \{m, b_1, b_2, ..., b_k\} (modp)$ περιεχει $ k \plus{} 1$ διαφορετικα ανα 2 και διαφορα του $ 1 (modp)$ στοιχεια ενω στην κατασκευη του χρησημοποιουνται μονο οι $ k \plus{} 1$ πρωτοι οροι της ακολουθιας $ a_i$ σε εκθετες < 2. Επαγωγικα λοιπον μπορουμε να κατασκευασουμε συνολο με $ p \minus{} 2$ διαφορετικα ανα δυο και διαφορα του $ 1 (modp)$ στοιχεια, χρησημοποιοντας μονο τους αριθμους $ a_1, a_2, ... a_{p \minus{} 2}$ σε εκθετες < 2, οποτε τελειωσαμε.[/hide]

Νικο μπραβο! Αυτη ειναι η ιδεα!  :)   :wink:
[hide]Mπορεις βεβαια να αρχισεις με το $ \{1,a_1\}$ για να μην παει με περιπτωσεις το αρχικο σταδιο αλλα μικρη η διαφορα![/hide]


Wat's the probability that wen 2 randomly selected 2-digit whole #'s are multiplied, the result is between 100 and 200, exclusive. Express answer as common fraction

[quote="espark52"]:bye:    :bye:[/quote]
What does that mean?

[quote="superwizard188"]hmm... whole numbers can be positive or negative...

so shouldn't the denominator be 180*179 or 180*180, depending on if you can choose the same number...[/quote]
Um whole numbers can't be negative.

The definition of whole number includes its being positive.

I learned that whole numbers are nonnegative integers, while natural numbers are positive integers. But it's debatable.

[quote="daermon"]but 0 isn't "whole"  :roll:[/quote]
Yes, It is :roll:.

[quote="espark52"][quote="meenamathgirl"]i think the definition of a whole number is a non-negative integer and a natural number is a positive integer.[/quote]

I thought whole numbers were: {0,1,2,3...}[/quote]
your definitions are the same and correct.

can you use the same number twice?
also, is $(13,14)$ the same as $(14,13)$?

Oddly, no one pointed out that It should've been $90(90)$ for the denominator. No one else has attempted this one either. Also, to daermon if you count them differently then the denominator is double of what it is if they are the same. The answer is then same either way.

[quote="pascal_1623"][quote="bpms"][hide]They both neem to be small. If 10 is a number then it could be 10-19. If 11 is selected it could be 10-18. If 12 is selected it could be 10-16. If 13 is selected it could be 10-15. If 14 then 10-14. If 15 then 10-13. If 16 then 10-12. If 17 then 10-11. If 18 then 10-11. If 19 then 10.
there are $90 \cdot 90$choices.
$\frac{49}{8100}$
[/hide][/quote]

Umm... I think there are $91\cdot91$ choices, and if 10 is selected, then 20 works too.[/quote]
He said exclusive.
Also 99-10+1=90.

[quote="bpms"][quote="Ignite168"]Do the two numbers have to be distinct?
[hide="Not distinct"]$\boxed{\frac{12}{2025}}$ 10 works with 11 through 19. 11 works with 10 through 18. 12 works with 10 through 16. 13 works with 10 through 15. 14 works with 10 through 14. 15 works with 10 through 13. 16 works with 10 through 12. 17 works with 10 and 11. 18 works with 10 and 11. 19 works with 10.[/hide]
[hide="Distinct"]$\boxed{\frac{22}{4005}}$ There were 48 ways for non distinct, 4 of those were squares. (11, 12, 13, 14) So discount those 4.[/hide][/quote]
Ignite is right. I counted 10x10 for my solution which threw it off.[/quote]

no, they don't have to be distinct


Find the roots $r_i$ of the polynomial \[p(x) = x^n + n \cdot x^{n-1} + a_2 \cdot x^{n-2} + \ldots + a_n\] satisfying $\sum^n_{k=1}r^{16}_k = n.$

[quote="Myth"]Maybe you lost the condition that all $r_i$ are real? :?[/quote]

It is not specified. Maybe Pascal knows.  :D So as long you state what you assume, you can assume whatever you want...  ;)


linear algebra
matrix
algebra
polynomial
complex numbers
linear algebra unsolved

$ A\in M_{2n\plus{}1} (R)$ s.t $ A^3\equal{}A^2\plus{}xI_{2n\plus{}1}$.Prove that if $ x \le \minus{}\frac{4}{3}$then $ detA \le 0$

All complex eigenvalues of $ A$ satisfy the equation $ y^3\minus{}y^2\minus{}x\equal{}0$ and $ \det A$ is the product of those complex eigenvalues.

Let's look at $ f(y)\equal{}y^3\minus{}y^2\minus{}x$ on the real line. It has a local maximum at $ y\equal{}0$ and a local minimum at $ y\equal{}\frac23.$

The local minimum, $ f\left(\frac23\right)\equal{}\minus{}\frac4{27}\minus{}x$ is positive if $ x\le\minus{}\frac43.$

Note that $ x\le\minus{}\frac43$ is overkill, and we could have said $ x<\minus{}\frac4{27}.$

So this equation has one real root $ a<0$ and a pair of complex conjugate roots $ b\pm ci.$

For this real matrix, all of the eigenvalues are either real or come in complex conjugate pairs. Hence $ a$ appears $ j$ times as a root of the characteristic polynomial and the pair $ b\pm ci$ appears $ k$ times. But then the total number of roots is $ j\plus{}2k\equal{}2n\plus{}1 .$ We conclude that $ j$ is odd.

As I said, the determinant is the product of the complex eigenvalues with their algebraic multiplicities:

$ \det A \equal{} a^j[(b\plus{}ci)(b\minus{}ci)]^k\equal{}a^j(b^2\plus{}c^2)^k<0$ since $ j$ is odd.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Problem: Let G be a group containing a normal subgroup H such that H is isomorphic to D8, the dihedral group of order 8. Show that |Z(G)|>1.
Here, Z(G) is the center of G.

My query:
H has center order 2, but what is it about D8 that allows this to carry through to the whole group?

(Note: p-group theory is not assumed material)

Thanks

Can you think of how a normal subgroup of order 2 and the center are related? Do you see how this solves your problem?

Steve

[quote="jamesc"]A normal subgroup of order 2 of a group K lies in the center of K. Here, we only know that H has a normal subgroup J order 2 (it has center order 2 anyhow), but I don't see what information this gives us about the normality of J in G.[/quote]

If you conjugate J by an element of G, it must still lie in H, since H is normal.  However conjugation by an element of g is an automorphism of H, and where must an automorphism of H send Z(H)?


This is the only problem I can't solve in my contest.
   a_1, a_2, ..., a_n are positive integers and (a_i <> a_ j) satisfying the conditions:
                \sum (1/a_i)=1 and Max{a_i}= 2p with p is a prime.
Find all the numbers a_1, a_2, ..., a_n !!!
   Please help me !!!

Wlog, we may assume that a_1 < a_2 < ... a_n = 2p.
Note that a_i > 1 for each i and that n \leq 2p-1.

If we reduce the numbers 1/a_i to a common denominator d, since a_n = 2p, we deduce that p divides d, so that there must be another one of the numbers a_i which has to be a multiple of p, that leads to a_q = p for some q.
It follows that no other of the a_i is divisible by p.

Now, let's reduce the fractions to the common denominator d = prod{i =/= q} a_i, and let P_k = d/a_k.
Thus :
1 = (P_1 + ... + P_(q-1) + P_(q+1) + ... + P_(n-1) + 3P_n)/d.

Note that p divides d, which divides the numerator, and that p divides each of the numbers P_i for i = 1,...,q-1, q+1,...,n-1, and p does not divides P_n.
Thus p divides 3, so that p = 3.

From above, we have a_n = 6 and a_q = 3 for some q. It follows that the sum of the other 1/a_i is 1/2. It leads easily to the only solution :
a_1 = 2, a_2 = 3 and a_3 = 6.

Pierre.

Thanks for your help.
Yesterday I found a sol which is very similiar to yours. So it isn't a difficult problem.


[quote="joejia"]Twenty-five percent of the entrants in a marathon did not finish the race. If 375 finished, how many entrants were there?[/quote]

[hide]
If 375 is 75%, then you divide it by three to get 25%. 375/3=125. 375+125=500.[/hide]

[hide]375/.75=500 runers. [/hide]

[quote="shobber"][quote="math92"]we have the same avatar[/quote]
Isn't that an embarassment?[/quote]

yeah it is  :rotfl:

[hide][quote="joejia"]Twenty-five percent of the entrants in a marathon did not finish the race. If 375 finished, how many entrants were there?[/quote]

\[\text{I'll set up a proportion}\]
\[\frac{375}{x} = \frac{75}{100}\]
\[75x = 37500\]
\[x=\frac{37500}{75}\]
\[x=500\]
\[\therefore  \text{There was 500 runners total.}\][/hide]

[quote="shobber"][quote="math92"]we have the same avatar[/quote]
Isn't that an embarassment?[/quote]
i don't really care


In $\Delta ABC$, points $E$ and $F$ divide sides $AC$ and $AB$ respectively so that $AE/EC=4$ and $AF/FB=1$. Suppose $D$ is a point on side $BC$, let $G$ be the intersection of $EF$ and $AD$, and suppose $D$ is situated so that $AG/GD=3/2$. FInd the ratio $BD/DC$.

[hide="Solution"]For some $\lambda\in(0,1)$ we have $\overrightarrow{AG}=\lambda\overrightarrow{AE}+(1-\lambda)\overrightarrow{AF}={4\lambda\over 5}\overrightarrow{AC}+{1-\lambda\over 2}\overrightarrow{AB}$

For some $\mu\in(0,1)$ we have $\overrightarrow{AD}=\mu\overrightarrow{AC}+(1-\mu)\overrightarrow{AB}$

Since $\overrightarrow{AG}={3\over 5}\overrightarrow{AD}$, we get

${4\lambda\over 5}\overrightarrow{AC}+{1-\lambda\over 2}\overrightarrow{AB}={3\mu\over 5}\overrightarrow{AC}+{3(1-\mu)\over 5}\overrightarrow{AB}$

Vectors $\overrightarrow{AC}$ and $\overrightarrow{AB}$ being linearly independent, that yields the system

${4\lambda\over 5}={3\mu\over 5}\land{1-\lambda\over 2}={3(1-\mu)\over 5}$

whose solution is $\lambda={1\over 3},\mu={4\over 9}$

Now ${BD\over DC}={\mu\over 1-\mu}={4\over 5}$[/hide]

I have a different, albeit more messy, solution.

