For $m,n\in\mathbb{N}$ and $a_{i}>0$ where $1\leq i\leq m$ Prove that:

$\sqrt[n]{a_{1}+a_{2}+\cdots+a_{m}}+\sqrt[n]{a_{2}+\cdots+a_{m}}+\cdots+\sqrt[n]{a_{m}}$

$\geq\sqrt[n]{a_{1}+2^{n}a_{2}+\cdots+m^{n}a_{m}}$

It`s well. my calculator was setted in other base :blush: I think i`m going to kill him:D

${\sum_{s=1}^{m}\left(\sum_{i=s}^{m}\alpha_{i}\right)^{1/n}\geq \left(\sum_{j=1}^{m}j^{n}\alpha_{j}\right)^{1/n}}$

just rewriting. dont know how to solve :)

Nice problem!

Let $b_{k}: = (a_{k}+\ldots+a_{m})^{1/n}$, then $b_{1}\ge \ldots \ge b_{m}$ and it suffices to show that
$(b_{1}+\ldots+b_{m})^{n}\ge \sum_{1 \le k \le n}b_{k}^{n}(k^{n}-(k-1)^{n})$.

But this can be shown just be expanding the LHS: By the multinomial expansion
the LHS can be written as a sum of monomials $b_{1}^{n_{1}}\ldots b_{m}^{n_{m}}$.
If $n_{m}= \ldots = n_{k+1}= 0$ and $n_{k}\neq 0$ we estimate this by
$b_{1}^{n_{1}}\ldots b_{k}^{n_{k}}\ge b_{k}^{n}$. Hence the result follows from
$\sum_{n_{1}+\ldots+n_{k}= n, n_{k}\neq 0}\binom{n}{n_{1}\ldots n_{k}}$
$= \sum_{n_{1}+\ldots+n_{k}= n}\binom{n}{n_{1}\ldots n_{k}}-\sum_{n_{1}+\ldots+n_{k-1}= n}\binom{n}{n_{1}\ldots n_{k-1}}$
$= k^{n}-(k-1)^{n}$.


Given real $ a, b, c$. Prove that:
$ \frac {(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)^3}{(a\minus{}b)(b\minus{}c)(c\minus{}a)}[\frac {1}{(a\minus{}b)^3} \plus{} \frac {1}{(b\minus{}c)^3} \plus{} \frac {1}{(c\minus{}a)^3}] \leq\ \frac {9}{4}$

[quote="nguoivn"]Given real $ a, b, c$. Prove that:
$ \frac {(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)^3}{(a \minus{} b)(b \minus{} c)(c \minus{} a)}[\frac {1}{(a \minus{} b)^3} \plus{} \frac {1}{(b \minus{} c)^3} \plus{} \frac {1}{(c \minus{} a)^3}] \leq\ \frac {9}{4}$[/quote]
I have obtained that a maximal value of
$ \frac {(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)^3}{(a \minus{} b)(b \minus{} c)(c \minus{} a)}\left(\frac {1}{(a \minus{} b)^3} \plus{} \frac {1}{(b \minus{} c)^3} \plus{} \frac {1}{(c \minus{} a)^3}\right)$
 is $ \minus{} \frac {405}{16}$ and occurs for example for $ a\equal{}0,$ $ b\equal{}2$ and $ c\equal{}1.$ :maybe:

@arquady: Yes, I checked and I also think Max is $ \frac { \minus{} 405}{16}$
Can you check for me my solution with $ \frac{9}{4}$ , is it true?
Let$ x \equal{} \frac {(a \minus{} c)(a \minus{} b)}{(b \minus{} c)^2}$ ; $ y \equal{}$...;$ z \equal{}$....
We can prove easily that: $ xyz \equal{} 1$ and $ xy \plus{} yz \plus{} zx \equal{} 3$
$ \Rightarrow (1 \plus{} x)(1 \plus{} y)(1 \plus{} z) \equal{} x \plus{} y \plus{} z \plus{} 3$
$ \Rightarrow (x \plus{} y \plus{} z)(1 \plus{} x)(1 \plus{} y)(1 \plus{} z) \equal{} (x \plus{} y \plus{} z \plus{} 1.5)^2 \minus{} 2.25 \geq\ \minus{} 2.25$
$ \Leftrightarrow \frac {(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)^3}{(a \minus{} b)(b \minus{} c)(c \minus{} a)}[\frac {1}{(a \minus{} b)^3} \plus{} \frac {1}{(b \minus{} c)^3} \plus{} \frac {1}{(c \minus{} a)^3}] \leq\ \frac {9}{4}$

[quote="nguoivn"]
Let$ x \equal{} \frac {(a \minus{} c)(a \minus{} b)}{(b \minus{} c)^2}$ ; $ y \equal{}$...;$ z \equal{}$....
We can prove easily that: $ xyz \equal{} 1$ and $ xy \plus{} yz \plus{} zx \equal{} 3$
[/quote]
I have obtained that $ xyz \equal{} \minus{}1$ and $ xy \plus{} yz \plus{} zx \equal{} \minus{}3.$ :wink:

Sr I posted wrong. Exactly, we have: $ xyz\equal{}\minus{}1$ and $ xy\plus{}yz\plus{}zx\equal{}3$.
And the solution above still true. I think equality can occurs.  :wink:

[quote="nguoivn"]
$ xy\plus{}xz\plus{}yz\equal{}3.$
[/quote]
Yes you are right!
[quote="nguoivn"]
And the solution above still true. I think equality can occurs.  :wink:[/quote]
If do you mean the following inequality:

$ \frac {(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)^3}{(a\minus{}b)(b\minus{}c)(c\minus{}a)}[\frac {1}{(a\minus{}b)^3} \plus{} \frac {1}{(b\minus{}c)^3} \plus{} \frac {1}{(c\minus{}a)^3}] \leq\ \frac {9}{4},$

what your example when equality occurs? Thank you!

Indeed, I also don't know equality occurs or not. But the proof above seems to be true.
So, I think you can post your proof with $ \frac {\minus{}405} {16}$.
We are going to discuss about it  :)

[quote="nguoivn"]Indeed, I also don't know equality occurs or not. But the proof above seems to be true.[/quote]
Your proof of your inequality is true and very nice!
But the equality does not occur because not exists real numbers $ x,$ $ y$ and $ z$ such that $ xyz\equal{}\minus{}1,$ $ xy\plus{}xz\plus{}yz\equal{}3$ and $ x\plus{}y\plus{}z\equal{}\minus{}1.5.$ :wink:
[quote="nguoivn"]
So, I think you can post your proof with $ \frac { \minus{} 405} {16}$.
We are going to discuss about it  :)
[/quote]
Here is my proof:
Let $ a\equal{}\min\{a,b,c\},$ $ b\equal{}a\plus{}x$ and $ c\equal{}a\plus{}y.$ Then $ x$ and $ y$ are positive different numbers.
Let assume also $ \frac{x}{y}\plus{}\frac{y}{x}\equal{}u.$
Hence,
$ \frac {(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)^3}{(a\minus{}b)(b\minus{}c)(c\minus{}a)}\left(\frac {1}{(a\minus{}b)^3} \plus{} \frac {1}{(b\minus{}c)^3} \plus{} \frac {1}{(c\minus{}a)^3}\right)\equal{}$
$ \equal{}\frac{(x^2\minus{}xy\plus{}y^2)^3((\minus{}x)^3(x\minus{}y)^3\plus{}(x\minus{}y)^3y^3\plus{}y^3(\minus{}x)^3)}{x^4y^4(x\minus{}y)^4}\equal{}$
$ \equal{}\minus{}\frac{(x^2\minus{}xy\plus{}y^2)^3((x^2\minus{}2xy\plus{}y^2)^2(x^2\plus{}xy\plus{}y^2)\plus{}x^3y^3)}{x^4y^4(x^2\minus{}2xy\plus{}y^2)^2}\equal{}$
$ \equal{}\minus{}\frac{(u\minus{}1)^3((u\minus{}2)^2(u\plus{}1)\plus{}1)}{(u\minus{}2)^2}\equal{}\minus{}\frac{(u\minus{}1)^3(u^3\minus{}3u^2\plus{}5)}{(u\minus{}2)^2}.$
Let $ f(u)\equal{}\minus{}\frac{(u\minus{}1)^3(u^3\minus{}3u^2\plus{}5)}{(u\minus{}2)^2}.$
Hence, $ f'(u)\equal{}\minus{}\frac{(u\minus{}1)^2(2u\minus{}5)(2u^3\minus{}6u^2\plus{}3u\plus{}4)}{(u\minus{}2)^3}.$
Easy to show that $ 2u^3\minus{}6u^2\plus{}3u\plus{}4>0$ for all $ u>2.$
It gives $ \max_{(2,\plus{}\infty)}f\equal{}f(2.5)\equal{}\minus{}\frac{405}{16}.$ :)

Yes, thank you arquady. I also think about it but I haven't patient enough to complete :D
And with $ \frac{9} {4}$, the equality is not occurs. Your explaination is very exactly :D


geometry probability AMC AIME HMMT Princeton college \/closed

The Art of Problem Solving Spring Schedule is available [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]here[/url].

Here are the classes we will offer this spring:

[b]Introductory level; Grades 6-10[/b]
*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=35]Introduction to Algebra[/url].  Wednesdays, Mar 4 - Sept 2, 7:30 PM - 9 PM ET (4:30 PM - 6 PM PT).
*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=10]Introduction to Geometry[/url].  Tuesdays, Mar 3 - Aug 11, 7:30 PM - 9 PM ET (4:30 PM - 6 PM PT).
*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=1]Introduction to Counting & Probability[/url].  Mondays, Mar 2 - May 18, 7:30 PM - 9 PM ET (4:30 PM - 6 PM PT).

[b]Intermediate level; Grades 8-12[/b]
*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=3]Intermediate Counting & Probability[/url].  Mondays, Mar 9 - Jul 13, 7:30 PM - 9 PM ET (4:30 PM - 6 PM PT).
[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=8]Special AIME Problem Seminar[/url].  Sat + Sun, Mar 7 - 8, 3:30 PM - 6:30 PM ET (12:30 PM - 3:30 PM PT).

[b]High school olympiad level[/b]
[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=7]Olympiad Geometry[/url]. Wednesdays, Feb 4 - Apr 22, 7:30 PM - 9:30 PM ET (4:30 PM - 6:30 PM PT).

Classes marked with a (*) have an accompanying textbook.  If you plan to enroll in one of these courses, you should do so soon, so that you receive the text before the class starts.

All of our courses offer full transcripts for each class, so if you're going to miss a class or two, you'll still have access to everything that happened in class.

We will host FREE Math Jams to discuss the classes.  The Math Jam schedule can be found [url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]here[/url].

If you have any questions about a specific class, please contact us at classes@artofproblemsolving.com.

Woot!  I'm taking Intermediate C&P, what are y'all taking?

Interm C&P isn't strictly a "contest prep" class like the Special AIME Seminar is.  It does have problems from the AIME, from HMMT, and from various olympiads and the Putnam.  There are Challenge Sets for the Intermediate Counting class, so you'll have the chance to have your work on these problems evaluated.

[quote="simo14"]For people who have taken the AIME seminar before, how much have you improved?
Is it worth it to cram?

EDIT: Also, for Interm C&P, is it more focused on AIME, or will we also get to do Olympiad proofs? are the proofs (if any) graded?
Thanks[/quote]

I took it last year in 5th grade and improved my score from practices (0,0,0,1,1,2) to a relatively good score (5)

i've taken intro to c + p, and studied from intro to geometry and number theory, and currently studying intermediate algebra. when i finish that, where should i go from there? (besides intermediate c + p) i'm considering taking one of the contest/problem series course.

if you've mastered the intros (do the "do i need this" tests), then take all of the "are you ready" tests for the intermediates. for the ones that you pass (if any), take the "do you need this". that should help you zero in on your weaknesses.

qwerty - AMC 12 or AIME Problem Series could well be a fit for you.

If you have taken Discrete Mathematics, should you take the Intermediate course for Counting & Probability???

That depends on the emphasis (and quality) of the discrete math class you took.  The one I took at Princeton 21 years ago would have been an adequate warm-up for the Intermediate Counting class, but not a replacement for it -- it maybe covered half the material we cover in the Interm Counting class.  You can probably best judge for yourself by looking at the [url=http://www.artofproblemsolving.com/Classes/IntermCounting/PostTest.pdf]Post Test[/url] for the class.

Okay, so I guess if I get into Princeton, I should get a math degree there, and then after I finish enroll in all the intermediate and olympiad classes here at AoPS the following year. I'll be "well-prepared".

About the olympiad geometry class, if we can answer most geometry problems about the basics and stuff, and let's say AIME 1-12, but are not so good at locus problems and have hardly ever been able to solve an olympiad geometry problem because our brains overload and commence a shut-down sequence when a diagram has too many lines and angles, would you say this class is about right, or might it be too much and just make us go crazy and have a lot of trouble following in class?

[quote="Duelist"]Okay, so I guess if I get into Princeton, I should get a math degree there, and then after I finish enroll in all the intermediate and olympiad classes here at AoPS the following year. I'll be "well-prepared".[/quote]
To be fair, the class was taught in the CS department, and was probably an easier class than it should have been.  The prof was teaching it for the first time and probably didn't have his difficulty calibration quite right yet.

[quote="Duelist"]
About the olympiad geometry class, if we can answer most geometry problems about the basics and stuff, and let's say AIME 1-12, but are not so good at locus problems and have hardly ever been able to solve an olympiad geometry problem because our brains overload and commence a shut-down sequence when a diagram has too many lines and angles, would you say this class is about right, or might it be too much and just make us go crazy and have a lot of trouble following in class?[/quote]

Sounds like you're in the target audience for the class.  Occasionally you'll suffer the shut-down sequence in the class, but that's a good thing -- if you're not pushing your limits, you're not learning anything.  (Of all the subjects I studied for math contests, I think geometry was the most helpful -- not for the subject matter, but for the problem solving skills.  As you note, the complexity of the problems can threaten overload because there's just so much going on.  It's not at all unlike trying to get a small business off the ground.)

Is there any financial aid for the classes?

