abstract algebra
function
modular arithmetic
group theory
number theory
relatively prime
superior algebra

Go here ([url]http://en.wikibooks.org/wiki/Math[/url]) and click on Abstract Algebra. Here is some stuff that might help:

[hide="Group"]
A group is a set $ G$ paired with a binary operation $ *$. We can denote this $ (G,*)$. The group must satisfy the following axioms:

Closure Under $ *$: If $ a,b\in G$, then $ a*b$ is also in $ G$.

Associativity of $ *$: If $ a,b,c\in G$, then $ (a*b)*c\equal{}a*(b*c)$.

Existence of an Identity: There exists an element $ e$ in $ G$ such that for any $ a\in G$, $ a*e\equal{}a$ ($ e$ is equivalent to $ 1$ under multiplication).

Existence of Inverses: For any $ a\in G$, there exists an element $ q$ in $ G$ such that $ a*q\equal{}q*a\equal{}e$ where $ e$ is the identity.

If a set and an operation fulfill these axioms, then they form a group.

Also, if a group is communative, then it is called an Abelian Group.
[/hide]

[hide="Homorphisms"]
If we have to groups, $ (G_1,*)$ and $ (G_2,\delta)$, a homomorphism is a mapping or function $ f$ from $ G_1$ to $ G_2$. Also, given elements $ x,y$ in $ G_1$,

$ f(x*y)\equal{}f(x)\delta f(y)$
[/hide]

This is just an extremely brief overview, but I hope it will get you started. Good luck. By the way, there are some very nice proofs using group theory, one being a proof of Fermat's Little Theorem.

Er,... actually I wanted learn about the application, how to apply it , become comfortable with it ....etc. etc..I already know this definitions.Could you name me such books?I would really appreciate it.
Thanks.

Could any of u guys help me?It's urgent, and helping would'nt hurt u guys,would it?

As an example of application, here's the following proof of Fermat's Little Therorem which states:

$ a^{p-1}\equiv 1\pmod{p}\forall a\in\mathbb{Z}^+: a\le p-1$

[hide]Assume the set $ G=\{1,2,3,...,p-1\}$ with the operation of multiplication taken modulo $ p$ forms a group.

Assume $ a\in G$ and $ k$ is the order of $ a$ such that,

$ a^k\equiv 1\pmod{p}$

By Lagrange's Theorem, $ k$ divides the order of $ G$, $ p-1$.

Therefore, $ p-1=km$ for some $ m\in\mathbb{Z}$.

Then, $ a^{p-1}\equiv a^{km}\equiv (a^k)^m\equiv 1^m\equiv 1\pmod{p}$.

To show that $ G$ is a group we must show invertibility as follows,

Assume $ b$ is relatively prime to $ p$,

Then by Bezout's Identity there exists $ x,y\in\mathbb{Z}$ such that,

$ bx+py=1$

Taking this equation modulo $ p$ we see,

$ bx=1$

Therefore, $ x$ is the inverse of $ b$ and all elements of $ G$ have inverses.

Q.E.D.
[/hide]

[quote="Cronus"]Hello, I want to learn more about group theory . Can you guys help me by giving me the names of some books that can help me learn it?
Thanks .[/quote]

Birkhoff/ MacLane: A Survey of Modern Algebra
http://www.amazon.com/Survey-Modern-Algebra-AKP-Classics/dp/1568810687/ref=sr_1_1?ie=UTF8&s=books&qid=1211877288&sr=1-1

Herstein: Topics in Algebra
http://www.amazon.com/Topics-Algebra-I-N-Herstein/dp/0471010901/ref=pd_bbs_sr_1?ie=UTF8&s=books&qid=1211877344&sr=1-1

Thanks guys (especially to ottem).I really appreciate it. :)

There is also the website of Peter Cameron:
http://www.maths.qmul.ac.uk/~pjc/
You will find nice things there,not only in group theory.

sir ,

  I want to know about the relation of order of an group and  order of an element...
generator of subgroup means what...

Read a textbook. Several have been recommended already here.


You are blindfolded and told there are at least 128 pennies on a table. You know there are 128 heads, and some number of tails. Without removing any pennies, looking at the coins, and feeling the pennies' tops to see if they are heads or tails, create an algorithm that partitions all the pennies into two sets of pennies using all the pennies such that each set has an equal number of heads in it? This algorithm must work for any number of pennies, as long as the total number of pennies is greater than or equal to 128.

[hide]I have seen this problem before, but I can't seem to remember the answer. All I remember is that it requires flipping of coins to get a certain arrangement.[/hide]

I have seen this problem somewhere too. All I can remember is that the solution was confusing.

i agree. btw, on a cloudy day, around 80% of the radiation still seeps through. :D

[quote="computer_nuke"][quote="durt"][hide]
I haven't seen this before. It's a bit tricky.  :)

Put $128$ of the coins in one pile and the remainder in another. The first pile will have $a$ heads and $b$ tails such that $a+b=128$. The other pile will contain $128-a$ heads and some unimportant number of tails. Now flip every coin in the first pile, so it has $b$ heads. But $b=128-a$, so both piles have the same number of heads.
[/hide][/quote]

But I thought you weren't allowed to look at the coins, so how do you know when you will have the desired amount of heads (In this case, b amound of heads)?[/quote]

$a+b=128$ You don't know what b is.
That means the number of tails in the pile of 128 is the same as the number of heads in the other pile. So if you turn over all the coins in the first pile, all the heads turn tails and the tails turn heads.

Hmm, I got it now  :D  :blush:

Srry I thought he meant by flipping until you get the number. Thanks alot. This truelly is pretty tricky but now I see it.


How is the Canadian IMO Team selected?
(And are Grade 12's allowed?)

Strong Canadian math students write the Canadian MO, USAMO, and the Asian-Pacific MO (of course, there are other lesser contests to identify the strong students).  A comittee looks at the papers and uses a top secret formula/process/vote/bribe-system to select the members of the team.  They've pretty much finished that by now, but nobody's allowed to publicly announce who will be on the team.  Grade 12s are allowed... in fact, I went to my first and only IMO as a grade 12.  Last year's IMO team had 4 grade 11s, and all of them are good enough to make it back this year as grade 12s.  In fact, I wouldn't be too surprised if this year's team has 5 or more grade 12s.

[quote="otaku"]well, i was just wondering if doing extremely well on the COMC and CMO next year is enough...
and if any record of previous achievements is necessary...

thanks for the quick reply, lol, i just joined the forum so didn't know what to expect
anyway, since you guys don't know me,

my name's Greg, Grade 11, found out about AoPS last summer, qualified for USAMO this year, not much inclined to sit and do math in my free time (yet), live in Toronto[/quote]

Ahh ok then.  If you own COMC, you'll get invited to the winter training camp and you'll be given a chance to write APMO and USAMO.  Doing very well on this year's CMO will also get you an invitation.

[quote="otaku"]not much inclined to sit and do math in my free time (yet), live in Toronto[/quote]

be gone heathen! :!:

 :rotfl:  :rotfl:

When will they tell us the results of CMO? Last year, there was a media release on May 12 for CMO results, so does anyone know when there will be that media relase this year??????? :?

Is anybody allowed to write CMO, or it can be written through invitation only?

I really really really want to be selected to write IMO. Getting good marks on COMC will probably earn me the invitation to write the IMO, but I am not sure if I can get good marks on COMC. This year I got 48 out of 80 and next year I expect to get something like 60. Does anyone know any good books to prepare for the COMC? University of Waterloo published a number of books to help students prepare for math contest, but I heard that those book are only good if you are perparing for Euclid, Fermat, Cayley, etc., but not good enough for COMC.

[quote="satyam"]
I really really really want to be selected to write IMO. Getting good marks on COMC will probably earn me the invitation to write the IMO, but I am not sure if I can get good marks on COMC.[/quote]

Hahaha careful, getting good marks will let you write the CMO, not the IMO.

CMO is invite only.  If you want to prepare, use the waterloo books first.  If you can't do the euclid, fermat problems, when you'll have lots of trouble on CMO, and trouble with the last half of the COMC.

[quote="satyam"]Does anyone know any good books to prepare for the COMC? University of Waterloo published a number of books to help students prepare for math contest, but I heard that those book are only good if you are perparing for Euclid, Fermat, Cayley, etc., but not good enough for COMC.[/quote]

A good way to prepare would be to join the Math Olympiad Correspondence Program (OLymon) where problems are posted on the website, you write them up and send it to them, and then they are marked and sent back. It's a great way to get feedback, and it forces you to practice writing up solutions, which people don't normally do for themselves. The site is http://www.math.ca/Competitions/MOCP/. Good luck! I hope you get on the IMO team someday.

[quote="soapbubble"]usually how long does it take to prepare for comc/cmo?[/quote]

It depends on the experience you already have, on how often you practice, etc. I've never paced myself in that sense, so I don't know. It seems more natural not to target yourself to a particular contest, but just to do various contest problems regularly. Problems on the level of the last COMC questions, CMO questions, etc, don't differ that much in style, so you would be preparing for all of those contests at once. Good luck, and have fun practicing!


What exactly is the Scotopic sensitivity curve ??

don't  quote me on this, but i believe that is has something to do with sensing different wavelengths of light better than others.

Yes, I also concur that scotopic sensitivity would be defined as having a super-sensitivity to certain colors over others.