[hide]
I will use capital letters for vectors. By our definitions,
$F = \frac{A+B}{2}\\ E = \frac{4}{5}C+\frac{1}{5}A \\ D = \frac{a}{a+b}C+\frac{b}{a+b}B \\$
Now we find the point of intersection $G$:
$G-F = c(E-F) \\ G-\frac{A+B}{2}= c\left(\frac{4}{5}C+\frac{1}{5}A-\frac{A+B}{2}\right) \\ G = \frac{1}{2}A+\frac{1}{2}B+c\left(\frac{4}{5}C-\frac{3}{10}A-\frac{1}{2}B \right) = \frac{4c}{5}C+\frac{5-3c}{10}A+\frac{1-c}{2}B$
Also,
$G = \frac{3}{5}D+\frac{2}{5}A = \frac{2}{5}A+\frac{3b}{5(a+b)}B+\frac{3a}{5(a+b)}C$
Comparing "coefficients":
$\frac{5-3c}{4}= \frac{2}{5}$
Thus $c=\frac{1}{3}$. Then substitute to find that
$\frac{4}{15}= \frac{3a}{5(a+b)}\\ \frac{15}{4}= \frac{5}{3}+\frac{5b}{3a}\\ a = \frac{4}{5}b$[/hide]


The sequence

$1,2,1,2,2,1,2,2,2,1,2,2,2,2,1,2,2,2,2,2,1,2,...$

consists of 1's separated by blocks of 2's with $n$ 2's in the $n$th block. The sum of the first 1234 terms of this sequence is

$\text{(A)} \ 1996 \qquad \text{(B)} \ 2419 \qquad \text{(C)} \ 2429 \qquad \text{(D)} \ 2439 \qquad \text{(E)} \ 2449$

[hide="Answer"]Each set beginning with a $1$ increases in number by one each time, so we have $\frac{49(50)}{2}=1225$, so we add one set to include $1234$ numbers, but subtract one because the first is not included, so we must be on the $49^{th}$ such set. Therefore, we have 49 $1$'s and 1185 $2$'s, so our sum is $49+2\cdot 1185=2419\Rightarrow \boxed{B}$.[/hide]

Can someone explain why my reasoning is wrong??

Since the each $n$ ones include $\frac{n(n+1)}{2}$ terms, we see that $n$ must be $49$ since we don't want to go over $1234$, so we have $49$ ones and ${\frac{49(50)}{2}}-49 \Longrightarrow 1176$ twos, so their sum is $2401$ :?

EDIT: Found my mistake, forgot to add the $9$ twos left over which would make $2419$ :blush:

[hide]We need to find the point at which the number of terms exceeds 1234. After block 1, there are 2 terms. After block 2, there are 5 terms. After block 3, there are 9 terms. etc...

Let $f(n)$ be the number of terms after block $n$, so $f(n+1)-f(n)=n+2$. Thus $f(n+1)=2+(1+2) + (2+2) + \cdots + (n+2) = 2(n+1) + n(n+1)/2=(n+1)(2+n/2)$, substitute $n-1$ to find $f(n)$.

If $f(n)=(n)(2+(n-1)/2) > 1234$, then $n > 48$. After 48 blocks, there are 1224 terms. The sum of each block is $1+2n$. Thus we need the sum \[ \sum_{n=1}^{48} (1+2n) = 48 + 2\sum_{n=1}^{48} n = 2400, \] and the sum of the 10 remaining terms is $1+2(9) = 19$, so the total is $2419$.[/hide]


A man drives 150 miles to the seashore in 3 hours and 20 minutes. He returns from the shore to the starting point in 4 hours and 10 minutes. Let  be the average rate for the entire trip. Then the average rate for the trip going exceeds , in miles per hour, by:
(A) 5
(B) 4.5
(C) 4
(D) 2
(E) 1

The average rate going there is 45, the average rate coming back is 36. The total average for the entire trip is 40, so the answer is $5$.


inequalities
quadratics
inequalities open

Let $x,y,z$ be real nonnegative numbers such that
\[ \frac1{x+1}+\frac1{y+1}+\frac1{z+1}=2.  \]
Prove that
\[ \frac1{xy+2}+\frac1{yz+2}+\frac1{zx+2}\geq\frac43.  \]

hey buddy, your inequality is wrong.......
you multiply the inequality by (xy+2)(yz+2)(zx+2).
then you get

4*(xyz)^2+5xyz(x+y+z)+4(xy+yz+zx)-4<=0
we use the condition in form : 2xyz +xy+yz+zx=1 and then we get
4*(xyz)^2+5xyz(x+y+z)-8xyz<=0
it's obvious that this works for x=0 or y=0 or z=0. so we assume that x,y,z are nonzero, and dividing by xyz we get
4xyz+5(x+y+z)<=8, which obvious doesn't work for enough large x,y or z

it doesn't work for $x=y=\frac15$, $z=2$

Ok, expanding the given we get $2xyz+xy+xz+yz=1$, and by AMGM we have $1=2xyz+xy+xz+yz\geq2xyz+3(xyz)^{\frac{2}{3}}$, so solving that inequality in $(xyz)^{\frac{1}{3}}$ we get $(\sqrt[3]{xyz}+1)^2(2\sqrt[3]{xyz}-1)\leq0$, and since squares are non negative than by our given we have $\sqrt[3]{xyz}\leq\frac{1}{2}$.

Next, we multiply through our inequality by $3(xz+2)(yz+2)(xy+2)$ and moving everything together we have $0\geq4(xyz)^2+5xyz(x+y+z)+4(xy+xz+yz)-4$. Since $xy+xz+yz=1-2xyz$ by our given, we have $0\geq4(xyz)^2 +5xyz(x+y+z) - 8xyz$. If any of the variables are 0, then the inequality is an equality. Otherwise, assume they are all positive, and we can divide by $xyz$, yielding $0\geq4xyz+5(x+y+z)-8\geq4yz+15\sqrt[3]{xyz}-8$ by AMGM again. Using the left and right part of that inequality chain, we can factor in $\sqrt[3]{xyz}$ to get

\[ (2\sqrt[3]{xyz}-1)(2(xyz)^{\frac{2}{3}}+\sqrt{xyz}+8)\leq 0  \]

The quadratic is unfactorable over reals, so it's always positive, thus our inequality is true when $\sqrt[3]{xyz}\leq\frac{1}{2}$, which is true by our given.

don't you think you're using your last AMGM on the wrong side?

and also there can not be a possible proof because there is a counterexample as adalbert has written above

The GM is on the less than part, I believe it's on the wrong side

though his counterexample is right, so I dunno

[quote="pvthuan"]What a pity, it is not true. It is curious to know that the equality occurs when $a=b=c=\frac12$ or $a=0$, $b=c=1$.

Here is another related inequality by me. It's solved.

Let $a,b,c>$ and
\[ \frac1{a+1}+\frac1{b+1}+\frac1{1+c}=1.  \]
Prove that
\[ \frac1{ab-1}+\frac1{bc-1}+\frac1{ca-1}\geq1.  \][/quote]

Hi pvthuan!

Nice to meet you again.  :)

I haven't found your ineq so let me write my solution.
Well here goes

We put   $a=\frac{x+y}{z}$,  $b=\frac{y+z}{x}$,  $c=\frac{x+z}{y}$.

so we need to prove that $\frac{1}{x+y+z} \cdot \sum \frac{xz}{y} \geq 1$

However this is true because $\sum \frac{xz}{y} \geq \sqrt{3(x^2+y^2+z^2)}$ (well known ineq)

and $3(x^2+y^2+z^2)\geq(x+y+z)^2$ (Cauchy - Schwarz)

Hope it's true!

 :)

[quote="pvthuan"]My solution is different. I used substitutions:
\[ \frac1{1+a}=x,\;\; \frac1{1+b}=y,\;\;\; \frac1{1+c}=z.  \][/quote]

It's almost the same  :P


The nth triangular number is $T_n=\frac{n(n+1)}{2}$.  We can see algebraically that $T_n=\binom{n+1}{2}$, but is there also a counting argument for the identity?

[hide="Solution"]
Consider a set of $n+1$ elements.  We will count the number of 2-element subsets, ${n+1 \choose 2}$, as follows:

Order the elements with indices.  Take the element of index $1$.  There are $n$ other elements, hence $\boxed{n}$ subsets that have the element of index $1$ as their least index member.

Take the element of index $2$.  There are $\boxed{n-1}$ subsets that have the element of index $2$ as their least index member.

And so forth.
[/hide]


I am shocked at the mumbai attacks.
One terrorist was very young. they showed the picture on the news.
The ATS chief died which has made me very sad. :(
Hope the police take control of this situation.
This act by the terrorists is simply unacceptable. :mad:

[quote]Hope the police take control of this situation. [/quote]
i HOPE so too
[quote]This act by the terrorists is simply unacceptable. Mad  [/quote]
they wanted it to be far more and it was

[quote="befuddlers"]This act by the terrorists is simply unacceptable. :mad:[/quote]
The problem is the don't need somebody's acceptance :(

I second Akashnil......
This is outrageous.... But wat can be done??? :(
There is nothing which we can do as common civilians...... :maybe:

dont underestimate yourself...
its pretty well known.... but very often forgotten...
the only thing thats makes a difference... in an individual all by himself..

i totally agree with keshav on this point

u will loath me 4 this :maybe:
but i really don't care!!! :ninja:

i just go were im lead :)

One of the worst consequences after this will be no foreign big ppl wanting to come to our country.

Like the situation in Pakistan. I hear some billionaire got killed in Taj. And many of the casualties were VVIPs of different nations. Look at the English cricket team. I'm not saying I'm particularly worried about cricket after so many ppl being killed but it shows a typical foreigner VVIP's impression of the country. No Bill gates (he came here a while ago) or ppl like him will come here which will affect the country.

[quote="lordvoldemort"]anurags92 , my dear , are not the army police and everyone else out there common people? they just look different in their uniforms  that's all .
We should atleast be happy that the Indian govt has not asked for a talk and then released  some terrorists in exchange for the civilians . that's going to be a bad , no worst  option if ever considered.

Now is not the time for Gandhian policy we've seen that they are very very mean . It's better to follow the Bhagat policy , kill or be killed. easy to say but not easy to do .
THINK OF THE BRUTAL NATURE OF THESE TERRORISTS , A NINE YR OLD BOY GOT THE FULL BLAST OF THE BOMB THROWN BY THE TERRORISTS , WHAT A PITY . WHAT I BELIEVE ARE FORGETING IS THAT WE ARE ALL INDIANS , THEN ONLY WE BELONG TO DIFFERENT STATES. THE FACT THAT ALL DON'T FELL EQUALLY FOR EVERY INDIAN LOST IS A CAUSE FOR CONCERN . I MEAN LOOK AT THACKERAY .
AND OTHER POLITICIANS . THEY'RE DIVIDING THE COUNTRY , TIME FOR R.K.MISHRA I THINK BUT THE DISADVANTAGE IS THAT I FEEL HE'S VERY TIMID NOT VERY BOLD , SORT OF LIFE LOVER , ANY TRUE CITIZEN OF INDIA WILL WISH TO DIE SERVING HER THAN DIE LIKE A COWARD ,BEING SHOT WHILE RUNNING AWAY .  WE MUST ALL NOW TAKE TIME TO REMEMBER THE SERVICE PROVIDED BY THE GREAT BRAVE-HEARTED MEN AND WOMEN WHO HAVE BEEN KILLED DURING THE PAST FEW DAYS AND ALSO THOSE WHO HAVE BEN KILLED IN THE PREVIOUS BLASTS. MOURNING ALL THOSE AFFECTED AND ALSO THOSE WHO HAVE LAID DOWN THEIR LIVES IN THE SERVICE OF THE NATION AND CONDEMNING ALL THE ACTS OF THE TERRORISTS , I SIGN OFF.

MAY THE CAPTIVE BE RELEASED AND THE TERRORISTS BE KILLLED , JAI HIND. HERE IS A SONG FROM THE ORDER OF THE PHOENIX WHICH IS QUITE RELEVANT
The warning history shows,
For our Hogwarts(INDIA) is in danger
From external, deadly foes
And we must unite inside her
Or we'll crumble from within
I have told you, I have warned you . . .

I FELT IT AS MY DUTY TO DO SO MY DEAR FRIENDS.