No - rather than structure a financial aid program, our classes are significantly less expensive than other online classes for gifted students offered by various "gifted ed" organizations.  (Not to mention all the free resources on the site.)

How do you do #1 on the Are you ready? test for olympiad geo?

Hints: [hide]Find CP.[/hide]

And

[hide]
There's a 60 degree angle.  What kind of triangles should you look for?[/hide]

Are we going to have other olympiad classes like Olympiad Combinatorics or something? Because in the WOOT lectures we didn't have much material because it was only one lecture every two weeks so like we didn't cover more difficult material (not necessarily advanced theory) that well. Also combinatorics is really popular and maybe you should conduct a poll or something because sometimes there are people that would sign up for a class if they knew about it. Also although you don't support learning advanced theory some of our WOOT problems could not be solved without it.
If your not going to have the class can you just sell the transcripts and problem sets or something?

[quote="hello123"]Are we going to have other olympiad classes like Olympiad Combinatorics or something? Because in the WOOT lectures we didn't have much material because it was only one lecture every two weeks so like we didn't cover very advanced material that well.[/quote]

We don't have any others planned besides Olympiad Geometry.

(I'll also note that mastering olympiad-level problem solving is somewhat orthogonal to mastering "very advanced material".  It's easy to learn very advanced tools.  It's hard to learn how to solve very hard problems.)

[quote="hello123"]Are we going to have other olympiad classes like Olympiad Combinatorics or something? [/quote] You should check my comment in [url=http://www.mathlinks.ro/weblog_entry.php?t=252502#comments]Richard Rusczyk's blog entry on programming.[/url] Programming is useful to have motivation to learn [url=http://en.wikipedia.org/wiki/Discrete_mathematics]discrete math[/url] which covers topics combinatorics, graph theory, algorithmics (useful as proof technique to show a process terminates under certain conditions with some final result), counting and information theory (basically just a different view on probability, statistics as it provides information how to efficiently compress, encode etc. information and transmit it over a possibly noisy channel with some bandwith) among other things. [url=http://www.artofproblemsolving.com/AboutUs/AoPS_A_Company.php]ML/AoPS[/url] staff members have crazy skills and an amazing experience in problem-solving issues that the disadvantages of taking their courses are neglectable I feel.


real analysis
search
topology
calculus
complex analysis

can anyone please post an e-book version of W L RUDIN'S mathematical anlysis

is it Principles of mathematical analysis, W.Rudin ?

yes it is principles of mathematical analysis by W L RUDIN

If so, here is it http://djvu.504.com1.ru:8019/WWW/a34c119e403e86bc213659e2e714ff19.djvu

from http://www.lookforbook.com/cgi-bin/search.cgi?lang=en&st=Rudin+&network=1

[quote="pardesi"]thanks again[/quote]
No, problem . You can find a lot of books from site http://lookforbook.com/  :wink:

well i can  search for many books then what i get is an dialog asking for password(in case of  [url]http://lib.homelinux.org[/url])
futher in other sites everything is russian ...can anyone help me with the password or the russian :(

Me too, but my friends can send to me those books. What is book's name ? Maybe I will help you  :maybe:

How do you get the book from that point?  I get some website with crazy characters on it.

[quote="agolsme"]how about this book:

3. Barbeau E. Pell`s Equation Springer, 2003 0387955291 225 p. English pdf, 1103 KB 4.9 KB/p

I found that on lookforbook.com but can't access because of that password.[/quote]
Please send e-mail of you to me.

[quote="pardesi"]well can you get me a copy of al fohr's complex analysis or any other really good book on complex analysis in pdf(if possible)  and a good book on topology please. :maybe:[/quote]
I haven't got those books, but I will try request to my friends. What is your e-mail?  :wink:

well sorry N.T.TUAN i have pm'ed u twice my id . one of them is right (sorry for the mistake  :oops: )

i got it thanks N.T.TUAN :lol:
[url]http://www.lookforbook.com/cgi-bin/search.cgi?lang=en&st=ahlfors&network=1[/url]

[quote="pardesi"]
[url]http://www.lookforbook.com/cgi-bin/search.cgi?lang=en&st=ahlfors&network=1[/url][/quote]
Thanks  :oops: I download this book.

Hi, I've been really interested in this site. Could I have the username and password as well? These books can be pretty expensive here in the states!

Most of these sites are run by the russian mafia.

two things
1. These sites might be illegal  :o (in usa atleast if you are in india don't worry)
2. How is the book analysis? I mean it seems like its just calculus 1


calculus
integration
trigonometry
function
derivative
calculus computations

every time i try to integrate cos(4x) over dx i get (-sin(4x))/4. can someone please list a step by step method of doing it correctly?

edit:nvm, I was differentiating when i was supposed to be integrating. lol

Given $ I \equal{} \int \cos 4x dx$. Set $ 4x \equal{} t$ so that you get $ dx \equal{} \frac{dt}{4}$.

Then $ I \equal{} \frac{1}{4}\cdot \boxed{\int \cos t dt}$. Can you find a function of t whose derivative is $ \cos t$ ? That function goes in the boxed portion and make sure to replace t with 4x in that function.


geometry
circumcircle
incenter
geometric transformation
reflection
perpendicular bisector
power of a point

Dear Mathlinkers,
I recall the Feuerbach's theorem (wrt triangle ABC, the incircle is tangent to the nine point circle at Fe)
and the Feuerbach's tangent (the tangent to the incircle at Fe).
The problem :
let 0 be the circumcircle of ABC, I the incenter of ABC, A' be the midpoint of arc BC which doesn't contain A, (F) the Feurbach's tangent, X the intersection of (F) with BC and I' the symmetric of I wrt BC.
Prouve : X, I' and A' are collinear.
Sincerely
Jean-Louis

$ O$ is the triangle circumcenter, $ M$ is midpoint of the segment $ BC.$ $ D, D_a$ are tangency points of the incircle $ (I)$ and A-excircle $ (I_a)$ with $ BC.$ $ (M)$ is a circle with center $ M$ and radius $ MD \equal{} MD_a.$ The 9-point circle $ (N)$ tangent to $ (I), (I_a)$ at $ F, F_a$ goes through $ M$ and $ (I), (I_a)$ are perpendicular $ (M)$ $ \Longrightarrow$ inversion in $ (M)$ takes $ (I), (I_a)$ to themselves and $ (N)$ to the common internal tangent $ F^*F_a^*$ of $ (I), (I_a)$ other than $ BC \equiv DD_a.$ By the basic properties of inversion, $ \triangle MFD \sim \triangle MDF^*$ are similar, $ \angle MFD \equal{} \angle MDF^* \equal{} \angle AA'O,$ the last equality because $ DF^* \perp AA'$ and $ MD \perp A'O.$

Let $ p$ be perpendicular bisector of the segment $ ID.$ Reflect $ M, F$ in $ p$ into $ M', F'.$ Also reflect the incircle $ (I)$ into a congruent circle $ (D) \cong (I)$ centered at $ D.$ Obviously, $ II'$ is a diameter of $ (D),$ $ F' \in (D)$ and $ DF, IF'$ intersect at a point $ R \in p.$ Let $ (Q)$ be a circle with diameter $ IA'.$ The angle $ \angle IM'A'$ is right, $ M' \in (Q).$ The angles $ \angle M'F'I \equal{} \angle MFD \equal{} \angle AA'O \equal{} \angle IA'M'$ are equal $ \Longrightarrow$ $ F' \in (Q).$ Thus $ F'$ is intersection of $ (D), (Q)$ other than $ I.$ Radical axis $ p$ of $ (I), (D)$ and radical axis $ IF'$ of $ (D), (Q)$ meet at the radical center $ R$ of $ (I), (D), (Q).$ Radical axis $ q$ of $ (Q), (I)$ also goes through $ R.$ Since the circles $ (D), (Q)$ with diameters $ II', IA'$ pass through the center $ I$ of $ (I),$ $ I'$ is the pole of $ p$ and $ A'$ is the pole of $ q$ with respect to $ (I).$ Obviously, $ X$ is the pole of $ DF$ with respect to $ (I).$ Since $ p, q, DF$ concur at $ R,$ their poles $ I', A', X$ are collinear.

$(N)$ is the 9-point circle of $\triangle ABC,$ $M$ is the midpoint of $BC,$ $V \equiv AA' \cap BC$ is the foot of the internal bisector of $\angle BAC,$ $D$ is the tangency point of $(I)$ with $BC$ and the tangent from $V$ to $(I),$ different from $BC,$ touches $(I)$ at $F.$  Since $ID$ and the A-altitude of ABC are parallel, then their reflections about $AV$ are also parallel $\Longrightarrow$ $IF \parallel OA \parallel MN$ $\Longrightarrow$ $MF$ goes through the exsimilicenter $F_e$ of $(I) \sim (N).$ If $U \equiv F_eX \cap VF,$ then $(I)$ becomes the U-excircle of $\triangle UXV$ $\Longrightarrow$ $(D,M,X,V)=-1.$ But, if  $A'I'$ cuts $BC$ at $X',$ we have

$\frac{\overline{X'D}}{\overline{X'M}}=\frac{\overline{DI'}}{\overline{MA'}}=-\frac{\overline{DI}}{\overline{MA'}}=-\frac{\overline{VD}}{\overline{VM}}\Longrightarrow$ $(D,M,X',V)=-1,$

which means that $X \equiv X',$ i.e. $A',I',X$ are collinear, as desired.


function calculus derivative algebra domain

For values of $ x$ less than $ 1$ but greater than $ \minus{}4$, the expression
\[ \frac{x^2 \minus{} 2x \plus{} 2}{2x \minus{} 2}
\]
has:

$ \textbf{(A)}\ \text{no maximum or minimum value}\qquad \\
\textbf{(B)}\ \text{a minimum value of }{\plus{}1}\qquad \\
\textbf{(C)}\ \text{a maximum value of }{\plus{}1}\qquad \\
\textbf{(D)}\ \text{a minimum value of }{\minus{}1}\qquad \\
\textbf{(E)}\ \text{a maximum value of }{\minus{}1}$

[hide="Solution"]Taking the first derivative, we get $ f'(x) \equal{} \frac {x^2 \minus{} 2x}{2x^2 \minus{} 4x \plus{} 2}$.

Solving $ f'(x) \equal{} 0$ for $ x$, we get $ x \equal{} \left\lbrace0,\,2\right\rbrace$. Since $ x\equal{}2$ isn't in the interval $ \left[ \minus{} 4,\,1\right]$, we only need to consider $ x \equal{} 0$.

Taking the second derivative, we get $ f''(x) \equal{} \frac {1}{x^3 \minus{} 3x^2 \plus{} 3x \minus{} 1}$.

At $ x \equal{} 0$, $ f''(0) \equal{} \frac {1}{ \minus{} 1} \equal{} \minus{} 1$. Since $ \minus{} 1 < 0$, the function has a local maximum at $ x \equal{} 0$.

Evaluating $ f(x)$ at $ x \equal{} 0$ yields $ f(0) \equal{} \frac {0 \minus{} 0 \plus{} 2}{0 \minus{} 2} \equal{} \frac {2}{ \minus{} 2} \equal{} \minus{} 1$.

Thus, the function has a local maximum value of $ \minus{} 1$ at $ x \equal{} 0$, so the answer is $ \text{(E)}$.[/hide]

[hide="without calculus"]

$ \frac{x^2\minus{}2x\plus{}2}{2x\minus{}2}\equal{}\frac{x\minus{}1}{2}\plus{}\frac{1}{2x\minus{}2}\equal{}\minus{}\frac{1}{2}[(1\minus{}x)\plus{}\frac{1}{1\minus{}x}]$

Let $ y\equal{}1\minus{}x$, then $ y \in (0,5]$.

$ y\plus{}\frac{1}{y} \ge 2$ by AM-GM, equality holds $ \Leftrightarrow y\equal{}1$, which is within the domain of $ y$.

So $ y\plus{}\frac{1}{y} \ge 2 \Leftrightarrow \minus{}\frac{1}{2}(y\plus{}\frac{1}{y}) \le \minus{}1$

So the answer is $ \text{(E)}$
[/hide]


Let n be a positive integer such that $ 2\plus{}2\sqrt{28n^{2}\plus{}1}$ is an integer. Show that $ 2\plus{}2\sqrt{28n^{2}\plus{}1}$ is the square of an integer.

$ 28n^2\plus{}1$ must be a square, and it is 1 modulo both 4 and 7, so it must be the square of a number of the form $ 14k\pm 1$: $ 28n^2\plus{}1\equal{}(14k\pm 1)^2\equal{}196k^2\pm 28k\plus{}1\equal{}28k(7k\pm 1)\plus{}1$. then $ n^2\equal{}k(7k\pm 1)$. since $ k$ and $ 7k\pm 1$ are relatively prime, they must both be squares: $ k\equal{}x^2, 7k\pm 1\equal{}y^2$. in the "$ \minus{}$" case, we get $ 7x^2\minus{}1\equal{}y^2$, which has no integer solutions. so the "$ \plus{}$" case must hold. then the expression equals $ 2\plus{}2(14k\plus{}1)\equal{}4(7k\plus{}1)\equal{}4y^2$, which is a square.

[quote="theSA"][b]yildobekir[/b] why are you posting PEN problems???
Go and see PEN section.[/quote]

yes, WHY?  :P


Show that any subset of $ V\equal{} \{ 1,2,...,24,25 \}$ with $ 17$ or more elements contains at least two distinct numbers the product of which is a perfect square.

[hide]The subset must not have more than one of $ 1, 4, 9, 16, 25$.
From the following pairs of numbers, $ (3,12),(2,18),(4,24),(5,20),(2,8)$,
the subset can at most have one number from each pair.
Since we are allowed to take away only $ 8$ numbers, such a subset is impossible.[/hide]


1) Find the minimum value for real number x:
sinx + cosx + tanx + cotx + secx +cscx

2) Prove that 1- cot 23 degrees = 2/1 - cot 22 degrees

Not sure. These were given by our teacher as extra credit problems.


Let $ ABC$ be an non-issosceles triangle.The points $ A_1$ and $ B_1$ are the feet of bisectors from the vertexes
$ A$ and $ B$,respectively. The line passing through the points $ A_1$ and $ B_1$ intersects the line $ AB$ at
the point $ D$. Show that $ m(\angle ACD)\plus{}(\angle BCD)\equal{}180^\circ$.