This in humans is known as Scotopic Sensitivity Syndrome or Irlen syndrome, a type of dyslexia.

Symptoms are categorized, under depth perception problems, discomfort with bright, vibrant patterns or extreme conditions of contrast (like light and dark), reading abilities, or text that seems to swirl. [1]

Helen Irlen stated in her book Reading By the Colors-- "Scotopic Sensitivity Syndrome--and the sensitivity to full-spectrum light that forms the basis of the syndrome--possibly involves a structural brain deficit involving the central nervous system. If so, then signals sent to the brain would be inappropriately processed, resulting in perceptual problems. For those with SSS, full-spectrum light would distort what is perceived and processed by the brain" [2]

I believe Helen Irlen developed a method of improving the vision of those with this syndrome, with the use of tinted glasses to aid reading. One of her first tests, was taking an SSS subject, and assesing his vision after placing a colored plastic over the text being read. It seemed to have improve.

Though many scientists believe that SSS exists, the actuality of this syndrome has received some skeptcism as follows:
[list]Whether SSS exists as a distinct, predictably identifiable disease with a reasonable pathophysiologic mechanism;
Whether SSS is causally or incidentally related to dyslexia, autism, or other conditions; and
Whether existing methods of SSS treatment are appropriate and effective. [1] [/list]

The above is the background behind scotopic sensitivity, the sources shall be listed below, if you wish to read over them.

Now my assumption, is an actual scotopic sensitivity curve is the one shown in the graph:
[img]http://www.ultimateavmag.com/images/archivesart/204eye.4.jpg[/img] [3]

Sources:
[1] http://en.wikipedia.org/wiki/Scotopic_Sensitivity_Syndrome
[2] http://findarticles.com/p/articles/mi_m2843/is_4_28/ai_n6100519
[3] http://www.ultimateavmag.com/images/archivesart/204eye.4.jpg


Anybody has ever done this? More precisely, I want my Java applet to keep some information (high scores chart) on the server but I can't seem to find any way to write to files.

The password part is a bit problematic though. I don't want to hard code the password into the Java applet nor provide users with the username / password. I don't want the people using my applet to be able to send any files they wish to the server, only the data that my applet produces.

Actually, I realized that what I want is probably impossible. I'll have to think of some other approach to make my hs chart secure.

Applets are client-side, not server-side. In order for applets to get to read/write stuff to a server (or even to the client for that matter) you need signed applets. Which are a pain :P

I believe you can use some server-side script that the Java applet will communicate to in order to store high scores. For example, you can make a PHP script and then pass the information through GET or POST. It may be, however, that even this is not allowed.

Yes, I know, that's the whole problem. I'll have to switch to a server-side application. I know it [i]is[/i] possible to communicate with a server-side daemon which will process the requests and update my hs chart but then anybody could do it and I really don't want to deal with any advanced secure communication techniques :). But hey, thanks for your input!


When I tell TeXnic Center to generate a PDF file, it will tell me about the "bad boxes" that I have in my file.  What is a bad box, and how is it significant?  I haven't noticed any problems with my PDF files, but maybe I'm just missing something.

it may occur when there's a too long formula and so on...

Does this mean that there is something inherently wrong with a bad box?

I get those all the time and largely ignore them.  Often this happens because something extends beyond a margin or something (I think).  There are probably many other causes.  If it looks the way you want, you're ok.

OK, thanks, that's all I wanted to know.

I think (not sure though) that if you extend a line beyond the margin, it's called an overfull box. A manual line break (\\) will typically result in an underfull box. I believe that each of these is considered a bad box. I tend to get hundreds of them and don't bother to get rid of them. (I wouldn't know how anyway.) Maybe stevem will enlighten us.

[quote="ComplexZeta"]I believe that each of these is considered a bad box. I tend to get hundreds of them and don't bother to get rid of them.[/quote] That's probably the sensible approach  ;)  The key here is to look at the resulting output and think about whether it looks right. If it does then don't worry about it.

Yes it is to do with [tex]\LaTeX[/tex] overflowing a line into the margin, maybe because the hyphenation rules don't allow it. You could try using a hyphen with \- or use \linebreak (which unlike \\ keeps the justification) or even \begin{sloppypar} paragraph text \end{sloppypar} which effectively turns off the hyphenation and can give different spacing between the words. You can learn more about what's going on at the [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=overfull]TeX FAQ[/url]; at which point you might decide to just ignore the problem  :)

Again with underfull just see how it looks. It occurs with left and right justified text when the spacing between the words is bad. This is measured by the 'badness'. [tex]\LaTeX[/tex] tolerates badness up to 200, up to 2000 is probably acceptable but 10 000 is 'infinite badness'.


Let $\vartheta'\geq\vartheta''$ be positive real sequences. Then for all $a_{1},\ldots,a_{n},b_{1},\ldots,b_{n}$ we have \[\left(\sum_{k=1}^{n}\vartheta'_{k}a_{k}^{2}\right)\left(\sum_{k=1}^{n}\vartheta'_{k}b_{k}^{2}\right)-\left(\sum_{k=1}^{n}\vartheta'_{k}a_{k}b_{k}\right)^{2}\geq\left(\sum_{k=1}^{n}\vartheta''_{k}a_{k}^{2}\right)\left(\sum_{k=1}^{n}\vartheta''_{k}b_{k}^{2}\right)-\left(\sum_{k=1}^{n}\vartheta''_{k}a_{k}b_{k}\right)^{2}\geq0.\]


calculus
integration
algebra
function
domain
polynomial
linear algebra

[i]Show that the integral operator defined by $ (Tf)(s): \equal{}\int_0^1(s\minus{}t)^2f(t)\text{d}t$ is self-adjoint and calculate the necessary ingredients of the spectral representation.
Hint: first determine $ \text{im }T$ and then make an 'ansatz' for possible eigenfunctions.[/i]

First of all, the exercise doesn't say what the domain and codomain of $ T$ are, so I guess I'll just take it to be $ T: C[0,1]\to C[0,1]$. Since $ k: [0,1]\times[0,1]\to\mathbb{R},\ k(s,t)\equal{}(s\minus{}t)^2$ is continuous and satisfies $ k(t,s)\equal{}k(s,t)\equal{}\overline{k(s,t)}$ for all $ s,t\in[0,1]$, it follows that $ T$ is (compact and) self-adjoint.

I have no idea how to determine $ \text{im }T$... any ideas?

\\edit: hmm...since $ (Tf)(s)\equal{}s^2\int_0^1 f(t)dt\minus{}2s\int_0^1tf(t)dt\plus{}\int_0^1t^2f(t)dt$, it seems I have to think about polynomials of degree 2. But if I propose an $ f$ of the form $ f(t)\equal{}at^2\plus{}bt\plus{}c$, then $ (Tf)(s)\equal{}\left(\frac{a}{3}\plus{}\frac{b}{2}\plus{}c\right)s^2\minus{}\left(\frac{a}{2}\plus{}\frac{2b}{3}\plus{}c\right)s\plus{}\left(\frac{a}{5}\plus{}\frac{b}{4}\plus{}\frac{c}{3}\right)$.
Solving $ Tf\equal{}\lambda f$ for such an $ f$ gives a=b=c=0...

[quote="BlueVelvet"]Solving $ Tf \equal{} \lambda f$ for such an $ f$ gives a=b=c=0...[/quote]

Without any calculation: This statement has to be wrong if you think about it. The operator $ T \vert\mathbb{C}[t]_2 : \mathbb{C}[t]_2 \to \mathbb{C}[t]_2$, where $ \mathbb{C}[t]_2 \supset \rm{im}\,T$ are the polynomials of degree (at most) $ 2$, is an operator between [i]finite dimensional[/i] spaces. To calculate the point spectrum of $ T$, just calculate the (non-empty!) spectrum of $ T \vert\mathbb{C}[t]_2$, for example using a matrix representation of that operator (which you already deduced implicitly).
And don't forget to treat the eigenvalue $ 0$ separately.

And by the way, talking about a self-adjoint operator, you should at least choose a Hilbert-space for your domain. $ L^2(0,1)$ will do here. Self-adjointness then follows from Fubini and compactness from the fact that the image of $ T$ is finite dimensional.

Thank you very much. Indeed, of course I should have chosen a Hilbert space. So let me choose $ T: L^2[0,1]\to L^2[0,1]$. Since $ \text{im } T\subseteq \mathbb{C}_2[t]$, the eigenvectors have to be such polynomials. Hence we consider $ T\vert\mathbb{C}[t]_{2}: \mathbb{C}[t]_{2}\to\mathbb{C}[t]_{2}$. With respect to the basis $ (1,x,x^2)$, the matrix representation is
$ A\equal{}\left(\begin{array}{ccc}
\frac{1}{3} & \frac{1}{4} & \frac{1}{5} \\
\minus{}1 & \minus{}\frac{2}{3} & \minus{}\frac{1}{2} \\
1 & \frac{1}{2} & \frac{1}{3}
\end{array}\right)$

There are three eigenvalues $ \lambda_1\equal{}\frac{5\plus{}3\sqrt{5}}{60},\lambda_2\equal{}\minus{}\frac{1}{6},\lambda_3\equal{}\frac{5\minus{}3\sqrt{5}}{60}$. Corresponding eigenvectors are
$ f_1\equal{}\left(\frac{5\plus{}\sqrt{5}}{20},\minus{}1,1\right),f_2\equal{}\left(\frac{\minus{}\sqrt{1}}{2},1,0\right),f_3\equal{}\left(\frac{5\minus{}\sqrt{5}}{20},\minus{}1,1\right)$.