@ rituraj , If I've understood u right ? U meant that u'll carry on where ever u are(i.e. consolidate on your strenghts , don't get disturbed ) neglecting this as a mionr hurdle , but this is serious dude . hopefully u'll atleast serve INDIA not go abroad and serve there .U'll never feel anything unless it affects u , my dear , hope I've driven home a point , else , njoy what fate has in store for you . can't say anything more.Atleast think of what INDIA is for you . without it you are nothing(we all are nothing!!!!) , REMEMBER.[/quote]
i wish all people thought like you yes i firmly believe in Bhagat Singh policy than the gandhian
the best thing you can do (wihout actually going to mumbai) is not to be terrorized (THAT IS EXACTLY WHAT THEY WANT FROM US)
continue our normal lives without letting these tut tut  manipulating our lives (THIS DOSENT MEAN WE STOP HELPING THE MUMBAI BUT MEANS THAT WE REPLACE OUR FEAR WITH ANGER)
it is time we unite once again as we did againt the britishers
yesterday i saw more than half of the news channels busil criticizing political parties and hence trying disntegrate us (WE WONT LET THAT HAPPEN)
they have no brains what so ever when are such opinions to be expressed
all this utterly dirty buisness its been almost 3 hours et the police are not able to stop them
now i dont think i have a reason to condemn the russin commandoes for their immensely successful operation atleast the succeeded in limiting their civilian + police casualties to arround 90
 :furious:  its high time we must give all possible moral support to the police and commandoes
this casualty i firmly believe that could have been avoided if we were not so engaged in inter religion fighting

but wat the f*** can we do?

any ideas?////////////////
give a practical approach to it ?

it depends upon what do you call practical
at the time of independence idea of having an ATS was not practical
(we had no information available about them)
but it gained info as it got some older

@ gauravpatil . well said brother, thanks for your support .

@rt_08
You want one ? I'm no good at setting up a website, but if any one could (and if they have free time?), they should try to setup a portal on which citizens can express their views abt the political system and also corruption and many other social topics(ALMOST FORGOT TERRORISM), U'll be surprised to know the number of hits that portal will get , U'll also be surprised to know that we can get many new ideas very quickly . the only problem in INDIA is that , people are afriad to speak out as they fear what'll happen to their kith and kin. once they have the liberty to freely express their feelings , then look out for the number of replies , u'll be very very surprised.

BELIEVE ME : THE TRUTH IS THIS , WHETHER YOU AGREE OR NOT. INDIANS ARE A SCARED LOT , BUT GIVEN A CHANCE  THEY'LL DEFINITELY SPEAK OUT THEIR HEART.

SO IF POSSBLE TRY THIS OPTION AND MAKE SURE THAT THE PERSON CANNOT BE IDENTIFIED , BUT IT'S A TOUGH JOB , WE'VE GOT TO PROTECT IT FORM HACKERS (THEY'LL BE MANY TRYING TO DESTROY IT , MIND YOU ) AND ALSO THERE'LL BE SOME TRAITORS WHO'LL  TRY TO CREATE DISHARMONY. AND MANY OTHER PROBLEMS ,FOR THAT I DON'T THINK WE AS STUDENTS ARE UP TO THE TASK . SO WE SHOULD OPEN THIS FORUM AND ALLOW OTHER TRUST WORHTY AND ALSO PATRIOTIC ELDERS [hide]( WHEN I  SAY THIS , ELDERS MEAN , NOT MORE THAN 45 YRS OLD. ABOVE THIS THEY PREFER GANDHI DUDE. NOBODY THINKS OF BHAGAT )[/hide] TO CONTINUE IT.

ATLEAST WE SHOULD START IT , AND HOPE THIS AWAKENS THE SLUMBERING NATION . WE'VE ALSO GOT TO BRIDGE THE SOUTH-NORTH DIVIDE. [hide]I'M A SOUTH INDIAN MIND YOU AND I'VE  HEARD THAT , IN THE IIT'S ITSELF  SOUTH AND NORTH PPL DON'T MIX MUCH , DON'T KNOW HOW FAR IT'S TRUE ,YET , I'VE HEARD. AND IT'S NOT GOOD IS IT?[/hide]

THIS THACKERAY BUSINESS IS EMBARRASSING  :oops: AND IS A BLACK MARK ON THIS WONDERFUL NATION.

LETS BUILD A NEW INDIA , JAI HIND. EVEN IF IT MEANS LAYING DOWN OUR LIVES , IF PPL ARE AFRIAD TO DIE FOR THEIR MOTHERLAND , WE SHOULD ATLEAST SET AN EXAMPLE , NOT FORCE PPL TO DIE , THAT WOULD BE GOING TOO FAR. HUMANENESS LOST. NO... NO....  NOT THAT.

SAYING THIS I SIGN OFF . LET OUR SLOGAN BE .

[size=200][color=orange]AAO [/color][color=darkblue]NAYA [/color][color=green]BHARAT BANAYE[/color] [/size]

HOPE IT'S FINE  , AND BY THE BY WHAT SCORE SHOULD WE GET IN JEE TO JOIN COMPUTER SCIENCE ???? IMPORTANT INFO. PLEASEEE........

THANK YOU .
[size=200][color=orange]JA[/color][color=darkblue]I HI[/color][color=green]ND[/color][/size]

Well said.
But I don't understand one thing.
Nothing against Bhagat really, but why do you not regard Netaji as a non-Gandhian Idol?

@rituraj some people are best ignored in everyones interest
@peteryellow you seem to be one of the first people to take interest to Indian forums not many non Indians take any particular interest if you dont like it don't watch it also i think we need intra India sympathy for the peoplewho wee trapped

peteryellow :if u dont know anything then it is better to keep you mouth shut.

[quote="ernie"]THANK GOD ITS OVER.

And one question that sounds intriguing to me: Why did the terrorists not attack the commercial area of Mumbai?[/quote]
May be the reason is they want people outside India to think that India is not a safe for any of  them as Oberio Trident Taj all were famous among foreigners .
Anyways whether it is commercial or not , we have lost so many lives and that's the big thing.
Let us just pray to Jesus to give strength to all of us so that we can come out from all this.

hey man i didn't mean it really!!!  :)

it isn't that i wasnt upset or sth

and @patteryyellow----do u want 2 get beaten :mad: ma ka dudh pia hai toh samne aa!!! :P

peter yellow will be the new target for NSG AND BLACK CAT COMMANDOS
 :P

The terrorists should be executed by drawing and quartering.

THESE ARE NOT PRACTICAL REASONS MATHLETE ALTHOGUH THEY ARE SUPPORTED BY EVRYONE

IF U THINK CLOSELY WE HAVE TO NIP THIS HABIT FROM ITS BUD

THINK CLOSELY AND ULL GET THE BUD
 :|

who the hell is that peteryellow ? crazy fellow . you idiot ! would you ever kill your own country men ? and what would you mean by SYMPATHY  :huuh: , we indians don't need sympathy . we have all that we want. Thank you. we are one of the fastest progressing economy and industrial nation ( I don't care whether you belonged to a developed nation or not, but this is the fact .)so mind your tongue. you are a disgusting human. Believe me it's true. You'll never realise how it pains us ,until you receive a taste of it.Beautiful , isn't it ?the sight of 200 dead bodies  :furious:  ?

I also agree with Arvind Srinivas L .well said buddy. :thumbup: You ( I mean peteryellow)   have no right whatso ever to question us. It's a general discussion  INDIAN forum .

Keep your crooked nose out of other people's business.( You may be a briany person but, humanity wise . a 3-yr old child is far better. atleast the child asks what ahppened when somebody falls, while you say . That guy's fallen to get my sympathy.) You're not fit to live in human society.That's the truth.

But do you'll think that the media has a positive or a negative role to play in all this?
I seriously would like to know your opinions..
On one hand they're making film stars and gliteratti come and say "enough is enough, politicians wake up, india is angry" coz that's what the ppl want to hear...but they don't do anything about it..
and on the other hand they give live coverage like- "as you can see, there is a commando hiding here, there is a commando hiding there, they have such and such weapons and are coming through such and such entrance.." and then the head of NSG issues a statement saying that if it wasn't for the media advertising their locations, they would have finished 12 hrs earlier and could have saved about 50 lives more..
But then why does everyone glorify them when they r the real market-force driven terrorists?
Geniune query, no rhetoric intended  :lol:

@peteryellow-If I may say so-get lost...you don't deserve to have even ur name in a forum related to India but true Indian hospitality demands that we put up with your sh**t till a limit-n I think ur very close to crossing it...

according to me.. the media has a negative role to play...
infact... the news we have nowadays... is srsly pathetic....

I dunno  whether it's good or abd. you know the should give out info but atleast hold tongue till they get a green signal from the government , esp in matters such as this which can cause panic and aslo unnessecary rumours . after everything is over , they can let out the details only as a superficial info . not all details. that means the other ppl who   shouldn't get to know all this also are aware of the tactics of the army and others. they'll be better prepared.

The main problem is in Indain media . making money is thier main goal . The paper or channel which first publishes these info is the greatest , that's what they feel as do the common ppl. so , to get better money for ads they do this .

If this was stopped , then our security would be more stronger . I feel.

i just dont like any media thats it :huh:  :|

specially INDIA TV :rotfl:  :rotfl:  :rotfl:


pigeonhole principle
inequalities
search

The sum of 5 integer numbers are 100. Show that there are at least two of them whose difference <= 10

maybe i understood something wrong but if we have integers a and b and

a-b>10 then b-a<-10<10 ....

 :huh:

Well, I think that he wanted $ 0\leq a \minus{} b\leq 10$
We don't have to use pigeonhole principle here.
First we assume $ a_{1} \leq a_{2}\leq a_{3}\leq a_{4}\leq a_{5}$
Then we assume the opposite, we have:
$ a_{5} \minus{} a_{4}\geq 10 \\
a_{4} \minus{} a_{3}\geq 10 \\
a_{3} \minus{} a_{2}\geq 10 \\
a_{2} \minus{} a_{1}\geq 10$
They're all $ \geq 0$ Nuw we sum all 4 of them, 3 of them, 2 of them and one last of them
We have a new set of inequalities, which is:
$ a_{5} \minus{} a_{1}\geq 40 \\
a_{4} \minus{} a_{1}\geq 30 \\
a_{3} \minus{} a_{1}\geq 20 \\
a_{2} \minus{} a_{1}\geq 10$
Now, summing up we have:
$ a_{2} \plus{} a_{3} \plus{} a_{4} \plus{} a_{5} \minus{} 4a_{1}\geq 40 \plus{} 30 \plus{} 20 \plus{} 10 \equal{} 100$ but on the other hand we have:
$ a_{2} \plus{} a_{3} \plus{} a_{4} \plus{} a_{5} \minus{} 4a_{1} \equal{} (\sum_{i \equal{} 1}^{5}a_{i}) \minus{} 5a_{1} \equal{} 100 \minus{} 5a_{1}\geq 100$ which gives us that $ a_{1} \equal{} 0$ is required, thus we have a contradicyion :wink:

I think the numbers must be nonnegative, but not necessarily integers.

And everyone must be made aware that this is HSB forum. So, 'proof' problems like this one belongs to Intermediate Topics forum.

[quote="zephyredx"]I think the numbers must be [b]nonnegative[/b], but [b]not necessarily integers[/b].[/quote]

zephyredx is right about the numbers being nonnegative but not working with integers... that I would like to see a proof.

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=541130490&t=187770

What are you both cheating off of? :P

[hide]Proof by contradiction:

Assume $ n$ as the first number and assume that four of them have a differnece greater than 10. Then you have $ n\plus{}(n\plus{}11)\plus{}(n\plus{}22)\plus{}(n\plus{}33)\plus{}(n\plus{}44)$.