[color=darkblue]Suppose $ AC \ge BC$, then $ B \in AD$. By Menelaus theorem in $ \triangle{ABC}$ for transversal $ \overline{DA_1B_1}$ we have $ \dfrac{AB_1}{B_1C}\cdot \dfrac{CA_1}{A_1B}\cdot \dfrac{DB}{AD} \equal{} \dfrac{AB}{BC}\cdot \dfrac{CA}{AB}\cdot \dfrac{DB}{AD} \Longleftrightarrow \dfrac{DB}{DA} \equal{} \dfrac{CB}{CA} \Longrightarrow CD$ is exterior bisector of angle $ C$. So  $ \angle{DCB} \equal{} 90^\circ \minus{} \frac {1}{2}\angle{C}$ and $ \angle{ACD} \plus{} \angle{BCD} \equal{} 2(90^\circ \minus{} \frac {1}{2}\angle{C}) \plus{} \angle{C} \equal{} 180^\circ$.[/color]

[quote="Ahiles"][color=darkblue]$ \dfrac{DB}{DA} \equal{} \dfrac{CB}{CA} \Longrightarrow CD$ is exterior bisector of angle $ C$. [/color][/quote] I got up to here, can you show why this is true? Sorry if it is elementary.

[b][color=darkred]Exterior bisector theorem.[/color][/b]
[color=darkblue]Let $ ABC$ be a triangle with $ AB \not\equal{} AC$. If $ AD$ is exterior bisector of $ \angle{A}, D \in BC$, then $ \dfrac{DB}{DC}\equal{}\dfrac{AB}{AC}$.
[hide="Proof"]
Let parallel to $ AE$through $ B$ intersects $ AC$ at $ N$. Then $ \angle{ABN}\equal{}90^\circ\minus{}\frac{1}{2}\angle{A}, \angle{BAN}\equal{}\angle{A} \Rightarrow \triangle{BAN}$ is isosceles, $ AB\equal{}AN$, Also by Thales $ \frac{EC}{EB}\equal{}\frac{AC}{AN}\equal{}\frac{AC}{BC}$.
[/hide][/color]


geometry
geometric transformation
calculus
calculus computations

The rate if change of the volume V of water in a tank with respect to time T is directly proportional to the square root of the volume.  Which of the following is a differential equation that describes this relationship?

A.  V(t) = k * square root of T
B.  V(t) = k * square root of V
C.  dv/dt = K * square root of T
D.  dv/dt = k / square root of V
E.  dv/dt = k * square root of V

I think the answer is E, but I'm not too sure.  It'd be great if someone can explain this to me.  Anyway, thanks for the help.

It's (E): $\frac{dV}{dt}=k\sqrt{V}.$

There really isn't anything to explain; this is simply a direct translation into symbols of the words of the problem. There's nothing to work, even though this separable equation is pretty easy to solve.


partition
number theory
Ramsey Theory
Hi
Coloring
Extremal combinatorics
IMO Shortlist

Assume that the set of all positive integers is decomposed into $ r$ (disjoint) subsets $ A_1 \cup A_2 \cup \ldots \cup A_r \equal{} \mathbb{N}.$ Prove that one of them, say $ A_i,$ has the following property: There exists a positive $ m$ such that for any $ k$ one can find numbers $ a_1, a_2, \ldots, a_k$ in $ A_i$ with $ 0 < a_{j \plus{} 1} \minus{} a_j \leq m,$ $ (1 \leq j \leq k \minus{} 1)$.

Call a set $A \subset {\mathbb N}$ [i]good[/i] if it has the desired property, i.e. there is an $m$ such that for every $k$ there are $a_{1},a_{2},...,a_{k} \in A$ s.t. $0<a_{i+1}-a_{i} \leq m$, for $i=1,2,...,k-1$. Call a sequence $x$-bounded if it is strictly increasing and the difference between any two consecutive numbers is at most $x$.

[b]Lemma[/b]. I'll prove that if $A,B \subset {\mathbb N}$ such that $A \cap B =\emptyset$ and $A \cup B$ is good, then at least one of $A$ and $B$ is good.

For any $k$, take an $m$-bounded sequnce $c_{1},c_{2},...,c_{k} \in A \cup B$ and let $l_{k}$ be the largest number for which there is a $j$ such that all $c_{j}$, $c_{j+1}$, ..., $c_{j+l_{k}-1}$ belong to the same set ($A$ or $B$). If the sequence $l_{k}$ is unbounded, then we are done, so suppose it is bounded and take $L>l_{k},\forall k$. In any sequence $m$-bounded sequence from $A \cup B$, the two subsequences formed from numbers belonging to the same set are $(L+1)m$-bounded, so at least one set must contain arbitrarily long $(L+1)m$-bounded sequences.

Back to the problem, let $A_{1} \cup A_{2} \cup ... \cup A_{n}$ be a partition of the set of positive integers. If $A_{1}$ is not good, then $A_{2} \cup A_{3} \cup ... \cup A_{n}$ is good. If $A_{2}$ is not good, then $A_{3} \cup ... \cup A_{n}$ is good etc. In the end, you get that if $A_{n-1}$ is not good, then $A_{n}$ is good.

Solution by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=6798]Severius[/url]:

For $ r \equal{} 2$, if some subset contains arbitrarily long blocks of consecutive numbers, we're done with $ m \equal{} 1$. Otherwise they have maximal block lengths $ M_1, M_2$, so that any block of consecutive numbers in the same subset is of length at most $ M$ for some fixed $ M$, and any two consecutive elements of a subset differ by no more than $ M \plus{} 1$. So we're done with $ m \equal{} M \plus{} 1$.

Now suppose the problem is true for $ r$, and take it for $ r \plus{} 1$. Among the $ r \plus{} 1$ subsets, imagine that two of them are the same and apply the induction hypothesis. If the resulting good subset is one of the singles, we're done; otherwise, it must be the "double" subset, say, the union of $ A$ and $ B$. For each $ k$ we take a sequence $ a_1 < a_2 < ... < a_k$ contained in $ A \cup B$ and with $ \minus{} m \leq a_{i \plus{} 1} \minus{} a_i \leq m$, and we consider the collection of these sequences. If among them there are arbitrarily long subsequences of consecutive elements $ a_{i \plus{} 1}, a_{i \plus{} 2}, ..., a_{i \plus{} d}$ in the same subset, we're done. Otherwise, as above, the maximal length of such a block is bounded by some fixed $ M$, and in a sequence $ a_1 < a_2 < ... < a_k$, for any element $ a_i$, the smallest $ j > i$ such that $ a_j$ is in the same subset as $ a_i$ satisfies $ j \leq i \plus{} M \plus{} 1$, so it is easy to see that we are done.

Call a sequence of $n$ consecutive integers an $n$-string. Call a subset $A_i$ [i]bad[/i] if it does not satisfy the property: namely, that for any $m$, there exists a $k$, such that in any $k$ sequential elements of $A_i$, at least one adjacent pair differs by more than $m$. You can verify the following consequence: for any $m$, there exists a $p$ (namely $p=mk$) such that any $p$-string contains an $m$-string with no elements from $A_i$.

Now assume all $A_2$ through $A_r$ are bad. We claim $A_1$ is good: namely, we prove that for any $k$, we can find a $k$-string contained in $A_1$ (the case where $m=1$ ).

Given any $k$, define the numbers $p_1,p_2, ... p_r$ as follows: $p_1=k$, and for $i>1$, $p_i$ is the number such that any $p_i$-string contains a $p_{i-1}$-string with no elements from $A_i$. The existence of these numbers is ensured by: (1) the existence of $p_1$, (2) the badness of $A_2$ through $A_r$, and (3) the consequence of badness described earlier.

It follows that any $p_r$-string contains a $p_{r-1}$-string with no elements from $A_{r}$, and this string contains a $p_{r-2}$-string with no elements from $A_{r-1}$ or $A_r$, and this string contains ... etc. .... and this string contains a $p_1$-string with no elements from any of $A_2$ through $A_r$. This $p_1$-string must then be contained in $A_1$, so we are done.


function
logarithms
inequalities
real analysis
real analysis unsolved

$1.$ $(r_{n})^{\infty}_{n=1}$ is a decreasing sequence of real numbers. If $lim_{n\rightarrow \infty}r_{n}=0$ Prove that:
$\left|\sum^{\infty}_{k=n+1}(-1)^{k}r_{k}\right|<r_{n}$

$2.$ All of the roots of the equation $ax^{4}-bx^{3}+cx^{2}-x+1=0$ are positive. Prove that: $c\geq80a+b$.

$3.$ Find all continuous functions $f$ such that for real numbers $a_{1},a_{2},\cdots,a_{n}$ that $a_{1}a_{2}\cdots a_{n}=1$ we have : $a^{f(a_{1})}_{1}a^{f(a_{2})}_{2}\cdots a^{f(a_{n})}_{n}\geq1$

I got these three beautiful problems froma book, but i can't completely understand the solutions, please give $cute$ solutions....thanks

#1 is a standard lemma used to prove the alternating series test.
#2 is wrong.

#3 is is relatively simple with the substitution $g(\ln t)=f(t)$; the inequality becomes $\sum x_{i}g(x_{i})\ge 0$ whenever $\sum x_{i}=0$. This is a convexity condition; we have the desired result whenever $xg(x)$ lies above some supporting line through the origin. Conversely, we can construct a counterexample if some secant line has negative $y$-intercept.
The function $f$ works if and only if there exists a constant $c$ such that $f(x)\le c$ when $x<1$ and $f(x)\ge c$ when $c\ge 1$. This $c$ must be $f(1)$ by continuity at 1.

[quote="jmerry"]$(x-1)^{4}$. 6 is not greater than 81.[/quote]

$(x-1)^{4}$ would give $x$ a coefficient of $4$, where it is 1 in the given problem.  Then, if you try $(x/4-1)^{4}$ instead to take care of that, you end up with $a = \frac 1{256}$, $b = \frac 4{64}= \frac 1{16}$, and $c = \frac 6{16}= \frac{3}{8}$.

Thus, $80a+b = \frac{80}{256}+\frac 1{16}= \frac 5{16}+\frac 1{16}= \frac 6{16}= \frac{3}{8}= c$.

So your example was close, but is only the case where equality holds.

[quote="jmerry"]#1 is a standard lemma used to prove the alternating series test.[/quote]

You mean it is ok to use #1 as a theorem in solving problems...? :maybe:


Prove that for $ n>1$
\[ \frac{1}{n\plus{}1}\plus{}\frac{1}{n\plus{}2}\plus{}\dots\plus{}\frac{1}{2n}>\frac{13}{24}\]

[quote="Yuriy Solovyov"]Prove that for $ n > 1$
\[ \frac {1}{n \plus{} 1} \plus{} \frac {1}{n \plus{} 2} \plus{} \dots \plus{} \frac {1}{2n} > \frac {13}{24}\]
[/quote]

Let $ f(n)\equal{}\frac {1}{n \plus{} 1} \plus{} \frac {1}{n \plus{} 2} \plus{} \dots \plus{} \frac {1}{2n}$. We get $ f(n\plus{}1)\equal{}f(n)\minus{}\frac{1}{n\plus{}1}\plus{}\frac{1}{2n\plus{}1}\plus{}\frac{1}{2n\plus{}2}$ $ \equal{}f(n)\plus{}\frac{1}{2n\plus{}1}\minus{}\frac{1}{2n\plus{}2}$ $ >f(n)$ and so $ f(n)$ is strictly increasing.

So $ f(n)\geq f(2)\equal{}\frac{7}{12}>\frac{13}{24}$ $ \forall n>1$

We have $ \sum_{i \equal{} 1}^{n} \frac {1}{n \plus{} i} \ge \frac {7}{12} > \frac {13}{24}$.

edit: too late^^

[quote="PhilG"]f(n) -> ln(2)[/quote]

Right, but it does not help to answer the required question.


algebra
polynomial
function
number theory
prime numbers

How is expressed a prime number? I mean how is the comon form of a prime number?

Your questions is ambiguous.  If you are asking for a formula that gives you the n-th prime number, then I believe that there is none.  However, you could, for example, say that all primes $ p>3$ can be expressed in the form $ 6n\pm 1, n\in\mathbb{N}$.

You could say $ \mathbb{P} : \equal{} \{ p \in \mathbb{N}, p > 1 : \forall d \in \mathbb{N}, 1 < d < p : d \nmid p \}$.... However, it has been proved[1] that there exists no polynomial function that gives prime numbers for all integers.

[1] http://en.wikipedia.org/wiki/Formula_for_primes#Prime_formulas_and_polynomial_functions


Find the number of solutions of the equation
$ 16(x^{2} \plus{} 1) \plus{} \pi^{2} \equal{} |tanx| \plus{} 8\pi x$ in the
interval $ [ \frac{\minus{}\pi}{2},\frac{3\pi}{2}]$

Has anyone got this one???
I tried it hard but not able to get the solution!!!

Try moving $ 8\pi x$ over and completing the square on the LHS, then sketch a mediocre graph of both sides within the given range and look for intersections. There isn't really a good way to find the solutions, so don't try to.


a) Describe completely the locus of points 2 units from the line whose equation is x = 5.

b) Describe completely the locus of points n units from
point P(3,4).

c) What is the TOTAL number of points that satisfy the conditions in parts a and b above at the same time for n = 4?

[hide="answer"]a) The lines $x=7$ and $x=3$

b) The circle $(x-3)^{2}+(y-4)^{2}=n^{2}$

c) Line $x=3$ creates a chord but line $x=7$ is tangent to the circle.  3 points.
[/hide]

This is great but did you find the answers?

What are the steps?

Can you show me the steps ONE BY ONE?

++++++++++++++++++++++++++++++

What about this question below?

Lines AB and CD are parallel to each other and 8 inches apart.
Point P is located BETWEEN the two parallel lines and 2 inches from line AB.

a) Describe the locus of points that are:

1) equidistant from lines AB and CD

2) 3 inches from point P

b) How many points satisfy BOTH conditions in part a above?


superior algebra
superior algebra unsolved

Check whether <2+3i> is a prime ideal or maximal ideal in Z[i]


Let $f(x)=2x-sinx$. Prove that : there exists a real b and functions f(x) and g(x) satisfies the coditions
1. g(x)=bx+h(x)
2. h(x) is periodic
3. f(g(x))=x for all x

too many typo's... I can't understand a thing!

