Point $ P$ is selected at random from the interior of the pentagon
with vertices $ A\equal{}(0,2)$, $ B\equal{} (4,0)$, $ C \equal{} (2\pi \plus{}1, 0)$, $ D\equal{}(2\pi
\plus{}1,4)$, and $ E\equal{}(0,4)$. What is the probability that $ \angle APB$ is
obtuse? Express your answer as a common fraction.

[asy]pair A,B,C,D,I;
A=(0,2);
B=(4,0);
C=(7.3,0);
D=(7.3,4);
I=(0,4);
draw(A--B--C--D--I--cycle);
label("$A$",A,W);
label("$B$",B,S);
label("$C$",C,E);
label("$D$",D,E);
label("$E$",I,W);[/asy]

[quote="football017"]
P.S.: how do you put a line over $AB$?[/hide][/quote]

Click this ---> $\overline{AB}$ (the command is basically \overline)

[quote=jbatterson]Beautiful problem...  great work making the answer a common fraction![/quote]

I thought answers that are fractions are always common fractions...

[quote=Eugenis]How did you figure out is it right iff it lies on the circle, and obtuse iff it lies inside the semicircle? Is there some geometric principle?[/quote]

Yeah, how?

Basically cuz any triangle inscribed in a semicircle with longest side as the diameter has a right angle opposite to the diameter. In other words, in a right triangle, the median to the hypotenuse is equal to half of the hypotenuse.

thanks, that explains how the problem is solved, but why is that fact true?

The angle that the two end points of the diameter of a circle and ANY POINT on the circumference is a right triangle.

How did you guys know for sure that the shape made where $\angle APB= 90 ^ o$ was a semicircle

for what?
the question?
didnt post 3 explain?

[quote=football017][hide="Solution"]We know that $\angle{APB}$ is obtuse if $P$ is in the interior of circle $O$, where $O$ is the midpoint of diameter $AB$ ($\angle{APB}$ would be right if $P$ was on circle $O$). However, since $P$ must be in the interior of the pentagon, we are dealing a semicircle because only half of circle $O$ will be in the pentagon. We can find the area of the semicircle and divide that by the total area of the pentagon.

[hide="Area of semicircle"] By distance formula, the diameter of the semicircle ( $AB$) is $\sqrt{(2-0)^2+(0-4)^2}=2\sqrt5$; thus, the radius is $\sqrt5$. Therefore, $A=\frac{r^2\pi}{2}=\frac{\pi(\sqrt5)^2}{2}=\frac{5\pi}{2}$[/hide]

[hide="Area of pentagon"] The pentagon is a $2\pi+1$ by $4$ rectangle minus a $2$ by $4$ triangle in the bottom left corner. Therefore, we have $(2\pi+1)(4)-(\frac{1}{2})(2)(4)=8\pi$[/hide]

[hide="Final Step"]Finally, we have $\frac{\frac{5\pi}{2}}{8\pi}=\frac{5\pi}{2}\times\frac{1}{8\pi}=\boxed{\frac{5}{16}}$[/hide]

P.S.: how do you put a line over $AB$?[/hide][/quote]

THANK YOU VERY MUCH!!!

I was SO CONFUSED! :D :D :D :D :D

[quote=GAMER100]I am confused too. How do we find the semicircle?[/quote]

Hope this is NOT TOO BIG OF A BUMP, but

If you remember the equation x^2+y^2=z^2, it is a circle. Again, it is also the formula for a right triangle's sides. So the angle must be 90 degrees. Within it, it is obtuse. Out of it, it is acute. So we find the area outside of it.

[quote=GuoGuoBellevue][quote=ThriftyPiano][quote=GAMER100]I am confused too. How do we find the semicircle?[/quote]

Hope this is NOT TOO BIG OF A BUMP, but

If you remember the equation x^2+y^2=z^2, it is a circle. Again, it is also the formula for a right triangle's sides. So the angle must be 90 degrees. Within it, it is less. Out of it, it is obtuse. So we find the area outside of it.[/quote]

Correct me if I'm wrong, but doesn't the problem use that within it it is obtuse, and out of it it is acute?[/quote]

Yes. I was dumb back them.

Fixed.


que te gusta hacer, ustedes?

Me gusta esuchar la musica en mi tiempo libre. :)

[quote="Silverfalcon"][b]¿[/b]Qu[b]é[/b] [b]les[/b] gusta hacer, ustedes?

Me gusta es[b]c[/b]uchar la m[b]ú[/b]sica en mi tiempo libre. :)[/quote]

No se puede usar 'te' y 'ustedes' juntos como así :)

Pienso que a toda la gente aquí le gusta las matemáticas, y probablemente un instrumento de música.. y quizás el ajedrez, como yo.

Wow.. ;)

Didn't really get much word out of it..

I think I got the part you like about music and math though :D

No se puede usar 'te' y 'ustedes' juntos como así :)
[b]You can't use 'te' and 'ustedes' together like that :)[/b]

Pienso que a toda la gente aquí le gusta las matemáticas, y probablemente un instrumento de música.. y quizás el ajedrez, como yo.

[b]I think that everyone here likes math, and probably a musical instrument.. and maybe chess, like me.[/b]

(about that te + ustedes thing, if you didn't realise: ustedes is the polite/formal form of 'you (plural)', wheres 'te' is in the casual form of 'you' - you have to make sure you use the same style throughout.)

I got where I screw up now.

I like chess too. Are you good at it?

Silverfalcon, a mi me gusta mucho la musica y llevo estudiandola desde hace tiempo ademas de las matematicas.
Que tipo de musica te gusta?

Me gusta jugar futbol, cantar, y escuchar musica.

Me gustan muchas cosas, por ejemplo, jugar deportes, jugar videojuegos, escucha la musica y matematicas.

(sorry, i don't know how to put accents on. my computer's sort of screwed up that way.)

A mi me gusta la musica, el ajedrez y en las matematicas me gusta la teoria de numeros y algebra.
Tambien me gusta mucho estudiar sobre la inteligencia artificial. No tengo musica favorita, ni grupos favoritos, ultimamente me esta gustando Green Day. Y ustedes?

"En Mexico y en el mundo, la cerveza es.... Corona!"

[quote="mcalderbank"]a mi, me gusta cerveza :P[/quote]

Calderbank,

No sos un poco joven para andar tomando cerveza? (aun cuando yo tambien pienso que es muy sabrosa :P )

Hmm...
I can only understand "tequila"  :lol:

[quote="DemisM"]Yo no le encuentro el gusto a la cerveza, pero hablando de mexicanos, me encanta la tequila.  :)[/quote]

Bueno, en general el tequila bueno, el que de verdad es de Agabe, a mi me encanta, sin embargo, la cerveza Corona no es nada especial (aunque mucho mejor que la gran mayoria de las cervezas estadounidences! :P )

Ciao,

[quote="Dave Vather"]La próxima vez que tomen tequila, recuerden que están ayudando a los matemáticos.
La industria tequilera (En particular la marca Sauze, creo) ha sido de las primeras industrias mexicanas en contratar matemáticos. No entiendo todavía qué hacen, pero tengo entendido que lo han hecho bien.[/quote]

 :huh:  :what?: Que rayos puede hacer un matematico en una fabrica de licor???

En fin, pues espero que el agave nunca se acabe, por que la verdad a mi si me gusta un Tequilita de vez en cuando!

Pero en serio, que pueden poner a hacer a un matematico en la industria tequilera.... Un misterio mas para la humanidad  :alien:


superior algebra
superior algebra unsolved

let $ a,b\in G$ such that $ a^2b^2\equal{}b^2a^2\ ,\ a^3b^3\equal{}b^3a^3$ prove $ ab\equal{}ba$

This is part of a more general result. $ m, n \in N, gcd(m,n) \equal{} 1, a^{m}b^{m} \equal{} b^{m}a^{m}$ and $ a^{n}b^{n} \equal{} b^{n}a^{n}$ implies $ ab \equal{} ba$. For a proof of this fact look at http://www.mathlinks.ro/viewtopic.php?p=274033#274033

Below you'll find a result that elaborates on the one mentioned by [b]alejnaser[/b]. Couldn't resist to bring it up, you know!  :D

[url]http://ndikandi.utm.mx/~lm2002070425/CHIDO.pdf[/url]

Best of Luck.


group theory
abstract algebra
superior algebra
superior algebra solved

Please tell me how to show that there are no simple groups of order $3875=31. 5^3$

The Burnside theorem states that no group of order $p^aq^b$ with $a,b\geq 1$ and $p,q$ prime is simple.  But I assume you don't know representation theory. :)

Anyways, less generally (but more generally than your problem), if $|G|=p^3q$, then either $G$ has a normal Sylow $p$ or $q$ subgroup or $|G|=24$.  This, at least, is not difficult to prove.  Put $n_p(G)$ to be the number of Sylow $p-$subgroups of $G$.  Proceed by contradiction.  We need to show that the cases $n_q(G)=p,p^2,p^3$ all lead to contradictions.  In all cases we must have $n_p(G)=q$ and $q\equiv 1\bmod p$ by the Sylow theorems, so that $q>p$.

$n_q(G)=p$.  $q>p$ so $p\equiv 1\bmod q$ is absurd.  It follows that $n_q(G)\not=p$

$n_q(G)=p^2$.  Then $p^2\equiv 1\bmod q$, so that $q|(p+1)(p-1)$, and since $q>p$, we must have $q|p+1$, which implies $q=p+1$ (this leads to the case $|G|=24$)

$n_q(G)=p^3$.  There are $p^3(q-1)$ elements of order $q$, so let $X$ be the set of elements with order different than $q$.  Clearly $|X|=p^3$.  So any Sylow $p-$subgroup of $G$ must be all of $X$, and so it is unique.  It follows that $n_p(G)=1$, which is impossible.

This proves your result.

Could u show it in a general way for the order of 3875 which I might mean by using $3875=31.5^3$ instead of using the $p,q$, I really wanna know how it goes to the result. Thx

By the way, how would show that there are no simple groups of order of $3874= 2.13.149$, similar way or??

The number of $149$-Sylowsubgroups is $\equiv 1\mod 149$ and divides $3874$, so there is just one. So this subgroups is normal.

Here is my idea to show that $|3874|$....

$|G|$=$|3874|= 2.13.149$
Suppose that G is simple. Then G has 26 sylow 149-sungroups and 298 sylow 13-subgroups. Therefore, there are 12.298=3576 element of G of order 13 and 148.26=3848 elements of G of order 249, and it comes total 3576+3848=7424 elements. But G has only 3874 elements, so contadiction.
Thus, G is no simple.

Is this right?

[quote="Disco Very"]Refers to "blahblahblah"'s post, where does that $G=|24|$ come from and could I put $31, 5$ into the theorem instead of $p, q$ and get the same result?

Could I apply the same method if I wanna show there are no simple groups of order of $3874= 2.13.149$[/quote]

Yes, you can plug in $31,5$.  The $|G|=24$ comes from the case in which you get $q=1+p$.  Since $2,3$ is the only pair of consecutive primes, this leads to $|G|=24$ (of course we can prove non-simplicity here via other methods)

Is my answer right? Could someone check it?

I do  not know

it is overkill though, you get contradiction earlier

remember sylow theorem gives you that the number of sylow p groups is 1 mod p

koola: I don't see how you come up with your numbers of Sylow subgroups (why the simplicity implies that numbers.
But look at post #4 ;)


Hey there, i'm icx from Romania and about a month ago i had the interesting idea of preparing for the romanian olympiads and perhaps for the international one. I don't know where it came from. (actually, the fact that i scored about 9.5 out of ten at a test gave me a small impulse).

I am 16 years old and just landed in a college :) (9th grade). I consider myself as 'ok' in maths, but definitely need training, because my level in this science is slightly above the stuff we do at school/high school, and that won't help me. I have some questions regarding this competition, in case i won't participate in 2007 or 2008:

- what's the average age of the competitors?
- what's the maximum age you can participate?
- why do some students get so little points? is this an emotional problem, or lack of preparation?
- are the subjects less difficult than they were in the 60s-70s? (i find them less difficult for our national olympiads, at least)

Hope my desire of participating at such an olympiad will be constant :D

Best wishes, icx.

ps: more questions still to come :)

[quote="icx"]
- why do some students get so little points? is this an emotional problem, or lack of preparation?[/quote]

I think that this happens because there a lot of "weak" countries taking part (and for me this is OK), and they don't have enough experience in this kind of competitions, so they have poor results. However, they will improve with time and practice. If the problems are easier than their are now, there would be 20 perfect scores, and that's not the idea perhaps, so maybe getting 4 or 5 points is not so bad, just with the fact of going to an IMO is great: then, what happens at the competition doesn't matter a lot (talking about weak countries)


Find the best constant $c$, such that the following inequality is valid for all $a,b>0$

$\frac{a^2}{b^3}+\frac{b^2}{a^3}\le c$

For $b$ constant $\lim_{a \to \infty} \left(\frac{a^2}{b^3}+\frac{b^2}{a^3} \right)= \infty$     :)


function
Functional Analysis
real analysis
real analysis unsolved

Let $F$ be a closed subspace of the space of continuous functions defined on $[0,1]$ with real values such that all elements of $F$ are continuously differentiable. Prove that $F$ has finite dimension.

This is a bit off topic, but I'm just curious how much time they allow you for solving a problem like this. Also, were you assumed to take a course in Functional Analysis before this examination?

Assume the space is infinite-dimensional. It's closed unit ball $\mathcal B(0,1)$ is then a non-compact closed bounded subset of $C(I,\mathbb R)$ ($I$ is the unit interval), so, by the Arzela-Ascoli Theorem, it cannot be equicontinuous. This means that there is an $x\in I$ and a $t>0$ such that no matter how small $\delta>0$ is, we can always find $|y-x|<\delta$ with $|f(x)-f(y)|\ge t$ for some $f\in\mathcal B(0,1)$. We may take $x=0$ WLOG.