This results in $ 4n\plus{}110\equal{}100$. Since $ n$ is negative, it is not possible.

Now try three of them having  a difference greater than 10.

$ n\plus{}(n\plus{}11)\plus{}(n\plus{}22)\plus{}(n\plus{}33)\plus{}(n\plus{}33\plus{}k)$. This comes to $ 4n\plus{}99\plus{}k\equal{}100$, and even if $ k$ is 0, then $ n$ is $ 1/4$, which is not an integer.

Therefore, less than 3 of them must have a differnece of greater than 10, showing that at least 2 must have a difference of $ \leq 10$.[/hide]


find the maximum value of thef\expression
$A= \frac{(2004x^2 +6006x+6\sqrt{(x^3-2x^2+x-2)}-8003)}{(x^2+3x-4)}$

why ? i think that  this problem  isn't hard .
         $A = 2004 + \alpha ................$

$6\sqrt{x^3-2x^2+x-2}=2\sqrt{9(x-2)(x^1+1)}=2\sqrt{(9x-18)(x^2+1)}\le x^2+1+9x-18=x^2+9x-17$
Then it is easy to get the result  :)
Hmmm this problem look familiar :|


limit
integration
real analysis
real analysis unsolved

Let $ f_n: {R}\to{R}$
i)$ f_n(t)\equal{}\frac {1}{{(1\plus{}\frac {t}{n})^n}.t^{\frac {1}{n}}}$    for all $ t \in(0,1)$
ii)$ f_n(t)\equal{} 0$ for all  $ t \not\in (0,1)$
       Determine $ \lim_{n\to\infty} \int_{ \minus{} \infty}^{ \plus{} \infty}f_n(t) \, \mathbf{d}t$

Look : $ \lim_{n\to\infty}\int_{\minus{}\infty}^{\infty}{f_{n}(x)dx}\equal{}\int_{\minus{}\infty}^{\infty}{\lim_{n\to\infty} f_{n}(x)dx}\equal{}$
$ \equal{}\int_{\minus{}\infty}^{0}{0\; dx}\plus{}\int_{0}^{1}{e^{\minus{}x}dx}\plus{}\int_{1}^{\infty}{0\; dx}\equal{}0\plus{}\frac{e\minus{}1}{e}\plus{}0$

$ \lim_{n\to\infty} f_{n}(x)\equal{}\frac{1}{e^x}\equal{}e^{\minus{}x}$

[quote="Yuriy Solovyov"]Look : $ \lim_{n\to\infty}\int_{ \minus{} \infty}^{\infty}{f_{n}(x)dx} \equal{} \int_{ \minus{} \infty}^{\infty}{\lim_{n\to\infty} f_{n}(x)dx}$[/quote]
why ?don't understand !


algebra
difference of squares
special factorizations

is there a quick way to do a prob like

501^2-499^2

without actually calcin it.

if so show me the steps in detail

Using the difference of squares factoring method...
$ a^2\minus{}b^2\equal{}(a\plus{}b)(a\minus{}b)$
So $ 501^2\minus{}499^2\equal{}(501\plus{}499)(501\minus{}499)\equal{}1000(2)\equal{}2000$.

here is how i do it: I MADE this formula (i dont know if it has already been invented though) that if you have teo consecutive numbers (3^2-2^2), u can put the bigger number as n in this formula: ;2n-1

so 501(2)-1=1001

then 500(2)-1=999

so 1001+999=2000

but i wold do the method posted above because it workss for anything

[quote="Brut3Forc3"]Using the difference of squares factoring method...
$ a^2 \minus{} b^2 \equal{} (a \plus{} b)(a \minus{} b)$
So $ 501^2 \minus{} 499^2 \equal{} (501 \plus{} 499)(501 \minus{} 499) \equal{} 1000(2) \equal{} 2000$.[/quote]

Thank u!

ajain, can you explain how you got those numbers, basically, what is your formula?

if you're just looking for ANOTHER trick to memorize, don't.  it's better to use difference of squares.  especially since it only takes 3-7 seconds.

ajain's trick:
Find $ (x\plus{}1)^2\minus{}(x\minus{}1)^2$. Let's expand and see what we get: $ 4x\equal{}2(x\plus{}1)\plus{}2(x)\minus{}2$. Which is pretty much his formula lol =). But this way is easier  :)

:P

Find $ (x\plus{}y)^2\minus{}(x\minus{}y)^2\equal{}4xy$.

oooo. convenient. lol =).


given a set A which contains 9 elements .Two subsets are taken from this set , what is the probability of having two subsets with no common elements

[hide]
We just count up the number of ways to count of 2 sets one after another, in the condition that the second one doesn't intersect with the first:

$ \binom{9}{0}(\sum_{i\equal{}0}^{9}\dbinom{9}{i})\plus{}\binom{9}{1}(\sum_{i\equal{}0}^{8}\binom{8}{i})\plus{}\binom{9}{2}(\sum_{i\equal{}0}^{7}\binom{7}{i})\plus{}.....\plus{}\binom{9}{9}(\sum_{i\equal{}0}^{0}\binom{0}{i})\equal{}\binom{9}{0}(2^9)\plus{}\binom{9}{1}(2^8)\plus{}\binom{9}{2}(2^7)\plus{}..\binom{9}{9}(1)\equal{}(2\plus{}1)^9$

Therefore, $ \frac{(2\plus{}1)^9}{(2^9)^2}\equal{}\frac{3^9}{2^{18}}$[/hide]

[hide]Consider two jars, one for holding elements of the first set (denote it as $ A$) and the other for holding elements of the second set (denote it as $ B$).

For each element of the subset, it could be in $ A$ but not in $ B$, in $ B$ but not in $ A$, or in neither, if the subsets are to have no element in common.

So for each element, there are exactly three ways to allocate it to the subsets such that there are no common elements.

For 9 elements, there are $ 3^9$ ways to do so. But the question implies the order of the subsets doesn't matter, so there are $ \frac {3^9 \plus{} 1}{2}$ subsets (we add one as the situation in which both subsets are zero sets doesn't have a complement). There are $ \binom {2^9}{2} \plus{} 2^9$ ways to choose two subsets such that the order doesn't matter. So the final probability should be $ \frac {3^9 \plus{} 1}{2^{18} \minus{} 2^9 \plus{} 2^{10}}$?[/hide]


linear algebra
matrix
search
graph theory

If each person, in a group of n people, is a friend of at least half the people in the group, then it is possible to seat the n people in a circle so that everyone sits next to friends only.

there is a generalization of this in graph theory:
[b]let $ G$ be a simple n-vertex graph,where $ n\geq 3$,s.t $ deg(x)+deg(y)\geq n$ for each pair of non-adjacent vertices $ x,y$,then $ G$ is hamiltonian.[/b]
[b]proof:[/b]

by way of contradiction,assume that the theorem is false,and let $ G$ be a maximal counterexample.that is,$ G$ is non-hamiltonian and satisfies the condition of the theorem,and the addition of any edge joining two non-adjacent vertices of $ G$ results in a hamiltonian graph.
let $ x,y$ be two non-adjacent vertices of $ G$ ($ G$ is not complete, since n>=3). to each a contradiction,it suffices to show that $ deg(x)+deg(y)\leq n-1$.
since the graph $ G+xy$ contains a hamiltonian cycle,$ G$ countains a hamiltonian path whose endpoints are $ x,y$.let $ <x=c_{1},c_{2},...,c_{n}=y>$ be such a path
for each $ i=2,...,n-1$,at least one of the pairs $ v_{1},v_{i+1}$ and $ v_{i},v_{n}$ is non-adjacent,since otherwise $ <v_{1},v_{2},...,v_{i},v_{n},v_{n-1},...,v_{i+1},v_{1}>$ would be a hamiltonian cycle in $ G$.this means that if $ (a_{i,j})$ is the adjacency matrix for $ G$,then $ a_{1,i+1}+a_{i,n}\leq 1$ for $ i=2,...,n-2$.
thus we have:
$ deg(x)+deg(y)=\sum_{i=2}^{n-1}a_{1,i}+\sum_{i=2}^{n-1}a_{i,n}$
$ =a_{1,2}+\sum_{i=3}^{n-1}a_{1,i}+\sum_{i=2}^{n-2}a_{i,n}+a_{n-1,n}$
$ =1+\sum_{i=2}^{n-2}a_{1,i+1}+\sum_{i=2}^{n-2}a_{i,n}+1$
$ =2+\sum_{i=2}^{n-2}(a_{1,i+1}+a_{i,n})$
$ \leq 2+n-3=n-1$
which establishes the desired contradiction...
as a corollary of this theorem we get that:
[i]let $ G$ be a simple n-vertex graph,where $ n\geq 3$,s.t $ deg(v)\geq \frac{n}{2}$ for each vertex $ v$,then $ G$ is hamiltonian.[/i]

could you please explain what does <x=c_{1},c_{2},...,c_{n}=y>  mean? At the end of line 4 of the proof..

and <v_{1},v_{2},...,v_{i},v_{n},v_{n-1},...,v_{i+1},v_{1}> at the end of line 5?[hide][/hide]

varsha, being new to this type of forum - take a closer look at the post-date of the material you want to comment or ask questions - in this case it is 2007 !

As to your questions, those sequences describe [color=blue]paths[/color] in the graph (sequences of non-repeating vertices). You need to be familiar with at least basic terminology, in order to understand a (slightly) technical proof.

You may search on Google "Dirac theorem" or "Ore theorem" on "hamiltonian circuits".

[quote="varsha"]How can I find the number of hamilton cycles in a complete graph, with n vertices?[/quote]

The number of distinct Hamiltonian cycles on $ K_n$ with labeled vertices is equal to the number of free circular permutations of $ \{1,2,\ldots,n\}$, which is $ (n \minus{} 1)!/2$.

The proof is simple.  Since $ K_n$ is complete, every vertex is incident to every other vertex.  Therefore, a Hamiltonian cycle is possible if and only if each vertex is visited exactly once, which in turn corresponds to an ordering of the vertices.  There are $ n!$ such orderings, but for each distinct cycle, there are $ n$ ways to select the starting/ending vertex, and 2 ways to select the traversal direction.  Therefore, the number of such cycles is $ (n \minus{} 1)!/2$, as claimed.

Here is a related question:  How many distinct Hamiltonian cycles are possible on the complete bipartite graph $ K_{n,n}$?


[img]http://img256.imageshack.us/img256/6195/47807816.jpg[/img]

I tried to solve the first one and I got something like an= 3n+1
Please correct me if i'm wronge and help me with the second one.  Note that the second sequence is alternating.

Thank you

$ \left( \left( \minus{}1\right) ^{n\plus{}1}\frac{n!}{2^{n}}\allowbreak \right) _{n\geq 1}$

Second one is $ ( \minus{} 1)^{n} \frac {(n \plus{} 1)!}{2^{n \plus{} 1}}$, starting from $ n \equal{} 0$.

How exactly do you go about solving it? These types of exercises are okay for simple pattern recognition, but I wouldn't say that their solutions are always unique. Given a finite amount of terms you can make a wide variety of general ``formulae'' for sequences which all agree on these first couple of terms, making these exercises a bit silly at times.

edit:

For example, rewriting those terms as

$ \frac24, \minus{} \frac24, \frac34, \minus{} \frac64$

We can describe the same sequence by $ \frac {( \minus{} 1)^n}4\left(2n! \plus{} n(n \minus{} 1)\left( \frac{(n \minus{} 3)}2 \minus{} (n \minus{} 2)\right) \right)$ or something even more elaborate.


category theory
superior algebra
superior algebra unsolved

Prove that there exists an equivalence of categories between $ PreSet$ and $ PoSet$.