I am sorry
Let $f(x)=2x-sinx$. Prove that : there exists a real b and functions h(x) and g(x) satisfies the coditions
1. g(x)=bx+h(x)
2. h(x) is periodic
3. f(g(x))=x for all x


algebra
function
domain
analytic geometry
combinatorics unsolved
combinatorics

An 8x8 chess board is filled in with a number in each square in such a way that each number is equal to the average of all the numbers adjacent to it. A square is adjacent to another square if they both share an entire side (simply touching at the corner doesn't count). If the lower left hand corner square has the number 10 in it, then what are all the possible values of the upper right hand square?

can you generalize for an n x n board? $2^{n}$ x $2^{n}$?

I think that it can be proved that all numbers are
 equal ? i hope this is correct, but if we assume that there
 exist a smallest element say $S$ , then since the average of
 the surrounding numbers is $S$, it follows that either all
 are $S$ or atleast one is smaller, which is not allowed.
 So, all are $S$. Hence the entire board consist only this
 number, and it follows that the top-right square is also
 a $10$.  since we have not talked about $8X8$ boards
 anywhere, i believe this is a generalisation in itself??

Woot you solved it. Good job Karunya. Your proof looks correct to me.

[quote="Ish"]Woot you solved it. .[/quote]

what's "woot"??

"Woot" is something like "WOW" but sounds more cool and trendy  :P

Could I say that the numbers of the pieces are the same for an infinite sized board?

This is classic and has been posted before.  The answer is no, but it becomes yes if you also impose the condition that the set of numbers used is bounded (and the proof is the same).

Cool. I got two proofs saying the answer is no. I fear though that they are not elegant. Can one of you guys right a really elegant proof?

The simplest proof would simply be an example, which are not too difficult to produce.

lol. I cant seem to get one  :blush: . can you help me with this?

i don't get it, why shouldn't the numbers be same even for an infinite board?

Ok, take an infinite chessboard, with squares denoted by a pair of integers $(m,n)\in \mathbb{Z}^{2}$ so, that two cells are adjacent iff $|m_{1}-m_{2}|+|n_{1}-n_{2}|=1$. Let the cells be filled with numbers $A_{m,n}$. Then the following holds:

If there exists $(m_{0},n_{0})$ such, that $A_{m_{0},n_{0}}$ is minimal or maximal among all $A_{m,n}$, then all $A_{m,n}$ are equal.

The proof is clear from above :roll:
However, if we only demand that $A_{m,n}\in \mathbb{Z}$, then you can check that $A_{m,n}=n$ is a non-constant valid "filling".

Just a remark... This has something to do with the maximum principle for Laplace operators. If the solution has a minimal or maximal value in the interior of a domain, then the solution is constant. Indeed, a valid filling of the chessboard must satisfy constraints, which are equivalent to a finite difference discretization of $\triangle u=0$ in $\Omega = \mathbb{R}^{2}$ :wink:

EDIT: There is another kind of infinite squared chessboards, where you have a lower left corner, but no upper right corner - in that case, the squares are  $(m,n)\in \mathbb{N}^{2}$. Then the aforementioned example does not work, and the analogy with the elliptic operators does not seem to help either, for we must control the boundary conditions...

EDIT2: In this situation, we start with $A_{1,1}=a$.
Put $A_{1,2}=b \neq a$; then $A_{2,1}=2b-a$.

Recursion step: Let the values of $A_{n,m}$ for $n+m\leq k$ be filled in. Consider the unknown values $A_{n,m}$ on the next diagonal $n+m=k+1$. These must satisfy $k$ linear constraints, but we have $k+1$ degrees of freedom :) - Let us choose $A_{1,k+1}\in \mathbb{R}$, now we can calculate the rest of the values in a unique way.

Now we are done. The chessboard is filled iff all the diagonals are filled. The constraints are satisfied by construction. Since $a\neq b$, we have $A_{1,1}\neq A_{1,2}$, therefore the filling is not constant. QED  :P

So, the answer is: on infinite squared chessboards, there exist non-constant fillings.

This was my counterexample.

consider the example such that a row is all of the integers from -inf to
inf, and the row above it is also the integers but with a shift to the left.
The row below is the integers with a shift to the right.

..-1 0 1 2 3...
..-2 -1 0 1 2...
..-3 -2 -1 0 1...

thus there exists at least two pieces that are not the same

Yes, in an infinite chessboard with INTEGER coordinates, you can put in $A_{nm}=an+bm+c$ for any real $a,b,c$, and it will be a valid filling. However, if you consider the limit case of the $n\times n$ chessboard (imagine yourself sitting on the square $a1$ - as the board expands, $a1$ will still remain the corner), you obtain a chessboard with NATURAL coordinates, and there you have a boundary. The corner would start something like

$\begin{array}{cccccc}\dots & \dots &\dots &\dots\\ 2 & 1 & 0 & \dots \\ 1 & 0 &-1& \dots \\ 0 & *-1* &-2& \dots, \end{array}$

and you see that the marked $-1$ is not the average of its neighbours $\{0,0,-2\}$, so this case deserves separate attention, IMHO :oops:

If you start with the same bottom row, and set $A_{2,1}=1$, then by the recursive constuction, you get (the first five diagonals)

$\begin{array}{cccccc}72 & \dots & \dots & \dots &\dots &\dots\\ 17 &-25 & \dots & \dots &\dots &\dots\\ 4 &-6 &-2 &\dots &\dots & \dots \\ 1 &-1 &-2 &-3&\dots & \dots \\ 0 &-1 &-2&-3&-4& \dots, \end{array}$

and you complete it diagonal by diagonal. In fact, we have shown the following. Given the sequence $A_{1,k}$ for $k=1, \dots, n, \dots$, then there exists one and only one valid filling with the given first row.

For a finite board, the problem is that the diagonals start getting shorter, therefore the only valid fillings are constant ones.


abstract algebra
superior algebra
superior algebra unsolved

maybe wellknown :prove that $\prod _{1}^{\infty}\mathbb{Z}$ is not free $\mathbb{Z}$-module.

Maybe well-known, but very hard.

http://www.maths.tcd.ie/pub/ims/bull40/coleman.ps


Can anyonen else explain to me how the virus works?

I wonder if it is possible to use one kind of virus to attack another kind of virus in order to keep our bodies healthy....... ( making use of the 陰陽五行 in Chinese culture.)
 :idea:

Maybe it is just a silly question.  :blush:

In most basic terms, a virus is a bag of DNA surrounded by some proteins that allow it to penetrate a cell.  Once it gets in there, it co-opts the cell's reproductive functions to copy its own DNA - to make copies of itself.  When it's forced the cell to make enough copies of itself, the cell bursts, allowing all of the new viruses to infect new cells.

This mechanism is particular to cells, so it would not be possible for viruses to attack each other.  It is conceivable, however, that viruses can attack bacterial cells.

Remember that a virus is only "alive" if it is functioning in a host cell. These inactive viruses are called virions.

It might possible to alter a virus's DNA so that when one does infect a cell, it is used to attack other viruses.

Viruses are actually called "obligate" parasites because of their dependence on a host for replication. Viruses have both living and non-living characteristics. For instance the ability to replicate (inside a host) and the ability to move. However, it's ability to from a crystaline structue (or protein capsid) is typical of non-living material. All viruses contain some nucleic acid core (either DNA or RNA) surrounded by this protein coat. A virus moves up the concentration gradient towards a suitable host cell (detecting chemicals released by the cell). There are two ways in which the nucleic acid can be introduced into the host cell. Some viruses latch onto the membrane then inject the nucleic acid through a tail. Others move into the cell by phagocytosis, where the membrane invaginates and "swallows" the viral particle. Once inside the host, the nucleic acid, or viral particle could remain in a state of dormancy for several years. If the viral nucleic acid is RNA, then the enzyme reverse transcriptase is used to form viral DNA. This DNA then begins to take control of the hosts metabollic processes, and uses its ATP, enzymes and nucleotides to replicate the nucleic acid cores. The protein capsids are manufactured using the host's amino acids. Assembly of the nucleic acid cores and protein capsids occurs forming new viral particles. The viral particles then either burst out of the cell (lysis), or can actually leave by a process called budding. This is exocytosis, but this way, the viral particles actually gain a further protein coat, which has been pinched off from the membrane. The new viruses are free to infect new host cells, leaving the old host cell damaged or destroyed. In fact viruses are evolving so that they become less harmful to their hosts. If a virus is more damaging, it is more likely to kill it's host organism, thus vastly reducing the chances of meeting another host organism to infect. Those viruses which are less harmful have a selective advantage.

Thank you very much for all of your explanations!
Probably it is hopeless to control a virus by another one.
Maybe we can attack the virus in other ways round like stopping the process of duplication or invade the virus before it really operates.......   :)


trigonometry
calculus
calculus computations

maybe some body don't know how to solve this simplest equation:
$ \sqrt{\frac{x^{2}\minus{}3x\plus{}2}{x^{2}\plus{}2x}}\equal{} x\plus{}1$

$ (x\minus{}1)(x\minus{}2) \equal{} (x^{2}\plus{}2x\plus{}1)(x^{2}\plus{}2x)$, $ A \equal{} x\minus{}2$, $ B \equal{} x^{2}\plus{}2x$, we have
$ (A\plus{}1)A \equal{} (B\plus{}1)B\rightarrow (B\minus{}A)(B\plus{}A\plus{}1) \equal{} 0$, so
$ (x^{2}\plus{}x\plus{}2)(x^{2}\plus{}3x\minus{}1) \equal{} 0$.


inequalities
induction
real analysis
real analysis unsolved

The elements of a [i]sequence[/i] {a_n} are given by:

a_(n+1) = ka_n(1 - a_n)

Prove that if k belongs to [0,1) and a_0 belongs to [0,1] then a_n -> 0 as n -> infinity.

[quote="bharathhm"]By AM-GM inequality, a_0(1-a_0)<=1/4, since, a_0<=1.

hence, a_1<=k/4 <1. So, we can apply AM-GM again, and get, a_2<1
By induction, we can prove that, a_n<=k/4 <1.

Observe that a_n+1 /a_n = k(1-a_n) <k<1. So, there exists, b, where k<b<1.

Hence, a_n+1 /a_n <b <1.

so, a_n+m < b^m-->0, as m-->infinity.[/quote]
fucker


So, ab jetzt ist die vollständige Liste unter [url=http://www.noname.schlussda.com/IMO/list.html]www.noname.schlussda.com/IMO/list.html[/url] erreichbar. Bitte Fehler sofort melden!

Die dritte Klausur wurde soeben in Bad Homburg geschrieben.

Wichtigste Änderungen: Annika Heckel und Martin Herdegen nehmen nicht mehr teil.

[quote="NoName"]

Wichtigste Änderungen: Annika Heckel und Martin Herdegen nehmen nicht mehr teil.[/quote]

Warum ?

[quote="orl"]Warum ?[/quote]

sie müssen (?) sich fürs abi vorbereiten... übrigens kann es sein dass sie die klausuren unabhängig nachschreiben; das einzige was sicher ist ist dass sie nicht zu den seminaren kommen.

  grüße,
  darij

Soeben haben wir die dritte Klausur zurück bekommen.

Die Punkteliste wurde aktualisiert.

Check the results of Christopher Wulff for the third exam again.  ;)

Jetzt ist auch die vierte Klausur geschrieben.

Es gab ziemlich viele Überaschungen...

so, das müsst ihr auch mal lernen, banaaaaaale aufgaben zu verscheißen: eine monotone funktion ist stets injektiv, gell? auch die nullfunktion natürlich. ansonsten viele grüße und das mit der seltsamen schätzung hat sich erledigt. btw meine 3 7 7 hat mir peter geschätzt, ich persönlich würde 1 7 7 erwarten.

  dg

Die vierte Klausur wurde zurückgegeben...


So the question is how many ways are there to paint each face of a cube with the colors Red, White and Black? I don't understand group theory, so I am not able to figure out Burnside's lemma. Can someone suggest an alternate solution which circumvents group theory?

Yes: list them all.

The whole point of Burnside's lemma is that it greatly lessens the amount of work you have to do.

I would suggest Polya's Enumeration Theorem, but that involves group theory.
On a side note, Polya's Enumeration Theorem looks eerily similar to Burnside's lemma; can someone explain the difference between them to me?
Thanks!

Benjamin,

The difference is that Burnside's lemma is just a special case of Polya's enumeration Theorem. I can't venture to explain how it is a special case because I don't know group theory.

Burnside's lemma is a key tool in the proof of the PET; I've written all of this up [url=http://qchu.wordpress.com/2009/06/13/gila-i-group-actions-and-equivalence-relations/]here[/url] if you're interested.


geometry
perimeter
inequalities
trigonometry
circumcircle
cyclic quadrilateral
geometry unsolved

How can we prove that, out of all convex polygons with n ≥ 3 sides and a same perimeter, the regular one has the maximum area?

This is a forum.  You can't expect responses this quickly.  See [url=http://en.wikipedia.org/wiki/Isoperimetric_inequality]isoperimetry[/url].

[b]Lemma1:[/b] given $ 4$ real numbers $ a,b,c,d$ able to form a quadrilateral, they can also form a cyclic quad. This is a known lemma. An intuitive demonstration is as follows: Consider one of the quads that can be formed and attach to each vertex a rigid stick all of them sharing a common vertex and all of equal length. Drop this figure into the plane, it will form a cyclic quad.

[b]Lemma2:[/b] From all quadrilaterals that can be formed using the four sides $ a,b,c,d$ the cyclic quad has the greatest area.
[b]Demonstration:[/b] Using Brahmagupta formula: Given a quadrilateral $ PQRS$ with side lengths $ a,b,c,d$, $ \alpha = \angle{PQR}$, $ \gamma = \angle{RSP}$ then the area $ F$ of $ PQRS$ is:

$ 16F^2 = (a + b + c - d)(a + b - c + d)(a - b + c + d)( - a + b + c + d) - 16abcd\cos^{2}({\frac {\alpha + \gamma}{2})}$ the maximum of $ F$ is obtained when $ \alpha + \gamma = \pi$ $ \Rightarrow$ $ PQRS$ is cyclic.