Thus the three eigenspaces are 1-dimensional: $ E_1\equal{}<f_1>,E_2\equal{}<f_2>,E_3\equal{}<f_3>$. Since $ T$ is compact and self-adjoint, by the spectral theorem we have the decomposition
$ L^2[0,1]\equal{}\ker T\oplus \overline{\text{im }T}\equal{}\ker T\oplus E_1\oplus E_2\oplus E_3$, and
$ T\equal{}\sum_{\lambda\in\sigma(T)}\lambda\pi_\lambda$ with $ \pi_\lambda$ the projection on the eigenspace $ E_\lambda$ (and $ E_0: \equal{}\ker T)$.

Now I'm a bit confused: apparently every $ g\in L^2[0,1]$ can be written (uniquely) as $ g\equal{}g_0\plus{}g_1\plus{}g_2\plus{}g_3$, where $ g_i\in E_i$? How is that possible, given that there are lots of weird functions living in $ L^2[0,1]$? For example $ g(t)\equal{}0$ if $ t<1/2$ and $ g(t)\equal{}1$ if $ t\geq 1/2$.
As you said, treating the eigenvalue $ 0$ seperately, I want to find $ 0\neq f\in L^2[0,1]$ such that $ 0\equal{}\int_0^1 f(t)dt\equal{}\int_0^1 tf(t)dt\equal{}\int_0^1 t^2f(t)dt$. I can take e.g $ f(t)\equal{}1$ if $ t\equal{}\frac{1}{n}$ ($ n\in\mathbb{N}$), and 0 everywhere else.

Now, I don't see how to calculate the projections explicitly, since there is no 'general form' of an $ g\in L^2[0,1]$. Could you/someone clarify this? Thanks again!


Beautifull!

It can be shown even that
$2n-\frac{4}{3}-\frac{1}{n}\le u_{n}\le 2n-\frac{4}{3}+\frac{1}{n}$.

[quote="cckek"]Be $u_{n+1}=n+\frac{2n^{2}}{u_{n}},u_{1}=1$. Is the sequence $u_{n}-\sqrt{n(4n+1)}$ convergent? If so find its limit.[/quote]

I attempted to use the technique in http://www.mathlinks.ro/Forum/viewtopic.php?t=126724 to solve your problem. However, it comes to the step that I have to prove:
$u_{n}\leq 2n$ by induction. Well, can you help me to prove that step?
Many thanks! :)

[quote="didilica"]Maybe this would work.

Observe that $\sqrt{n(4n+1)}-(2n+\frac{1}{4})\rightarrow 0$, hence it is sufficient to calculate $\lim (u_{n}-2n)$.

On the other hand, note that:

$\frac{u_{n+1}}{n}=1+\frac{2n}{u_{n}}$, hence $\lim \frac{u_{n}}{n}=2$.

Now, $u_{n+1}-2(n+1)=\frac{2n^{2}}{u_{n}}-n-2$.

Let $z_{n}=u_{n}-2n$. Then we get that

$z_{n+1}=\frac{n(2n-u_{n})}{u_{n}}-2=-2-\frac{z_{n}}{\frac{u_{n}}{n}}$.
Letting  $n$ converge to $\infty$ in the preceding recurrence relation we get that:$L=-2-\frac{L}{2}$ which implies that $L=-\frac{4}{3}$.

I denoted $L=\lim z_{n}$.[/quote]

how did you get $\lim\frac{u_{n}}{n}=2$?


Scheduling for me will start very soon, and for my senior year, I'm not sure what to take... I'm thinking between AP Biology or AP Physics, but I'm not very sure... For those who took both classes, please reply giving your comments on both classes and if you don't mind, your AP Scores for both of them.  For those that only took AP Biology, please describe the course content and how difficult it is.  For AP Physics, please describe the course content as well, and how difficult...

please reply whenever you can :) thank you

I took AP Bio and I would recommend taking it.  Physics B has more topics in it but Physics C is calculus-intensive.

Have you already done AP Chemistry? If not, that may be more important to biology/biochem/chem/premed majors than AP Biology.

Depends on your interests and goals.  As far as your interests go, Physics B consists of a lot of topics, but basic principles in each should guide you well.  Physics C has fewer topics but harder problems, and as JRav said is calculus-intensive.  Biology is also a lot of topics and is probably the most fact-based of the three (as in, there are many things that can't be re-derived and must simply be known).

yeah I'm currently taking AP Chemistry right now :P

and unfortunately I'm not sure what to major in too =/

The worst (or best) it could be ... taking all the AP sciences as I did :P

Also, it depends on what you're interested in and the universities you will apply to.  If you like thinking and mass-intensive work, pick physics.  If you like reading and learning about really intricate processes that make life go, then take bio.  Also, some schools may not even offer credit for physics B.  Some may only offer 3-4 for C but 6-8 for bio.  It depends.

Physics B is probably not given credit by top-notch colleges, but I don't know, I could be wrong  :roll:

The underlying principle here is you can't be undecisive. Look at your interests and the challenges ahead. If you are not interested but want to take them because you just want to be 'famous' or 'notorious' or 'looks good on resume' or 'mom said so' or 'my friend is taking it too (although this can be an advantage :D)', please don't.

My own experience: my friend and I are good students. We get good grades. We both took the same AP classes and tests. He liked math, computer science, biology and I like math, physics, chemistry. We studied together and look at the different AP tests scores:

$ \begin{array}{|l|l|l|} \hline \text{AP Exams} & \text{My score} & \text{Him} \\
\hline \text{Calc BC} & 5 & 5 \\
\hline \text{Physics C: both} & 5,5 & 3,3 \\
\hline \text{Chem} & 5 & 4 \\
\hline \text{Bio} & 4 & 5 \\
\hline \text{Stat} & 5 & 4 \\
\hline \text{Comp Sci AB} & 2 & 5 \\
\hline \end{array}$

Your interest is definitely the #1 drive for success in a class.

Eh... I didn't know what I was doing for Comp Sci  :rotfl:

[EDIT]
Big PS: AP Physics C is calc-intensive, even though you may not be doing heavy calculations. That's because vector integrals are so important in E&M and just plain AP Calc can be really helpful for understanding Mechanics as well. If math is a problem, that leaves you AP Bio as a pick. Do not take AP Bio lightly, it is just as tough as AP Phys, more fact oriented and a lot of conceptual/terminology understanding. A good grasp of chemistry is needed for AP Bio, you can't expect to succeed AP Bio without a good knowledge of the basics of chemistry.

To succeed one or both classes: good studying habits and hardwork... and must be interested ;)

:) ok thanks :) i'll consider about the college that I would like to attend... I'm currently taking many honors and AP Courses, and everything seemed very interesting to me, especially music and science/math.... so really, I'm not sure... thanks

The classes' difficulty vary by school as well so it's kind of hard to gauge the level of the two. However, I've taken AP bio and am currently taking AP physics (C). AP bio was pretty easy for me mainly because I'm really interested in it. (I got a 5). However, AP physics has been somewhat of a struggle. However, keep in mind that this is THE hardest class at my school. Also, around 95% of the AP physics students get 5s at my school due the extreme level of difficulty of the class.

It might depend on the teacher as well. I'm currently taking AP bio as a freshman in high school and the teacher is pretty cool. I plan to take AP chem next year but I heard the teacher wasn't as good.

You're pretty fortunate that you can take AP Bio as a freshman.  In my HS, you can't take an AP science until junior year.  You'd better make sure you know what you're doing.

[quote="JRav"]You'd better make sure you know what you're doing.[/quote]

Yeah, I'm part of the second generation AP Freshman Bio program. I have the first generation to look for help as well as some experienced councelers to plan my future HS classes.


A person buys 20%, 40%, and 40% of books respectively from the three vendors A B C, 1%, 3% and 3% respectively of the products are defective. A given book is defective. Find the probability it was from B.

Please help me ....

This is a classic Baysian Probability question. As a prerequisite to this you should already be familiar with conditional probablity.

There is a formula for computing Baysian Probability which you can directly apply to this problem (I'm sure it can be found in any textbook covering the topic), but I usually advise my students against memorizing this formula as it is easy to misremember.

Instead I advise you to make a probability tree. First you should have three branches, one for the options A, B, and C respectively. Then each of those branches should split into two branches - for Defective (D) and Not Defective (D').

Now use what you know (or hopefully know) about conditional probability.

$ P(B|D) \equal{} \frac {P(B \cap D)}{P(D)}$

The probability of $ B \cap D$ can be found by multiplying the probability along branch B to the probability along the subsequent branch D, i.e. (.4)(.03). The probability on the denominator, $ P(D)$, can be found by adding the probabilities leading to each end result of D.

Try it and see if this makes sense to you. Your final answer should come out to be between 40% and 50%.


Solve if $ x,y \in \mathbb {N}$

$ x^{2} \minus{} 3y^{2} \equal{} \minus{}1$

$ 3y^2\equal{}x^2\plus{}1$, so $ x^2\equiv2\pmod3$. That's impossible. No solutions.

Then what about Pell's equation. Does that mean Pell is wrong?