That would require that every PreSet (Preordered set) is isomorphic to some PoSet. But already the PreSet $ \{x,y\}$ with $ x \leq y , y \leq x$ isn't (the isomorphic PoSet would have two elements, too, but there surely can be no isomorphism).

Why is that, exactly? You could hypothetically construct functors each way that don't come from monotone morphisms.

I see what you are pointing to: I was only considering the canonical functor $ PoSet \to PreSet$ (which by the above isn't an equivalence)  :blush:


geometry
circumcircle
rectangle
vector
IMO Shortlist

Let $ S$ be any point on the circumscribed circle of $ PQR.$ Then the feet of the perpendiculars from S to the three sides of the triangle lie on the same straight line. Denote this line by $ l(S, PQR).$ Suppose that the hexagon $ ABCDEF$ is inscribed in a circle. Show that the four lines $ l(A,BDF),$ $ l(B,ACE),$ $ l(D,ABF),$ and $ l(E,ABC)$ intersect at one point if and only if $ CDEF$ is a rectangle.

The official solution uses complex number :(

Isn't there any synthetic proof??? I am really interested to see a very elegant one...

Denote $ \mathcal N_{PQR}$ the 9-point circle of a $ \triangle PQR.$ Consider quadrilaterals $ ABCE, ABDF.$ The 9-point circles $ \mathcal N_{ABC}, \mathcal N_{BCE}, \mathcal N_{CEA}, \mathcal N_{EAB}$ concur at a point $ X$ and the 9-point circles $ \mathcal N_{ABD}, \mathcal N_{BDF}, \mathcal N_{DFA}, \mathcal N_{FAB}$ concur at a point $ Y$ [1].

$ ABCE, ABDF$ are cyclic with a common circumcircle $ \mathcal O \Longrightarrow$ all 9-point circles are half the size of $ \mathcal O$ and congruent. Moreover, $ X, Y$ are anticenters of $ ABCE, ABDF,$ respectively, i.e., reflections of the circumcenter $ O$ in their respective centroids $ G, H.$ In addition, Simson lines $ l(E, ABC), l(A, BCE), l(B, CEA), l(C, EAB)$ concur at $ X$ and Simson lines $ l(F, ABD), l(A, BDF), l(B, DFA), l(D, FAB)$ concur at $ Y$ [2].

Let $ M, K, L$ be midpoints of $ AB, CE, DF.$ The centroids $ G, H$ are midpoints of $ MK, ML.$ The anticenters $ X, Y$ are identical $ \Longleftrightarrow$ the centroids $ G, H$ are identical $ \Longleftrightarrow$ the midpoints $ K, L$ of the chords $ CE, DF$ are identical $ \Longleftrightarrow$ the chords $ CE, DF$ are identical, in which case $ CDEF$ is a rectangle degenerated into a line segment, or $ CE, DF$ are both diameters of $ \mathcal O,$ in which case $ CDEF$ is a rectangle. In the second case, $ X \equiv M \equiv Y.$

References:
[1] [url]http://www.mathlinks.ro/viewtopic.php?t=85989[/url]
[2] [url]http://www.mathlinks.ro/viewtopic.php?t=179089[/url]
(and elsewhere)

Perhaps I am missing something, but it seems to me that if $AD$ and $BE$ are diameters, then $l(A,BDF)=l(D,ABF)=BF$, and $l(B,ACE)=l(E,ABC)=AC$ and there is a point of intersection of all four lines, without $CDEF$ needing to be a rectangle.

Anyway, assume that neither $AD$ nor $BE$ are diameters.  Let $X$ be intersection of the four Simpson lines.  Let $A_1, A_2$ be the altitudes from $A$ to $BD,DF$ and let $D_1, D_2$ be the altitudes from $D$ to $AB,AF$.  Then $A,A_1,A_2,D,D_1,D_2$ all lie on the circle with diameter $AD$.  So by Pascal's theorem applied to $A_1A_2AD_1D_2D$, we have that $X,B,H_{AFD}$ are collinear, where $H_{AFD}$ is the orthocenter of $AFD$.  Similarly we have that $X,F,H_{ABD}$ are collinear.

Let $a,b,c,d,e,f,x$ denote the vectors from the center of the circle to $A,B,C,D,E,F,X$.  Then $x=r(a+f+d)+(1-r)b=s(a+b+d)+(1-s)f$, for some $0 \le r,s \le 1$.  We, (or at least I) have assumed that $AD$ is not a diameter, so $l(A,BDF)$ and $l(D,ABF)$ are distinct lines with a unique point of intersection, so we have that $x=(a+b+f+d)/2$.  Similarly, working on the other two Simpson lines yields that $x=(a+b+c+e)/2$, which in turn means that $CE$ and $FD$ have the same midpoint.  Each non-diameter chord of the circle has a unique midpoint, which means that $CE$ and $FD$ are diameters, which in turn means that $CDEF$ is a rectangle.


Εχουμε [b]p , q, r[/b] θετικα με το αθροισμα τους 1.Πινακας [b]Α[/b] με 1σειρα=p q r
                                                                                2σειρα=r p q
                                                                                3σειρα=q r p

Δειξτε οτι το οριο του γινομενου του στο απειρο( Α^ν ) ισουται με τον πινακα που σε καθε θεση εχει 1/3.

Αρκετά σημαντική η μέθοδος της απόδειξης εδώ.

[hide]Διαγωνοποιήστε τον πίνακα![/hide]

Πραγματι σημαντικη μεθοδος καικ αυτη που συνηθως χρησιμοποιειται σε αντιστοιχα προβληματα. Αν το προσεγγισω με οριζουσα τοτε ευκολα μετα απο πραξεις λυνεται. Αν κανω ομως τον πνακα 4*4 με την ιδα φιλοσοφια και ξανακανω προοσεγγιση με οριζουσα τοτε δε βγαζω ακρη γιατι δεν μπορω να προσδιρισω τις ριζες του πολυωνυμου.

Το ερωτημα ειναι μπορω να αποδειξω το οριο χωρις χρησιμοποιηση της οριζουσας ? Ας κρατησουμε τον πινακα 3*3και το οριο γνωστο(χεχε μη μου πειτε οτι θα μαντεψετε τις ιδιοτιμες  και ας αποφυγουμε τα συστηματα γιατι τοτε θα τον κανω 6*6...)


Is there any fromula for this ?  Counting is  a big headache! :?

The value of the expression 1/2/3/5/7/11/13/15 can be altered by including parentheses.  If we are allowed to plasce as many paretheses as we want, how many differect distict values can be obtained for this expression?

[quote="falconwing64"]...the expression 1/2/3/5/7/11/13/15...[/quote]
There is a MATHCOUNTS problem like this, in which every divisor is a prime number.  Are you sure the final divisor of this problem isn't intended to be a prime number (such as "17", instead of "15")?

Does the observation that all of the divisors are prime help you think about this problem?

[hide][i]How many ways are there of arranging n pairs of parenthesis?[/i] I got the answer from [u]The Book of Numbers[/u] by John Conway (who you may be familiar with) and Richard K. Guy. The answer is the nth Catalan number, where Catalan numbers are defined as the middle elements of Pascal's triangle divided by their position in this list. In other words, they are 2n choose n over n+1 equals 2n+1 choose n over 2n+1. Since there are 8 elements, you need 6 pairs of parenthesis (I think), so the answer is 13 choose 6 / 13 = [b]42[/b]. Don't take my word for it, though.[/hide]

Also, I don't think primacy matters in this case and there is a maximum number of pairs of parenthesis you can put in (I'm pretty sure it's 6) such that some represent unnecessary parenthesis, but all answers are distinct.

if the numbers were all prime, then you would know that the 1 has to be in the numerator, the 2 has to be in the denominator, and the other numbers could be either in the numerator or the denominator if you just put the right number of parenthesis. thus, the answer would be 2^6 or 64

however, since they arent all prime, first, find the number of ways the 3, 5, and 15 can be arranged to be in the numerator or denominator. There are eight ways (2^3) this can happen, but one must be subtracted from this because the number 1 appears twice (15/(3*5) or (3*5)/15). now, you find there are eight ways (still 2^3) to arrange the other three numbers (one and two are still fixed). thus, the answer is 7*8= 56.

not sure this is right, but its all i could come up with. :roll:

Basically, you can either multiply or divide a number when you can use parentheses.  Of course, there are still a few wrinkles left.

Aargh! Everybody has different answers (mine's probably the most far-fetched though...)!

[quote="dwx314"]Assuming the 15 was meant to be a 17...
[hide]So, when you divide with parentheses like this, you can either have each # on the numerator or the denominator when you simplify to just x/y
The one doesn't affect the equation, and the 2 is always on the denominator...
but all the other primes have 2 places to go. Top or bottom
3,5,7,11,13,17
2^6
>>32<<[/hide]
[hide="formula"]If you ever have something like this, count the number of terms, make that n, then 2^(n-2)[/hide]
I may be wrong though...(on both)[/quote]

2^6 is 64, not 32...

[hide]All of those nimbers will be in the numerator or the denominator of the fraction when expressed as a/b and prime factorized, each of the numbers besides 1 have 2 placement possibilities and 1 has only 1. so the answer is 1*2^7 = 128[/hide]

2^7 is 64...
and its suppose to be 2^6 which is 64 but that is assuming they are all prime
if the 15 is supposed to be a 15, then its 56 i think

whoops...
i meant 2^7 is 128
and according to the problem, you should use 2^6 to get 64


How many inches are in the area of a circle inscribed in a regular hexagon with side length 12 inches? Express your answer in terms of $ \pi$.
[asy]size(75);
draw(dir(0)--dir(60)--dir(120)--dir(180)--dir(240)--dir(300)--cycle);
draw(circumcircle((dir(0)+dir(60))/2,(dir(60)+dir(120))/2,(dir(120)+dir(180))/2));[/asy]

Using 30-60-90 ratios, we find the apothem of the hexagon which, conveniently, is the radius of the circle.  THe apothem is 6sqrt3, so the area of the circle is 36*3pi=108pi.

Finding the height of one of the 6 equilateral triangles is the same thing as the radius of the circle and because the side length of the equilateral triangle is 12, our area is 36Sqrt3 after using the formula (s^2sqrt3)/4 whereas s=side of equilateral triangle. Now we divide 12 to 36sqrt3 which we get 3sqrt3 and multiply by 2 since the height and length would equal a square not a triangle. The radius is 6sqrt3 thus the area is 36x3xpi=108pi.


I think (1) should be the formula
\[\cos^{2}A+\cos^{2}B+\cos^{2}C+2\cos A\cos B\cos C=1\]
[hide="proof"]
We have $\cos A=\cos (180^\circ-B-C)=-\cos (B+C)$, or $\cos A+\cos (B+C)=0$. But also, $2\cos B\cos C=\cos (B+C)+\cos (B-C)$ from product-to-sum formula. So we try to prove that
\[\cos^{2}A+\cos^{2}B+\cos^{2}C+2\cos A\cos B\cos C-1=\left(\cos A+\cos (B+C) \right) \left( \cos A+\cos (B-C) \right)\]
It only remains to show that
\[\cos (B+C) \cos (B-C) =\cos^{2}B+\cos^{2}C-1\]
which follows from product-to-sum and double angle formulas:
\[2\cos (B+C) \cos (B-C)=\cos 2B+\cos 2C=(2\cos^{2}B-1)+(2\cos^{2}C-1)\]
\[=2(\cos^{2}B+\cos^{2}C-1)\]
[/hide]


How many square units are in the area of the convex
quadrilateral with vertices (0, 0), (3, 0), (2, 2) and (0, 3)?