[b]General demonstration:[/b] Consider our polygon $ P = P_1P_2\dots P_n$ of maximal area and fixed perimeter $ p$. If it has $ 3$ sides it is known the solution is an equilateral triangle. (I will omit this). The first four vertexes of $ P$ form the quad $ P_1P_2P_3P_4$ with sides $ p_1,p_2,p_3,q_1$ with $ q_1 = P_1P_4$ if it is not cyclic, then we can construct the cyclic quad with the same sides $ p_1,p_2,p_3,q_1$ as $ P$ paste this quadrilateral by its $ q$ side and get a $ P'$ with greater area than $ P$, contradiction. This means $ P_1P_2P_3P_4$ is cyclic. Every four consecutive points lie in a circle. This means all polygon vertexes must lie in the same circle.

Consider the same polygon $ S = P_1P_2P_3P_4$ such that $ S$ has sides $ l_1,l_2,l_3,s_1$ ($ s_1 = P_1P_4$) because of Brahmagupta, $ S = \sqrt {\frac {l_1 + l_2 + l_3 - s_1}{2}\cdot\frac {l_1 + l_2 - l_3 + s_1}{2}\cdot\frac {l_1 - l_2 + l_3 + s_1}{2}\cdot\frac { - l_1 + l_2 + l_3 + s_1}{2}}$ let $ d = \frac {l_1 + l_2 + l_3}{3}$ because of [b]Lemma1[/b] there is a cyclic quad $ S'$ of sides $ d,d,d,s_1$ (with perimeter $ p$ as $ S$) using Brahmaguptha again $ S' = \sqrt {\frac {d + d + d - s_1}{2}\cdot\frac {d + d - d + s_1}{2}\cdot\frac {d - d + d + s_1}{2}\cdot\frac { - d + d + d + s_1}{2}}$,
$ S' = \sqrt {\frac {3d - s_1}{2}\cdot(\frac {d + s_1}{2})^3}$ , let $ l_1 + l_2 - l_3 = x_1$, $ l_1 - l_2 + l_3 = x_2$, and $ - l_1 + l_2 + l_3 = x_3$ then:
$ S = \sqrt {\frac {3d - s_1}{2}\cdot\frac {x_1 + s_1}{2}\cdot\frac {x_2 + s_1}{2}\cdot\frac {x_3 + s_1}{2}}$,
lets see $ S \leq S'$. This happens $ \iff$  $ \sqrt {\frac {3d - s_1}{2}\cdot\frac {x_1 + s_1}{2}\cdot\frac {x_2 + s_1}{2}\cdot\frac {x_3 + s_1}{2}} \leq \sqrt {\frac {3d - s_1}{2}\cdot(\frac {d + s_1}{2})}$
we know that: $ x_1 + x_2 + x_3 = 3d$
conclusion $ \iff (d + s_1)^3\geq (x_1 + s_1)(x_2 + s_1)(x_3 + s_1)$
$ \iff d^3 + 3d^2s_1 + 3ds_1^2 + s_1^3\geq x_1x_2x_3 + s_1(x_1x_2 + x_2x_3 + x_1x_3) + s_1^2(x_1 + x_2 + x_3) + s_1^3$
$ \iff d^3 + 3d^2s_1\geq x_1x_2x_3 + s_1(x_1x_2 + x_2x_3 + x_1x_3)$
$ \iff (\frac {x_1 + x_2 + x_3}{3})^3 + 3(\frac {x_1 + x_2 + x_3}{3})^2s_1\geq x_1x_2x_3 + s_1(x_1x_2 + x_2x_3 + x_1x_3)$
Which follows easily from the known inequalities:
$ \frac {x_1 + x_2 + x_3}{3}\geq \sqrt [3]{x_1x_2x_3} \Rightarrow (\frac {x_1 + x_2 + x_3}{3})^3\geq x_1x_2x_3$
and $ 3(\frac {x_1 + x_2 + x_3}{3})^2\geq x_1x_2 + x_2x_3 + x_1x_3$
this means $ S'\geq S$ with equality $ \iff x_1 = x_2 = x_3 = d, \iff l_1 = l_2 = l_3$

Then, if the polygon has two consecutive different sides (Say $ S$), we can construct $ S'$ and paste it to the side $ s_1$ (not of consecutive vertexes) of the polygon $ S$ and obtain a polygon $ P'$ with greater area than $ P$ contradiction. This means $ P$ is cyclic and has equal sides. Therefore it must be regular.

A direct consequence of this result is the inequality of this topic:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=204265[/url]

[quote="Nemion"][b]Lemma1:[/b] given $ 4$ real numbers $ a,b,c,d$ able to form a quadrilateral, they can also form a cyclic quad. This is a known lemma. An intuitive demonstration is as follows: Consider one of the quads that can be formed and attach to each vertex a rigid stick all of them sharing a common vertex and all of equal length. Drop this figure into the plane, it will form a cyclic quad.[/quote]
Another:  consider the cyclic quadrilateral with side lengths $ a, b, c$ in a circle of radius $ R$ as $ R$ varies from the circumradius of a triangle with side lengths $ a, b, c$ to infinity.  The side length of the fourth side varies continuously from $ 0$ to infinity, so it must take the value $ d$.


[hide]we can make a nice explicit sequence for this

d = 24 - 7n for all saturdays.
we want d = 1

1 = 24 - 7n
-23 = -7n
3.28 ~ n
Meaning the day was 3 weeks and .28 days from saturday.
.28[7] = x
1.96 = x
1.96 is so close to 2 that we can round up[also we cant have .96 when we are talking about weekdays].
Meaning the day was 2 days from saturday making it a Thursday.[/hide]

[hide="We can"]Subtract 7 from the 24 since there are 7 days in a week. We subtract it until we see that on the 3rd it was also Saturday. The first day was Thursday.[/hide]

[hide][quote="anirudh"]The 24th day of a particular month is a Saturday. On what day of the week was the first day of that month?[/quote]

thursday. check the calendar (thats cheating)[/hide]

thursday:

track back weeks. we know that the 24th was a saturay and so the 17th, 10th, and the 3rd were all saturdays as well. then we backtrack up to the first day of the month to get that the month began on a thursday.

[hide] you could just realize that every week has 7 days. then, you can take 24, and find the closest multiple of 7. which is 3. $7*3=21$. therefore, we know that the 3rd was a saturday.
$24-21=3$. then, two days before was the 1st, so friday being the 2nd, thursday as the first.
[b] THURSDAY[/b]  :lol:  [/hide]


function
algebra
floor function
Additive Number Theory
pen

Let's assume for the sake of contradiction the set \[ S:=\left\{n \in \mathbb{N}:n=\sum_{i=1}^5{x_i^{2i+1}}\text{ for some }x_i \in \mathbb{N}\right\} \] is not finite. Then $|\mathbb{N} \setminus S|=n_0$ for some $n_0 \in \mathbb{N}$. But for all reals $x>0$ and positive integers $h$ we have \[ | \{n^h:n \in \mathbb{N}\}\cap [0,x] | \le  (x+1)^{h^{-1}} = \mathcal{O}(x^{h^{-1}}) \]
It means that, given positive integer $h_1,h_2,\ldots,h_m$, in the best case we'll have: \[ \left|\bigcup_{i=1}^m{\{n^{h_i}:n \in \mathbb{N}\}} \cap [0,x]\right|=\mathcal{O}(x^{\sum_{i=1}^m{h_i^{-1}}}). \]
But in our case $\sum_{i=1}^5{(2i+1)^{-1}}<1$, so it implies that \[ |S\cap [0,x]|=\mathcal{O}(x^{\sum_{i=1}^5{(2i+1)^{-1}}})=o(x) \].
In other words $\mathbb{N} \setminus S$ cannot be finite, and that's a contradiction.

[quote=AndrejK]Pick a big number N. The number of 5-tuples (x_1,x_2,...,x_5), such that x_1^3+x_2^5+...+x_5^11 is smaller or equal to N is not more than f(N^(1/3))*f(N^(1/5))*...*f(N^(1/11))<=N^(1/3+1/5+...+1/11)< N^0.9 (where f is floor function), but N/N^0.9=N^0.1, which tends to infinty for big N, so there are infinitely many numbers which are not representable as this power sum.[/quote]

This solution is good but let me rewrite it better:
Start counting number of numbers which are less than N and they can be written in this form:($=K$)
$$K =\sqrt[3]{N}\times\sqrt[5]{N}\times\sqrt[7]{N}\times\sqrt[9]{N}\times\sqrt[11]{N}$$ $$=N^{\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}}\approx N^{0.87}$$
Now pick N a big number like $N=x^{100}$
$\Rightarrow$number of numbers which can not be written in that form=$N-K =x^{100}-x^{87}$
And this polynomial ( $x^{100}-x^{87}$ ) for big $N$s $\rightarrow \infty$ which means there are infinite N


A game is played with tokens according to this rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token into a discard pile. The game ends when some player runs out of tokens. Players A, B, and C start with 17, 16, and 15 tokens, respectively. How many rounds will there be in the game?

Further: Is there a general forumla for this type of problem?

I'm not sure how to generalize this problem,
but I've solved this particular game ... well, sort of.[hide]
$.\qquad\quad A\quad\; B\quad\; C\quad\text{discard}$

$\text{Start}\quad17\quad16\quad15\qquad0$
$.\;\;\;1\;\;\quad14\quad17\quad16\qquad1$
$.\;\;\;2\;\;\quad15\quad14\quad17\qquad2$
$.\;\;\;3\;\;\quad16\quad15\quad14\qquad3$

After a cycle of 3 rounds, each player has lost exactly one token.

Hence, in 14 cycles (42 rounds), we have:
$.\;\;42\;\;\;\quad3\;\quad2\;\quad1\;\qquad 42$

And the 43rd round will be:
$.\;\;43\;\;\;\quad0\;\quad3\;\quad2\;\qquad43$

$\text{Therefore, the game will have 43 rounds.}$[/hide]

Don't look unless you want to spoil all the fun...

[hide]Or: You start with 48, each time one  is thrown out , game ends when you are left with  5 piles ( After 1-2-3 you get to throw away  1)... Yes just pay attention to the [b]total[/b] number of  tokens... and net effect is one token is thrown out  in each round. [/hide]


There are 30 senators in a senate. For each pair of senators, the two senators are either friends or enemies. Every senator has exactly six enemies. Every three senators form a committee. Find the total number of committees whose members are either all friends or all enemies of each other.
[/hide]

[hide]
Call a committee [i]typical[/i] if there is a pair of enemies and a pair of friends in it ( :P ). In each typical committee, two of the senators have both a friend and an enemy, the third does not. Therefore, the number of typical committees is equal to half the number of ways of choosing a senator, a friend of that senator, and an enemy of that senator. So there are
\[\frac{1}{2}\cdot 30\cdot 23\cdot 6=2070\]
typical committees. Since we are looking for the number of non-typical committees and there are a total of $\binom{30}{3}=4060$, we get $\boxed{1990}$. (I'm guessing I did not make a stupid mistake, just based on the answer.)
[/hide]


might be this is not intersting but that was just to get you in here... :D

[x] denotes the greatest integer value just below x and {x} its fractional value. The sum of [x]^3 & {x}^2 is -7.91. Find x.

Now i dont understand what [x] or {x} mean...can some one explain with an example... here is my limited understanding...

x = 1.21
[x] = 1
{x} = .21

x = -8.21
[x] = 8
{x} = .21

I have a problem understanding the question and [x] & {x}... could you please explain what they mean and give me some examples? :?:

$[x]$ is the greatest integer value just below (or equal to) $x$.
A simple way to find it: $[x]\leq x<[x+1]$ (put $x$ between consecutive integers and pick the integer below)
Some examples:
 $3\leq 3<4$ so $[3]=3$
 $3\leq 3.1<4$ so $[3.1]=3$
 $3\leq\pi<4$ so $[\pi]=3$
 $-3\leq-3<-2$ so $[-3]=-3$
 $-4\leq-3.1<-3$ so $[-3.1]=-4$
 $-4\leq-\pi<-3$ so $[-\pi]=-4$
Then we have always $\{x\}=x-[x]$,  $\{x\}\in [0,1)$.
Some examples:
 $\{3.1\}=3.1-3=0.1$
 $\{-3.1\}=-3.1-(-4)=0.9$


abstract algebra
integration
advanced fields
advanced fields unsolved

Let $u \in \mathcal{C} (\mathbb{R}^{n} \times \mathbb{R})$ such that
for every $\epsilon > 0$ there exists $C_{\epsilon} > 0$ such that
$|u(x,t)| \leq C_{\epsilon}\exp(\epsilon |x|^{2}) , x \in \mathbb{R}^{n}, t \geq 0$.

Also, let as usual, $K(x,t) = \frac{1}{4\pi t)^{n/2}}\exp(-|x|^2/4t), x \in \mathbb{R}^{n}, t \geq 0$

For a fixed $x \in \mathbb{R}^{n}$ show that

$\int_{\mathbb{R}^{n}} u(x-y,t)K(y,t)dy \longrightarrow u(x,0)$ as $t \searrow 0$.

Thanks!

May assume $u(x,0)=0$. Then $u(x-y,t)$ is small for small values of $|y|$ and $t$, and since $K$ integrates to $1$, such values of $y$ don't contribute much to the integral. The contribution from $|y|>\varepsilon$ is also small because $K$ kills $u$.


Erm, I don't think that's right.  First of all, you never said the line even had to go through the circle, so the minimum number ought to be 1.  Then, what about a tangent?  Does a tangent count?  You will need to define a "line thru a circle" as a line that forms a chord inside the circle or something of the sort.

And with one line, the minimum will be 2 and the maximum will be 4 according to your formula, whereas in reality, you can only form 2 regions no matter what you do.  When n is 2, max = 8.  But I can only make 4.  When n is 3, max = 14, but I can only make 7.  When n is 4, max = 22 but I can only make 11.