This specific Pell equation has no solutions. There exist plenty that do.

http://en.wikipedia.org/wiki/Pell's_equation

I always thought it could be $ \pm 1$...?

You are right, it's negative Pell's equation.
But $ x^2\minus{}dy^2\equal{}\minus{}1$ is solvable in first 10 natural numbers for 2,5 and 10 only.


Prove or disprove that  then $ x_i>0,i\equal{}1,2,...,n$,$ n\in N,n\geq 3$

$ \sum_{i\equal{}1}^n x_i \sum_{i\equal{}1}^n \frac{1}{x_i}\minus{}n^2\geq 4 \sum_{i\equal{}1}^n \frac{x_i}{x_{i\plus{}1}\plus{}x_{i\plus{}2}}\minus{}2n$

$ x_{n\plus{}1}\equal{}x_1,x_{n\plus{}2}\equal{}x_2$.

thank you! :lol:

[quote="hedeng123"]Prove or disprove that  then $ x_i > 0,i \equal{} 1,2,...,n$,$ n\in N,n\geq 3$

$ \sum_{i \equal{} 1}^n x_i \sum_{i \equal{} 1}^n \frac {1}{x_i} \minus{} n^2\geq 4 \sum_{i \equal{} 1}^n \frac {x_i}{x_{i \plus{} 1} \plus{} x_{i \plus{} 2}} \minus{} 2n$

$ x_{n \plus{} 1} \equal{} x_1,x_{n \plus{} 2} \equal{} x_2$.

thank you! :lol:[/quote]

Using the computer can verify that $ n\equal{}3,4,5,6$ the  inequality holds.

 $ n\geq 7$ ?

thank you.

[quote="hedeng123"]Prove or disprove that  then $ x_i > 0,i \equal{} 1,2,...,n$,$ n\in N,n\geq 3$

$ \sum_{i \equal{} 1}^n x_i \sum_{i \equal{} 1}^n \frac {1}{x_i} \minus{} n^2\geq 4 \sum_{i \equal{} 1}^n \frac {x_i}{x_{i \plus{} 1} \plus{} x_{i \plus{} 2}} \minus{} 2n$

$ x_{n \plus{} 1} \equal{} x_1,x_{n \plus{} 2} \equal{} x_2$.

thank you! :lol:[/quote]
The inequality holds. We can prove it as follows:

Notice that for each $ i,$
$ x_i\sum^n_{i\equal{}1}\frac{1}{x_i} \equal{}x_i\left(\frac{1}{x_{i\plus{}1}}\plus{}\frac{1}{x_{i\plus{}2}}\right)\plus{}x_i\sum_{j \not\in \{i\plus{}1,\; i\plus{}2\}}\frac{1}{x_j}.$
Therefore,
$ \left(\sum^n_{i\equal{}1}x_i\right) \left( \sum^n_{i\equal{}1} \frac{1}{x_i}\right) \equal{}\sum^n_{i\equal{}1} x_i \left(\frac{1}{x_{i\plus{}1}}\plus{}\frac{1}{x_{i\plus{}2}}\right) \plus{}\sum^n_{i\equal{}1}x_i\sum_{j \not\in \{i\plus{}1,\; i\plus{}2\}}\frac{1}{x_j}.$
By the Cauchy-Schwarz Inequality, we have
$ \sum^n_{i\equal{}1} x_i \left(\frac{1}{x_{i\plus{}1}}\plus{}\frac{1}{x_{i\plus{}2}}\right)  \ge 4\sum^n_{i\equal{}1}\frac{x_i}{x_{i\plus{}1}\plus{}x_{i\plus{}2}}.$
So it is enough to prove that
$ \sum^n_{i\equal{}1}x_i\sum_{j \not\in \{i\plus{}1,\; i\plus{}2\}}\frac{1}{x_j} \ge n^2\minus{}2n,$
which is clearly true according to the AM-GM Inequality.

[quote="can_hang2007"][quote="hedeng123"]Prove or disprove that  then $ x_i > 0,i \equal{} 1,2,...,n$,$ n\in N,n\geq 3$

$ \sum_{i \equal{} 1}^n x_i \sum_{i \equal{} 1}^n \frac {1}{x_i} \minus{} n^2\geq 4 \sum_{i \equal{} 1}^n \frac {x_i}{x_{i \plus{} 1} \plus{} x_{i \plus{} 2}} \minus{} 2n$

$ x_{n \plus{} 1} \equal{} x_1,x_{n \plus{} 2} \equal{} x_2$.

thank you! :lol:[/quote]
The inequality holds. We can prove it as follows:

Notice that for each $ i,$
$ x_i\sum^n_{i \equal{} 1}\frac {1}{x_i} \equal{} x_i\left(\frac {1}{x_{i \plus{} 1}} \plus{} \frac {1}{x_{i \plus{} 2}}\right) \plus{} x_i\sum_{j \not\in \{i \plus{} 1,\; i \plus{} 2\}}\frac {1}{x_j}.$
Therefore,
$ \left(\sum^n_{i \equal{} 1}x_i\right) \left( \sum^n_{i \equal{} 1} \frac {1}{x_i}\right) \equal{} \sum^n_{i \equal{} 1} x_i \left(\frac {1}{x_{i \plus{} 1}} \plus{} \frac {1}{x_{i \plus{} 2}}\right) \plus{} \sum^n_{i \equal{} 1}x_i\sum_{j \not\in \{i \plus{} 1,\; i \plus{} 2\}}\frac {1}{x_j}.$
By the Cauchy-Schwarz Inequality, we have
$ \sum^n_{i \equal{} 1} x_i \left(\frac {1}{x_{i \plus{} 1}} \plus{} \frac {1}{x_{i \plus{} 2}}\right) \ge 4\sum^n_{i \equal{} 1}\frac {x_i}{x_{i \plus{} 1} \plus{} x_{i \plus{} 2}}.$
So it is enough to prove that
$ \sum^n_{i \equal{} 1}x_i\sum_{j \not\in \{i \plus{} 1,\; i \plus{} 2\}}\frac {1}{x_j} \ge n^2 \minus{} 2n,$
which is clearly true according to the AM-GM Inequality.[/quote]

very nice...

[quote="can_hang2007"][quote="hedeng123"]Prove or disprove that  then $ x_i > 0,i \equal{} 1,2,...,n$,$ n\in N,n\geq 3$

$ \sum_{i \equal{} 1}^n x_i \sum_{i \equal{} 1}^n \frac {1}{x_i} \minus{} n^2\geq 4 \sum_{i \equal{} 1}^n \frac {x_i}{x_{i \plus{} 1} \plus{} x_{i \plus{} 2}} \minus{} 2n$

$ x_{n \plus{} 1} \equal{} x_1,x_{n \plus{} 2} \equal{} x_2$.

thank you! :lol:[/quote]
The inequality holds. We can prove it as follows:

Notice that for each $ i,$
$ x_i\sum^n_{i \equal{} 1}\frac {1}{x_i} \equal{} x_i\left(\frac {1}{x_{i \plus{} 1}} \plus{} \frac {1}{x_{i \plus{} 2}}\right) \plus{} x_i\sum_{j \not\in \{i \plus{} 1,\; i \plus{} 2\}}\frac {1}{x_j}.$
Therefore,
$ \left(\sum^n_{i \equal{} 1}x_i\right) \left( \sum^n_{i \equal{} 1} \frac {1}{x_i}\right) \equal{} \sum^n_{i \equal{} 1} x_i \left(\frac {1}{x_{i \plus{} 1}} \plus{} \frac {1}{x_{i \plus{} 2}}\right) \plus{} \sum^n_{i \equal{} 1}x_i\sum_{j \not\in \{i \plus{} 1,\; i \plus{} 2\}}\frac {1}{x_j}.$
By the Cauchy-Schwarz Inequality, we have
$ \sum^n_{i \equal{} 1} x_i \left(\frac {1}{x_{i \plus{} 1}} \plus{} \frac {1}{x_{i \plus{} 2}}\right) \ge 4\sum^n_{i \equal{} 1}\frac {x_i}{x_{i \plus{} 1} \plus{} x_{i \plus{} 2}}.$
So it is enough to prove that
$ \sum^n_{i \equal{} 1}x_i\sum_{j \not\in \{i \plus{} 1,\; i \plus{} 2\}}\frac {1}{x_j} \ge n^2 \minus{} 2n,$
which is clearly true according to the AM-GM Inequality.[/quote]

thank  you. :lol:


Να υπολογιστει το:
$ \int{\frac{dx}{2 \plus{} sinx}}$

Ηταν το μονο ερωτημα που δεν προλαβα να ολοκληρωσω λογο χρονου, η λυση που ξεκινησα ομως η οποια ειναι και η μονη που εχω στο μυαλο μου ακομα και τωρα βασιζεται σε μια αντικατασταση: $ t \equal{} \frac{dx}{2 \plus{} sinx}$ και τοτε το ολοκληρωμα μετατρεπεται ευκολα σε ολοκληρωμα αρρητης συναρτησης, επηδη ομως το αποτελεσμα που προκειπτει μου φαινεται καπως παραξενο, θα ηθελα να δω και τις δικιες σας λυσεις για να δω αν ειναι σωστη η δικια μου...