Graph the points and outline the quadrilateral.

Draw perpendiculars to each axis from (2,2), and draw a line from (2,2) to the origin.

These 4 triangles within the quadrilateral are easy to calculate areas of, and the quad. is just the sum of them. The four triangles are with areas 1, 1, 2, 2.
$ 2\plus{}2\plus{}1\plus{}1\equal{}6$


MATHCOUNTS HCSSiM AMC USA(J)MO USAMO geometry

rank the states that you think will get in the top 5 in order.

1)California
2)Virginia
3)Pennsylvania
4)New Jersey
5)Oklahoma

oklahoma wasnt really biased, i think that we can get the perfect team easily and then it really only depends on the individual scores of the other two people on our team. (the first two being martin and The Great Neal)

blah don't depend too much on your teammates. we got perfect team last year. and another guy got 11th i think, so pretty good. but we still - er - didn't do that great (5th i think)

HAHA...

oh well... i voted for oklahoma, because i do want it to win...

but i think California will win...

Ok...

1. California
2. New Jersey
3. New York
4. Illinois
5. Oklahoma

i voted for oklahoma too, though there really is no chance because our other teammates will get like a 30.

[quote="Syntax Error"]rank the states that you think will get in the top 5 in order.[/quote]

1)California
2)Virginia
3)New Jersey
4)Maryland
5)New York

I regret to say that Minnesota doesn't stand a chance for a while. Maybe in a few years . . . .

P.S. I don't have any inside knowledge that prompts my ranking above; I'm just replying as participants were asked to reply.

those are the states that usually do well, so youre pretty accurate. california has won 4 of the last 5 years.

where's the California option up on the poll?
i hope indiana could do better in mathcounts though i don't really think that's possible.

I'm not sure why New York is in the running.  I don't belive we've been in the top ten since 1999.

lots of states just jump up in random years though due to some random kids working stremely hard

or some smart kid just getting into his/her first year of mc.

ALABAMA!!!! :lol:
Even though I'm too old to participate ;).

Naw, not Nevada.  I say NV-#1 for sure!  </jkjkjk>

I think the highest we've ever gotten is 24th.  Sigh.  California has my bet.

Oklahoma actually got 3rd before, so there's hope.

California would be my bet. I saw the team at countdown and I know one of them. They'll have a harder time avoiding the championship than getting it.

[quote="tarquin"]It will definitely be Texas. The scores this year have always been really high. Two seventh graders were on Texas's national team last year. This year they are scoring even higher and I don't doubt there will be 2-3 perfect scores at state.[/quote]

The fact that two seventh graders made it last year doesn't mean that they are doing better this year...I know of several cases (including my state) where people who made it last year didn't even make it this year.  Then again, my friend and I competed for the first time and both made it to nationals.

My money is on California. THen again, why wouldn't it be. I made it in, barely in 4th. If CA wins, i get 2k. CAlifornia's strength this year is SPEED. Oh, and we're all asian too.

NO, i did not copy uberl33tmage's avatar. This is his new acct. I got tired of that name.

I think, from the people i know, that the national winner for this year will either be from CA, or PA.

hmmmm i wonder what the 31 people who voted for other thought which state was going to win?
edit: changed the wording of my question so that its not as ambiguous

It's just me, daiwei, patricia, and alex

GO TEXAS! isnt jeffery chen on the team? or was he 8th grade last year... cuz he's crazy good
i dont know who else is on the team though (14th last year, made a really stupid mistake on one target)

Well if you're going just by population then [b]of course [/b]it will be california (then texas, new york, florida, etc).

[size=125]Question: My school is in a different state then I live in.  If (when) I don't get to nationals then which state should I root for (or is it whichever will beat california?)[/size]

Jeffrey Chen isn't in 8th grade anymore, but the there were 2 7th graders on the team last year.

We were last in the nation last year: number 53, I think.  It won't happen again.  Watch Coach John Rushman lead SC to a Top 10 Finish in 2005.  Watch Phillip Cross in the Countdown.  This kid has won every Countdown at every level but National.  He has never lost on even one question.  He is AMAZING!

[quote="bob123"]CA will win because I'm on the team!!! :)[/quote]

I predict Alaska to finish number 1.  They have the strongest team I have seen in 15 years.

This year's CA team is weaker than the '02 team, when we took 1st. We only edged out 2nd place team by .25 points, and I believe we got extremely lucky as well...

Yeah, even though I want CA to win, I wouldn't put my money on us. The '02 team was composed of 4 Chinese 8th graders (including me), and we met for practice A LOT! We had a great coach, too (Darren's dad).


limit
real analysis
real analysis unsolved

A set is dense in the set of real numbers if for any two realnumbers a and b with a < b, there is an element s found in the mentioned set  such that a < s< b.  If S is dense, show that any real number x is an accumulation point of S.

just use the definition of accumulation point.

I am totally clueless... how do I start proving this....?     :(

An accumulation point is just a limit of some sequence contained in the set.

Basically, pick any sequence $a_i$ which limits towards $x$.  By density, there is always an element of the set in between $a_i$ and $a_{i+1}$, say $b_i$.  Then $\displaystyle \lim_{i\rightarrow\infty}b_i=\lim_{i\rightarrow\infty}a_i=x$, so $x$ is an accumulation point.


We say a number is ascending if its digits are strictly increasing. For example, 189 and 3468 are ascending while 142 and 466 are not. For which ascending 3 digit number n is 6n also ascending?

578 times six is 3468. I knew 8 was the units digit, and I just tried different numbers for the other two.

rigorous?

[hide]
well the units digit does have restrictions.

First, we prove that 6n has 4 digits. If not, then $ n < \frac {1000}{6}$
So, the hundreds digit of n is 1. Furthermore, we know that the hundreds digit of 6n cannot be 8 or 9 (otherwise it can't be ascending), so we have a tighter bound on n: $ n < \frac {800}{6} \implies n \le 133$
so only 123 fits this bound of all ascending numbers, but $ 123*6 \equal{} 738$ Not ascending.
Yeah..... so 6n has 4 digits.
The units digit of n must be at least 4. Because it is ascending, it cannot be 0, 1, or 2; and if the units digit is 3, then $ n \equal{} 123$ which we already know doesn't work.
The units digit of 6n cannot be 0, 1, 2, 3, or 4. The fact that it is ascending rules out 0, 1, 2, and 3; and if 6n ends in 4, then $ 6n \equal{} 1234$ but this makes n not an integer.
Also, the last digit of 6n must be even, making it either 6 or 8.
In both of these cases, this implies that the last digit of n is the last digit of 6n.
If the last digit of 6n is 6, then 6n equals one of the following
1236
1246
1256
1346
1356
1456
2346
2356
2456
3456

But alot of these are not multiples of 6.

the list of the multiples of 6 is
1236
1356
2346
3456
Dividing this list by 6 gives
206
226
391
576
each not ascending.
So, the units digit of n and 6n is 8.
Let's do casework on the tens digit of n.
It cannot be 0, 1, 8, or 9. The hundreds digit tells us that the digit is not 0 or 1, and the units digit tells us that it is not 8 or 9.
Now let's look at the possible last 2 digits of n.
28
38
48
58
68
78
Multiplying this list by 6 and taking mod 100 gives
68
28
88
48
08
68
The 2nd, 3rd and 5th residues do not accommodate an ascending number, so n ends in either 28, 68, or 78.
let's look at the '68' case.
6n ends in 48.
The only ascending 4 digit numbers ending in 48 are
1248
1348
2348
Only 1248 out of these is a multiple of 6, but this gives 208 as n which is not ascending.
So, the number ends in either 28 or 78.
128 is the only ascending number that ends in 28, but $ 6 \cdot 128 \equal{} 768$ which is not ascending.
So, n ends in 78.
Now, we multiply 78 by 6 and add 600 6 times. (this represents multiplying $ 178, 278, \ldots , 678$ by 6)
468
1068
1668
2268
2868
3468
4068
And we see that only 3468 works for 6n so n is 578.
[/hide]
darn this method could be troublesome without a calculator

I actually did it way faster than that. Let n =  xyz and 6n = mnow. If the units digit of n was 9, that would make the units digit of 6n, 4, which would have been unexceptable. If we had 8 as the units digit for n then 6n would have the same units digit. When we multiply the three digit number n--> xy8 by six, we carry over four, so in order to make o as large as possible, 6y + 4 has to have a units digit of 6 or 7, so that o will be six or seven. It is impossible for it to be seven, so it must be six. Which means that 6y has a units digit of 2. Which means if is 7 or 2. Seven leaves us with less restrictions so we have;

n = x78
6n = mn68

After multiplying the 78 by six, we carry over the four again. The units digit of 6x + 4 has to be from 2 - 5. We want it to be five, so we'll have less restrictions for m. It is impossible though, because 6x never has a units digit of one. So we try four. This works if x is 5. So we try it.

n = 578
6n = 3468

And it works. Does that make sense?

it's good except for 2 things
The first is rather easy to fix: you've used 'n' in 2 different ways!
The second is that you are using a sort of greedy algorithm.
If the only solution had 6 as the units digit, then you would have many more cases, but it still would not be too long.
Using your logic, you have not actually ruled out that the last digit of n is 6, you, in effect, have saved it for further casework, but you did not need to address that case.


The height of a right cylinder is 2.5 times its radius. If the surface area of the cylinder is $ 112\pi$ cm$ ^2$, what is the radius of the cylinder in centimeters?

Say the radius of the cylinder is $ r.$
The height, $ h$, is $ 2.5r$.

Then its surface area,
$ 2\pi r(h)$ + $ 2\pi r^2$ is equal to
$ 5\pi r^2 \plus{} 2\pi r^2$ which is
$ 7\pi r^2 \equal{} 112\pi$

Dividing by $ 7\pi$, we have

$ r^2 \equal{} 16$
$ r \equal{} \boxed{4}$

EDIT: oops forgot to delete it.

[quote="CA Math"]Say the radius of the cylinder is $ r.$
The height, $ h$, is $ 2.5r$.

Then its surface area,
$ 2\pi r(h)$ + $ 2\pi r^2$ is equal to
$ 5\pi r^2 \plus{} 2\pi r^2$ which is
$ 7\pi r^2 \equal{} 112\pi$

Dividing by $ 7\pi$, we have

$ r^2 \equal{} 16 r$
$ r \equal{} \boxed{4}$[/quote]

Uh... If $ r^2\equal{}16r$, then $ r\equal{}16$, not $ 4$...


Solve algebraically, showing work:

$ log_{x} (3x\plus{}2) \equal{} x\plus{}1$

It's easy to know when x=2 is an answer

let $ f(x)\equal{} log_{x}(3x\plus{}2)$

then f'(x)=(xlnx-(3x+2)ln(3x+2))/(lnx)^2

It's clear that f'(x)<0<1 so x=2 is the only answer

[quote="12markkram34"]Solve algebraically, showing work:

$ log_{x} (3x \plus{} 2) \equal{} x \plus{} 1$[/quote]

we have that $ x^{x \plus{} 1} \equal{} 3x \plus{} 2$,so $ 1 \equal{} \frac {3}{x^x} \plus{} \frac {2}{x^{x \plus{} 1}}$ .since RHS is decreasing,
there exists only one possible solution.easy to see that $ x \equal{} 2$ satisfies the equation. :wink:


$ \odot O_1$ externally-tangent $ \odot O_2$ at $ T$. $ PQ$ is a chord of $ \odot O_1$.