Clearly, you're off by a factor of 2.  I think the formula for maximum should read: "The maximum amount of regions using n lines is: [b](n^2 - n + 2)/2[/b]

[color=cyan]Induction does work for proving that formula.  However, it is a little bit tricky.  You have to consider intersections between lines and their relationship with the number of regions formed.

As for the various different figures, there really isn't a significant change caused by changing the circle to a triangle or quadrilateral or anything else.  In fact, it works just as well if you consider the entire plane instead of an enclosed region.  In fact, any convex region will give you that same formula.  To see why, just try zooming in very very much -- you can push the boundary far enough away that it effectively doesn't matter.  Then lay down your extra lines.  Then zoom out again.  Of course, adding in concave figures will add all sorts of problems.[/color]


geometry
circumcircle
modular arithmetic
geometry unsolved

Let $ABCD$ be a convex quadrilateral. Take an arbitrary point $M$ on the line $AB$, and let $N$ be the point of intersection of the circumcircles of triangles $MAC$ and $MBC$ (different from $M$). Prove that:

a) The point $N$ lies on a fixed circle;

b) The line $MN$ passes though a fixed point.

the circles MAC and MBC intersect at points M and C.
So maybe its MAD circle ?

a) Let $T={AC}\cap{BD}$ then $C, T, D, N$ are concyclic (angle chasing).

b) Suppose $MN$ intesects $(CTD)$ in point $X$ then $NC, NT, NX, ND$ form $r$-corelative
harmonic fascicle on a fixed circle (again angles), hence $X$ is a fixed point.

For b) you can just observe that $\angle{DNX}=\angle{DBA}$ and $\angle{CNX}=\angle{CAB}$ and thus $X$ is a fixed point.

For a)
$\angle DNC=\angle DNM+\angle MNC=180^o-\angle A+180^o-\angle B=\angle C+\angle D$
which is constant.
But $DTC\neq 180^o-\angle D-\angle C$ such that the quadrilateral $CNTD$ to be cyclic.. Why i'm wrong?

For b) On the specified fixed circle, denote by $T_M$ the second intersection of $MN$ with the circle. The arc $TD$ always have measure $180^o-\angle A$. Hence $T$ is a fixed point on $MN$.

This problem is nice but easy :D . I think to prevent from pointing out the case of the figure of this problem (At least there must be 2 cases), I would prefer using the directed angle to kill it. Here is my solution:
a) Let $ O$ be the intersections of the $ 2$ diagonals of $ ABCD$. $ (OA,OB)\equiv (OC,OD)\equiv \minus{} (DO,DC) \minus{} (CD,CA)\equiv (DM,DB) \plus{} (CA,CM)\equiv (NM,NB) \plus{} (NA,NM)\equiv (NA,NB)\pmod\pi$. Therefore, $ N$ blongs to $ (OAB)$, which is obviously fixed. (Done!)
b) Let $ P$ be the intersection of $ MN$ with $ (OAB)$. Then, we get, $ (PN,PO)\equiv (AN,AO)\equiv (AN, AC)\equiv (MN,MC) \pmod \pi$, which leads to the fact that $ PO//CD$, but $ CD$ is fixed, $ O$ is fixed, $ (OAB)$ is fixed. Hence, $ P$ is fixed too, which completes our proof. :D

@Mod please post the prob in it's correct form

Here is my solution for this problem
[b]Solution[/b]
a) We have: ($NC$; $ND$) $\equiv$ ($NC$; $NM$) + ($NM$; $ND$) $\equiv$ ($AC$; $AB$) + ($BA$; $BD$) (mod $\pi$)
Then: $N$ lies on fixed circle is the circle passes through $C$, $D$ which satisfies: ($NC$; $ND$) $\equiv$ ($AC$; $AB$) + ($BA$; $BD$) (mod $\pi$)
b) Let $P$ $\equiv$ $AC$ $\cap$ $BD$
Draw line through $P$ and parallel with $AB$ intersects ($PCD$) at $S$ ($S$ $\ne$ $P$)
We have: ($NC$; $ND$) $\equiv$ ($NC$; $NM$) + ($NM$; $ND$) $\equiv$ ($AC$; $AB$) + ($BA$; $BD$) $\equiv$ ($AP$; $AB$) + ($BA$; $BP$) $\equiv$ ($PA$ ; $PB$) $\equiv$ ($PC$ ; $PD$) (mod $\pi$)
So: $P$, $C$, $N$, $D$ lie on a circle
But: ($NM$ ; $NS$) $\equiv$ ($NM$ ; $ND$) + ($ND$ ; $NS$) $\equiv$ ($BM$ ; $BD$) + ($PD$ ; $PS$) $\equiv$ ($PS$ ; $PD$) + ($PD$ ; $PS$) $\equiv$ 0 (mod $\pi$)
Then: $M$, $N$, $S$ are collinear
Hence: $MN$ passes through fixed point $S$ when $M$ moves on $AB$


quadratics
linear algebra
matrix
function
symmetry
calculus
derivative

I am having some difficulties in understanding some concepts related to Quadratic Forms (apologize if any of the following questions is silly).

1) Need the matrix A associated with a quadratic form Q be symmetrical? Why?

2) How can a given Q be brought into a diagonal form?

3) Why the positive definiteness of Q implies that the eigenvalues of A are all positive? Is the converse also true?

4) Can someone explain the chain of reasoning that allow one to conclude that for Q to be positive definite, every principal minor of A must be positive?

If you can direct me instead to a link where all this is clearly explained, that would be also helpfull, since all internet pages I have consulted are not helping.

Thanks in advance.

We do need to specify what field we are dealing with. In my answers here, I will make the assumption that you are working over the real numbers. If that's not the case, a few things will need to be modified.

1. Any square matrix $ A$ defines a quadratic form by $ Q_A(x,y)\equal{}y^TAx.$ But this is only symmetrical - that is $ Q_A(x,y)\equal{}Q_A(y,x)$ - if $ A\equal{}A^T.$ However, we do have that $ Q_A(x,x)\equal{}Q_{A^T}(x,x).$ In fact, if we only focus on that, on $ Q_A(x,x),$ then any linear combination of the form $ (1\minus{}t)A\plus{}tA^T$ defines the same function on $ x.$ Note then that we can always let $ t\equal{}\frac12$ and thus assume that our matrix is symmetric.

Inspired by that, many authors choose to include that $ A\equal{}A^T$ in the very definition of quadratic form. I would choose that path myself - include symmetry in the definition.

2. The full theorem is this: if $ A$ is symmetric ($ A\equal{}A^T$), then there exists $ Q$ orthogonal ($ Q^TQ\equal{}I$) such that $ Q^TAQ\equal{}D,$ diagonal. There are many nice proofs of this. I'd probably move into the complex numbers and invoke Schur's Lemma, but there are other possibilities. We can break off parts of this theorem and prove them separately, such as all eigenvalues of $ A$ are real, or eigenvectors associated to different eigenvalues are orthogonal.

3. By #2, we do have that $ A$ is diagonalizable. Suppose $ Ax\equal{}\lambda x.$ Then $ x^TAx\equal{}x^T(\lambda x)\equal{}\lambda x^Tx\equal{}\lambda\|x\|^2.$ But the hypothesis of positive definite gives us that $ x^TAx>0$ for all $ x\ne 0,$ so we must have $ \lambda>0.$

As for the converse: again working through #2, we look at the diagonal matrix $ D.$ Suppose $ D$ has only positive entries on the diagonal, $ D\equal{}\text{diag}(\lambda_1,\lambda_2,\dots,\lambda_n)$ where $ \lambda_j>0.$ Then for all $ y\ne0,\ y^TDy\equal{}\sum_{j\equal{}1}^n\lambda_jy_j^2>0,$

Then $ x^TAx\equal{}x^TQDQ^Tx\equal{}(Q^Tx)^TD(Q^Tx)>0$ by the previous paragraph.

4. Write $ A$ in block-structured form as $ \begin{bmatrix}M&B\\B^T&C\end{bmatrix}.$

Let $ x\equal{}\begin{bmatrix}y\\0\end{bmatrix}$ for an arbitrary nonzero $ y.$ Then $ x^TAx\equal{}y^TMy.$ (Just carry out the block-matrix multiplication to see that.) Since $ A$ is positive definite, we must have $ M$ positive definite.

I did that argument on the assumption that your principal minor was located in the top left, but we can handle other locations by conjugating with permutations.

Professor Merryfield, thank you very much for your explanation, and sorry for only answering now, but I decided to read something about Matrix Algebra before posting further doubts. The problem which I am interested to analyse (regarding a text I am reading) is the following.

I have a function Z of several variables $ X_i$ and I am interested in the conditions for which Z is a minimum. The condition dZ = 0 was already analysed, so now the text skips to the next condition, the second order variation be positive, $ d^2 Z > 0$ (actually the text says $ \geq 0$: which is the correct sign?). But then we want

$ d^2 Z \equal{} \sum_i \sum_k \frac {\partial ^2 Z}{\partial X_i \partial X_k} dX_i dX_k > 0$

(another question: since we want to minimize a function of several variables, why not write directly that the Hessian determinant must be positive?)

This is of course a quadratic form whose matrix is the Hessian of Z (and which is a symmetric matrix due to Schwartz). Now the text says that, in order for a quadratic form to be positive (definite) the principal minors of the matrix must be positive. Then the following explanation is given as a note (together with some notes/questions from me):

A quadratic form $ Q \equal{} \sum_i \sum_k a_{ik} x_i x_k$ ($ \equal{} x^T A x$) can be brought into a diagonal form $ Q \equal{} \sum_i a_i y_i^2$ by an orthogonal transformation $ x_i \equal{} \sum T_{ik} y_k$.

(Although the notation presents some difficulty to me, I think I understood; from what I have read and you have written above, any symmetric matrix can be diagonalized to D by some orthogonal matrix T. Question: has the resulting matrix D always the eigenvalues of A on its diagonal, or can we always choose T such that D has the eigenvalues of A on the diagonal?)

The positive definiteness of Q means that the eigenvalues $ a_i$ of matrix A are all positive.

(From your explanation above, I have understood this.)

Since we have $ det\,D \equal{} a_1 a_2 ... a_n \equal{} det\,T^{ \minus{} 1}AT \equal{} det\,T^{ \minus{} 1}\,\,det\,A\,\,det\,T \equal{} det A$ (as $ det\,T \equal{} det\,T^{ \minus{} 1} \equal{} 1$), then det A must be positive.

[Just a question: I know that given two matrices n x n A and B, det(AB) = det(a)det(B); therefore, for three (or more) matrices we can say that det(ABC) = det(AB)det(C) = det(A) det(B) det(C), such that the determinant of a product of matrices is the product of the determinants. Is this correct?]

A principal minor of the matrix A is ... (definition). The corresponding quadratic form is a special case of Q in which only the corresponding x's are allowed to take finite values. Thus it must be positive.

(This part I didn't understand.)

By applying the previous argument, one finds that the principal minor itself must be positive.

(Another doubt here: Now are considering only the eigenvalues of the principal minor? Are they part of the eigenvalues of matrix A or are different ones?)

Therefore, for a positive definite quadratic form Q, every principal minor must be positive.

(Question: by definition of principal minor, all the diagonal elements of A are principal minors, and so they must be positive, right? At least the text assumes this in the next part of it).

Any help on these questions would be appreciated. Thanks in advance.

In order to have a sufficient condition for a smooth function of several variables to have a local minimum, the Hessian quadratic form must be positive definite, not merely positive semidefinite. (That is, $ >0$ rather than $ \ge 0.$)

That's the sufficient condition; having the Hessian be positive semidefinite would be a necessary condition for a local minimum. What if the Hessian is nonzero and positive semidefinite but not positive definite? Then we can rule out the possibility of a local maximum, but the second derivatives alone are not enough to decide what kind of critical point we have.

Here's my favorite family of examples. Each of these has a critical point at the origin, and each has the same nonzero positive semidefinite but not positive definite Hessian:

$ f(x,y)\equal{}x^2.$ Non-isolated local minimum.

$ f(x,y)\equal{}x^2\plus{}y^3.$ Not any kind of extreme point.

$ f(x,y)\equal{}x^2\minus{}y^4.$ Saddle point (minimax).

$ f(x,y)\equal{}x^2\plus{}y^4.$ Local minimum.

As for some of your other comments:

Yes, when you diagonalize such a matrix: $ D\equal{}T^TAT,$ then the diagonal entries of $ D$ are the eigenvalues of $ A.$ They must be. In other words, $ D$ is determined up to permutation - the only choice you have is what order to write the eigenvalues in.

Yes, $ \det(ABC)\equal{}\det(A)\det(B)\det(C).$ In particular, $ \det(P^{\minus{}1}AP)\equal{}\det(A).$

[quote]A principal minor of the matrix A is ... (definition). The corresponding quadratic form is a special case of Q in which only the corresponding x's are allowed to take finite values. Thus it must be positive.

(This part I didn't understand.) [/quote]
This is essentially what I wrote in #4 in my first post. But the eigenvalues of the minor $ M$ need not be eigenvalues of $ A.$

Let's take a very simple case. $ A\equal{}\begin{bmatrix}2&1\\1&2\end{bmatrix}.$

By the test you've been looking at, this is clearly positive definite. Its characteristic polynomial is $ x^2\minus{}4x\plus{}3,$ from which we get that the eigenvalues of $ A$ are $ 1$ and $ 3.$ Yet, both possible eigenvalues of principal minors are $ 2,$ which is not on the list of eigenvalues of $ A.$

Yes, all the diagonal elements are principal minors. Having them all be positive is necessary for the matrix to be positive definite, but not sufficient.


function
logarithms
algebra unsolved
algebra

Let $ h: \mathbb{R} \longrightarrow \mathbb{R}$ be a real function that fulfils the following conditions:

I) $ h(x) \plus{} h(y) \equal{} h(xy), \forall$ $ x, y \in \mathbb{R}$

II)$ h\left(\max\left(\frac{m}{2^p3^q}, \frac{n}{2^p}\right) \plus{} \min\left(\frac{m}{2^p}, \frac{n}{2^p3^q}\right)\right) \equal{} \log_2(q \plus{} 1), \forall$ $ m, n \in \mathbb{Z}$ and $ \forall$ $ p, q \in \mathbb{N}$.