[quote="NickNafplio"]Να υπολογιστει το:
$ \int{\frac {dx}{2 \plus{} sinx}}$[/quote]

Bleeeh, δεν είναι και το πιο όμορφο που έχω δει, αλλά εξετάσεις είναι αυτές... δοκίμασες το $ t \equal{} \tan\frac {x}{2}$?

Cheerio,

Durandal 1707

οχι δεν το κοιταξα, το εθεσα ολο t και μετα κατεληξα σε αρρητη συναρτηση που υπολογιζεται ευκολα θετοντας παλι $ sinu \equal{} at \plus{} b$ (δεν θυμαμε τα α,β ποσο ηταν), αλλα με παραξενευει το γεγονος οτι στο τελος προκειπτει μια μεγαλη τριγωνομετρικη παρασταση μεσα σε $ Arcsin$

Αν δεν έκανα κάποιο λάθος το έβγαλα με το χέρι. Είναι μια κλασσική τεχνική που την περιγράφει ο  Σπανδάγος στη σελίδα 123.

Εκφράζεις το $ sin{x}$ σα συνάρτηση του $ tan{\frac{x}{2}}$ και θέτεις $ tan{\frac{x}{2}}\equal{}y$. Λίγο πιο μετά κάνεις το ίδιο και σε ένα συνημίτονο που θα προκύψει.
Μετά πέφτεις στα γνωστά. Θέλει μια συμπλήρωση τετραγώνου και ένα ψιλομετασχηματισμό αλλά στο τέλος βγαίνει. Ωστόσο συμφωνώ με τον Παναγιώτη. Πράξεις πράξεις πράξεις. Δεν ξέρω τι νόημα έχει να τα μαθαίνεις απέξω όλα αυτά. Εγώ δε θυμάμαι ούτε τις τριγωνομετρικές ταυτότητες απέξω.

Το τελικό αποτέλεσμα θα μπορούσε να είναι και (πολύ) ασχημότερο:

$ \int\frac {dx}{2 \plus{} \sin x} \equal{} \frac {2\sqrt {3}}{3} \arctan\left(\frac {1 \plus{} 2 \tan\frac {x}{2}}{\sqrt {3}}\right)$

Meeeeeeh... :|

Cheerio,

Durandal 1707

Τέτοια ολοκληρώματα συνήθως λύνονται με την αντικατάσταση $ t \equal{} \tan{x/2}$ που πρότειναν ο Παναγιώτης και ο Γιώργος. Αν γνωρίζεις μιγαδική ανάλυση, τότε λύνονται και με την αντικατάσταση $ z\equal{}e^{ix}$.


Euler
integration
logarithms
calculus
limit
calculus computations

Prove that
$ \int_{0}^{\infty}e^{-x}\ln x dx=\int_{0}^{\infty}\frac{1}{x}(e^{-x}-\frac{1}{x+1}) dx=-\gamma$

The half of this equality can be shown using the relation

$ \int_{0}^{\infty}\frac{e^{-t}-e^{-xt}}{t}\; dt \; = \; \ln x.$

Then

\begin{eqnarray*}\int_{0}^{\infty}e^{-x}\ln x \; dx & = & \int_{0}^{\infty}e^{-x}\left( \int_{0}^{\infty}\frac{e^{-t}-e^{-xt}}{t}\; dt \right) \; dx\\ & = & \int_{0}^{\infty}\int_{0}^{\infty}\frac{e^{-(x+t)}-e^{-x(t+1)}}{t}\; dt \; dx\\ & = & \int_{0}^{\infty}\int_{0}^{\infty}\frac{e^{-(x+t)}-e^{-x(t+1)}}{t}\; dx \; dt\\ & = & \int_{0}^{\infty}\frac{1}{t}\left( e^{-t}-\frac{1}{t+1}\right) \; dt.\\ \end{eqnarray*}

I was not able to prove the other half. In fact, I saw the integral equality $ \int_{0}^{\infty}e^{-x}\ln x \; dx \; = \;-\gamma$ 3 years ago, but still I can't prove this.

Here's a low-tech solution for the first part:

We have that
\[ \int_{1/t}^{t}e^{-x}\ln x \, dx =-e^{-x}\ln x \Big|_{1/t}^{t}+\int_{1/t}^{t}\frac{e^{-x}}{x}\, dx . \]
Hence,
\[ \int_{1/t}^{t}e^{-x}\ln x \, dx-\int_{1/t}^{t}\frac{e^{-x}-\frac1{x+1}}{x}\, dx = \int_{1/t}^{t}\frac1{x(x+1)}\, dx-\left( e^{-1/t}+e^{-t}\right) \ln t . \]
That equals $ \left( 1-e^{-1/t}-e^{-t}\right) \ln t = \left( 1-e^{-1/t}\right) \ln t-e^{-t}\ln t$. Both parts go to zero as $ t \to \infty$ (for the first one I would use the substitution $ u = 1/t$, and then some Big $ O$ notation).

We're not quite done, since we have to prove that the original integrals are convergent. But I reckon nobody will mind if I leave out the details.

In aid of my friend, I found a proof.

Assume that the sequence $ \gamma_{n}=-\int_{0}^{n}\left( 1-\frac{x}{n}\right)^{n}\ln x \; dx$ converges to $ I =-\int_{0}^{\infty}e^{-x}\ln x \; dx$. Simple(really?) calculation shows that

$ \gamma_{n}= \frac{n}{n+1}\left( H_{n+1}-\ln n \right)$

where $ H_{n}= \sum_{k=1}^{n}\frac{1}{k}$. Since $ \gamma_{n}$ converges to $ \gamma$, $ I =-\gamma$ and we are done. Now all we have to do is to justify that $ \gamma_{n}$ really converges to $ I$.

Actually, I was thinking something along those lines... Anyway, here's a possible justification:
Split that integral into $ \int_{0}^{1}$ and $ \int_{1}^{n}$ (just to avoid sign changing problems).
Fix some $ k$. We have that $ \int_{1}^{n}= \int_{1}^{k}+\int_{k}^{n}\geq \int_{1}^{k}$. Use the monotone convergence theorem and let $ n \to \infty$; after that, let $ k \to \infty$ to get that $ \liminf \gamma_{n}\geq \int_{1}^\infty e^{-x}\ln x \, dx$.
Also, $ \left( 1-\frac{x}{n}\right)^{n}\leq e^{-x}$, from where we get that $ \limsup \gamma_{n}\leq \int_{1}^\infty e^{-x}\ln x \, dx$.

My main problem was computing an integral of $ \left( 1-\frac{x}{n}\right)^{n}\ln x$; I still can't see how to do it cleanly, but I trust you on it.

[quote="sos440"]
I was not able to prove the other half. In fact, I saw the integral equality $ \int_{0}^{\infty}e^{-x}\ln x \; dx \; = \;-\gamma$ 3 years ago, but still I can't prove this.[/quote]

$ \int_{0}^\infty\;e^{-x}\;\ln x\;dx\;=\;\lim_{k\to\infty}\;\int_{0}^{k}\;\left( 1-\frac{x}{k}\right)^{k}\;\ln x\;dx$

$ =\;\lim_{k\to\infty}\;\left( \;\sum_{j=0}^{k}\;\left\{ \frac{(-1)^{j}\;\dbinom{k}{j}}{k^{j}}\;\int_{0}^{k}\;x^{j}\;\ln x\;dx\;\right\}\;\right)$

$ =\;\lim_{k\to\infty}\;\left\{ \;k\;\left(\;\ln k\;\sum_{j=0}^{k}\;\frac{(-1)^{j}\;\dbinom{k}{j}}{j+1}\;-\;\sum_{j=0}^{k}\;\frac{(-1)^{j}\;\dbinom{k}{j}}{(j+1)^{2}}\;\right)\;\right\}$

$ =\;\lim_{k\to\infty}\;\left\{\;\frac{k}{k+1}\;\left(\;\ln k\;-\;\sum_{j=1}^{k+1}\;\frac{1}{j}\;\right)\;\right\}\;\;\;\quad (\;\;\;\textbf{why}\;? \;\;\; )$ :D

$ =\;\boxed{-\gamma}$

[ this one is due to didilica, he (or she) was the first to
show me how to do it! :D ]


I'm living in California and considering several UC schools...
Is one year of art required for UCs??

Thanks,

[url=http://www.ucop.edu/a-gGuide/ag/welcome.html]University of California -- a-g guide[/url]

A few years ago, that was one of the differences between the CSU system and the UC system - that we required a year of visual or performing arts and the UC didn't. But now the UC has added that requirement as well, and there are essentially no differences between UC course requirements and CSU course requirements.


geometry
3D geometry
calculus
calculus computations

I need to know what the volume in litres and gallons of a plastic tube  and the cube ( or part thereof ) that is 50 metres long and 100mm in circumference ..... what would be the amount used  if it was filled with water?

This does not belong in the calculus forum.

What's the volume of any object?  The area of the base times the height.  I don't know what shape the tube has but you mentioned a cube, so I'll assume that it is a rectangular tube (which doesn't really make much sense to me, but then again, neither does this problem).  Then the area of the base is $ .1\times .1\equal{}.01$ meters squared.  Multiply by the height to get $ 50\times .01\equal{}.5$ meters cubed.  You can do the conversions between liters and gallons yourself.