Take $f_n\in\mathcal B(0,1)$ with $f_n(y_n)-f_n(0)\ge t$ for some $y_n<\frac 1n$. Our aim will be to find a function of the form $f=\sum\alpha_kf_k\in F$ for $\alpha_k\ge 0$ and a properly chosen subsequence $(n_k)_k$ such that $f$ is not differentiable in $0$ (instead of $f_{n_k}$, I wrote $f_k$, and instead of $y_{n_k}$ I'll write $y_k$). Since I'm sure I'll get a bit lost in notation, I'll just sketch this part.

Let $\phi_j=\sum_{i=1}^j\alpha_if_i$. First of all, the $\alpha_k$ can be taken $<\frac 1{2^k}$, so we're sure that our series converges absolutely and uniformly. Now, on the one hand, we take $\alpha_{j+1}$ small enough that no matter which $n_{j+1}$ we choose, the values $\frac{\phi_{j+1}(y_i)-\phi_{j+1}(0)}{y_i},\ i=\overline{1,j}$ stay larger than $1,2,\ldots,j$ respectively, and, on the other hand, we choose $n_{j+1}$ large enough to have $\frac{\phi_{j+1}(y_{j+1})-\phi_{j+1}(0)}{y_{j+1}}>j+1$. We will then have $y_k\to 0$ and $\frac{f(y_k)-f(0)}{y_k}>k$, so $f$ is not differentiable in $0$.

[quote="grobber"] We may take $x=0$ WLOG.
[/quote]
Why? $x=x(\delta)$ depends on $\delta$, so it is a "wandering" point, not a fixed one.

Nevertheless, grobber's idea is very good and can be used to show much more: it suffices to assume that every function in $F$ is everywhere differentiable (not necessarily continuously). For harazi's original problem,  instead of doing things by hand, we can use a big cannon. Just consider the mapping $f\mapsto f'$ from $F$ to $C[0,1]$. It's graph is closed and, therefore, the mapping is bounded, so $\|f'\|_{C[0,1]}$ is bounded by some constant in the unit ball of $F$. That is more than enough for equicontinuity.

[quote="fedja"][quote="grobber"] We may take $x=0$ WLOG.
[/quote]
Why? $x=x(\delta)$ depends on $\delta$, so it is a "wandering" point, not a fixed one.

[/quote]

Actually, no, I believe we can find such a point which does not depend on $\delta$ :?. Here's the version of the Arzela-Ascoli Theorem that I am referring to:

We say that a family $\mathcal F$ of functions $f: X\to Y$ ($X,Y$ are metric spaces, and in order to keep things simple, the distance in both of them will be denoted by $|\cdot|$) is equicontinuous in $x_0$ if for every $\varepsilon>0$ we can find $\delta>0$ such that $|x-x_0|<\delta\Rightarrow |f(x)-f(x_0)|<\varepsilon$. The family is then said to be equicontinuous if it's equicontinuous in every point. The theorem states that if $X$ is compact and $Y$ is some $\mathbb R^d$, a family of functions from $C(X,Y)$ is compact iff it's closed, bounded, and equicontinuous.

The negation of the above (we do [i]not[/i] want the unit ball to be compact) gives us an $x$ which does not depend on $\delta$ (in fact, that's how I formulated it: "we can find a $t$ and an $x$ such that no matter how small $\delta>0$ is..").

Yes, you're right  :blush:. I just got used to the Arzela-Ascoli theorem  formulated with "uniform equicontinuity" and automatically assumed that you are using this version (I often remind myself that I should be more attentive when reading other people's posts :( ). Sorry for the first comment. Nice solution :).

The examinations in France are very special (especially for ens) because they give extremely difficult problems (at least for people like me), but the examination is in fact a long dialogue between the examinator and the student and the examinator often gives hints for problems like this. For example, since I didn't know the theorems of open application and closed graph, he asked me first to prove them (I only had 1 indication for these proofs) and after that he said that the problem is a consequence of the closed graph theorem. Then I knew what I had to do (hopefully, I knew Riesz theorem). And no, we didn't have courses of functional analysis.


[quote="daine"]What are your favorite authors and books?[/quote]

"El Último Teorema de Fermat", by Simon Singh.

"Geometry Revisited", by Coxeter and Greitzer.

"Harry Potter", by J.K. Rowling. (all HP's books).

My favoirite auther is Orson Scott Card!

"Harry Potter", by J.K. Rowling. (all HP's books)

El Último Teorema de Fermat", by Simon Singh(english though)

oh and with dain Eoin Colfer and guess what it is time to submit spells!!

And all artemis fowl books alive

I just finished reading "A Madman Dreams of Turing Machines," by Janna Levin. I enjoyed its prominent display math history and references. Has anyone else heard of/read this book? What do you think of it?

[quote="daine"]What are your favorite authors and books?

My favorite authors are:
Tamora Pierce (which leads to my screen name)
Orson Scott Card
Phillup Pullman
Garth Nix
Anne McCaffrey
Eoin Colfer
Lemony Snicket
Peter Dickinson
...[/quote]

I'm a little bit more classic than you :) but I also love fantasy or sci-fi novels. However I tend to read only books I know are going to be pretty good. For instance, I've only read Orson Scott Card's Ender's Game, but this is because when I read one scifi author for the first time I prefer the book to be somewhat awarded. Then, if I like it, I mark the author and start searching for other books. In this particular case I liked the book so I'm looking forward to read Speakers for the Dead, etc.
In sci-fi I think one should try P.K Dick(great atmospheres), Bradbury(wonderful writing style), Stanislav Lem(different, non-american), some Asimov(classics, very good stories, "Nightfall") and the Dune & Hyperion series. You [i]daine[/i] might like Le Guin which by the way is the reason of my nick, that's because I see you enjoy tamora pierce. A.C. Clarke is OK too

Non sci-fi I choose using the same principles. So I've read some Nobel Prizes and classics. I tell you, in general, all those books you hear about them being great masterpieces, they usually are. Maybe one needs to be a bit old to appreciate them but for instance I've read Tolstoi's "War and Peace" and "Anna Karenina" and they are both amazing books to read, I think. You also learn a lot of stuff. Somewhat modern authors, there is this german guy called Heinrich Böll, he is a nobel, read "Opinions of a Clown". Read Umberto Eco's "The Nameof the Rose". Short stories by argentinian  authors Jorge Luis Borges and Julio Cortazar. There is this spanish author Carlos Ruiz Zafon who wrote recently a book called "The Shadow of the Wind". Read it please, even if you don't feel atracted by the ones I've said before because it is beautiful I think for the broadest kind of reader.

Maybe I put some more later, I like this topic

Ursula

PD: HP books are getting worse dont' you think? Third was the greatest. Also everybody seems to have read the book about Fermat LT, I've never heard of it. Anyway one thing I don't do is read science-for-everyone kidn of books. Either I understand the science or the book seems absurd to me. For instance Stephen Hawking's books in my opinion suck. Things are not that simple and if you don't know the subject you will be asking things no one is going to answer at every line of the book. In the other hand are Roger Penrose books which you do understand if you are a mathematician or a physicist but probably don't if you belong to the general public.

I enjoy reading fantasy and I can't speak of it without mentioning Tolkien. I haven't read better fantasy novel then LOTR  :yup:  and it seems that I won't.
I don't like some author's tendency to simplify style, reduce the language and prolong the text just to put in 13 books something that would perfectly fit in 5.
Good example for this is Robert Jordan's Wheel of Time series. First 4 books had some dynamics, characters were nice shaped and distinctive, story had some development...Fifth book is bearable but for following books even the most devoted fan  doesn't have good opinion.  :furious:
Anne McCaffrey and Ursula Le Guin are very good writers. I must mention one more female writer ,Marija Semenova, in whose novel I really found some female and slavic touch which is rarely found in fantasy. I'm sorry but I don't know how to translate name of the book. In russian it's Volkodav.
One of my favoured books is Alienist by Caleb Car. It's amazing story about crime and human soul, but also about uncovering the secrets those around us and in us...

[quote="Nienna"]I enjoy reading fantasy and I can't speak of it without mentioning Tolkien. I haven't read better fantasy novel then LOTR  :yup:  and it seems that I won't. [/quote]

Lord of the Rings was the first book that really impressed me. I read it pretty young, I was 13 or 14 and I thought it was the best book I had read for a long time. Now it's been some time and I've  learned to value other stuff and I find pretty difficult to name one book as the best. Lord of the Rings nevertheless is still unique.
Nice Valar name by the way ;)

Ursula

[quote="kate25"]I like classics the best, with an occasional dash of fantasy.
Authors:
Jane Austen
Louisa May Alcott
Alexander Dumas
Charles Dickens
CS Lewis (Narnia!)

Books:
Pride and Prejudice
Count of Monte Cristo
Jane Eyre
HP
Alias Grace
Last and Future King
Northanger Abbey[/quote]

Of these I've read Count of Monte Cristo, Jane Eyre adn HP. I still think Dumas is better in The Three Musketeers series(is this how it is said?), including Twenty Years Later and the Viscount of Bragelonne than in CMC. Count of Monte Cristo is vicious in an unrealistic manner in my opinion. Well, maybe it's just im unable of putting myself in Dantes's place, but I think that at some point of his vengeance almost everyone would have realized its pointlessness.
Jane Eyre was a great book though one has to put it in perspective when judging Jane's actions. I liked the ending, cried a bit I guess.

HP up to the fourth book is excellent. I have this question because you see, one important part of the HP magic comes form the fact that there is a magical world running parallel to our lives of which we are not aware. Most things in this world function just like they do in our world, there are schools, government, shops, banks, pubs, post offices, and soccer world cups(well quidditch) everything but with the additional ingredient of magic. And my question is have you heard of someone to come up with this idea before Rowling? Because I see everybody inventing new worlds and putting magic there, typically Middle Age like worlds? But do you know of some other book where magic is integrated with modern society? I find this idea as brilliant as it is obvious and of course she can play a lot with this to produce funny situations like at the beggining of the fourth book when they arrive to the quidditch game and there is this muggle guy guarding the entrance because no wizard wants to miss the game but they must erase his memory every ten minutes or so. That was hilarious.

I will read Jane Austen some time this year, I have this list of books to read which I try to follow sequenntially and Pride and Prejudice is approaching. Of course it is not rigid and if somebody lends me a book for a short time it will go first but that way I don't forget the books that have interested me.

Oh and I didn't like Narnia at all. I don't believe in God so Narnia's obvious and ubiquitous metaphores related to christian beliefs kind of cool me down. I quitted after reading two of the books, also the naivety like the bad witch is very very bad and the good lion is purity itself, and this also happens with every other character. I don't think it is good not only from my perspective but from the perspective of a child to think in black and white like in Narnia.

Ursula

[quote="junggi"]"룬의 아이들"-전민희
THE best fantasy novel I've ever read. Too bad my mom won't buy me the new ones. =(
And I second "El Último Teorema de Fermat", aka "Fermat's Last Theorem."(In english, the book's called "Fermat's Enigma")[/quote]
thankyou junggi i totally forgot it's real name bt that is the whole reason i got into math by reading it

Eric Nylund
J.K. Rowling
Eoin Colfer
Mr. Clark (screen name but he writes a really good fanfic)

Does anyone like nobel prize books? i read some with the awards but they all seem to suck...

[quote]Oh and I didn't like Narnia at all. I don't believe in God so Narnia's obvious and ubiquitous metaphores related to christian beliefs kind of cool me down. I quitted after reading two of the books, also the naivety like the bad witch is very very bad and the good lion is purity itself, and this also happens with every other character. I don't think it is good not only from my perspective but from the perspective of a child to think in black and white like in Narnia[/quote]

I didn't like Narnia either. Book is full of logical mistakes-one situation is mentioned twice but with different explanations. Strictly black-white perspective really makes plot simple and style isn't sofisticated at all.

Ursual, I really enjoy reading your comments... :)

Am I the only one that never noticed any Christian metaphors in Narnia?

Brandon Sanderson's Mistborn and Stormlight trilogy, All of Brandon Mull's Fablehaven, Dragon watch, Five Kingdoms, and Beyonders

[quote=AureliaQueenOfScotland][quote=Felicityhe]I love the Percy Jackson books. I also really like the Grishaverse. My favorite stand alones are most of the Newbery books.[/quote]

YES PERCY JACKSON AND ALL OF THE RIORDANVERSE[/quote]

YEAH LET'S GO PERCY JACKSON FANS!

Also a book named "Speak" and "Thirteen Reasons Why" which talk about suicide and mental health

@abhavsar looks like he did, but in his post, he mentions @bove bumping the thread, meaning someone else bumped it but deleted the post lol

[quote]might just be me who reads it though :|[/quote] Oh no. I own the entire series and I've read it several times over since 3rd grade  :love:. It's an amazing series. All of Stuart Gibb's books are.

[quote=RoyalCactus][hide=SamuraiA (#189)][url=aops.com/community/user/380888][b]SamuraiA (Profile)[/b][/url] $\cdot$ 40 minutes ago $\cdot$ [aops]Y[/aops] 0 $\cdot$ [tip=Avatar][img]https://avatar.artofproblemsolving.com/avatar_380888.png?t=1618596579[/img][/tip]
 ----- [quote name="SuperTarantula" url="/community/p21637301"]
The Percy Jackson series is really good, but HoO has a lot of problems, mostly with Piper (i think so anyway)
[/quote]

wait what problems

HoO is better than the original Percy Jackson series imo $\phantom{modified by WyattB, do not take without permission. RoyalCactus has permission.}$[/hide]

HoO's plot is better than Percy Jackson
but Percy Jackson (the original series) gives off better vibes.[/quote]

What is the best HoO book in your opinion.