Demonstrate that if $ x \in \mathbb{Q}$, then $ h(x) \equal{} 0$.

[quote="VSelkoR"]Let $ h: \mathbb{R} \longrightarrow \mathbb{R}$ be a real function that fulfils the following conditions:

I) $ h(x) \plus{} h(y) \equal{} h(xy), \forall$ $ x, y \in \mathbb{R}$

II)$ h\left(\max\left(\frac {m}{2^p3^q}, \frac {n}{2^p}\right) \plus{} \min\left(\frac {m}{2^p}, \frac {n}{2^p3^q}\right)\right) \equal{} \log_2(q \plus{} 1), \forall$ $ m, n \in \mathbb{Z}$ and $ \forall$ $ p, q \in \mathbb{N}$.

Demonstrate that if $ x \in \mathbb{Q}$, then $ h(x) \equal{} 0$.[/quote]

That's wrong : choose $ q\equal{}1$ in the second equation and you get $ h\left(\max\left(\frac {m}{2^p3}, \frac {n}{2^p}\right) \plus{} \min\left(\frac {m}{2^p}, \frac {n}{2^p3}\right)\right) \equal{} 1$ and we have $ \max\left(\frac {m}{2^p3}, \frac {n}{2^p}\right) \plus{} \min\left(\frac {m}{2^p}, \frac {n}{2^p3}\right)\in\mathbb Q$

So $ \exists x\in\mathbb Q$ such that $ h(x)\neq 0$

:rotfl: The problem has been presented exactly as proposed. I noticed the inconcistency. How about solving it for $ h(x) \neq 0$?


Anyone going? I think it's on the 29th or 30th.

orlando. if you're going your school probably already planned the trip.


Μεταξύ των ορθών κυλίνδρων με σταθερο όγκο να βρεθεί το μικρότερο εμβαδόν επιφάνειας (συμπεριλαμβανομένων και το εμβαδόν των βάσεων)

Βρες το ελάχιστο της $ f(R)\equal{}\frac{2V}{R}\plus{}2\pi R^2 , R>0$

[quote="Durandal"][quote="r_boris"]Βρες το ελάχιστο της $ f(R) \equal{} \frac {2V}{R} \plus{} 2\pi R^2 , R > 0$[/quote]

Rudy, μη γεωμετρική λύση? Εκπλήσσομαι :)

Cheerio,

Durandal 1707[/quote]
Παναγιώτη
 δεν εννοούσα παραγώγους αλλά ανισότητα ΑΜ-ΓΜ πάντως αν θες βάλε εσύ τη γεωμετρική λύση  :)

[quote="r_boris"]
Παναγιώτη δεν εννοούσα παραγώγους αλλά ανισότητα ΑΜ-ΓΜ πάντως αν θες βάλε εσύ τη γεωμετρική λύση  :)[/quote]

Εγώ? Εγώ ένα μπακάλικο έχω...  :lol:

Cheerio,

Durandal 1707


Let $n$ be a given integer . Find the number of $(x,y)$ such that

$\frac{xy}{x+y}=n$

This one is well known!

Rewrite as $(n-x)(n-y) = n^2$ . We can see that the number of solutions is $\tau(n^2)$ (Assuming $x$ and $y$ are positive integers).


inequalities
triangle inequality
inequalities unsolved

[quote="mkrnic"]Prove that if a, b, c, d are positive real numbers for which the follow-
ing inequality is satisﬁed
$ (a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)^2 > 3(a^4 \plus{} b^4 \plus{} b^4 \plus{} c^4)$
then can be formed a triangle with sides equal to any three from the
numbers given.[/quote]
It 's ChinaMO2004 :)
By Cauchy-schwarz inequality,we have:
$ (a^2\plus{}b^2\plus{}c^2\plus{}d^2)^2\equal{}(\frac{a^2\plus{}b^2\plus{}c^2}{2}\plus{}\frac{a^2\plus{}b^2\plus{}c^2}{2}\plus{}d^2)^2 \le 3(\frac{(a^2\plus{}b^2\plus{}c^2)^2}{2}\plus{}d^4)$
But $ (a^2\plus{}b^2\plus{}c^2\plus{}d^2)^2> 3(a^4\plus{}b^4\plus{}c^4\plus{}d^4)$
So, $ (a^2\plus{}b^2\plus{}c^2)^2 > 2(a^4\plus{}b^4\plus{}c^4)$
$ <\equal{}> (a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b) > 0$
Then,we have Q.E.D :)

Note that the triangle inequality can be written as $ 2a^2b^2 \plus{} 2b^2c^2 \plus{} 2c^2a^2 > a^4 \plus{} b^4 \plus{} c^4$

Let $ x \equal{} a^2$, $ y \equal{} b^2$, $ z \equal{} c^2$, $ w \equal{} d^2$. Then we have to show $ 2(xy \plus{} xz \plus{} yz) > x^2 \plus{} y^2 \plus{} z^2$

$ xy \plus{} xz \plus{} yz \plus{} w(x \plus{} y \plus{} z) > x^2 \plus{} y^2 \plus{} z^2 \plus{} w^2$
$ 4(xy \plus{} xz \plus{} yz) \minus{} 4(x^2 \plus{} y^2 \plus{} z^2) \plus{} (x \plus{} y \plus{} z)^2 > 4w^2 \minus{} 4w(x \plus{} y \plus{} z) \plus{} (x \plus{} y \plus{} z)^2\equal{}(2w\minus{}(x\plus{}y\plus{}z))^2 \ge 0$

$ 6(xy \plus{} xz \plus{} yz) \minus{} 3(x^2 \plus{} y^2 \plus{} z^2) > 0$

$ 2(xy \plus{} xz \plus{} yz) > x^2 \plus{} y^2 \plus{} z^2$. Similarly for the other triples. Q.E.D.

If $ a_i$>0(1≤i≤n),n≥3,and $ (a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2)^2$≥$ (n\minus{}1)(a_1^4\plus{}a_2^4\plus{}...\plus{}a_n^4)$,then $ a_i,a_j,a_k$ are side lengths of a triangle for 1≤i<j<k≤n.


Hi,

We will be there too in Slovenia, bringing some of our best books with us.

Here is a list published in English by GIL Publishing House

1. T. Andreescu, D. Andrica - An Introduction to Diophantine Equations [ ISBN 973-9238-88-2 ]
2. T. Andreescu, D. Andrica - 360 Problems for Mathematical Contests [ ISBN 973-9417-12-4 ]
3. T. Andreescu, V. Cartoaje, G. Dospinescu, M. Lascu - Old and New Inequalities [ ISBN 973-9417-35-3 ]
4. Negut A. - Problems for the Mathematical Olympiads [ ISBN 973-9417-52-3 ]
5. Cirtoaje V. - Algebraic Inequalities. Old and New Methods [ ISBN 973-9417-66-3 ]

You can find more details on our website. For orders e-mail at sales@gil.ro .

Also send an e-mail there in case you want to buy some books there we we'll make notes and e-mail you where we can meet to give you the books.

Regards!

Hi  me and  my  friend   would like to buy some  books from you 3 in particular  old and  new inequalities, algebraic inequalities and introduction to diophantine equations. anytime you  want we  can meet i will even go to your hotel if necessary.

please  pm me to confirm. thank you very much  :lol:


Is the contest over? I kind of lost track since I'm not participating anymore, and I'm asking because one of the problems which are still unsolved in the NT section is strongly related to one of the problems from round 7, so maybe, in case the contest is over, those who have solved that problem from the contest can also solve this one (it's been bugging me for quite some time now  :D ).

the announcement in the first page was pretty clear grobber :D yes it is over. next one will be next spring. by now I have 2 days more of college and then I can find the time to finish the results, announce the winners, and post the solutions to the remaining rounds.

When I brows this website I connect directly to the forum, skipping the main page, so you can see why I've missed the announcement.

The contest being over, I can now speak freely about that particular problem (not that I have a soln or anything... :D).

yes of course. try to look on the main page also. some announcements are posted from time to time.

If there can be a direct link to the poll on the main page, is it possible to link a post in a forum thread to the main page too?  Then those who never bother to check the main page can just read the announcement here.


Functional Analysis
real analysis
real analysis unsolved

Let $ T$ be a bounded linear operator from the Banach Space $ X$ to $ X$,and $ X\equal{}Y\bigoplus Z$,where $ Y$ is closed and $ Z$ is finite dimension,If $ T(X)$ is closed,Prove that $ T(Y)$ is also closed.

[b]Theorem.[/b] Let $ X,E$ be Banach spaces, $ T: X \to E$ be a bounded linear operator, and let $ T(X)$ be closed. Then for every closed finite-codimensional subspace $ Y \subset X$ its image $ T(Y)$ is closed.

To prove this one needs a number of simple steps.

[b]Step 1.[/b] It is sufficient to prove the Theorem for 1-codimensional subspaces $ Y$: after that the general statement can be easily obtained by induction.

[b]Step 2.[/b] It is sufficient to consider surjective $ T$: $ T$ acts from $ X$ to $ T(X)$, so we can consider $ E \equal{} T(X)$.

[b]Step 3.[/b] It is sufficient consider the case of $ \mathrm{Ker} T \subset Y$. Othervise there is an element $ e \in \mathrm{Ker} T \setminus Y$, so $ X \equal{} \mathrm{Lin}(Y, e)$ (here we use the assumption that $ Y$ is 1-codimensional), so $ T(X) \equal{} T(Y)$ and we are done.

[b]Step 4.[/b] One can pass to injectivization $ \bar T$ of $ T$. Namely one can consider the quotient space $ X/\mathrm{Ker} T$ and operators $ q: X \to X/\mathrm{Ker}$ and $ \bar T: X/\mathrm{Ker} \to E$, where $ q$ is the natural quotient map (it maps every element of $ X$ to its equivalence class) and $ T \equal{} \bar T \circ q$. Since $ \mathrm{Ker} T \subset Y$ then $ q(Y) \equal{} Y/\mathrm{Ker} T$, so it is a complete space, and consequently it is a closed subspace of $ X/\mathrm{Ker} T$.

[b]Step 5.[/b] This last step is just an application of the Banach's Inverse Operator Theorem. Namely, $ \bar T: X/\mathrm{Ker} T \to E$ is a bijective continuous operator, so its inverse is continuous.Then $ \bar T$ maps closed sets to closed sets. So  $ T(Y) \equal{} \bar T (q(Y))$ is closed.


algebra
binomial theorem
binomial coefficients

$ (a\plus{}b)^n$  How do we prove something when $ n$ is simply $ n$ and is not given.  Say for example...

$ a$ is a positive even integer and $ b$ is a positive odd integer.  Show that $ (a\plus{}b)^n$ is a positive odd integer for all positive integers $ n$.  (I just made this up so if theres something wrong with the problem just give me a different example.)

...so $ a$ is in form of $ 2k$ and b is in form of $ 2k'\plus{}1$.

then $ a\plus{}b\equal{}2k\plus{}2k'\plus{}1\equal{}2(k\plus{}k')\plus{}1$, which is an odd number.

any power of an odd number is odd.

...Q.E.D.

The underlying principle here has nothing to do with the binomial theorem, so I'm not sure if this is the example you want.  Your statement relies on two basic facts:

- The sum of an even and odd number is odd.
- The product of two odd numbers is odd.

$ (a \plus{} b)^n$ is a product of $ n$ odd numbers.  Do you have another example in mind?

The hard part about writing a proof with binomial expansion is that theres always this ... in the middle of the expansion since was dont know the value of $ n$ in $ (a\plus{}b)^n$ so we cant expand the whole thing.  If thats the case, then would we usually not attempt to prove something by expanding the binomial but prove whatever is trying to be proven by the properties that $ a$ and $ b$ have as what was done with my odd/even number example?  Sorry if this is a confusing question.  But thanks to those who reply.

I still think you're talking about a different kind of question, and I'm not sure if it's the kind of question that focuses on the properties of the sum or on the properties of the binomial coefficients themselves, so I'm afraid you'll have to be more specific.

There are cases where we can solve problems by a "generic" expansion of the binomial theorem, even when we don't know what $ n$ is.  For example, it's possible to prove that for any $ n$ we have $ 2^{n \plus{} 2} | 5^{2^n} \minus{} 1$ by binomial expansion.

Edit:  It might be instructive to give a "binomial" proof of the original question.  If $ a$ is even and $ b$ is odd, then

$ (a \plus{} b)^n \equal{} a^n \plus{} {n \choose 1} a^{n\minus{}1} b \plus{} {n \choose 2} a^{n\minus{}2} b^2 \plus{} ... \plus{} {n \choose n\minus{}1} ab^{n\minus{}1} \plus{} b^n$.

Note that any term that contains $ a$ is even, which means that every term except $ b^n$ is even, and $ b^n$ itself is a power of an odd integer and therefore odd.  Thus $ (a \plus{} b)^n$ is odd.  (Note that the "..." doesn't matter.  Even though we haven't written down every term, we know what those terms look like, and we know those terms are divisible by $ a$.  This is the basic strategy when dealing with problems of this type.  If it really bothers you, we can use summation notation

$ \sum_{k\equal{}0}^{n} {n \choose k} a^{n\minus{}k} b^k$

and reason with a generic member of the sum.)

Actually I think through your explainations Im starting to understand.  Thanks for your help.


Find all real numbers a such that : the sequence {xn} ,n=0,1,2..
are defined by :
      x_0= \sqrt 1996,x_(n+1)=a/(1+x_n^2) ,n=0,1,2..
 has limit when n ->   \infty


What exactly is AMC and AIME, and where do you take it? What happens if you get in it?  :?:

You people must be thinking I'm an alien o.0

Thanks for your help

You seem like an alien.

AMC can refer to the AMC 8, AMC 10, or AMC 12; or the general AMC test progression of AMC 10/12/AIME/USAMO. AMC=American Mathematics Competitions. Website: unl.amc.edu/
AMC 8=25 questions,  40 min, 8th grade or younger
AMC 10=25 questions, 75 min, 10th grade or younger
AMC 12=same, except 12th and under

AIME is the American Invitational Mathematics Exam, which you take as a result of doing well on the AMC 10 and/or AMC 12. It is 15 questions, 3 hours, and relatively hard.