I've been trying to dervate the following expression but without obtaining the requiered answer I guess. It looks like:
$f'(x)=\left(\frac{(n!)(-1)^{n}}{\left(x-\frac{1}{2}\right)^{n+1}}\right)^{'}$  :| Please don't laugh  :D

I'm a little confused about what you're asking, do you want to differentiate $\frac{n!(-1)^n}{(x-1/2)^{n+1}}$?  If so, here's what I'd do:

$\frac{d}{dx}\left(\frac{n!(-1)^n}{(x-1/2)^{n+1}}\right) = n!(-1)^n\cdot\frac{d}{dx}\left((x-1/2)^{-n-1}\right)$$= -n!(-1)^n(n+1)(x-1/2)^{-n-2}$


geometry
parallelogram
trigonometry
trig identities
Law of Cosines

Given that the medians of a triangle are 6,8, and 10 find the sides of that triangle.
Explain it step-by step and very clear. Thanks! :lol:

I will not explain it step by step.

[hide="just a hint"]
$ a^2\plus{}b^2\equal{}2m_c^2\plus{}\frac{c^2}{2}$.
Where $ m_c$ is the the c-median.

Work analogously for the other cases.

Sum it up.....[/hide]

Ok length of a median is $ m\equal{} \frac{\sqrt{2a^2\plus{}2c^2\minus{}b^2}}{2}$ . Do that for all three medians and you will get

                              $ 144 \equal{} 2a^2\plus{}2c^2\minus{}b^2$

                              $ 256 \equal{} 2a^2\plus{}2b^2\minus{}c^2$

                              $ 400 \equal{} 2b^2\plus{}2c^2\minus{}a^2$

Solving gets $ a \equal{} \frac{20}{3}$ , $ b \equal{} \frac{4\sqrt{31}}{3}$ and $ c \equal{} \frac{8\sqrt{13}}{3}$

Can you explain me why the lenght of a median is $ m\equal{}\frac{\sqrt{2a^{2}\plus{}2c^{2}\minus{}b^{2}}}{2}$???

[quote="Chalao"]Can you explain me why the lenght of a median is $ m \equal{} \frac {\sqrt {2a^{2} \plus{} 2c^{2} \minus{} b^{2}}}{2}$???[/quote]
Consider the law of cosines(for the $ \angle ADC$ and $ \angle ADB$) in the triangles $ ABD$ and $ ACD$ where $ AD$ is the median and use the fact that $ cosx \equal{} \minus{} cos(\pi \minus{} x)$.

P.s.
It is actually a special case of stewart's theorem which someone can easily prove with the same way. :wink:

[geogebra]b9d6ff00cec09110634c736845c4b482e367ba47[/geogebra]

OK start off with a triangle $ ABC$. Then make a parallelogram like shown. The diagonals of a parallelogram bisect each other so $ CD$ is a median of $ ABC$.  Let $ m$ be the length of the median and $ O$ be the intersection of the diagonals. Let $ \angle BOC \equal{} q$.

Now you use the law of cosines with supplementary angles, $ Cos(q) \equal{} \minus{} Cos(180\minus{}q)$

$ a^2 \equal{} m^2 \plus{} \frac{c^2}{4} \minus{}mcCos(q)$

$ b^2 \equal{} m^2 \plus{} \frac{c^2}{4} \minus{}mcCos(180\minus{}q)$ or

$ b^2 \equal{} m^2 \plus{} \frac{c^2}{4} \plus{} mcCos(q)$

Add those two and get $ a^2 \plus{} b^2 \equal{} 2m^2 \plus{} \frac{c^2}{2}$

Multiply by two and solve for m,

$ m\equal{}\frac{\sqrt{2a^{2}\plus{}2b^{2}\minus{}c^{2}}}{2}$

Nice solution JITTT. Where did you find that formula with its demonstration? Was it on those AoPs books?

I don't know for sure. I don't remember where I learned that.

[hide]

[geogebra]132cb3059668d10c1aeb9fdf1e528d18ce6ee4a3[/geogebra]

We have triangle ABC with medians 6,8,10. Find D such that ABCD forms a parallelogram. Let E be the midpt of BC, and F be the midpt of CD.

Then, note that

$ AG \equal{} \frac{2}{3} \times 6$,
$ GH \equal{} 2GJ \equal{} \frac{2}{3} \times 8$
$ HA \equal{} \frac{2}{3} \times 10$

Noting that AGH is therefore a right triangle, we have

$ AJ \equal{} \frac{AC}{2}\equal{}$ median of $ AGH \equal{} \frac{2}{3} \sqrt{6^2 \plus{} 4^2}$, so
$ AC \equal{} \frac{8\sqrt{13}}{3}$.

The other two sides can be found similarly.
[/hide]


logarithms
calculus
calculus computations

[quote="mdk"]For $ x > 0$, show that $ e^{x}< (1+x)^{1+x}$.[/quote]
We need prove $ x<(1+x)\ln (1+x)\forall x>0$. It is easy.

@mdk: You have a lot of problems :P

[quote="N.T.TUAN"][quote="mdk"]For $ x > 0$, show that $ e^{x}< (1+x)^{1+x}$.[/quote]
We need prove $ x<(1+x)\ln (1+x)\forall x>0$. It is easy.[/quote]

Just use $ 1-\frac{1}{x}<\ln x<x-1\ (x>0)$. :D


trigonometry
geometry
circumcircle
geometry proposed

Given triangle $ ABC$ such that $ AB \minus{} BC \equal{} \dfrac{AC}{\sqrt 2}$. Let $ M$ be the midpoint of $ AC$, and $ N$ be the base of the bisector from $ B$. Prove that
\[ \angle BMC \plus{} \angle BNC \equal{} 90^{\circ}\]
[i]Author: A.Akopjan[/i]

[quote="April"]Given triangle $ ABC$ such that $ \dfrac{AB}{BC} \equal{} \dfrac{AC}{\sqrt 2}$...[/quote]
Dimensions do not compare $ \Longrightarrow$ unit length has to be selected. Selecting the unit length appropriately, this condition can hold for any triangle - it is not a constraint. By any chance, should not the condition be $ AB \plus{} BC \equal{} \frac{AC}{\sqrt{2}}$ :?:

[color=darkblue][b]I'm sorry April[/b], in my opinion here is the correct enunciation of this easy and nice problem :[/color]

[quote="A. Akopjan"][color=darkred]Let $ ABC$ be a triangle in which denote the  midpoint $ M$ of $ [AC]$ and the base $ N$ of the bisector

from $ B$. Prove that $ m(\angle BMC) \plus{} m(\angle BNC) \equal{} 90^{\circ}\ \Longleftrightarrow b \equal{} (c \minus{} a)\cdot\sqrt 2$ .[/color][/quote]

[color=darkblue][b][u]Proof 1 (metric)[/u].[/b] $ m(\angle BMC) \plus{} m(\angle BNC) \equal{} 90^{\circ}\Longleftrightarrow m(\angle BMC) \equal{}$ $ 90^{\circ} \minus{} \left(A \plus{} \frac B2\right)$ $ \Longleftrightarrow$

$ \frac {m_b}{\sin C} \equal{}$ $ \frac {a}{\cos\left(A \plus{} \frac B2\right)}$ $ \Longleftrightarrow m_b\cdot\cos\left(A \plus{} \frac B2\right) \equal{}$ $ a\cdot\sin C$ . Multiply by $ 2\sin\frac B2\ :$

$ m_b\cdot\left(\sin C \minus{} \sin A\right) \equal{}$ $ 2a\sin C\sin\frac B2$ $ \Longleftrightarrow$ $ m_b(c \minus{} a) \equal{} 2ac\cdot\sin \frac B2$ and $ \boxed {c > a}$ $ \Longleftrightarrow$

$ (a \minus{} c)^2\left[2\left(a^2 \plus{} c^2\right) \minus{} b^2\right] \equal{}$ $ 4ac(b \minus{} a \plus{} c)(b \plus{} a \minus{} c)$ $ \Longleftrightarrow$

$ 2(a \minus{} c)^2(a^2 \plus{} c^2) \minus{} b^2(a \minus{} c)^2 \equal{}$ $ 4ac\left[b^2 \minus{} (a \minus{} c)^2\right]$ $ \Longleftrightarrow$

$ 2(a \minus{} c)^2(a \plus{} c)^2 \equal{} b^2(a \plus{} c)^2$ $ \Longleftrightarrow$ $ (c \minus{} a)\sqrt 2 \equal{} b$ .[/color]

[color=darkblue][b][u]Proof 2 (synthetic)[/u].[/b] Denote the projection $ P$ of the vertex $ B$ on the sideline $ AC$ . Observe that

$ m(\angle BMC) \plus{} m(\angle BNC) \equal{} 90^{\circ}\Longleftrightarrow m(\angle BMC) \equal{}$ $ 90^{\circ} \minus{} \left(A \plus{} \frac B2\right)$ and $ \triangle PBM\sim \triangle PNB$ ,

i.e. $ \boxed {\ MP\cdot NP\equal{}h_b^2\ }$ . Show easily that $ MP\equal{}\frac {c^2\minus{}a^2}{2b}$ and $ NP\equal{}\frac{2p(p\minus{}b)(c\minus{}a)}{b(a\plus{}c)}$ . Therefore,

$ p(p\minus{}b)(c\minus{}a)^2\equal{}b^2h_b^2$ $ \Longleftrightarrow$ $ p(p\minus{}a)(c\minus{}a)^2\equal{}4p(p\minus{}a)(p\minus{}b)(p\minus{}c)$ $ \Longleftrightarrow$

$ 4(p\minus{}a)(p\minus{}c)\equal{}(c\minus{}a)^2$ $ \Longleftrightarrow$ $ b^2\minus{}(c\minus{}a)^2\equal{}(c\minus{}a)^2$ $ \Longleftrightarrow$ $ (c\minus{}a)\sqrt 2\equal{}b$ .[/color]

Oh my God, I am really sorry for my negligence. The correct enunciation of this problem must be $ AB \minus{} BC \equal{} \dfrac{AC}{\sqrt 2}$  :blush:  :blush:  :blush:  Please sympathize for my carelessness  :blush:  Thank you, [b]yetti[/b] and [b]Virgil Nicula[/b], for pointed out my mistake.