[quote=Felicityhe]Hoh or MoA. God the percabeth content fed me for generations.[/quote]

agree

HoH makes me cry twice, the first one is for Nico, and the second one is for Percy and Annabeth

[quote=Wukong]they're the same series to me
equally nice

although I stopped midway through Trials of Apollo since I had to wait for the books to come out and forgot the plot

---

also Spy School is nice
might just be me who reads it though :|[/quote]

lol ive read it so many times


[quote="wow346"]So...anyone else get a 23 on this USAMO? I really hate it when I miss a cutoff by a small amount, but I guess being a first-year USAMO qualifier, having never seen any of these proofs that you guys seem to know like the back of your hands, I should be proud :). Anyway senior year for me, so I hope they have good math competitions at Caltech. Thanks AoPS![/quote]
Well, you can start to prepare for Putnam competition. The next Putnam is in Dec, 2005. Great job on your USAMO!

Since you're going to Caltech, here are some results from the 2004 Putnam.
Caltech came in fifth. Four Caltech students, including the three members of the team determining school rank, were between 25th and 75th individually.

The problems and solutions for past years are available at the AMC website, as well as other places.
Putnam problems have a rather different style than the USAMO. They use advanced concepts, but that doesn't necessarily make them harder. Most importantly, it's twice as many problems in 2/3 the time. The problems tend to be less complicated to balance this, but the time pressure is very serious if you want a high score.
The grading is also very tough- a solution you thought was nearly correct is likely to get one or two points out of 10.

Wow, I hope I can get a 23 by my senior year! However, I have a lot of studying to do in order to achieve that, hehe. Anyways, good luck at CIT!

I got a 23 as well... it was my first year taking USAMO since it was the first year in a while that London (Ontario) offered AMC competitions.  I think I'd have done better if I'd done practise USAMO's in exam settings.  Also, my friends warned me that I might not be able to get >= 1 question, and so by the time I got 3, I figured I'd made my quota and I stopped working quite so hard.  And I don't think I should have taken that break on day 2 to write that limerick about question 4... although I expect it gave the markers a good laugh.

So close!!  Oh, well, I'm still pretty happy with my score - as I said elsewhere, I was not expecting any double-digits, and I ended up with a double-digit score and a double-digit placing.  :)  I wish I could do it again.


The length of rectangle $ABCD$ is $5$ inches and its width is $3$ inches. Diagonal $AC$ is dibided into three equal segments by points $E$ and $F$. The area of triangle $BEF$, expressed in square inches, is:

$\text{(A)} \ \frac 32 \qquad \text{(B)} \ \frac 53 \qquad \text{(C)} \ \frac 52 \qquad \text{(D)} \ \frac13\sqrt{34} \qquad \text{(E)} \ \frac13\sqrt{68}$

[hide]Area ratios will do the trick. We know that the altitude of triangle BEF is equal to that of triangle ABC. Additionally, its base is 1/3 of ABC, so that triangle BEF has 1/3 the area of triangle ABC.

Similarly, triangle ABC is half the rectangle. Thefore, triangle BEF is 1/6 the area of the rectangle, or $\frac{15}{6}=\frac52$.[/hide]


calculus
derivative
calculus computations

Hello everyone. I am double checking a derivative problem and would like a yea or nay please.

f(x) = ae^kx

pluging in equals

f(x) = 1.49e^.0099x

the derivate of the above
f'(x)= .014e^01x

is this correct at all

[quote="spix"]$f(x)=ae^{kx}$ so $f'(x)=k\cdot a\cdot e^{kx}$ so your derivative should be $f'(x)=0.0099\cdot 1.49 \cdot e^{0.0099x}$[/quote]

Thanks for the reply. Is it possible to multiple .0099 with 1.49 to get the answer of .014e^.0099x? I found some examples in my textbook which state this.


[quote="Minh Thang"]Your problem is a problem in international mathematics olympics 2001.Because if we give x=bc/a 2 ;y=ac/b 2 ;
z=ab/c 2 ,it becomes
 \sum a/ \sqrt (a 2 +8bc) \geq 1
Perhaps you proved it by the inequality a^(4/3) / \sum a^(4/3) \leq
a/ \sqrt (a 2 +8bc.It can be proved by Cauchy.
But it isn't national. But we can use BCS and Shur .
You should try.[/quote]
Two posts are nearly identical except the last post by you has one more sentence. There is a missing parenthesis after 8bc. 'national'? You mean 'rational'? Replace Schur for Shur.

[quote="Valiowk"]If x, y, z > 0 and xyz = 1, show that 1/\sqrt(1+8x) + 1/\sqrt(1+8y) + 1/\sqrt(1+8z) \geq 1.

I have a solution for this inequality - as a result of another, well-known inequality.  I'm wondering if it's possible to do this question without making use of it.[/quote]

Let's try: Put  a = 1/\sqrt(1+8x), b = 1/\sqrt(1+8y), c = 1/\sqrt(1+8z) then  (1/a^2 - 1)(1/b^2-1)(1/c^2-1) = 64 and we have to show that  a + b + c \geq 1.

This is equivalent to: if a+b+c = 1 then (1/a^2 - 1)(1/b^2-1)(1/c^2-1)  \geq 64. The last one can be proven easily by replacing 1 to (a+b+c)^2 then using AM-GM.

Namdung

Thanks Namdung! (You meant (1/a2 - 1)(1/b2 - 1)(1/c2 - 1) \geq 512, I believe?)

I noticed the similarity to the IMO 2001 inequality, but I didn't want to use that to get a solution for this one.  I remembered that the Cauchy-Schwarz inequality worked for that question, but it doesn't seem to work for this...


integration
geometry
rotation
geometric transformation
calculus
calculus computations

i keep getting the wrong answer for this one:

find the volume of the solid generated ne revolving the region bounded by the graphs of the equations about the line $x=6$
$x=y^2$, $x=4$

and i dont know how to solve this one:

Find the volume of a solid whose base is bounded by the circle $x^2+y^2=4$ with a cross section of a square perpendicular to the x-axis.

thanks

[hide="1st one"]
$x=y^2$, so $y=\pm\sqrt{x}$.

Using the method of "cylindrical shells",
$V=2\pi\int_0^4 (6-x)\sqrt{x} dx=\boxed{\frac{192\pi}{5}}$
[/hide]

[hide="2nd one"]
It's hard to explain without a picture, but basically you have to express the side length of the square in terms of $x$ and then "add up" the areas of the squares.

$x^2+y^2=4\Longleftrightarrow y=\pm\sqrt{4-x^2}$, so the side length of the square is $2\sqrt{4-x^2}$.

So $V=2\int_0^2 (2\sqrt{4-x^2})^2 dx=\boxed{\frac{128}{3}}$
[/hide]

I hope I didn't make any mistakes...

i just got the first one
i used the disk method:
rotate the graph so the equation is $y=x^2$ $y=4$ and the rotation line is $y=6$
the area of a disk is $(6-x^2)^2-2^2$ and all of that times pi
$\pi\int_-2^2 (6-x^2)^2-2^2 dx=\frac{384\pi}{5}$

the second one
and the socond one is also what i got

thanks

Could somebody draw a picture for the second (or explain better)... I don't understand how the solid looks like.


calculus
integration
trigonometry
logarithms
calculus computations

I noticed a pattern that

$\int_0^{\frac{\pi}{2}} \cos(nx)\cos^{n} x \ dx = \frac{\pi}{8\prod_{k=1}^{\frac{n}{2}-1} 4} = \frac{\pi}{2^{n+1}}$

Plugging in $n=2004$, we get $\frac{\pi}{2^{2005}}$

Did you create this problem based on that pattern, or do you know how to solve this with a clever substitution?

Yes.
Let $I_n=\int_0^{\frac{\pi}{2}} \cos(nx)\cos^{n} x \ dx$, $I_n$ satisfies the following recursive formula.

$I_n=kI_{n-1}\ (n=1,2,\cdots)$, where $k$ is a constant number.

Is this identity a specific case of a well-known identity or does it have its own name?

I don't know. I found that accidentally.

That's cool, most discoveries are by accident; I accidentaly discovered $\tan^{-1} z = \frac{1}{2i}\ln\left(\frac{1+iz}{1-iz}\right)$, when I was working with integrals. :)


Two mathematicians were both born in the last 500 years. Each lives to be 100 years old, then dies. Each mathematician is equally likely to be born at any point during those 500 years. What is the probability that they were contemporaries for any length of time?

Umm... I vote no. And, also, I'm too lazy to do the prob right now, so do it yourself ppls. Hint though- If you use probability with geometry and graph the two out, it becomes a pretty routine Alcumus problem.


I recently got a letter in the mail from "LeadAmerica", sponsored by the "Congressional Student Leadership Conference".  They apparently offer summer programs for high schoolers who are nominated by their teachers.  There's a couple I'm considering... the booklet says it's "for outstanding students"... I was wondering if anyone had heard of it before and could vouch for the prestige of the program.

Thanks.  :D

I don't believe these programs are very prestigious...but I really don't know enough about them to judge.


Ok. I got this right by calculators. :D But how do you do if you DON"t have calculator?

If A = 20 degrees, and B = 25 degrees, then the value of (1+tan A)(1+tan B) is:

A) $\sqrt {3}$
B) 2
C) $1+\sqrt {2}$
D) $2(\tan A + \tan B)$
E) None of these

[hide="Hint"]Look at the tangent addition formula[/hide]

I haven't learn it yet ;)

Expansion gives:

$1+\tan A + \tan B + \tan A \tan B$

And since the answer was [hide="a"]2[/hide]

$\tan A + \tan B + \tan A \tan B = 1$ also.

I just don't know how to do it. :?

[hide]
Expanding:

$1+tanA+tanB+tanA \cdot tanB$

Also:

$tan(A+B) = \frac{tanA + tanB}{1-tanA \cdot tanB}$

$A+B=45$, so $tan(A+B) = 1$

Thus,
$tanA+tanB = 1 - tanA \cdot tanB$
$tanA+tanB+tanA \cdot tanB = 1$

adding 1 to the above will give the answer (in the expansion)
so the answer is $2$
[/hide]

Wait. Your answer is confusing me.

The formula for addition of the tangent is:

$\tan {\alpha + \beta} = \frac {\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$

But your step's confusing me.

put tan(A+B)=1 into the addition formula and then multiply by the denominator on both sides

and then you will have $tan A \cdot tan B +tan A + tan B = 1$

pluggin that into the original gives 2


$(a)$ In interval $[0,10000]$ which naturel number have most divisor?
$(b)$ In interval $[0,10000]$ which naturel number have largest positive divisor sum?

For [b]a[/b], consult

http://en.wikipedia.org/wiki/Highly_composite_number


A circle on the coordinate plane is centered at the origin and has radius $ r$. What proportion of the circle is located below the line $ y \equal{} r/2$?

[hide="Solution"]The total area of the circle is $ \pi r^2$. The area below the line $ y \equal{} \frac {r}{2}$ is $ \frac {3}{4}$ of the circle plus a triangle whose area is $ \frac {r^2}{2}$.
\[ \frac {\text{below}}{\text{total}} \equal{} \frac {\frac {3\pi r^2}{4} \plus{} \frac {r^2}{2}}{\pi r^2} \equal{} \frac {\frac {3\pi}{4} \plus{} \frac {1}{2}}{\pi} \equal{} \boxed{\frac {2 \plus{} 3\pi}{4\pi}}\approx90.9\%
\]
[/hide]

[hide="Not quite"]

How did you get the area of the region to be 3/4th of the circle plus an additional triangle?[/hide]

actually i think it is $ \frac23$ of the cirle plus a triangle...
for my sakes, the radius will be 2
the line is y=1 so the line intercepts the circle at $ (\sqrt {3},1) and ( \minus{} \sqrt {3},1)$
radii connecting these points to the center of the circle intersect at a 120 degree angle
this means two thirds of the circle plus a triangle with height 1 and base $ 2\sqrt3$ lies below the line
im too lazy to calculate tho...
[asy]draw(Circle((0,0),2));
dot((0,0));
draw((1.732,1)--(0,0)--(-1.732,1)--cycle);[/asy]

[hide="Solution"]

Let the center of the circle be $ O$, the points of intersection of the secant line and the circle be $ A$ and $ B$, and the midpoint of $ AB$ be $ M$. So we have $ OA \equal{} OB \equal{} r$ and $ OM \equal{} \frac {r}{2}$. Thus $ \angle AOM \equal{} \angle BOM \equal{} 60^o$, and $ \angle AOB \equal{} 120^o$. We can now find the area of the circle under the secant line by adding the area of the triangle and the sector of angle $ 240^o$, and dividing by the area of the circle to find the desired percentage:

$ \frac {\left(\frac {r^2}{2}\right)\sin 120^o \plus{} \frac {2}{3}\pi r^2}{\pi r^2} \equal{} \boxed{\frac {3\sqrt {3} \plus{} 8\pi}{12\pi}}\approx 80.45\%$
[/hide]


function
real analysis
real analysis unsolved

function
let f:R to R be any function and g(x) = 1/f(x).then g is
a) one to one if f in one to one

b)onto if f is onto

c)continuous if f is continuous

d)differentiable when f is differentiable

please help me

Consider the function $ f(x)\equal{}x$.  Then $ g(x)\equal{}\frac{1}{x}$.  Clearly, $ f$ is one-one, onto, continuous, and differentiable.  How about $ g$?

but if f is one to one isnt g one to one

I responded to your question and I answered a good deal of it.  You can do some of it by yourself.  Try to find another counterexample or try to prove it.

Jrav's example gives you an instance where $ f(x)$ causes $ g(x)$ to be discontinuous (and also non-differentiable) at $ 0$. That should already hint that the answer is a: one-to-one.

Of course, we still need to prove that $ g$ is one-to-one if $ f$ is.

Let $ f(x)$ be defined such that $ f(x_{1}) \neq f(x_{2})$ where $ x_{1}$ and $ x_{2}$ are any 2 values in the function's domain. And so $ \frac {1}{g(x_{1}) } \neq \frac {1}{g(x_{2})}$ for non-zero $ x$. Cross multiplying gives $ g(x_{1}) \neq g(x_{2})$, as required.