$ PT$ and $ QT$ meet $ \odot O_2$ at $ R$,$ S$ respectively.

$ PAB$ is a tangent of $ \odot O_1$ and meets $ \odot O_2$ at $ A$ and $ B$

$ QDC$ is a tangent of $ \odot O_1$ and meets $ \odot O_2$ at $ D$ and $ C$

$ SA$ meets $ PQ$ at $ F$, $ RD$ meets $ PQ$ at $ E$

Prove that:

$ \angle EAF \equal{} \frac{1}{2} \angle BAC$

1.△PAR∽△QFS    PA/QF=PR/QS
△PER∽△QDS    PE/QD=PR/QS
∠APE=∠DQF
=>  △PEA∽△QDF   => ∠QFD=∠PAE=∠EAD
2.∠AFD=180-∠QFD-∠PFA=180-∠BAM-∠AMR=∠AKM
A,K,R,E are concyclic
3.It is easy to know RA=RB  SD=SC MD=MB
Let O be the midpoint of arcAD , then AO=OD  we will prove O is the circumcenter of AEFD
∠AKM=1/2(arcAM + arcRB )
∠AOD=1/2(arcAM + arcMB + arcRB + arcDR)=1/2(arcAM + arcRB + arcMD + arcDR )
=1/2(arcAM + arcRB + arcAM + arcAR )
So ∠AOD=2∠AKM=2∠AED=2∠AFD   => O is the circumcenter of AEFD
4. ∠ADE=∠ASR
  ∠AOE=2∠ADE=2∠ASR=∠ASB=∠AMB
E,O,B are collinear
Analogously, F,O,C are collinear
5. ∠EAF=1/2∠EOF=1/2∠COB=1/2∠CAB
DONE!

To plane geometry

$ \Longrightarrow \angle PAE \equal{} \angle EAD$

How can it be derived ? I think it is equivalent to " $ A$, $ E$, $ F$, $ D$ are concyclic "
But I don't know how to derive it . Please tell me a little in detail . Thank you .
Ashamed of my poverty at math  :blush:

A,S,R,D concyclic   =>   ∠ADE=∠ASR
We can easily prove  PQ∥SR  =>   ∠ASR=∠AFE
∠ADE=∠AFE  =>   A,D,F,E concyclic


geometry
incenter
circumcircle
cyclic quadrilateral
angle bisector

1)
ABCD is a square.
E is a point outside this square such that angle BEC = 90°.
The angle bisctors of angle EBC and angle ECB meet at F.
[b]Prove that angle AFB = angle DFC.[/b]
[i](Hint: From chapter on concyclic points)[/i]

2)
Two circles C1 and C2 touch externally at A.
P and Q are points on the circumference of C1.
PA is produced to meet C2 at S.
QA is produced to meet C2 at T.
PT cuts C1 at X and C2 at Y.
QS cuts C1 at U and C2 at V.
[b]Prove that angle XAY = angle UAV[/b]
[i](Hint: From chapter on tangency)[/i]

Please do help. Looking for all kinds of solutions to these two.  :)

[color=blue][b](1)[/b][/color] Since the angle $\angle BEC = 90^o$ is right, the point $E$ lies on a semicircle $(M)$ outside of the squere $ABCD$ centered at the midpoint $M$ of the diameter $BC$. The intersection $F$ of the bisectors of the angles $\angle EBC, \angle ECB$ is the incenter of the right angle triangle $\triangle BCE$. The angles $\angle EBC + \angle ECB = 90^o$ add up to a right angle. The angle $\angle BFC$ is equal to

$\angle BFC = 180^o - \frac{\angle EBC + \angle ECB}{2} = 180^o - 45^o = 135^o$

regardless of the position of the point $E$ on the semicircle $(M)$. Thus the point $F$ lies on a fixed circumcircle $(O)$ of the triangle $\triangle ABF$. The angles $\angle BFC$ spanning the minor arc $AB$ of the circle $(O)$ are equal to $135^o$, hence, the angles spanning the major arc $BC$ of the circle $(O)$ are equal to $180^o - 135^o = 45^o$. Since the angles $\angle BAC = \angle BDC = 45^o$, the remaining 2 vertices of the square $ABCD$ also lie on the circle $(O)$. i.e., this circle is the circumcircle of the square $ABCD$. The angles $\angle AFB = \angle DFC = 45^o$ are equal, because they span the major arcs $AB, DC$ of the circle $(O)$ with the equal chords $AB = DC$, 2 sides of the square $ABCD$.

[color=blue][b](2)[/b][/color] The tangency point $A$ of 2 touching circles $(O_1), (O_2)$ with radii $r_1 = O_1A, r_2 = O_2A$ is their (internal) center of similarity and their coefficient of similarity is $\frac{O_1A}{O_2A} = \frac{r_1}{r_2}$. To see that, note that the triangles $\triangle O_1AP \sim \triangle O_2AS$ are similar, because their are both isosceles with $O_1A = O_1P = r_1,\ O_2A = O_2S = r_2$ and the angles $\angle O_1AP = \angle O_2AS$ are vertical. Likewise, the the triangles $\triangle O_1AQ \sim \triangle O_2AT$ are similar, because their are both isosceles with $O_1A = O_1Q = r_1,\ O_2A = O_2T = r_2$ and the angles $\angle O_1AQ = \angle O_2AT$ are vertical. Consequently, $\frac{AP}{AS} = \frac{AQ}{AT} = \frac{r_1}{r_2}$. This means that the angles $\angle AXP = \angle AVS$ are equal, because they span the similar major arcs $AP, AS$ of the circels $(O_1), (O_2)$. Likewise,  the angles $\angle AUQ = \angle AYT$ are equal, because they span the similar major arcs $AQ, AT$ of the circels $(O_1), (O_2)$. Hence, the angles $\angle AUS = 180^o - \angle AUQ = 180^o - \angle AYT = \angle AYP$ are also equal. As a result, the triangles $\triangle AXY \sim \triangle AVU$ are similar, because they have 2 internal angles equal. It follows that their remaining angles $\angle XAY = \angle UAV$ are also equal.

1) Here is another way to prove that $ABFCD$ are concyclic:

$ABFCD$ is made up of 2 quadrilaterals $ABCD$ and $ABFC$.
A quadraliteral is a cyclic quadrilateral if the opposite angles of the quadrilateral are supplementary.
$ABCD$ is a cyclic quadrilateral because it being a square, its opposite angles are supplementary
As for $ABFC$, lets take a look at $\angle CFB$ and $\angle CAB$ (which are opp)
$\angle CFB = 135^o$ as proven by yetti
$\angle CAB = 45^o$ because the diagonal of a square is an angle bisector
Since $\angle CFB + \angle CAB$ = 135^o + 45^o = 180^o$, ABFC is a cyclic quadrilateral.
Since ABC are already points along one circle, F has to belong to the circle too.
Therefore, ABCD and ABFC are cyclic quadrilaterals with the same circumcircle.

Is this correct?

the question was very good and the answer was brilliant :D


HELP!

ok, so here's the problem:

Rectangle ABCD is folded in half to form rectangle AEFD, which is then folded in half again to form rectangle AGHD. The perimeter of AGHD is 23 cm and the perimeter of ABCD is 47 cm. What is the area of rectangle ABCD?

Sorry I don't have the picture, since I don't know how to get the picture on here, and I can't copy and paste. If you can post the picure, thanks.

ok, so's here's my answer: 30
(i'll show my work later if I have to)

ok, so's here's the answer they have: 120

is it just me? or is there a bug?

just wondering, please post both your answer AND YOUR SOLUTION.
Thanks,
LHM

It's just you :) Be attentive: what you most likely did was to compute the area of AGHD, the area of ABCD is 4 times as much.

The solution goes like this: the difference 24 in perimeters is 6 times $ |AG|$, so $ |AG|\equal{}4$. Hence $ |AD|\equal{}7.5$. Therefore the area of AGHD is 30 and the area of ABCD is 120.


[b]Mafia Game 39.[/b]

[size=200]Alternate Forum Mafia.[/size]

At the Awesome Forum, things have been going really well. Until the spammers and the hackers came... You are thirty members, some who want to preserve the Awesome Forum, others who want to ambush it. Here is a post in the Awesome Forum:

Subject: SOS
Our forum has received a large influx of new members, but most are bad guys trying to overload our servers. So, I've opened up a poll: [b]Who do you want to ban?[/b]

Remember, you may NOT unvote, and majority may not be the lynching criterion...
<end topic>

[size=200][b]The Rules[/b][/size](stolen from pacman2812 (stolen from perfect628 (stolen from Klebian (stolen from someone in MafiaScum))))
[list][*]Votes must be in [b]bold[/b].  If you do not bold your vote, it will not be counted.
[*][b]No unvoting. Seriously, that's part of the game.[/b]
[*][b]Executions may not require simple majority...  Once a player has reached lynching state,[/b] his pleas are useless and any attempts to unvote will be unheeded. You may [b]vote: no lynch[/b] – majority votes of this kind are necessary to end the day without a death.
[*]The game is not to be discussed outside of the thread unless your role specifically states that you may do so.
[*]Once your death scene has been posted, you’re dead.  Stop typing. The only thing you can type is a "bah" post.
[*]Don’t edit/delete previously submitted posts
[*]Don’t quote any PMs from me. Neither true, nor false. Seriously.
[*]If you have a night choice to make, it is due by the posted deadline.  I will not wait for you – if you do not submit a choice to me, tough!  No choice will be made.
[*][b]If you anticipate being unavailable for more than a 48-hour period, please post a notice to that effect in the thread.[/b]  Treat this game as a commitment.  Be considerate – don’t leave us hanging.
[*]Remember, it’s a game.  It’s supposed to played.  It’s supposed to be [i]fun[/i].  Do your part.[/list]

Changes from normal mafia in bold.

If your role PM has a bolded username, discussion with THAT PERSON ONLY is allowed. If your role PM has an unbolded username, you MAY NOT discuss with THAT PERSON. Thank you.

[b]Player List[/b]
01. crazychinesemaniac
02. cf249
03. mz94
04. Brut3Forc3
05. mewto55555
06. chenhsi
07. bigboy82892
08. mathelete9999
09. setheus_soul
10. erabel
11. 1=2
12. zapi2007
13. 325155319110133325116104
14. jibjab3
15. 142857
16. budi713
17. Wickedestjr
18. pythag011
19. alexhhmun
20. iphailedusamo
21. AIME15
22. Dojo
23. aleph0
24. dragon96
25. johnnybeard
26. Peter
27. isabella2296
28. eggylvl999
29. rt_08
30. mathblitz

Game begins now.

Interesting setup...

/confirm.

Wait, most are bad guys trying to overload our servers?

[quote="james4l"] [b]No unvoting.[/b][/quote]

Seriously.

[b]FoS: iphailedusamo[/b]

[quote="iphailedusamo"][b]VOTE: AIME15[/b] for being the first to post.

Random voting sucks.

[b]UNVOTE[/b]

This was basically a confirm post.[/quote]

Read the rules before posting. You can't unvote.

I don't think random voting is a good idea because we can't unvote...

[quote="iphailedusamo"][b]VOTE: AIME15[/b] for being the first to post.

Random voting sucks.

[b]UNVOTE[/b]

This was basically a confirm post.[/quote]
There is no unvoting. >.>

Mod: When a lynch occurs, are our votes reset?

[quote="AIME15"]Wow usamo...