@Virgil Nicula: I am sorry, but $ \angle BMC \plus{} \angle BNC \equal{} 90^{\circ}\iff\angle BMC \equal{} 90^{\circ} \minus{} \left(180^{\circ} \minus{} C \minus{} \frac {B}{2}\right)$ $ \iff \angle BMC \equal{} C \plus{} \frac {B}{2} \minus{} 90^{\circ}$ $ \iff m_b\color{red}\cos\color{black}\left(C\plus{}\frac{B}{2}\right)\equal{}a\cdot\sin C$

[quote="April"]...The correct enunciation of this problem must be $ AB \minus{} BC \equal{} \dfrac{AC}{\sqrt 2}$...[/quote] Fixed.
Let $ E$ be foot of B-altitude. Let external bisector of $ \angle B$ cut the circumcircle $ (O)$ again at $ K.$ $ B$ is the internal similarity center of the excircles $ (I_c), (I_a)$ $ \Longrightarrow$ $ BE \equal{} \frac {2r_cr_a}{r_c \plus{} r_a}.$ $ K$ is midpoint of $ I_cI_a$ $ \Longrightarrow$ $ KM \equal{} \frac {r_c \plus{} r_a}{2} \equal{} \frac {r_cr_a}{BE}.$ $ BNMK$ is cyclic with a circumcircle $ (P),$ on account of right angles at $ B, M.$

$ \frac {b}{\sqrt {2}} \equal{} c \minus{} a$ $ \Longrightarrow$ $ \frac {b^2}{2} \equal{} b^2 \minus{} (c \minus{} a)^2$ $ \Longleftrightarrow$ $ \frac {8[ABC]^2}{b^2} \equal{} \frac {[ABC]^2}{(p \minus{} c)(p \minus{} a)}$ $ \Longleftrightarrow$ $ 2BE^2 \equal{} 2h_b^2 \equal{} r_cr_a$ $ \Longleftrightarrow$ $ KM \equal{} 2BE$ $ \Longleftrightarrow$ $ (P)$ is tangent to $ BE$ at $ B.$ Together with $ c \minus{} a \equal{} AB \minus{} BC > 0$ $ \Longrightarrow$ $ \angle BMC \equal{} \angle EBN$ $ \Longleftrightarrow$ $ \angle BNC \plus{} \angle BMC \equal{} 90^\circ.$


Notice similar triangles.....

Thus, the area is in a $\frac{4}{25}$ ratio and the sidelengths are in a $\frac{2}{5}$ ratio.


In quadrilateral ABCD construct  point X symmetric to diagonals intersection point
     wrt  the line  m  thru midpoints of diagonals. Draw line thru X  parallel to  m.
     Show  that segments of this line (from side to diagonal) are equal.

      Cmdr.   Pestich

Note [i]T[/i] the intersection of the diagonals. Then let line [i]l[/i] through [i]X[/i] paralel to [i]MN[/i] intersect diagonals at points [i]U[/i] and [i]V[/i]. It's easy to see that [i]AT[/i]=[i]UC[/i] and [i]BT[/i]=[i]VD[/i]. Using [i]Ceva[/i] 2 times in triangle [i]TUV[/i] with lines [i]AD[/i] and [i]BC[/i] we get the desired result.
Can anyone find a beautiful solution using construction ?


abstract algebra
group theory
superior algebra
superior algebra unsolved

[b]Problem.[/b] Let $ G$ be an abelian group of order $ n$. Suppose there exist $ a,b\in G$ such that $ a\neq b$ and $ \text{ord }(a) \equal{} \text{ord }(b) \equal{} 2$. Show that $ 4|n$.

The following is my proof. But is there a simpler way, say, without using the quotient, to prove it?

[b]Proof.[/b] Since $ \text{ord }(a) \equal{} |\langle a\rangle| \equal{} 2$ and $ \langle a\rangle < G$, by Lagrange Theorem we have $ 2|n$.

Then, consider the quotient group $ G/\langle a\rangle$. Note that $ \langle a\rangle \equal{} \{e,a\}$. As $ b\neq a$ and $ \text{ord }(b) \equal{} 2$, we know $ b\notin \langle a\rangle$. In addition, we have $ \text{ord }(b\langle a\rangle) \equal{} 2$ since $ b\langle a\rangle\cdot b\langle a\rangle \equal{} e\langle a\rangle \equal{} \langle a\rangle$.

Now, since $ \langle b\langle a\rangle\rangle < G/\langle a\rangle$ and $ |G/\langle a\rangle| \equal{} |G|/|\langle a\rangle| \equal{} n/2$, again by Lagrange Theorem we obtain $ 2|(n/2)$, i.e. $ 4|n$. $ \Box$

If $ G$ is abelian then $ \left< a, b \right>$ is the Klein group.

[quote="ifai"][b]Problem.[/b] Let $ G$ be an abelian group of order $ n$. Suppose there exist $ a,b\in G$ such that $ a\neq b$ and $ \text{ord }(a) \equal{} \text{ord }(b) \equal{} 2$. Show that $ 4|n$.[/quote]

The set $ \{1, a, b, ab\}$ is a subgroup of G. Its order is four and the result follows now as an immediate consequence of Lagrange's theorem.

coquitao, you beat me to the punch! :P   I saw this problem yesterday, forgot about it, then figured out that solution while on a bus this afternoon.  I planned to post that solution but I was too late.


I'm new to FTW, and relatively new to AoPS and math contests, so please forgive my ignorance.

Why are 15, 10, or 5 second games popular at FTW?  Are there that many people who are that good?

It seems that FTW can be used to prepare for things besides MathCounts CountDown, like MC Sprint, Target, (Team?), and AMC8, in which one has at least 80 seconds per problem.  Why are games with 60 seconds or more per question and scored my responses so unpopular?

Most real life problems that are worth solving can wait for a minute.  Furthermore a solution obtained in 15 second with a significant chance of being wrong is usually worth far less than a solution obtained in a minute or more that is almost certainly correct.

As many FTW players have played thousands of games, have they seen so many questions multiple times so that they are no longer problem solving but rather practicing speed reading, memory recall, a few rote calculations, and answer entry?  The popularity of 5-15 second games leads me to suspect so.  I'm starting to see what some gifted math teachers say about how math contests can potentially lead to training that weakens creative problem solving and thinking.

I hope I'm wrong.  Please educate me.

Most people just use FTW to prepare for countdown, which stresses speed.  Additionally, in some chapters and states, cd is OFFICIAL, meaning in order to make states or nats, you HAVE TO DO WELL IN COUNTDOWN.

Most people use practice tests, Alcumus, and the forums to prepare for other types of problems.  I use the 90 second responses game occasionally, but I don't like it because it makes it really easy to lose rating.

In FTW, there are some people who have played enough games to be able to memorize some of the answers.
Most 5-second games are people competing to see who has memorized more.
Really quite pointless if you ask me.

10-15 second games practice speed.
In countdown rounds you should try to get answers as fast as you can, and this trains for that.
In sprint rounds, where question difficulty ranges, you should be able to get the easier ones quickly and have more time to spend on the tougher problems.

Longer games are better for learning how to solve the problem.
There is more time to think about the problem and find a solution.
Here is where you would get to practice something new that you just learned or refresh your memory of information learned before.

While longer games produce better solutions and an increased chance of getting correct answers, fast games train for speed and accuracy.
When in a competition, you would benefit from being able to solve questions quickly (and correctly), which is why some people play those games.

My suggestion is that if you are just starting out and don't know how to solve some types of problems, start with longer games.
It doesn't really matter if no one joins; just go ahead and play.
Get exposed to the problems and if you don't know how to solve one, ask about it on the forum.
This is the learning process.

Once you feel you know enough to try to solve the problems a little quicker, shorten the time limit.
Go for speed and accuracy.

If you get to the point where you start memorizing answers, I suggest using another tool to help you get better.
FTW will not help you anymore if all you are doing is memorizing the answers.

Actually, 5-second games are satisfying to win IF you don't memorize and you beat people who memorize.

Speed is key. The reason why 1+ minute / problem is unpopular is that people who take 5-10 seconds to solve problems will have to wait for people who may take the full minute (or even worse, be afk).

I am relatively slow; even so, I have abandoned 45 second games, as they are quite unpopular. For a compromise, I have found 25-30 second games are tolerable by many FTWers.