[quote="george_eleth"]Χρησιμοποιώντας την ανισότητα t^(2n+2)/(1+t^2)<=t^(2n+2), Θέλω να αποδείξω ότι lim(n->00)Rn = 0 αν -1<=x<=1,
όπου [img]http://imageshack.gr/thumbs/uz0jezxtzzotzwjmu1ik.gif[/img].
Καμια ιδέα από που να ξεκινήσω (κριτήριο Leibniz?), έχω κολλήσει απίστευτα... :blush:[/quote]

Katarxin xronia sou polla!

Nomizw oti o parakatw tropos einai swstos. Xrisimopoiw tin gnwsti anisotita pou leei oti ean $a<b$ kai $f$ sinexis sto $[a,b]$ tote \[\left|\int_{a}^{b}f(x)dx \right | \leq \int_{a}^{b}|f(x)|dx\]

Diakrinw tis eksis periptwseis

1) An $0 \leq x \leq 1$ tote (ap'tin parapanw anisotita) \[|R_{n}(x)| \leq \int_{0}^{x}\left | \frac{t^{2n+2}}{t^{2}+1}\right | dt \leq \int_{0}^{x}t^{2n+2}dt = \frac{x^{2n+3}}{2n+3}\]

Ara \[-\frac{x^{2n+3}}{2n+3}\leq R_{n}(x) \leq \frac{x^{2n+3}}{2n+3}\ \ \ (1)\] kai ta oria twn $2$ pleurikwn sinartisewn kathws $n \to \infty$ einai $0$ (afou $0<x<1$).

2)  An $-1 \leq x \leq 0$ tote $0\leq-x \leq 1$ opote h parapanw anisotita $(1)$ isxiei ean valw opou $x$ to $-x$. Etsi pairnw ($2n+3$ perittos)

\[\frac{x^{2n+3}}{2n+3}\leq R_{n}(-x) \leq-\frac{x^{2n+3}}{2n+3}\] Omws $R_{n}(-x)=-R_{n}(x)$ (H apodeiksi einai poli eukoli me mia allagi metavlitis ara h $R_{n}$ einai peritti sinartisi),

ara pairnoume telika tin anisotita $(1)$ h opoia telika isxiei gia kathe $x$, me $-1 \leq x \leq 1$. Sinepws epeidi kai pali ta oria twn pleurikwn sinartisewn tis $(1)$ pigainoun sto $0$ kathws $n\to\infty$, ara kataligoume telika sto epithimito: \[\lim_{n\to\infty}R_{n}(x)=0, \ \ \forall x\in[-1,1]\].

Alexandros

Ειλικρινά σας ευχαριστώ πολύ για την βοηθειά σας! Είμαι υπόχρεος.

Kalhspera , exw mia entelos diaforetiki lisi , me xrisi Pragmatikis analisis kai Theoria Metrou kai Oloklirosis Lebesque

 Katarxas epeidi exoume sinartisi orismeni apo olokliroma me x na kinite sto -1 , 1 tote kai to t me endiaferei pws  tha kinite safto to diastima.Exoume lipoion to esoteriko tou olokliromatos na einai mia akolouthia sinartisewn i opoia efkola dixnete pws sigklinei sxedon pantou , otan to t kinite metaksi tou -1 kai tou 1 sto 0 kai gia t=1 i f paramenei fragmeni me supremum to 1/2 , opote i akolouthia sinartisewn , mporei na kiriarxithei apo tin statheri sinartisi 1 pou to olokliroma tis einai enas pragmatikos arithmos gia kathe klisto kai fragmeno diastima(einai Lebesque oloklirosimi).Episis i akolouthia sinartisewn einai fanera Lebesque oloklirosimi.Ara apo Theorima kiriarxoumenis sigklisis mporoume na valoume to orio tou n mesa sto okliroma tou Rn , pou tinei sto 0 opote kai to Rn ---> 0

P.S. To t gia tin akrivia kinite sto [0,x] 'h sto [x,0] analogos an to x einai thetikos 'h arnitikos , me tin apodiksi afti pou parethesa , dixnoume kati pio evri , oti diladi i oloklirotea posotita paramenei 0 sto ipersinolo [-1,1] , ara kai sto [x,0] , [0,x].....


geometry
geometric transformation
rotation
algorithm

Really Interesting and COOL Problem :
Let $A,B$ be two strings of Real numbers with the lenght of $n$.
We say $A$ is a rotation of $B$ ,if and only if by shifting(just shifting not any arbitariry permutation) the elements of $A$ we can reach $B$.
We wana check if these two strings are rotation of each other or no.
In each step we can ask a question.
In each question you can ask if the $i_{th}$ element of $A$ is equal to the $j_{th}$ element of $B$.The answer is either yes or no,Of course.

Determine the minimum number of needed questions to find out if
$A$ and $B$ are rotation of each other.

PS.An ordered array of $n$ real numbers is called an string with the lenght $n$.

The whole IOI question :
http://www.ioi2006.org/practice/the_fifth_sun.pdf

A very nice interactive program indeed, but a bit too easy for IOI (I noticed this is only a PRACTICE problem, not actually an IOI problem.)  Probably this was just used to get the contestants familiar with how interactive programs will be judged.  Most IOI contestants would find this really easy... it's more of a textbook-level problem than a contest-level problem.

There is a well-known linear algorithm for determining if one string is a rotation of the other:

[hide]
Essentially, you write the first string twice in a row and determine if the 2nd string is a substring of it using the KMP algorithm.  Obviously you don't have to actually use up 2x the memory required, but that's the basic idea

If you've never seen KMP before, it's quite interesting:
[url]http://en.wikipedia.org/wiki/Knuth-Morris-Pratt_algorithm[/url]
[/hide]


$ \frac{2^{p\minus{}1}\minus{}1}{p}\equal{}n^{2}$    ---> $ p\in P$ and $ n\in N$ .   find all p .

we take $ m\equal{}\frac{p\minus{}1}{2}$
$ (2^{m}\minus{}1)(2^m\plus{}1)\equal{}pn^2$ and $ gcd(2^m\minus{}1,2^m\plus{}1)\equal{}1$
so $ \{2^{m}\minus{}1,2^{m}\plus{}1\}\equal{}\{x^2,py^2\}$ for some $ x,y$ such that $ xy\equal{}n$
if $ 2^{m}\plus{}1\equal{}x^2$ then $ 2^{m}\equal{}(x\minus{}1)(x\plus{}1)$ give $ x\equal{}3,m\equal{}3$ ==> $ \boxed{p\equal{}7}$
if $ 2^{m}\minus{}1\equal{}x^2\equiv 1(mod\ 4)$ then $ m\equal{}1$ ==> $ \boxed{p\equal{}3}$

$ p\equal{}3,7$

where did you get it from? it was taken in a TST for the iberoamerican olympiad in 2004 in Argentina but maybe there's an older source
A very nice problem...


summer program
MathZoom
AMC
AIME
USA(J)MO
USAMO
ARML

MathZoom is offering Math Contest Prep courses from Fall 2007 through Spring 2008.  Classes to prepare for AMC8/10/12, AIME, USAMO and ARML are held at various locations.  Here is a list of courses, the dates, time and locations for the classes.  Registration forms are available on [url]http://www.mathzoom.org[/url]

--------------------------------------------------------------------------------

Title: Math Contest II
This course is focused on problem solving skills and test strategies for students up to 12th graders to participate in the American Mathematics Competition (AMC) 10, 12 and ARML contests. Through effective guidance and direction by our experienced teachers, students develop strong problem solving skills that make them perform well in the AMC 10, AMC 12, and ARML Contests.
Dates: Saturdays 10/13, 10/27, 11/3, 11/10, 12/1, 12/15, 12/22, 1/5, 1/12
Time: 1pm-4pm
Location: Arcadia: 411 E Huntington Dr, Suite 309, Arcadia, CA 91006
OR
Dates: Saturdays 10/13, 10/27, 11/3, 11/10, 12/1, 12/15, 12/22, 1/5, 1/12
Time: 6pm-9pm
Location: Irvine: 4255 E Campus Drive, Suite 290, Irvine, CA 92612
To register, please download the registration form and follow the instruction on the form.
--------------------------------------------------------------------------------

Title: Math Contest III
This course is focused on problem solving skills and test strategies for students up to 12th graders to participate in the American Invitational Mathematics Examination (AIME) and the USA Math Olympiad (USAMO) contests. Through effective guidance and direction by our experienced teachers, students develop strong problem solving skills that make them perform well in the AIME and USAMO.
Dates: Saturdays 1/12, 1/19, 1/26, 2/2, 2/9, 2/16, 2/23, 3/1, 3/8
Time: 1pm-4pm
Location: Arcadia: 411 E Huntington Dr, Suite 309, Arcadia, CA 91006
OR
Dates: Saturdays 1/12, 1/19, 1/26, 2/2, 2/9, 2/16, 2/23, 3/1, 3/8
Time: 6pm-9pm
Location: Irvine: 4255 E Campus Drive, Suite 290, Irvine, CA 92612
To register, please download the registration form and follow the instruction on the form.

I just updated the course schedules.  The registration form is also attached.

Note: The schedule in Arcadia no longer conflicts with UCLA Math Circle.  Also, we don't have classes on Southern California ARML practices dates either.  MathZoom students are strongly encouraged to attend the So Cal ARML practices.

The latest class schedules of our Year-round programs can always be found at [url=http://www.mathzoom.org/mathzoom_yearround.html]Math Zoom website[/url].

We currently have:

MC I in Arcadia Saturdays 10am-12pm

MC I in Irvine Saturdays 6:30pm - 8:30pm

MC II in Irvine Fridays 7:30pm - 9:30pm

MC I in Cerritos (not started yet)


From the University of South Carolina High School Math Contest, 1992

20. Let x be the smallest number larger than 2 that leaves a remainder of 2 when divided by each of 3, 5, and 6.  What is the sum of the digits of x?

(A) 1
(B) 5
(C) 11
(D) 12
(E) 14

25.  A cylindrical open-topped can has radius 3 cm and height 10 cm.  If the can is tilted at a 45 degree angle, what volume of water will it hold?

(A) 45 :pi:
(B) 63 :pi:
(C) 70 :pi:
(D) 75 :pi:
(E) 80 :pi:

27. Which has the greatest value?  (All values in degrees.)

(A) sin(sin(20))
(B) sin(cos(20))
(C) cos(sin(20))
(D) cos(cos(20))
(E) cos(20)

(Edited to make the :pi:s look like :pi:)

Quote:20. Let x be the smallest number larger than 2 that leaves a remainder of 2 when divided by each of 3, 5, and 6. What is the sum of the digits of x?
[hide]
x :equiv: 2 (mod 3)
x :equiv: 2 (mod 5)
x :equiv: 2 (mod 6)

x-2 is divisible by 3, 5, 6
x-2 = 30
x = 32
3+2=5 (B)
[/hide]

Quote:25. A cylindrical open-topped can has radius 3 cm and height 10 cm. If the can is tilted at a 45 degree angle, what volume of water will it hold?
[hide]
I got (B) 63 :pi:
Just draw the diagram and use 45-45-90 triangles
[/hide]
Quote:27. Which has the greatest value? (All values in degrees.)

How would you do this without a calculator?

timhlu, for the last one, you just judge how close to zero and 1 cos20 and sin20, etc, are and go from there. remember that cos0=1 and sin0=0.

And, of course, you know that sin and cosine are either increasing or decreasing over a half-period, so you can use that as well.


3 words each in honor of the 3's  :rotfl:

Let's STart

He!

 :rotfl:  and junggi doesn't count if he says anything!

[quote="junggi"]NO[/quote]
[size=200][color=red][b]YES![/b][/color][/size]

Size 29 + red + bold = pwnage.

[quote="i_like_pie"][quote="junggi"]NO[/quote]
[size=200][color=red][b]YES![/b][/color][/size]

Size 29 pwns.[/quote]
OH CRAP I'LL GET OUT MY SECRET WEAPON

[u][i][b][color=violet][size=200]NO![/size][/color][/b][/i][/u]
PINK BEATS RED MWAHAHAHAHAHAHA

[quote="mathnerd314"]Unless you're going to post more than a word each time, this topic is about to get really :spam:my. Reminiscent of the $3$'s.[/quote]
lol yea, let's start over.
How bout we post 3 words each, in honor of the $3$'s?
I'll start.
He was a...

@ junggi - That's violet. :rotfl:

LET'S START OVER:

She...

he was a big fat woman

who lived in....

to kill them!

(I was leading up to that! :rotfl: )

:mad:  :mad:

New sentence:
Kenai and Koda ...

killed each other! :rotfl:  :rotfl:

ok, that was just a kidding one. Here is a genuine one:

went hunting for


A jar filled with water weighs 10 pounds. When one half of the water is poured out, the jar and the remaining water weighs 5.75 pounds. How much does the jar weigh? Hint hidden:

[hide][hide][hide]Hint: What was the weight of the water poured out?[/hide][/hide][/hide]

[hide]Let J denote the weight of the jar and W, the weight of the water. Since the jar and the water weigh 10 pounds, we can come up with the following equation:

J+W = 10

We know that when you take out half of the water, the jar weighs 5.75 lbs. This allows us to write the following equation:

J-.5W = 5.75

Then, we simply subtract the 2nd equation from the 1st equation and we get

.5W = 4.25

Then, we solve for W by dividing 4.25 by .5. Now we know that the water weighs 8.5 lbs. So we can subtract 8.5 from 10, and the jar weighs 1.5 lbs.

simple algebra :!:  :!:  :!:  :!:  :!:  :!:[/hide]

[hide]We can make the systems of equations:
$x+y=10$
$\frac{1}{2}x+y=5.75$,
Where x is the amount of water and y is the weight of the jar.
Solving it we get...
$x=8\frac{1}{2}, y=1\frac{1}{2}$
Thus the jar weighs $1\frac{1}{2}$ pounds[/hide]

[quote="kishiman1"]10- 5.75 = 4.25

4.25 x 2 = 8.5

10.5- 8.5 = 1.5 pounds[/quote]

Please, in the future hide your solutions....