Good luck :)

And for more information, you can go [url=http://www.artofproblemsolving.com/Wiki/index.php/AMC]Here for AMC info[/url] and [url=http://www.artofproblemsolving.com/Wiki/index.php/American_Invitational_Mathematics_Examination]here for AIME info[/url]

Hope this helps  :lol:

For even more information, refer to the official website of the AMC.

http://www.unl.edu/amc/

 :)

And for even MORE information, I think [url=http://www.artofproblemsolving.com/Wiki/index.php/AMC]AoPSWiki[/url] has a long description and some advices, too. AoPS books/classes are probably the best resources.


[quote="Mathfun062"]I am currently in middle school and am having problems with a pattern type question, The numbers are as follows: 125 135 145 155 163 168...i was told i had to find the pattern between each number. After a couple of days i am still clueless, any ideas on how this sequence works Thank You[/quote]

$125 (130) 135 (140) 145 (150) 155 (159) 163 (165.5) 168...$

does averaging work?

maybe the numbers are all "related" to 5. for example, the first 4 numbers are divisible by 5. if u add the digits of the 5th number, you get something that is divisible by 5. same with the 6th number.

this is hard...

exactly what i thought, i tried several things and nothing, anyone else have any ideas?

this is the exact question, just found it and it might help: what do you think the next 2 numbers in the following arrangement will be? 125, 135, 145, 155, _, _.  to find the solution use your inductive reasoning skills. you base your response on what you think will happen. (the  answer is not 165 and 175. it is 163 and 168. why?)

These kinds of questions are ill-conceived.  There are in fact infinitely many possible ways to create a pattern from the sequence you started.  The trick is to just try to find the simplest one.  If the common difference is 10, then keep on using that common difference, but if the problem claims some strange deviation from that pattern, then the writer is trying to be way too clever.

[quote="Ravi B"][quote="Mathfun062"]125 135 145 155 163 168[/quote]These numbers happen to be consecutive stops on the C line of the New York City subway.  Maybe that is what the question writer was looking for?

MCrawford is right: The question is not well posed.[/quote]

lol that's not math!

wow that's a strange sequence... I even tried the x and y method, sequence method, etc but none of them worked, couldn't really find much of a solid pattern....
and I am almost completely sure that the problem has nothing to do with the New York Subway's line C.

[quote="monoversketcher"][quote="MCrawford"]I bet Ravi is right.  The answer is too nonstandard to be a real math problem.  It's certainly not something that should show up on a math test.

Where did the problem come from?[/quote]

i don't think it was from anything related to math...maybe he got it from his social science class and thought it was math?[/quote]
LOL.  :rotfl:
No offense, but that would be kinda dumb. :wallbash_red:

The everage doesn't have anything in common or maybe I don't understand . Can anyone tell that again .
I mean the answer of the matter


If ABCD is a trapezoid with $ \overline{DC}$ parallel to $ \overline{AB}$, $ \angle{DCB}$ is a right angle, $ \overline{DC} \equal{}6$, $ \overline{BC} \equal{}4$, $ \overline{AB} \equal{} y$, and $ \angle{ADB} \equal{} x$ find $ y$ in terms of $ x$.

[size=150]Please provide HINTS ONLY[/size]

[hide]Represent the area of triangle $ ADB$ in two different ways (the super simple way, and then the way that uses sine)[/hide]


Prove that

$sen^2x+sen^22x+sen^23x+...+sen^2nx=\dfrac{n}{2}-\dfrac{sen(nx)cos[(n+1)x]}{2senx}$

Maybe this formula $sin^2x=\dfrac{1-cos2x}{2}$ be helpful.....
[color=darkred][i]or[/i][/color] induction......

just some first thoughts......... :)

By using Socrates' identity, you need to prove that

$\cos 2x+\cos 4x+\ldots+\cos 2jx=\dfrac{ \sin jx\cos (j+1)x } {\sin x}$

Try adapting one of the methds found in

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53164

to finish the problem.

L.H.S = $\dfrac{1}{2} + \cdots + \dfrac{1}{2} - \dfrac{1}{2}(cos 2x + cos4x + \cdots + cos 2nx)$

Let $S = cos2x + cos4x + \cdots + cos2nx \Rightarrow (2sin x)S = sin 3x - sin x + sin 5x - sin 3x + \cdots + sin(2n+1)x - sin (2n - 1)x$


Mary is $ 20\%$ older than Sally, and Sally is $ 40\%$ younger than Danielle. The sum of their ages is 23.2 years. How old will Mary be on her next birthday?

$ \textbf{(A) } 7\qquad \textbf{(B) } 8\qquad \textbf{(C) } 9\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11$

[hide]
Let Mary's age equal m, Sally's equal s, and Danielle's d.

Then,
$m=1.2s$
$s=.6d$

So,
$m=.72d$
$m+s+d=d+.6d+.72d=2.32d=23.2$

And $d=10$
$m=7.2$

So, Mary will be 8.

B[/hide]

I hated this one... I didn't know what they meant by "their" ages...they shouldve said "sum of all three of their ages"
i realized that "their" refers to ALL three of them when I came out of AMC

By substituting $s$ for $.6d$ into the equation $m=1.2s$, we have $m=1.2(0.6d)=0.72d$


function
induction
algebra unsolved
algebra

Find all functions $f\colon R \to R$ such that
\[f\left(x^{2}+yf(x)\right) = f(x)^{2}+xf(y)\]
for all reals $x,y$.

i think there is some mistake because it is trivial, you just put $x=0$ and you have $f(yc)=c^{2}$ so it's $const$, and if you put it back you will get $f(x)=0$  :D

(*) $f\left(x^{2}+yf(x)\right) = f(x)^{2}+xf(y)$
$x=1$ gives
(**) $f(1+yf(1))=f(1)^{2}+f(y)$ for all $y\in\mathbb{R}$.
Assume $f\not\equiv 0$.

(1) $f(1)=1$:
Assume $f(1)\not =1$.
Then $1+yf(1)=y$ for $y={1\over 1-f(1)}$, and so (**) gives $f(1)^{2}=0=f(1)$.
And then $0=f(1)=f(1+yf(1))=f(1)^{2}+f(y)=f(y)$ for all $y\in\mathbb{R}$ and $f\equiv 0$ using (**). Contradiction!

(2) $f(x+n)=f(x)+n$ and $f(n)=n$ for all $x\in\mathbb{R},n\in\mathbb{Z}$:
(**) and (1) give $f(y+1)=f(y)+1$ for all $y\in\mathbb{R}$.
Then induction gives $f(y+n)=f(y)+n$ for all $y\in\mathbb{R},n\in\mathbb{Z}$.
And (1) gives $f(n)=f(1+(n-1))=f(1)+(n-1)=n$ for all $n\in\mathbb{Z}$.

(3) $f(nx)=nf(x)$ for all $x\in\mathbb{R},n\in\mathbb{Z}$:
$x=n\in\mathbb{Z}$ in (*) gives
$n^{2}+f(ny)=f(n^{2}+ny)=f(n^{2}+yf(n))=f(n)^{2}+nf(y)=n^{2}+nf(y)$
and $f(ny)=nf(y)$ for all $y\in\mathbb{R},n\in\mathbb{Z}$ using (2).

(4) $f(x+r)=f(x)+r$ for all $x\in\mathbb{R},r\in\mathbb{Q}$:
Say $r={m\over n}$ for some $m,n\in\mathbb{Z},n\not =0$.
Then using (2) and (3) we get
$f(x+r)={n\over n}f(x+r)={1\over n}f(n(x+r))={1\over n}f(nx+m)={1\over n}(f(nx)+m)\\ \hspace*{0.72in}={1\over n}(nf(x)+m)=f(x)+r$ .

(5) $f(x)\geq 0$ for all $x\geq 0$:
$f(0)=0$ by (2), so $y=0$ in (*) gives $f(x^{2})=f(x)^{2}\geq 0$ for all $x\in\mathbb{R}$.

(6) $f(x)\geq x$ for all $x\in\mathbb{R}$:
For all $x\in\mathbb{R},r\in\mathbb{Q}$ we have
$x\geq r\Rightarrow x-r\geq 0\Rightarrow f(x)-r=f(x-r)\geq 0\Rightarrow f(x)\geq r$ using (4) and (5).
So $f(x)\geq \sup_{r\in\mathbb{Q}, r\leq x}r=x$ for all $x\in\mathbb{R}$.

(7) $f(x)=x$ for all $x\in\mathbb{R}$:
(3) and (6) give $f(x)\geq x$ and $-f(x)=f(-x)\geq-x$ for all $x\in\mathbb{R}$.

So there are $2$ solutions $f\equiv 0$ and $f(x)=x$ for all $x\in\mathbb{R}$.


induction
function
algebra
polynomial
system of equations
combinatorics unsolved
combinatorics

Can anyone solve any of these problems? Click on the picture to enlarge it if you can't read it.

[img]http://www.anova-design.com/images/1.jpg[/img]

Much Thanks,

[url=http://www.anova-design.com]Austin[/url]

All of this problems are well known, for 47 is just guessing and induction would work:
You get
$f_0+f_2+...+f_{2n}=f_{2n+1}$
For the second you see that
$f_1+f_3+...+f_{2n-1}=f_{2n}-1$
So what you want is the first minus the second. We have to cases, when $n$ is odd and when is even:
$f_0-f_1+...-f_{2n-1}+f_{2n}=f_{2n+1}-f_{2n}+1=f_{2n-1}+1$
$f_0-f_1+...+f_{2n}-f{2n+1}=f_{2n+1}-f_{2n+2}+1=1-f_{2n}$
And for the last, you get
$f_0^2+...+f_n^2=f_nf_{n+1}$

For the next one, right now I dont remember a way to do it nicely, I think there was a neat solution using generating functions, but I dont remember the proof exactly.
For the last one, we proceed as follows
Call the nth term of the sequence, $a_n$. I dont remember exactly how was the example with the rabits, but in general suppose that $a_1=a$ and $a_2=b$. Since the recurrence relation $a_{n+1}=a_n+a_{n-1}$ holds, we get the caracteristic polynomial
$x^2-x-1=0$
with roots $x_1=\frac{1+\sqrt{5}}{2}$,$x_2=\frac{1-\sqrt{5}}{2}$. so we get that
$a_n=c_1\left(\frac{1+\sqrt(5)}{2}\right)^n+c_2\left(\frac{1-\sqrt(5)}{2}\right)^n$

Where $c_1$ and $c_2$ are numbers determined by the system of equations

$a=c_1+c_2+\sqrt{5}(c_1-c_2)$
$b=\frac{3}{2}(c_1+c_2)+\frac{\sqrt{5}}{2}(c_1-c_2)$

And thats it  :lol:


Two circles with spinners at their centers are divided into
three equal regions as shown. When both spinners are spun,
what is the probability that the product of the two values is
negative? Express your answer as a common fraction.
[asy]size(140);
void spinner(picture pic=currentpicture, real rotate, int a, int b, int c)
{
dot(pic,(0,0),linewidth(4));
draw(pic, unitcircle ^^ (0,0)--dir(90+rotate) ^^ (0,0)--dir(210+rotate) ^^ (0,0)--dir(-30+rotate));
label(pic, "$"+string(a)+"$",dir(30+rotate)/2,fontsize(10pt));
label(pic, "$"+string(b)+"$",dir(150+rotate)/2,fontsize(10pt));
label(pic, "$"+string(c)+"$",dir(-90+rotate)/2,fontsize(10pt));
}
picture p;
spinner(0,-1,-3,5); spinner(p,180,2,4,-6);
draw((0,0)--dir(-10)/2,EndArrow(4)); draw(p,(0,0)--dir(-75)/2,EndArrow(4)); add(shift(2.5*right)*p);[/asy]

negative and positive

$ \frac{2}{3}*\frac{2}{3}\equal{}\frac{4}{9}$

positive and negative

$ \frac{1}{3}*\frac{1}{3}\equal{}\frac{1}{9}$

answer : $ \frac{5}{9}$


Fie trapezul $ABCD$ $(AB\parallel CD)$ si punctele $E\in{(BC)}$, $F\in{(AD)}$ astfel incat $AE\parallel CF.$
Sa se arate ca $BF\parallel DE.$

[color=blue]Comentariu. Problema apare intr-un manual de clasa a VII-a, inainte de asemanare. Ce solutie putem da?[/color]

[quote="mchis68"]Folosind Thales.
Daca $O\in AD\cap BC$, atunci
\[(1)\qquad\frac{OD}{OA}=\frac{OC}{OB}\]
si
\[(2)\qquad\frac{OF}{OA}=\frac{OC}{OE}\,. \]
Impartind cele doua relatii, obtinem
\[\frac{OD}{OF}=\frac{OE}{OB}\,, \]
de unde rezulta paralelismul dorit.[/quote]

Solutia dumneavoastra este buna.
Autorul manualului a pus insa problema inainte de capitolul "Asemanare", asa ca trebuie gasita o solutie fara Thales sau TFA.

O soluţie simpatică foloseşte următoarea lemă, uşor de demonstrat:
Fie $O$ intersecţia diagonalelor patrulaterului convex $XYZT$. Atunci $XY$ e paralel cu $ZT$ dacă şi numai dacă $aria(XOT)=aria(YOZ)$ (vezi trapez1.jpg).
Astfel, deoarece $ABCD$ este trapez, avem (vezi trapez2.jpg)
\[[4]+[7]+[10]+[13]=[6]+[9]+[12]+[16]\]
(am notat cu $[x]$ aria regiunii marcate cu $x$ pe figură)
iar deoarece $AECF$ e trapez, avem şi
\[[10]+[13]=[6]+[8]+[9].\]
De aici rezultă
\[[4]+[7]+[8]=[12]+[16],\]
adică $DEBF$ e trapez  :)

Da, solutia este frumoasa; tot cam asa am dat-o si eu azi elevilor mei. Dupa ce am terminat solutia, un elev mi-a atras atentia ca problema apare in manual inainte de "Ariile suprafetelor poligonale". Ca argument, i-am spus ca ariile au fost facute in clasa a VI-a. I-am promis ca voi cauta si o solutie care nu foloseste ariile. Mai caut. Prezint in attach expunerea mea de azi.