[b]5 billion tons of FoS: iphailedusamo[/b][/quote]

I doubt iphailedusamo is mafia...

@iphailedusamo: Please read the rules next time.

[quote="AIME15"][quote="james4l"]
Our forum has received a large influx of new members, but [b]most are bad guys[/b] trying to overload our servers. [/quote]

Chances are for it.[/quote]

Multiple mafias?

Or is this some crazy setup...

I bet on crazy setup, because then mafia would have already won...

I guess random-voting might not be such a bad idea.

[quote="isabella2296"]Confirm.

I think usamo just made a mistake.

I'm a newbie, so what does FoS mean, exactly?[/quote]

Finger of Suspicion. Indicates that someone is suspicious, but your not going to vote for him.

Me too.

[b]Vote 142857 [/b]

Come on. Lets end this game.

if my math is right, the otherforum faction won

[quote="budi713"]I think so too.

4>3[/quote]

Unneccesarily... we'll see how this game plays out.

Ok NEVERMIND.

At ARML it has been proven that, so long that everyone who is cult knows that they are cult, it is...

dun dun dun
[size=200][b]
Other Forum (Cult) Win![/b][/size]

lol is this the first cult win in AoPS Mafia?

with there be analysis or at least basic info on everyone?

OK.

I vigged rt_08. I didn't kill myself, I know this as a fact. Yay I vigged mafia.

Also, the cult was an UNINFORMED CULT. That means that the cult members didn't know they were cult. (Unless james4l told me something false at mc nats.)

Finally, your INSANE COP got an innocent on a certain person and you never noticed. Hmm.

Alexhhmun and I were like one-way lovers (if he died, I'd sitll be fine).

Also, we could investigate someone and find their level of technical expertise or something. I can't remember the results of that.

Roles I know:

budi-Uninformed Cult Recruiter+Vig
pythag011-Improbably Vig, every night can kill someone using either gun A, gun B, or gun C. gun A has 4/6 chance of killing. gun B has 5/6 chance of killing but also 1/6 chance of killing yourself. gun C has 6/6 chance of killing but 1/3 chance of killing yourself.
somebody who i forgot-Dreaming God, Also the Deal or No Deal host, could choose a briefcase every night, would change the voting.

This game proves that cults are swingy. Even Uninformed cults.

Comments:

If everyone had reread, then I think that at least one person would have noticed that an INSANE COP got an INNOCENT on BUDI=> BUDI IS SCUM => DIE SCUM DIE/DIE UNINFORMED CULT RECRUITER DIE.

Flipping Insane Cop gives you a LOT of info, so you should have reread the insane cop's posts....

Also, I don't understand why everyone was like "oh pythag011 couldn't have killed rt_08."

Mafia kills were weird. james4l guessed that dragon96's kills were ftw-related, but why did the hackers kill me...?

Maybe you were half-experienced? (wickedestjr was Hacker Disabler and thus was in control of killing on the behalf of the hackers)

Mini-FAQ:

1) When will we see the role list?
Soon. Very soon.

2) How did that lynch criterion thingy work?
Ok, so there was Hacker Elite (eggylv999) and Game Mod: Deal or No Deal (142857). eggy gets to choose a number from 26-50 and 142857 gets to choose a number from 1-25. Add them up and that is the percentage of votes (rounded to the nearest integer) needed to lynch. I'm sorry if I was one vote off, but remember this is VOTES and not PEOPLE (ccm was a doublevoter, both multiaccounts are doublevoters (as long they both are alive), rt_08 and alexhhmun had no votes)

3) About the uninformed cult?
No, well they were notified but as with, say, Dojo, they might have missed the notification.

4) And the cultist being "SK/Vig"?
Well they got the role ONLY when they beat out the other three groups.

[quote="pythag011"]WHAT?

You told me they were uninformed cult...[/quote]

No, misquote: "Well, if they forget that they were cult then the are uninformed... lol, it's like an uninformed majority versus an uninformed minority"

[quote="1=2"][b]do not replace: 1=2[/b]

COUNTERROLECLAIM!

[b]I[/b] am the inspector. [b]vote: wickedestjr[/b]

I inspected budi729 on night 2 and found that he was innocent. I don't know what kind of cop I am, though...[/quote]

I know I'm reviving, but this is for stressing how important it is to read the topic.

lastpost (this is used to preserve order of locked topic)


I think $\frac{2}{9}$ is right I've tried a proof by contradiction but can't seem to make it work.

[hide]Let $AQ=x.$ Because triangles $APQ$ and $BPR$ both contain a right and a $45^\circ$ angle, they are isosceles-right. Hence, $PQ=RC=x$ and $QC=PR=BR=1-x.$

Now let's consider when $\frac13 <x<\frac23,$ or else one of triangles $APQ$ and $PBR$ will automatically have area greater than $\frac29.$ In this case, $[PQCR]>[ABC]-[APQ]-[PBR]\frac29.$ Therefore, one of these three figures will always have area greater than $\frac29,$ regardless of where $P$ is chosen.[/hide]

[quote][color=darkred]Let $ABC$ be the $C$-right-angled isosceles triangle whose equal sides have length $1$ . For $P\in [AB]$ denote the feet of the

perpendiculars from $P$ to the other sides are $Q\in CA$ and $R\in CB$ . Consider the areas of the triangles $APQ$ and $PBR$

and the area of the rectangle $QCRP$ . Prove that regardless of how $P$ is chosen, the largest of these three areas is at least $\frac 29$ .[/color][/quote][b][u]Proof[/u].[/b] Denote $\left\{\begin{array}{c}
QA=QP=CR=x\\\\
RB=RP=CQ=1-x\end{array}\right\|$ and $\left\{\begin{array}{ccc}
m=[AQP] & = & \frac {x^2}{2}\\\\
n=[BRP] & = & \frac {(1-x)^2}{2}\\\\
p=[CQPR] & = & x(1-x)\end{array}\right\|$ . Prove easily that $\max\{m,n,p\}=$

$\left\{\begin{array}{ccc}
n & \mathrm{if} & 0\le x\le \frac 13\\\\
p & \mathrm{if} & \frac 13\le x\le \frac 23\\\\
m & \mathrm{if} & \frac 23\le x
\le 1\end{array}\right\|$ and $\left\{\begin{array}{ccccc}
n\ge \frac 29 & \iff & 9x^2-18x+5\ge 0 & \iff & x\le\frac 13\\\\
p\ge\frac 29 & \iff & 9x^2-9x+2\le 0 & \iff & \frac 13\le x\le \frac 23\\\\
m\ge \frac 29 & \iff & 9x^2-4\ge 0 & \iff & \frac 23\le x\end{array}\right\|$ . Thus,  $\max\{m,n,p\}\ge \frac 29$ .

[quote=chess64]Let $ABC$ be the right-angled isosceles triangle whose equal sides have length 1. $P$ is a point on the hypotenuse, and the feet of the perpendiculars from $P$ to the other sides are $Q$ and $R$. Consider the areas of the triangles $APQ$ and $PBR$, and the area of the rectangle $QCRP$. Prove that regardless of how $P$ is chosen, the largest of these three areas is at least $2/9$.[/quote]

There is some ambiguity here, (in this diagram D and E) Q and R can be on either sides BC and AC or AC and BC, in that order. We look at the cases separately. In the first case, the areas of the two triangles combined are 2(PQ*PR)/2, plus the rectangle is 2PQ*PR=1/2, so the rectangle that has the largest area =PQ*PR=1/4=2/8>2/9.

Let PQ=x and PR=y. In the second case, we have APQ and PBR are isosceles right triangles (similar to ABC), so 1=AQ+QC=PQ+PR=x+y, or y=1-x. Clearly x and y must be between 1/3 and 2/3, since if not then either x^2/2 or y^2/2 would be at least (2/3)^2/2=4/9/2=2/9. In this case, x(1-x)=-x^2+x=-(x-1/2)^2+1/4 has minimum -(1/3-1/2)^2+1/4=2/9 and maximum 1/4, which makes the rectangle always have area at least 2/9.

Can somebody tell me how to rigorously say that x(1-x) takes on values from 2/9 to 1/4 for 1/3<x<2/3 without graphing? The best way i think of it is just to graph the graph and that’s really easy to prove, either on paper or on desmos by showing that for 1/3<x<2/3 the graph is increasing from 1/3<x<1/2 and decreasing from 1/2<x<2/3.


What is the length of the diagonal of a square with side length
$ 50\sqrt{2}$ cm? Express your answer in simplest form.

Expanding on AIME's idea.

We have a square.

[asy]draw((0,0)--(10,0)--(10,10)--(0,10)--cycle);[/asy]

We can form an isosceles right triangle with side $ 50\sqrt2$.

[asy]draw((0,0)--(10,0)--(0,10)--cycle);[/asy]

The hypotenuse of a 45-45-90 triangle is $ \text{leg}\sqrt2$, so the answer, as AIME stated, is $ \boxed{100}$.


geometry
3D geometry
tetrahedron
parallelogram

Let $E$ be a finite set of points in the space such that $E$ is not contained in a plane and no three points of $E$ are collinear. Show that $E$ contains the vertices of tetrahedron $T = ABCD$ such that the projection of $A$ on the plane $BCD$ is inside the triangle which is similar to the triangle $BCD$ and whose sides have mid-points $B,C,D,$ and only the vertices of $T$ belong to $E.$

What exactly do you mean by "only the vertices of $T$ belong to $E$?"

I think he wanted to say that the vertices of the ortopedal triangle are not in $E$. So it starts like this. We have 4 non-coplanar points. We know that in every tetrahedron there is a vertice from which the altitude on the other face falls in the interior of the triangle. So let's take $A$ to be this vertice. If there is one of the vertices of the ortopedal triangle in $E$, we can't choose $A,B,C,D$. There can't be 2 of these vertices in the set (collinearity) so we work with this 5-th point F, and A,B,C, where the parallelogram formed on the base is $DBFC$.  Also in this triangle there is a vertice from which the altitude falls in the interior of the other face (and this vertice can't be $A$ again, right?). So we now find another point on one of the faces of this tetrahedron, a vertice of an ortopedal triangle. If we show that this doesn't end somewhere(so become cyclic), we will always find another point in $E$, which is finite. Further more I coudn't find an argue of this.


In how many ways can two integers be selected from the integers 1, 2, 3, …..100 so that their difference is exactly 7?

a. 23 b. 47 c. 93 d. 100 e. NG

[hide] We can select any two differences from $ 8 \minus{} 1$ to $ 100 \minus{} 93$. Counting this up as either the number of integers from $ 8$ to $ 100$ or from $ 1$ to $ 93$, we see that there are a total of $ \boxed{93}$ ways to select two integers, so C.[/hide]


calculus
integration
inequalities proposed
inequalities

Are you sure? I think it must is \[\sum_{i,j=1}^{n}\frac{a_{i}a_{j}}{1+|i-j|}\geq 0.\]

[quote="chien than"]
I think this ineq not true with $a_{1}a_{j}$ negative[/quote]
Why you think like that? :D It is true  :wink:

and what about if one is $+$ and the other is $-$
and $i=j$

[quote="chien than"][quote="N.T.TUAN"]Are you sure? I think it must is
\[\sum_{i,j=1}^{n}\frac{a_{i}a_{j}}{1+|i-j|}\geq 0. \]
[/quote]
I think this ineq not true with $a_{1}a_{j}$ negative[/quote]

OK. How to solve the above problem? Could you post a solution now? Integral?

It is too advance and unthinkable. I remember that an elementary solution is existed.