Let f(0) = 1 and f(x) = cos(1/x) for 0 < x <= 1.
f is not continuos on 0 <=  x <= 1.
Is f Riemann-integrable on 0 <=  x <= 1?

It is Riemann-integrable because it is continuous almost-everywhere (except at point 0 actually).

Here's a direct argument, using the Darboux approach (upper and lower sums).

Let $\epsilon>0$ be arbitrary.

$f(x)$ is continuous on the interval $\left[\frac{\epsilon}4,1\right].$ Since a continuous function on a closed bounded interval is integrable, there exists a partition of that interval on which the upper and lower sums differ by less than $\frac{\epsilon}2.$

For the interval $\left[0,\frac{\epsilon}4\right]$, just take a partition with one interval. The sup of the function is 1 and the inf is -1, so the difference between the upper sum and lower sum is $(1-(-1))\frac{\epsilon}4=\frac{\epsilon}2.$

Put the two intervals together and we have a partition on which the upper and lower sums differ by less than $\epsilon.$ Since $\epsilon$ was arbitrary, this proves the function is Riemann integrable.

Of course, what julien_santini gave is the reason you should know the answer immediately upon seeing the problem.

Another problem to review the concept:

Let $g$ be the characteristic (indicator) function of the usual Cantor set. That is, let $g(x)=1$ for $x$ in the Cantor set, and $g(x)=0$ otherwise.

Is $g$ Riemann integrable on $[0,1]$?

Yes, actually I saw the solution, but I did not find a satisfactory way to show it...


I would post this in getting started, but I don't think trig is allowed there.

I don't know why I am not getting the answer to this problem... maybe I am having some sort of mental block.

If $f(x)-\sin x = \sec x (\cos x - \tan x)$ for all $x$, then $f(x)$ equals:

$\text{A. } 1$
$\text{B. } -\tan x$
$\text{C. } \cos x$
$\text{D. } \tan x$
$\text{E. } -\cos x$

Are you sure you typed the problem correctly?
I don't think any of the choices are correct.

[hide="Maybe the problem meant"]
$f(x)-\sin{x}= \cos{x} (\sec{x} - \tan{x})$?

Then the answer is simply $A$.[/hide]

Yeah!!!! That's what I thought too! But that's the way the problem showed up.  Silly test writers.


[quote="ragingbullrocky"]Lets say we have an economy with 2 industries: $A$, and $B$. Assume that the consumer demand for their products is, respectively, 1000 and 780, in millions of dollars per year. Industry $A$ produces electricity. Suppose that industry $B$ needs 10 cents worth of electricity for each 1 dollar of output B produces and that industry $A$ needs 20 cents worth of $Bs$ products for each 1 dollar of output $A$ produces. Find the outputs $a$ and $b$ needed to satisfy both consumer and interindustry demand.

Can somebody help me set up the linear equations?

Thanks[/quote]

Well, but written

A produses a=b/0.2=5b (estimated in dollars, if fixed b)
B produses b=a/0.1=10a (estimated in dollars, if fixed a)

I do not get in what it was called "consumer demand", first? At the end of sentence it's written separately, to satisfy "BOTH consumer AND (also) demand"! Economic tasks tied often with linear inequalities, to satisfy them (not only with equations, a>0, b>0 naturally, for example). It's more problem of terminology (not very good written), have to be something very simple.


function
inequalities
inequalities proposed

Let $ a,b,c$ be nonnegative real numbers such that $ a\plus{}b\plus{}c\equal{}1$. Prove that
\[ \sum a\sqrt{8b^2\plus{}c^2} \le 1\]
\[ \sum a\sqrt{4b^2\plus{}c^2} \le \frac{3}{4}\]
:)

PS: A good application of Schur. :)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that
\[ \sum a\sqrt {8b^2 \plus{} c^2} \le 1
\]

\[ \sum a\sqrt {4b^2 \plus{} c^2} \le \frac {3}{4}
\]
:)

PS: A good application of Schur. :)[/quote]
See http://www.mathlinks.ro/viewtopic.php?t=150089 for the first one. (no nice solution receive)
I will post another solution of mine :)
Note that
\[ 3b \plus{} c \minus{} \frac {3bc}{2b \plus{} c} \ge \sqrt {8b^2 \plus{} c^2}
\]
Hence, it suffices to show that
\[ (\sum a)^2 \ge \sum a(3b \plus{} c \minus{} \frac {3bc}{2b \plus{} c})
\]

\[ \Leftrightarrow 3abc\sum \frac {1}{2b \plus{} c} \plus{} \sum a^2 \minus{} 2\sum ab \ge 0
\]
The Cauchy Schwarz Inequality gives us
\[ \sum \frac {1}{2b \plus{} c} \ge \frac {3}{a \plus{} b \plus{} c}
\]
It suffices to show that
\[ \frac {9abc}{a \plus{} b \plus{} c} \plus{} \sum a^2 \minus{} 2\sum ab \ge 0
\]

\[ \Leftrightarrow \sum a^3 \plus{} 3abc \ge \sum bc(b \plus{} c)
\]
which is true by Schur. :)

I am really happy to see this very nice solution. :)

Thank you very much.

We can also prove the following by the same way
\[ \sum \sqrt{\frac{a}{b\plus{}3c}} \ge \frac{3}{2}\]
PS. It's VasC's Inequality. :)

Another is
\[ \sum a^2 \plus{}\sum ab \ge \sum a\sqrt{3b^2\plus{}c^2}\]
:)

[quote="can_hang2007"]Another is
\[ \sum a^2 \plus{} \sum ab \ge \sum a\sqrt {3b^2 \plus{} c^2}
\]
:)[/quote]
May VasC or aqardy have a nice solution for this? :)

[quote="manlio"]Sorry can_hang2007, I cannot find the equivalent of
\[ 3b \plus{} c \minus{} \frac {3bc}{2b \plus{} c} \ge \sqrt {8b^2 \plus{} c^2}
\]
for

[quote="can_hang2007"]We can also prove the following by the same way
\[ \sum \sqrt {\frac {a}{b \plus{} 3c}} \ge \frac {3}{2}
\]
PS. It's VasC's Inequality. :)[/quote]

Can you please help me.

Thank you very much.[/quote]
Dear manlio, I just said "the same way" (it means the similar way, sorry for my bad E :(), and nothing is equivalent here. :)

Another question, please. :wink:

I see that
\[ 3b \plus{} c \minus{} \frac {3bc}{2b \plus{} c} \ge \sqrt {8b^2 \plus{} c^2}
\]

is AM-GM

 $ (2x\plus{}1)^2 \plus{} (8x^2\plus{}1) \geq 2(2x\plus{}1) \sqrt {8x^2 \plus{} 1}$

but for the other inequality how to estabilish the right factor $ Ax \plus{}B$ to write

$ (Ax\plus{}B)^2 \plus{} (4x^2\plus{}1) \geq 2(Ax\plus{}B) \sqrt {4x^2 \plus{} 1}$

Thank you very much. :)

[quote="manlio"]Another question, please. :wink:

I see that
\[ 3b \plus{} c \minus{} \frac {3bc}{2b \plus{} c} \ge \sqrt {8b^2 \plus{} c^2}
\]
is AM-GM

 $ (2x \plus{} 1)^2 \plus{} (8x^2 \plus{} 1) \geq 2(2x \plus{} 1) \sqrt {8x^2 \plus{} 1}$

but for the other inequality how to estabilish the right factor $ Ax \plus{} B$ to write

$ (Ax \plus{} B)^2 \plus{} (4x^2 \plus{} 1) \geq 2(Ax \plus{} B) \sqrt {4x^2 \plus{} 1}$

Thank you very much. :)[/quote]
Well, for the second inequality, I don't establish the inequality (since it reverse...)
\[ (Ax \plus{} B)^2 \plus{} (4x^2 \plus{} 1) \geq 2(Ax \plus{} B) \sqrt {4x^2 \plus{} 1}\]
I used another. :)

[quote="can_hang2007"]Another is
\[ \sum a^2 \plus{} \sum ab \ge \sum a\sqrt {3b^2 \plus{} c^2}
\]:)[/quote]
We have

$ \sum a\sqrt{3b^2\plus{}c^2}\le \sum a(2b\plus{}c\minus{}\frac {2bc}{b\plus{}c})\equal{}3\sum ab\minus{}2abc\sum\frac 1{b\plus{}c}\le 3\sum ab\minus{}\frac{9abc}{a\plus{}b\plus{}c} \le \sum a^2\plus{}\sum ab.$

There is a nice solution for the following statement:

Let $ a,b,c$ be nonnegative real numbers such that $ a\plus{}b\plus{}c\equal{}3$. If $ p\ge 7$, then

$ \sum a\sqrt{pb^2\plus{}c^2} \le 3\sqrt {p\plus{}1}$.

[quote="Vasc"][quote="can_hang2007"]Another is
\[ \sum a^2 \plus{} \sum ab \ge \sum a\sqrt {3b^2 \plus{} c^2}
\]
:)[/quote]
We have

$ \sum a\sqrt {3b^2 \plus{} c^2}\le \sum a(2b \plus{} c \minus{} \frac {2bc}{b \plus{} c}) \equal{} 3\sum ab \minus{} 2abc\sum\frac 1{b \plus{} c}\le 3\sum ab \minus{} \frac {9abc}{a \plus{} b \plus{} c} \le \sum a^2 \plus{} \sum ab.$[/quote]
Thanks, VasC. It's also my solution :)

[quote="can_hang2007"]Another is
\[ \sum a^2 \plus{} \sum ab \ge \sum a\sqrt {3b^2 \plus{} c^2}
\]
:)[/quote]
Actually, for any positive number $ p$, the inequality holds
\[ \frac {\sqrt{p\plus{}1}}{2}(\sum a^2 \plus{} \sum ab) \ge \sum a\sqrt {pb^2 \plus{} c^2}.
\]

[quote="can_hang2007"][quote="manlio"]Dear can_hang2007, can you please post your solution for
\[ \sum a\sqrt {4b^2 \plus{} c^2} \le \frac {3}{4}
\]
Thank you very much. :)[/quote]
I used two lemmas in my solution
1)
\[ 2b \plus{} c \minus{} \frac {2bc(2b \plus{} c)}{4b^2 \plus{} 3bc \plus{} c^2} \ge \sqrt {4b^2 \plus{} c^2}
\]
2)
\[ 8\sum \frac {2b \plus{} c}{4b^2 \plus{} 3bc \plus{} c^2} \ge \frac {27}{a \plus{} b \plus{} c}
\]
:)[/quote]

Thank you very much, can_hang2007......but I cannot prove  :blush:

\[ 8\sum \frac {2b \plus{} c}{4b^2 \plus{} 3bc \plus{} c^2} \ge \frac {27}{a \plus{} b \plus{} c}
\]

I tried by Cauchy-Schwarz but I failed. Can you please help me.

[quote="can_hang2007"]We can also prove the following by the same way
\[ \sum \sqrt {\frac {a}{b \plus{} 3c}} \ge \frac {3}{2}
\]
PS. It's VasC's Inequality. :)[/quote]

I would like to see your solution, can_hang2007.

Thank you very much.

[quote="manlio"][quote="can_hang2007"]We can also prove the following by the same way
\[ \sum \sqrt {\frac {a}{b \plus{} 3c}} \ge \frac {3}{2}
\]
PS. It's VasC's Inequality. :)[/quote]

A hint, please, can_hang2007.

Thank you very much.[/quote]

\[ \sum \sqrt {\frac {a}{b \plus{} 3c}} \ge \frac {3}{2}
\]

I am struggling with this problem for a lot of days without success. :(

Can_hang2007, can you please give me a hint?

Thank you very much

[quote="Vasc"]There is a nice solution for the following statement:

Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. If $ p\ge 7$, then

$ \sum a\sqrt {pb^2 \plus{} c^2} \le 3\sqrt {p \plus{} 1}$.[/quote]
We can prove it easily with Cauchy Schwarz, VasC. :)
Also with this: :)
[quote="Vasc"][quote="can_hang2007"]Another is
\[ \sum a^2 \plus{} \sum ab \ge \sum a\sqrt {3b^2 \plus{} c^2}
\]
:)[/quote]
Actually, for any positive number $ p$, the inequality holds
\[ \frac {\sqrt {p \plus{} 1}}{2}(\sum a^2 \plus{} \sum ab) \ge \sum a\sqrt {pb^2 \plus{} c^2}.
\]
[/quote]

[quote="can_hang2007"][quote="manlio"][quote="Vasc"]There is a nice solution for the following statement:

Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. If $ p\ge 7$, then

$ \sum a\sqrt {pb^2 \plus{} c^2} \le 3\sqrt {p \plus{} 1}$.[/quote]

Can you please post your solution, Vasc?

Thank you very much. :)[/quote]
Cauchy Schwarz, my friend. :)[/quote]

My friend, I used Cauchy Schwarz in this way

$ (\sum a\sqrt{p^2b^2\plus{}c^2})^2 \leq \sum a(pb\plus{}c) \sum \frac{a(p^2b^2\plus{}c^2)}{pb\plus{}c} \equal{} (p\plus{}1)\sum ab[ \sum a(pb\plus{}c) \minus{}2abcp\sum \frac{1}{pb\plus{}c}] \leq (p\plus{}1)\sum ab[(p\plus{}1) \sum ab \minus{}18\frac{pabc}{(p\plus{}1)\sum a}]$

and I obtained it is sufficient to prove

$ \frac{9(p^2\plus{}1)}{q} \plus{}6pr \geq q(p\plus{}1)^2$

with $ q\equal{}ab\plus{}bc\plus{}ca$ and $ r\equal{}abc$

But how to prove this inequality for $ p \geq \sqrt{7}$ as Vasc said ? :blush:

[quote="manlio"][/quote][quote="can_hang2007"][quote="manlio"][quote="Vasc"]There is a nice solution for the following statement:

Let $ a,b,c$ be nonnegative real numbers such that $ a + b + c = 3$. If $ p\ge 7$, then

$ \sum a\sqrt {pb^2 + c^2} \le 3\sqrt {p + 1}$.[/quote]

Can you please post your solution, Vasc?

Thank you very much. :)[/quote]
Cauchy Schwarz, my friend. :)[/quote][quote="manlio"]

My friend, I used Cauchy Schwarz in this way

$ (\sum a\sqrt {p^2b^2 + c^2})^2 \leq \sum a(pb + c) \sum \frac {a(p^2b^2 + c^2)}{pb + c} = (p + 1)\sum ab[ \sum a(pb + c) - 2abcp\sum \frac {1}{pb + c}] \leq (p + 1)\sum ab[(p + 1) \sum ab - 18\frac {pabc}{(p + 1)\sum a}]$

and I obtained it is sufficient to prove

$ \frac {9(p^2 + 1)}{q} + 6pr \geq q(p + 1)^2$

with $ q = ab + bc + ca$ and $ r = abc$

But how to prove this inequality for $ p \geq \sqrt {7}$ as Vasc said ? :blush:[/quote]
You can read at the document I sent you, my friend. :)

Dear can_hang2007, your proof is very nice but hard to find as you prove the inequality for a stronger number, that is $ p \geq (9\plus{}4\sqrt{2})/7 \equal{}2.0938$ while Vasc limit is weaker $ p \geq \sqrt{7}\equal{}2,64...$ so I think for a weaker limit there is a simpler way to prove it, but I cannot see it  :oops:

Manlio I proved the inequality of Vasc for the cases $ p\geq 8$ .
If this is the case then put $ \sqrt {p \plus{} 1} \equal{} u$ and use the inequalities :

$ ub \plus{} c \minus{} \frac {ubc}{(u \minus{} 1)b \plus{} c}\geq\sqrt{(u^2 \minus{} 1)b \plus{} c}$ and also

$ \sum\frac {1}{(u \minus{} 1)b \plus{} c}\geq\frac {9}{u(a \plus{} b \plus{} c)}$

[quote="silouan"]Manlio I proved the inequality of Vasc for the cases $ p\geq 8$ .
If this is the case then put $ \sqrt {p \plus{} 1} \equal{} u$ and use the inequalities :

$ ub \plus{} c \minus{} \frac {ubc}{(u \minus{} 1)b \plus{} c}\geq\sqrt {(u^2 \minus{} 1)b \plus{} c}$ and also

$ \sum\frac {1}{(u \minus{} 1)b \plus{} c}\geq\frac {9}{u(a \plus{} b \plus{} c)}$[/quote]

Thank you very much, silouan.  :)

I think the case $ 7 \leq p \leq 8$ is harder :wink:

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that
\[ \sum a\sqrt {8b^2 \plus{} c^2} \le 1
\]

\[ \sum a\sqrt {4b^2 \plus{} c^2} \le \frac {3}{4}
\]
:)

PS: A good application of Schur. :)[/quote]

Very nice solution, my Bro  :) . And your ineq is stronger than Vasc' one:
$ \sum\ \sqrt[]{\frac {a}{8b \plus{} c}} \geq\ 1$

@Vasc: Can U explain for me.
I also try with your ways but...  :huh:
Example,: $ \sum\ a\sqrt[]{8b^2\plus{}c^2} \leq\ (a\plus{}b\plus{}c)^2$
Using Am-Gm, we' ll have to prove:
$ \sum\ \frac{a(8b^2\plus{}c^2)}{8b\plus{}c} \leq\ (a\plus{}b\plus{}c)^2$
But it isn't trues (examle, $ a\equal{}b\equal{}1; c\equal{}0$)


Let $a, b, c$ be positive real numbers such that $a+b+c=1$. Prove that

          \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\frac{25}{1+48abc}.\]

Post here before

http://www.mathlinks.ro/Forum/viewtopic.php?t=148331

But the extended for 4 variable of mine is still unsolved

For all positive real numbers a,b,c,d with sum 1 then
\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge \frac{63}{2(1+47abcd)}.\]

[quote="hungkhtn"]

For all positive real numbers a,b,c,d with sum 1 then
\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge \frac{63}{2(1+47abcd)}. \]
[/quote]
Maybe $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge \frac{63}{2(1+248abcd)}$ $?$ :wink:
If so it's stronger than
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=62669

[quote=sqing]Irish Mathematical Olympiads 2006:
Let $a, b$ be positive real numbers such that $a+2b=1$. Prove that\[\frac{1}{a}+\frac{2}{b}\geq\frac{25}{1+48ab^2}.\][/quote]

we can use $AM-HM$  here

$\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\geq\frac{9}{a+2b}=9\leq\frac{25}{1+48ab^2}$
and equality occurs when $a=b=\frac{1}{3}$

[quote=crezk][quote=sqing]Irish Mathematical Olympiads 2006:
Let $a, b$ be positive real numbers such that $a+2b=1$. Prove that\[\frac{1}{a}+\frac{2}{b}\geq\frac{25}{1+48ab^2}.\][/quote]

we can use $AM-HM$  here

$\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\geq\frac{9}{a+2b}\geq9\geq\frac{25}{1+48ab^2}$
and equality occurs when $a=b=\frac{1}{3}$[/quote]
$\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\geq\frac{9}{a+2b}=9\leq\frac{25}{1+48ab^2}$


Baltic Way 2004 problems can be found [url=http://bw2004.olimpiados.lt/?x=12]here.[/url] Previous years are provided in German [url=http://www.hh.schule.de/ifl/mathematik/baltic-way.htm]here.[/url]

Solutions are there already as well. Read the notes, and note the encouragement to send solutions if required.


real analysis
integration
advanced fields
advanced fields unsolved

Suppose $E$ measurable, $me<\infty$, prove that
$f \in L(E)$ iff $\displaystyle \sum_{n=1}^{\infty}mE(|f|\geq n) < \infty$.

Is this right? How about this: consider E=[0,3], let
\[f = \left\{
\begin{array}{ll}
\infty,&x \in (1,2) \\
x,& {\rm otherwise}
\end{array}
\right.
\]

This problem is from an entrance exam, so it would be my mistake. Help please, thanks!

You don't have a counterexample. Note that $\infty>n$ for any $n.$

Hence for each $n$, $m(\{|f|\ge n\})\ge 1,$ and thus

$\sum_{n=1}^{\infty}m(\{|f|\ge n\})\ge \sum_{n=1}^{\infty}1=\infty.$

Since $\int_E|f|=\infty$ in your example, there is no contradiction.

Now for a proof:

For notation, let's let $E_n=\{x: |f(x)|\ge n\}.$

Let $h(x)=\sum_{n=1}^{\infty}\chi_{E_n}(x).$

Suppose $x\in E_n \setminus E_{n+1}$, which means $n\le|f(x)|<n+1.$

Then $h(x)=\sum_{k=1}^n1=n.$

Therefore, $h(x)\le|f(x)|<h(x)+1.$

But $\int_Eh=\sum_{n=1}^{\infty}\int_E\chi_{E_n}= \sum_{n=1}^{\infty}m(\{|f|\ge n\})$

(That claim uses the Monotone Convergence Theorem.)

Hence the sum we are interested in is $\int_Eh$ and we have

$\int_Eh\le\int_E|f|\le\int_E(h+1)=\int_Eh+m(E).$

Since we are assuming that $m(E)$ is finite, the sum is finite if and only if $f$ is integrable.

Note that we did explicitly use the finite measure of $E.$ The theorem would be false otherwise. Use $f(x)=\frac1x$ on $[1,\infty)$ for a counterexample.


geometry
3D geometry
sphere
symmetry
geometric transformation
geometry solved

Here goes:
Because of the symmetry the plane containing the centers of the spheres is the bisector plane of the angle you want to calculate. The triangle formed by the centers of the spheres (call them A,B,C) is projected on the table as
the triangle formed by the points of tangency to the table (call them A1,B1,C1). It's not hard to calculate the sides of the 2 triangles, so you can easily get their areas. But you know that S(A1B1C1)=S(ABC)*cos(the angle between the planes ABC & A1B1C1), so you can find this angle, which is half the angle you are looking for.
I sure hope it's right.

[quote="Cristi Sav"]Fie s1,s2,s3 3 sfere "lipite"(tangente) 2 cate 2. Aceste sfere sunt asezate pe o masa, si au razele 1,2,3. Se pune peste sfere o placa (un plan) ce va fi tangenta fiecarei sfere. Ce unghi face planul placii cu planul mesei ?[/quote]

please use English on the forum - or at least try to provide an translation thank you!

the translation [quote] Let s1, s2, s3 be 3 tangent spheres each with each. These spheres are placed on a table, and their radii are 1,2,3 respectively. A plane tangent to all these spheres is considered. What is the angle made by this plane with the plane of the table (on which the 3 spheres lie)?[/quote]


number theory
relatively prime
superior algebra
superior algebra unsolved

Hello, I think it's very easy but I don't see the solution !!
Let be $n\in\mathbb{N}$ and $\alpha, \beta \in\mathbb{F}_{2^n}$ such that $\mathbb{F}_2(\alpha)\cap\mathbb{F}_2(\beta)=\mathbb{F}_2$. Show that $\mathbb{F}_2(\alpha,\beta)=\mathbb{F}_2(\alpha+\beta)$

hi,
see lang pg 267,by that u can conclude that the field on the left side of ur equation has order =product of degrees of alpha and beta,on the other hand the degree of "alpha + beta" is certainly product of degrees of alpha and beta,so both fields  have same degree of the base field and since we r in finite fields they r isomorphic,even otherwise as one is containned in the other and has same degree they r isomorphic,sorry for such bad explanation,i do'nt know latex.
regards
Rohith

[quote]"alpha + beta" is certainly product of degrees of alpha and beta[/quote]

Why ?!!
I knew that we just have to show that.
I think that there is a "trick" using that we are in $\mathbb{F}_2$ in order to find the order of $\alpha+\beta$

By general properties of finite fields, $\alpha$ and $\beta$ must have relatively prime degrees.
In general, if $a$ and $b$ are algebraic over $F$ with relatively prime degrees $m,n$, some linear combination $c_1a+c_2b$ (with $c_1,c_2\in F$) has degree $mn$ over $F$. In this case, with $F$ having only two elements, this linear combination must be $a+b$.

[quote]some linear combination c_1a+c_2b (with c_1,c_2\in F) has degree mn over F[/quote]
Could you prove it please ?!

That's a good question. The primitive element theorem relies on an infinite field to get a version of this conclusion in general...

I think I should just drop that and look for another way. How about the multiplicative order of $\alpha+\beta$? The field generated by $\alpha+\beta$ has order $2^k$ with $k|m$ if and only if $(\alpha+\beta)^{2^m}=\alpha^{2^m}+\beta^{2^m}=\alpha+\beta$.

Since the fields generated by $\alpha$ and $\beta$ have trivial intersection, this requires $\alpha^{2^m}=\alpha$ and $\beta^{2^m}=\beta$. If $r$ and $s$ are the degrees of the extensions generated by $\alpha$ and $\beta$ respectively, this gives $r|m$ and $s|m$. Since $r$ and $s$ are relatively prime, $rs|m$ and $\mathbb{F}_2(\alpha,\beta)=\mathbb{F}_2(\alpha+\beta)$.

Replacing $2$ with $p$ everywhere, the same is true for extensions of other prime fields. We don't even need an arbitrary linear combination.

hi,
but the  fields generated by alpha and beta has {0,1} as its intersection and not the trivial one.
wo'nt u have to  deal with then one more case.
rohith


number theory
greatest common divisor
algebra
polynomial

How do you prove:
$(a^{n}-1,a^{m}-1)=a^{(n,m)}-1$, where $(x,y)$ represents the gcd of $x$ and $y$?

Thanks. If the formula is not correct, please help me make any modifications to it.

\[a^{n}-1=(a-1)(a^{n-1}+a^{n-2}+\ldots+1)\]  \[a^{m}-1=(a-1)(a^{m-1}+a^{m-2}+\ldots+1)\] Let $g=\gcd(n,m)$. There are $n$ terms in the second product of the first line and $m$ terms in the second product of the second line. Hence, by grouping, their greatest common divisor is \[a^{g-1}+a^{g-2}+\ldots+1,\] so then \[(a^{n}-1, a^{m}-1)=a^{(n,m)}-1,\] as desired.

[quote="towersfreak2006"]
by grouping, their greatest common divisor is \[a^{g-1}+a^{g-2}+\ldots+1,\] [/quote]

I don't really get this part... :help:

[color=darkred]For $\{m,n\}\subset N^{*}$ given are the polynomials $F=X^{m}-1$ and $G=X^{n}-1\ .$
Denote $D=X^{d}-1$, where $d=(m,n)\ .$ Then $D=(F,G)\ .$[/color]

[color=darkblue][b]Proof.[/b] The polynomials $F,G$ [u]have only simple roots (essentially !).[/u] There is $\{u,v\}\subset Z^{*}$ so that $d=um+vn\ .$ Therefore,
$\ F(\alpha )=G(\alpha )=0\Longrightarrow \alpha^{m}=\alpha^{n}=1$ $\Longrightarrow$ $\alpha^{d}=\alpha^{um+vn}=\left(\alpha^{m}\right)^{u}\cdot \left(\alpha^{n}\right)^{v}=1$ $\Longrightarrow$ $D(\alpha )=0\ ,$
i.e. the polynomial $D$ has the all common (simple) roots of the polynomials $F$ and $G\ .$ In conclusion, $D=(F,G)\ .$[/color]


find all real valud functions $f,g,h$ such that for $x\not=y$:
$\frac{f(x)-g(y)}{x-y}=h(x+y)$

$\frac{f(x)-g(y)}{x-y}=h(x+y)=\frac{f(y)-g(x)}{y-x}\Longrightarrow f(x)+f(y)=g(x)+g(y).$ Therefore $g(x)=f(x)+const, const=f(y)-g(y)$. Choose 3 distinct $x_{1},x_{2},x_{3}$ from this we have const=0, it mean g(x)=f(x). Let $x=a+z,y=a-z$, then $\frac{f(a+z)-f(a-z)}{2z}=h(2a)$
It give $h(x)=const=h$ and $f(x)=g(x)=hx+d, d=const$.

[quote="Rust"] Let $x=a+z,y=a-z$, then $\frac{f(a+z)-f(a-z)}{2z}=h(2a)$
It give $h(x)=const=h$ and $f(x)=g(x)=hx+d, d=const$.[/quote]
How It give $h(x)=const=h$?
you lose the solution$f(x)=x^{2},g(x)=x^{2},h(x)=x$


factorial
number theory unsolved
number theory

For what $k$ is it true that $\exists n \in \mathbb{N} : \frac{ (2n)!}{(n+k)!^2} \in \mathbb{N}$?

For $k = 1$ we find $n = 5$ and I suspect (and believe I can prove) there are infinite $n$, but for $k \ge 2$ I remain at a loss :(


How many distinct triangles can be formed by selecting three vertices of a regular octagon?

There are eight possible ones for the first vertex, then there are seven possible ones for the second vertex, and then there are six possible for the third vertex. This is $ 8$ times $ 7$ times $ 6$. Then because the answer is 56, that means you have to divide by 6. Why do you have to divide by 6?

Say there is one triangle $ \triangle ABC$.

With your method, you are also counting $ \triangle ACB,\triangle BAC,\triangle BCA,\triangle CAB,\triangle CBA$.

So you also have to divide by $ 6$, since there are $ 6$ permutations of the order that the vertices go in for each distinct triangle.

Instead of saying "divide by 6", it should be "divide by 3!".  That should help you understand better.

$ \binom{8}{3}\equal{}\dfrac{8!}{3!(8\minus{}3)!}\equal{}\dfrac{8!}{3!5!}\equal{}56$

It is a combination not permutation because the order of the triangles do not matter.

Um...you know that I already posted that a while ago, don't you? :wink:  :P


Does there exsist such  $f$  that  $f$ is continuous all rational points and  discontinuous all irrational points?


If the terms in the successive row of a Pescal's Triangle are written successively, we form the sequence 1,1,1,1,2,1,1,3,3,1,1,4,6,4,1,1,5,10,10,.... If the sum of the first 212 terms of this sequence is $ 2^k \plus{} k,$ compute the integer $ k$.

[hide="Solution"] Recall that the sum of the terms in the $ n^{th}$ row of Pascal's triangle is $ 2^n$ (starting from $ 0$).  It follows that the sum of the terms in the $ 0^{th}, 1^{st}, 2^{nd}, 3^{rd}, ... (n\minus{}1)^{th}$ rows (in other words, the sum of the first $ \frac{n(n\plus{}1)}{2}$ terms of the given sequence) is $ 2^{n} \minus{} 1$.

Let $ n \equal{} 20$.  We conclude that the sum of the first $ 210$ terms is $ 2^{20} \minus{} 1$.  The $ 211^{th}, 212^{th}$ terms are the first two entries in the $ 20^{th}$ row, which are $ 1, 20$, so our desired sum is $ 2^{20} \plus{} 20$ and the value of $ k$ is $ \boxed{20}$. [/hide]

Yup. That's what I got as well (I don't have an answer key for this year).

[hide="Just a note"]
If anyone wondered why this works, try to prove it using induction ($ 2^k \minus{} 1 \equal{} 2^0 \plus{} 2^1 \plus{} \cdots \plus{}  2^{k\minus{}1}$). [/hide]


find all positive integers $a,b,m,n$ such that

$a^2+b^2 = 5mn$

$m^2+n^2 = 5ab$

I have one solution seeing a lot of cases.

Please try this.  :)


geometry
integration
perimeter
calculus
calculus computations

i was not able to solve those problems ,so i would appreciate it if someone gives me a hint of light for me to be able to understand . thanks  :lol:  :blush:

Problem 2 seems to be missing a constraint. Is the perimeter supposed to be fixed? Maximizing the area for a fixed perimeter (or the dual problem of minimizing the perimeter for a fixed area) is a classic textbook problem, but what you have here doesn't seem to be fully stated.


$p,n\in N^*,\ p\ge 2,\ 2^n\le p\Longrightarrow\left[ \sum_{k=0}^n \sqrt [p] {1+{n\choose k}}\right]=n+1.$

(If $\ x\in R,$ then $\ [x]=z\Longleftrightarrow z\in Z,\ z\le x<z+1$.)

Well,it's kind of late(and I'm lazy to write all details) but this problem, Levi, follows imdediately from the definition of integer part,from Bernoulli's inequality and finally from the wellknown identitty: $C_{n}^{0}+C_{n}^{1}+....+C_{n}^{n}=2^n$ ;)

  Well,I think this is right.Isn't it,Levi? ;)


logarithms
combinatorics unsolved
combinatorics

For $n\geq k\geq 3$, let $X=\{1,2,...,n\}$ and let $F_{k}$ a the family of $k$-element subsets of $X$, any two of which have at most $k-2$ elements in common. Show that there exists a subset $M_{k}$ of $X$ with at least $[\log_{2}{n}]+1$ elements containing no subset in $F_{k}$.

[quote="Megus"]I'm quite sure it was posted before - have you searched for it?[/quote]

I can't it  :(

Choosing a subset of $X$ with $k-1$ elements, we know that it can be a subset of at most one of $F_k$.
Since each $F_k$ has $k$ subsets with $k-1$ elements, we can deduce that $|F_k| \le \frac{1}{k}\binom{n}{k-1}$
Let $t=[\log_{2}{n}]+1$. The problem is trivial if $t<k$, so assume that $t \ge k$.
Let us pick a $M_k$, a subset of $X$, with $t$ elements.
Then, the probability that an element in $F_k$ is also in $M_k$ is $\frac{\binom{n-k}{t-k}}{\binom{n}{t}}$.
Let $T$ be the the number of elements in $F_k$ that is also in $M_k$.
We have that $E(T) = |F_k|\frac{\binom{n-k}{t-k}}{\binom{n}{t}} \le \frac{\binom{t}{k}}{n-k+1}$.
Using $\binom{t}{k} \le \frac{2^t}{\sqrt{t}} \le \frac{2n}{\sqrt{t}}$, we can get that $E(T) < 1$ and then we are done. $\blacksquare$


search
induction
combinatorics unsolved
combinatorics

The natural numbers are written in sequence, in increasing order, and by this we get an infinite sequence of digits. Find the least natural $k$, for which among the first $k$ digits of this sequence, any two nonzero digits have been written a different number of times.

[i]Aleksandar Ivanov, Emil Kolev [/i]

[quote="imortal"]...any two nonzero digits have been written different times.
 [/quote]

What does this mean?

[quote="imortal"]Sorry, I now edited to make it more clear.
I meant different [b]number of[/b] times.
If you have the first k digits of

123456789101112131415161718192021........$d_k$ there are $a_1$ digits 1, $a_2$ digits 2, ..., $a_9$ digits 9,

this means that $a_i\neq a_j$ for $i\neq j$[/quote]

So a trivial upper bound is 2345678

For a number that is all nines, all things have obvious equality.

If we put a 1 before it there are more 1's, a 2, there are more 1's and 2's, etc.

I dunno, right after having written 2345678 seems to me like the first time but I don't have proof.

assume that we have written al the numbers up to $m=\overline{b_nb_{n-1}\ldots b_0}$ in base $10$. We may assume that all the numbers have $n+1$ digits by placing enough zeroes at the beginning since they don't affect the conditions.

Now let's compute how many times $a$ appears as $k$'th digit in a number(denote this number by $f_k(a)$).

If $a<b_k$ then any number $\overline{c_1c_2\ldots c_{k-1}ax_1x_2\ldots}$ satisfying $\overline{c_1c_2\ldots c_{k-1}}\leq \overline{b_1b_2\ldots b_k}$ is good, so $f_k(a)=10^{n-k}( \overline{b_1b_2\ldots b_k}+1)$.

If $a=b_k$ then we must either have $\overline{c_1c_2\ldots c_{k-1}}< \overline{b_1b_2\ldots b_k}$ or $\overline{c_1c_2\ldots c_{k-1}}=\overline{b_1b_2\ldots b_k}$ and $\overline{x_1\ldots} \leq \overline{b_{k+1} \ldots}$ so we deduce $10^{n-k}\overline{b_1b_2\ldots b_{k-1}}+\overline{b_{k+1} \ldots b_0}$.

Finally if $a>b_k$ the condition is $\overline{c_1c_2\ldots c_{k-1}}< \overline{b_1b_2\ldots b_k}$ so $f_k(a)=10^{n-k} \overline{b_1b_2 \ldots b_{k-1}}$.

So we deduce that $f_k(a)$ depends only on the ordinal relation between $k$ and $a$.

If the occurences of any two distinct digits $a,b$ are distinct then for some $k$ we must have $f_k(a) \neq f_k(b)$ which by the ordinal dependence noticed implies $b_k$  is between $a$ and $b$ (including $a,b$).

Now taking $b=a+1$ we see that from any two consecutive digits, one is a digit of $m$.

So now it's clear that the smallest possible number is $2468$ and if we compute $f_k(a)$ we can see that it indeed is good.

well, actually 2468 is incorect answer by two reasons:
1)(trivial) you need the number of digits in the sequence, not the natural number n that you reach, consider also that you may have only part of some number.
2)(non-trivial)
in the sequence of digits
123456...(2467)(2468) you have equal number of 7's and 8's.

Hint: The answer of the problem (as it should be if you search for the natural n)is bigger than 2468 about 5 times.

hmm, yes i realized my mistake.

By a similar reasoning we deduce that one of $k,k+1$ is a digit of $n$, where $n$ is the last number whose digits are not used (maybe not all).

Suppose the last digit of $n$ is $l$. Then if $l$ is used, it can not make the difference between the number of digits $l$ and the number of digits $l-1$, so one of the digits $l-1,l$ is used one more time. If $l$ is not used,  again.

Now we deduce that $n$ has at least $5$ digits, and the smallest example is $13578$. Is it correct now?

Yes, Iura, now it is correct. :)
The sequence is
123456........1357713578 and the number of its digits is [b]56784[/b].

I did it and my result is also 13578.
n-digit numbers can make at most 2n kinds of number of different digit.(use induction to prove it,and it is easy)


1. Given positive real numbers $ A,B,C$ prove the inequality $ \max(A^2 \minus{} B, B^2 \minus{} C,C^2 \minus{} A)$ $ \ge$ $ \max(A^2 \minus{} A, B^2 \minus{} B,C^2 \minus{} C)$

2. Find the number of ordered triplets of non-negative integers $ (m,n,p)$ such that $ m \plus{} 3n \plus{} 5p\le 600$

3. Prove the identity
\[ \phi(ab)\phi(gcd(a,b)) \equal{} \phi(a)\phi(b)gcd(a,b).\]

EDIT: Thanks for catching the typo.

[hide]For the third take (a,b)=d and then the prime factorisaton of the numbers that appear.[/hide]

I believe you mean \[ m\plus{}3n\plus{}5p\le600\] for the second one.

C'mon nobody? Here are two more problems (I don't want to post this in a separate thread to occupy more room)

4. How many integers $ n$ in the interval $ 1\le n\le 1000000$ are there such that the equation $ n \equal{} x^{y}$ has solutions in integers $ x$ and $ y$ with $ x\neq n$?

5. Let $ a,b \in \mathbb{N}_0$ and $ p \ge 5$ a prime. Show that $ \binom{pa}{pb} \equiv \binom{a}{b} \pmod{p^3}$
By the way we can start a pre-olympiad NT marathon if you like... :D


Let G be a group, and let a be in G. The set C(a)= { x in G| xa=ax} of all elements of G that commute with a is called the centralizer of a.

a)  show that C(a) is a subgroup of G.
b) show  that <a> is a subset of C(a).

I don't know if you've noticed, but this forum is for "Puzzles and Brainteasers."  There are 4 entire subfora devoted to abstract algebra, in the College section of this site.  Please try to post your questions in the proper section of the website.

Just use the definition of a subgroup.

$e\in H$

$a\in H\implies \overline{a}\in H$

$a,b\in H\implies a\circ b\in H$

It should be really straightforward after that.

For the second, you need to show that $x\in\left<a\right>\implies x\in C\left(a\right)$, so you should pick an element and then work from there...


Let be $ a,b,c\ge 0$ . Prove that :

$ \sqrt{a(b\plus{}c)}\plus{}\sqrt{c(a\plus{}b)}\plus{}\sqrt{b(c\plus{}a)}\le \sqrt{2}(a\plus{}b\plus{}c)$

Cauchy-Schwarz;

$ (a\plus{}b\plus{}c)((a\plus{}b)\plus{}(b\plus{}c)\plus{}(c\plus{}a)) \ge \left(\sum \sqrt{a(b\plus{}c)}\right)^2$

$ \therefore \sqrt{2}(a\plus{}b\plus{}c) \ge \sum \sqrt{a(b\plus{}c)}$

A bashing way:
Let $ a\plus{}b\plus{}c\equal{}1$, then the inequality is $ \sum \sqrt{a(1\minus{}a)}\leq \sqrt{2}$.
If we can prove $ \sqrt{x(1\minus{}x)}\leq mx\plus{}b$ for some $ m\plus{}3b\equal{}\sqrt{2}$, then we will be done. Equality should happen when $ x\equal{}\frac{1}{3}$, so we take the derivative of the LHS when $ x\equal{}\frac{1}{3}$ to guess what $ m$ should be. This gives $ m\equal{}b\equal{}\frac{\sqrt{2}}{4}$.
Now we prove $ \sqrt{x(1\minus{}x)}\leq \frac{\sqrt{2}}{4}(x\plus{}1)$, which, after squaring, multiplying by 8, and rearranging, is equivalent to $ 0\leq(3x\minus{}1)^2$, so we are done.


[quote="archimedes1"]Any tips for preventing stupid mistakes?

I hate being able to re-check each problem twice and still get 3 wrong. :wallbash_red:[/quote]

yea i made this mistake in school competition:

scanning the question too fast and missing a sentence....it was critical too and i did it on 3 different questions  :ewpu:

Work out problems neatly.  At first I thought this was a joke, but it [b]really [/b]has helped a lot, because you see what you are doing, and catch your mistakes easily.

http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php

I really can't prevent my mistakes, they always slip through.
That's probably why I can never get a good score in MathCounts. :(

Try to read the problem word by word... even when you're checking them. Better to get a problem wrong because it takes too much time than to get a problem wrong because you read the problem wrong, and changed it into a wrong answer when you are checking. You won't be [i]as[/i] mad at yourself. I've done that before  :oops: ... getting a question right, but changing it to a different answer when I check it. Also, be careful when you are writing down your answers. You could be thinking one thing, but writing down something else. A former MC National Countdown round participant's strategy is to circle or underline the important parts of a question, so it forces you to see that part and read a little slower. Hope that helps! :)
~*Star*~

The day before any major competition, make sure you're well rested, fed, and prepared. :lol:

It definitely does not help your concentration if you're tired, hungry, or the like.

I agree, get well rested, eat healthy, eat well, and go over the problem, but always say it in a whisper, sometimes, it makes sense in your brain but it doesn't when you speak it out loud.  Also, make sure the answer makes sense and the problem solving you did relates to the problem.  Just go in there and make sure you can keep all of these things in mind.  Have fun too, don't be under pressure.

[quote="jhredsox"]poopyloopy2003 (05:50 pm): 1. DID I READ THE PROBLEM?
poopyloopy2003 (05:50 pm): 2. DID I PUT DOWN WHAT THEY ASKED FOR?
poopyloopy2003 (05:51 pm): 3. SLOW DOWN

-jorian[/quote]

woohooo, i've never said anything before worth being quoted (yes, I stole that).

For me, at least 50% of my mistakes come from not reading the problem.  The other 50% are just dumb.  We fix the reading errors using methods (1) and (2) described above.  The dumb errors should be lessened by (3).

when checking, do it a different way. if you do it the same way, you likely make the same mistake.

well just keep ur head...the last thing u need is getting sick on chapters....

anyway do what i just did....

1. read each question and make sure you understand it....
2. skip questions if necessary if you don't understand the material (or have no way to tackel it)
3. draw diagrams (hint hint) and they will help in chapters

i did the above three and i got to states  :lol:


MATHCOUNTS
geometry
3D geometry
sphere
trigonometry
Pythagorean Theorem

A ball was floating in a lake when the lake froze. After the ball was removed, it left a hole 8 cm deep and 24 cm wide. What was the radius of the ball?

This was the Problem of the Week for last week.

This was also an AMC problem, if I remember correctly.

I believe the answer is [hide]13[/hide].  i got it in a really weird way like i always do [hide]i used PYTHAGOREAN THEORUM!!![/hide]  Do not ask how i derived it off that.

Yes that would be correct - and there is nothing wrong about using the Pythagorean theorem there. Just x^2 + 12^2 = (8+x)^2. and add x to 8.

I looked at this a different way.  There is a theorem, rule, formula.. that states when two chords of a circle intersect, the product of the parts of each are equal.  So in this case if you draw the diameterof the sphere, it will bisect the 24 in chord.  So (12)(12) = 8( the rest of the diameter).  The rest of the diamter is thn 18, making the whole diameter 8+18 = 26, and the radius 13.

I like it because there is very little algebra really involved.

Interesting...my friend who also competes in this used trigs and stuff, very amateur, but yours is quite elegant...good job.

By the way I believe that may be power of point but not sure...

wow.... i was actually CORRECT!!!!!!!!!!!! (first time :mrgreen:) if frost was right...

[color=white]I dont get it. HELP ME!!!!!! [/color]


What is the equation of the plane containing point (20,0,3) that is parallel to the plane containing the points (2,1,1),(4,4,-1), and (2,-1,2)

Call the plane we already have $ \Pi_{1}$.  We need two vectors that span $ \Pi_{1}$ so using the points we already have, let $ \vec{a}\equal{}(2,3,\minus{}2)$ and $ \vec{b}\equal{}(2,5,\minus{}3)$.  Now, find the normal vector, which is given by the cross product.  So we have $ \vec{n}\equal{}(19,\minus{}10,4)$.  For two planes to be parallel, their normal vectors must be parallel as well, meaning their cross product is 0.  Unfortunately, my mind is drawing a blank at the moment and I can't make the numbers work out.

[hide]
The plane containing 3 points(vectors) is given by (r-a).((b-a)*(c-a)).
Then find its normal and the normal of the reqd plane will be lambda times that normal
So then u have a point and the normal so solve by using (r-a).n = 0
:)
[/hide]


inequalities
number theory unsolved
number theory

find all [b]x,y [/b]in integer number such that [b] x 2-y 2 =3 y [/b]

Then $y \geq 0$ and $0 < x-y = 3^a$ and $0 < x+y = 3^b$ where $a+b = y$.
Thus $y = \frac {3^b - 3^a} 2$ so that $3^b = 3^a + 2a+ 2b$.
Clearly, if $b=a$ then $b=a=0$ and $y=0,x=1$ or $x=-1$, which give two solutions.

Thus, we assume that $b > a$, so that $b \geq a+1$.
But, it is easy to verify that if $n \geq 2$ then $3^n > 4n$.
Thus, if $a \geq 2$ we have :
$3^b = \frac 1 2 3^b + \frac 1 2 3^b > 2b + \frac 1 2 3^{a+1}  = 2b + 3^a + \frac 1 2 3^a \geq 2b + 3^a +2a.$ A contradiction.
Thus $a=0$ or $a=1$.
If $a=0$ the inequality (1) leads to $b=1$ (since $b > a$). Thus $y=1$ and $x = 2$ or $x=-2.$
If $a=1$ then as above we deduce that $b=2$ and $y=3$ so that $x=  6$ or $x=-6$.

Conversely all these pairs are solutions.

Pierre.


conics
parabola
function
LaTeX
absolute value

I think you mean two roots that are the same, so there's actually only one root.
[hide]To have only one root (when graphed, the parabola has its vertex on the x-axis), you have to have a perfect square. The only way the function can be a perfect square is if m is [b]12. [/b][/hide]
BTW, please refrain from posting the same problem twice

Your function has one zero <=> $\Delta=0$ <=> $m^2-4\cdot 4\cdot 9=0$ <=> $m^2=144$ <=> $m=\pm 12$, so that's why it's 12. :)

[hide]I used the discriminant, set it equal to zero and I got 12.

$12$[/hide]

$x_1;x_2=\frac{-m \pm \sqrt{m^2-4*4*9}}{2*4}$

We want $x_1=x_2$, which is the same as $\frac{-m+\sqrt{m^2-144}}{8}=\frac{-m-\sqrt{m^2-144}}{8}$, which is true $\Longleftrightarrow$ $\sqrt{m^2-144}=0$, from where $m=\pm 12$. Then, the answer is $\boxed {12}$

[hide]Determinant=m^2-4*4*9=0 since there is only one root. |m|=12.[/hide]

[quote="silver"]What is the absolute value of m for which the equation 4x2 - mx + 9 = 0 has equal roots?

[color=blue][size=75]kyyuanmathcount: Make sure you erase the problem from the right thread  :P .[/size][/color][/quote]

You should learn $\LaTeX$, because it makes equations such as that much easier to read.

$4x^2-mx+9=0$

[hide]Yes, as it has already been stated, the answer is indeed 12.[/hide]


Find the least positive integer, which may not be represented as ${2^a-2^b\over 2^c-2^d}$, where $a,\,b,\,c,\,d$ are positive integers.

I think it is 11. Clearly powers of 2 can be written since $2^k=\frac{2^{k+2}-2^{k+1}}{2^2-2}$ and of course, if $x$ can be written, so does $2^px$ for all $p$. Next, 1 can be written, so does $3=\frac{2^3-2}{2^2-2}$; $5=\frac{2^5-2}{2^3-2}$, $7=\frac{2^4-2}{2^2-2}$ and $9=\frac{2^7-2}{2^4-2}$. Suppose that we can write 11 like that and take $a>b$. Then $c>d$ and $ 2^b(2^{a-b}-1)=11\cdot 2^d\cdot (2^{c-d}-1)$. Thus $b=d$ and since $11$ divides $2^{a-b}-1$ we must have $a-b$ multiple of 10. Write the equation as $ 2^{a-b}+10=11\cdot 2^{c-d}$ and deduce that $ c-d=1$ since otherwise 4 divides 10. But then $a-b$ is not a multiple of 10, false.

Well you cant just reason by trial and error. How can you be sure that  you have exausted EACH AND EVERY possiblity for the required $a,b,c,d$ ???? A mathematical proof needs to be water tight. I must say, your solution has TOO many loopholes . Please post a proper solution if you have one!

Let's looking at this problem.
  Prove that in the interval$(2^{p+1}-1;2^p+1)$,wher $p$
  is a prime numbre,don't existe odd numbre that  be represented in the form  $\frac{2^{a} - 2^{b}}{2^{c} - 2^{d}}$ ;)

hey its easy to prove that nos. of the form 2^k +1 and 2^k -1 can be represented in this form.
further we can prove that 2^s(2^k +1) and 2^s(2^k -1) can also be represented in this form.
then we can prove that no other nos. can be represented in this form.
its easy to note that 11 is the 1st no that cannot be represented in the form 2^k -1 & 2^k +1.
this is how i solved it.hope it works ur purpose.

[quote="venkata"]then we can prove that no other nos. can be represented in this form[/quote]
Please use LaTeX, and please give your proof. :)

[quote="fleeting_guest"][quote="Rushil"]least positive integer that cannot be represented in the form $\frac{2^{a} - 2^{b}}{2^{c} - 2^{d}}$ [/quote]

A representable number is of the form $2^p (2^n - 1) / (2^m - 1)$.
Lemma: this is integral if and only if $m | n$.
Conclusion: a number is representable if and only if the positions of the 1's digits in its binary representation, form an arithmetic progression.
Famous exercise: what is the GCD,  $(2^m - 1, 2^n - 1)$?[/quote]
If $m,n$ are positive integers and $a>1$ is also an integer, then $gcd(a^m-1,a^n-1)=a^{gcd(m,n)-1}$

In fact, if $a$ and $b$ are coprime, then $\left(a^m - b^m,a^n - b^n\right) = a^{(m,n)} - b^{(m,n)}$ :)

Arne, yours is correct
To prove this Lemma: $\left(a^m - b^m,a^n - b^n\right) = a^{(m,n)} - b^{(m,n)}$
assume $gcd$ of $(m,n)$ is $k$
$\left(a^m - b^m,a^n - b^n\right) = (a-b) * gcd(\sum a^ib^{m-i-1}, \sum a^ib^{n-i-1}) = (a-b)*sum a^ib^{m-i-1}, \sum a^ib^{n-i-1})$
$a^{(m,n)} - b^{(m,n)} = a^k-b^k = (a-b)(\sum a^ib^{k-i-1})$
$\text{ The value of } \sum a^ib^{m-i-1} \text{ is always a prime number when } m \text{ is odd, and so does n}$
If one of $m$ and $n$ is odd, then $k$ is odd, the lemma is obviously correct
If both of them are even, then assume $m=2^{\alpha}m', n = 2^{\beta}n'$where $m', n'$ are all odd, $\alpha,\beta \in N$
then we can prove that the lemma is still true.

You can look here too:

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=pgcd&t=36731

[quote="Kalimdor"]
$\text{ The value of } \sum a^ib^{m-i-1} \text{ is always a prime number when } m \text{ is odd, and so does n}$
If one of $m$ and $n$ is odd, then $k$ is odd, the lemma is obviously correct
If both of them are even, then assume $m=2^{\alpha}m', n = 2^{\beta}n'$where $m', n'$ are all odd, $\alpha,\beta \in N$
then we can prove that the lemma is still true.[/quote]

What? :huh: I don't think this works.

it was easier to proof c-d|a-b
but writing them as binaries was a great idea!!

[hide=Solution + Storage]Firstly, if $b=d$, we get :

$$\frac{2^a-2^b}{2^c-2^d} = \frac{2^{a-b}-1}{2^{c-d}-1}$$

Hence, $2^{a-b} \equiv 2^{c-d} \pmod {2^{c-d}-1} \implies 2^{{c-d}}(2^{(a-b)-(c-d)}-1)  \equiv 2^{(a-b)-(c-d)}-1 \pmod {2^{c-d}-1}$. So, $\textbf{ord}_{2^{c-d}-1}2 \mid (a-b)-(c-d)$. But $\textbf{ord}_{2^{c-d}-1}2$ is precisely $c-d$. Hence, $c-d \mid a-b$. In this case, we observe that $\frac{2^a-2^b}{2^c-2^d} = \frac{2^{a-b}-1}{2^{c-d}-1}$ is odd.

If we set, $b > d$ and $c-d = a-b$. We get $\frac{2^a-2^b}{2^c-2^d} = 2^{b-d}\frac{2^{a-b}-1}{2^{c-d}-1} = 2^{b-d}$. Hence, all powers of $2$ can be represented. Further, $2^m \cdot N$ can also be represented where $m \in \mathbb{N}$ and $N$ is of the form $\frac{2^{a-b}-1}{2^{c-d}-1}$. So, combining all these, we can check by hand that all numbers $1$ to $10$ can be represented.

Now, we check for $11$. Since, $11$ is odd, it can only be represented in $\frac{2^{a-b}-1}{2^{c-d}-1}$. Also, $11$ must divide $2^{a-b}-1$. Hence, by FLT, $a-b$ is a multiple of $10$. As the above posts mentioned, this cannot happen since $10$ doesn't divide $a-b$, contradiction. So, we conclude that smallest such integer is $11$, as desired $\blacksquare$

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