I think 10-15 second games are reasonable (but 10 is still pushing it) and 5 sec are just insane.
i did about 20 5 sec games and i was only able to read 50% of problems and successfully answer 7 q's...
it was sad

cuz i'm a ninja... thats right.... :ninja:

If you aren't comfortable playing them, then just ignore them.  I find 5-second games useful, because they test your computation speed.

On FTW, I find that there are two main reasons users make 5 second games:

1. They want to play a 'mem' game, where all the problems of the game must be solved my memory or VERY fast computation.

2. They are some low-rated player seeking to a higher rated player, so the low-rated player will gain points when the game ties most of the time. (Don't do this...It is very cheap)

Thanks!

5 second games, unless based on response, is normally won by one simple question and being faster by half a second.

Well, if you're like 8th in CD in an official MathCounts competition, the first few questions you see can probably be answered in 5 seconds; that may be why 5 second games are useful. I for one prefer 15 sec - 30 sec games.

5 is one extreme... Some people have response games on 2 minutes. :huh:

Personally, I never join 2 minute games... they're too long and annoying.
[quote="k00lperson"]people mem a lot too, but how is memming going to help in mc..
they're obviously not going to give the same questions in mc[/quote]
Some people mem because they are only focusing on getting their rating up, rather than using FTW as an aid to help with Mathcounts.

On the other hand, though, there are people who mem after seeing the same question multiple times. (For example, I had the same question 3 times in 5 games, so I couldn't help memming it).

Ya, i know many people mem to get their ratings up, but I mean...you mem but why do people PURPOSELY MEM to get their rating up....

I agree, it totally ruins the purpose of the game.

Please don't revive threads without an addition of new information. Also, it is unclear what you mean by your post.


Let be $ x,y,z>0$ . Solve this system :

$ \{\begin{aligned}x+\lfloor y\rfloor +\{z\}&=3,4\\
y+\lfloor z\rfloor +\{x\}&=4,5\\
z+\lfloor x\rfloor +\{y\}&=5,3\end{aligned}$

Sum up: $ 2x\plus{}2y\plus{}2z\equal{}13.2$, $ x\plus{}y\plus{}z\equal{}6.6$. Substract the first equation from this, $ \{y\}\plus{}\lfloor z\rfloor\equal{}3.2$, we get $ \{y\}\equal{}0.2,\lfloor z\rfloor\equal{}3$. Similarly, we have $ \lfloor x\rfloor \equal{} 2,\{z\}\equal{}0.1$ and $ \{x\}\equal{}0.3,\lfloor y\rfloor\equal{}1$. Therefore $ x\equal{}2.3,y\equal{}1.2,z\equal{}3.1$.


An 80  spacewalking astronaut pushes off a 640  satellite, exerting a 100  force for the 0.50  it takes him to straighten his arms.  How far apart are the astronaut and the satellite after 1.0 ?

A 1000  car pushes a 2000  truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 4500 . What is the magnitude of the force of the car on the truck?  What is the magnitude of the force of the truck on the car?

These problems mean absolutely nothing without units.

Thank you for making the exact same comment that I made. :mad:

[quote="amangupta2"]Please Mention the Units!![/quote]
You Welcome !!


More than 20 and less than 50 people went to a concert. The price of each ticket was the same whole number of dollars, and the total cost was $ \$377$. How many dollars did each individual ticket cost?

Please make your answer clear when you are posting a solution.
Yes, $ 13\cdot29\equal{}\$377$, but for someone who does not know the answer, this would be confusing. The answer is $ \boxed{\$13}$.


symmetry
algebra
system of equations
algebra unsolved

Three real numbers are given. Fractional part of the product of every two of them is $ 1\over 2$. Prove that these numbers are irrational.

[i]Proposed by A. Golovanov[/i]

Let numbers are $ a,b,c$, then $ 2ab \equal{} k_1,2ac \equal{} k_2,2bc \equal{} k_3$, were $ k_1,k_2,k_3$ are odd integers.
Therefore $ c \equal{} \sqrt\frac {k_2k_3}{2k_1},b \equal{} \sqrt {\frac {k_1k_3}{2k_2}},a \equal{} \sqrt {\frac {k_1k_2}{2k_3}}$ are irrationals.

Sorry for my stupid question,
can someone explain why is $ \sqrt{\frac{k_2 k_3}{2 k_1}}$ irrational?
Thx..

Alternately, if the three are $ a,b,c$, then $ (ab)(bc)(ca) \equal{} (abc)^2 \equal{} \frac{k}{8}$, for some $ k$ not divisible by $ 8$; then $ abc$ is irrational, since $ 8$ is not a square. Then $ a(bc)$ is irrational, while $ abc$ is rational; therefore $ a$ is irrational. Likewise $ b$ and $ c$ are irrational.

[quote="TFPNH"]Sorry for my stupid question,
can someone explain why is $ \sqrt {\frac {k_2 k_3}{2 k_1}}$ irrational?
Thx..[/quote]

Assume that is rational: $ \sqrt {\frac {k_2 k_3}{2 k_1}}\equal{}\frac{x}{y}$ then
$ 2k_1 x^2\equal{}k_2 k_3 y^2$ and remember that $ k_1,k_2,k_3$ are odd.
then if $ 2^k | LHS$ then $ k$ is odd and if $ 2^k | RHS$ then $ k$ is even. Contradiction

[quote="mszew"][quote="TFPNH"]Sorry for my stupid question,
can someone explain why is $ \sqrt {\frac {k_2 k_3}{2 k_1}}$ irrational?
Thx..[/quote]

Assume that is rational: $ \sqrt {\frac {k_2 k_3}{2 k_1}} \equal{} \frac {x}{y}$ then
$ 2k_1 x^2 \equal{} k_2 k_3 y^2$ and remember that $ k_1,k_2,k_3$ are odd.
then if $ 2^k | LHS$ then $ k$ is odd and if $ 2^k | RHS$ then $ k$ is even. Contradiction[/quote]

Or just $ \sqrt {\frac {k_2 k_3}{2 k_1}}$ is $ \sqrt2$ times something which cannot cancel $ \sqrt2$, so it's irrational.

Thanks for the explanation, I got it now  :)

Does this work? The part which I am unsure about is whether the following statement is true (I think it is).

A real number $a$ is rational if and only if the product of the denominator and numerator in any representation of $a^2$ is a perfect square.

[b]Solution[/b]

Let $a,b,c$ denote the three numbers. Then for some $x,y,z \in \mathbb{Z}$ we have that $2ab=2x+1$, $2bc=2y+1$ and $2ac=2z+1$. Solving the system of equations yields that,
\[a^2 = \frac{(2x+1)(2z+1)}{2(2y+1)}\]

If $a$ were rational, then it would follow that the product of the numerator and denominator in the above fraction would be a perfect square. However the largest power of $2$ that divides this product is $2$. Hence $a$ is irrational. Symmetry yields the same result for $b,c$.


counting
derangement
combinatorics unsolved
combinatorics

consider the traditional derangements problem.
is there any analytical reasoning* available to compute the numbers of ways in which 1 letter alone is put into the correct envelope,

*i mean without calculations.

If only one of the $ n$ letters is in the correct envelope then none of the remaining $ n \minus{} 1$ letters are in the correct envelopes. And the number of ways to put the $ n \minus{} 1$ letters in wrong envelopes is $ D_{n \minus{} 1}$. Hence choosing first the letter that gets in the correct envelope in one of $ n$ possible ways our answer should be $ nD_{n \minus{} 1}$, where
\[ D_k \equal{} \frac {k!}{0!} \minus{} \frac {k!}{1!} \plus{} \dots \plus{} ( \minus{} 1)^k\frac {k!}{k!}
\]
Were you asking for this? :maybe:


calculus
integration
search
number theory
number theory solved

show that $\frac1{15} < \frac12\cdot\frac34\cdots\frac{99}{100} < \frac1{10}$

yes. a bit of workaround, but it's correct.

But you also need to prove that it's larger than $\frac 1{15}$ ;)

Of course, liyi's method is straigt forward .. same strategy and  all you   you can see it is  > (1/sqrt(200).

Obviously it is between  sqrt(2/(pi*100)  and sqrt(2/(pi*101)..or somewhere between (1/12.5 and 1/12.6))
..(Simple  infinite product of sin x , and  put x=pi/2)

More accurate value would be somethinng like 1/12.564512 just adding few more terms in

Of course, if you have a pencil and paper, and a minute or so to spare:
12611418068195524166851562157/158456325028528675187087900672
and that is between 1/10 and 1/15.

I cant  resist a nice way to show  this Wallis formula: Hope it is helpful for some.

We define   I(n) =  (integral  0  to (pi/2)(  (sin(x))^n) dx )

One can easily show I(n) =  (n/((n-1))  I(n-2)
                                          = (n/(n-1))((n-2)/(n-3)) I(n-4)

Recursively you will see that  it produces the product like the one given above..

Last term (depending on n is even or odd) will be I(0) = pi/2 or I(1) = 1.
In fact  if we   put
K =   (1/2)(3/4). (99/100)

One can see I(99) =  1/(100 k)
 I(100) =    k*(pi/2)
And I(101) =  1/(101.k)
Obviously I(99)>I(100)> I(101)
I(99)/I(100) > 1 and I(100)/I(101) < 1
We see K^2  is between      2/(100*pi) and  2/(101*pi)
This , of course, is a neat way to prove Wallis formula.

You call this the "Wallis formula", is this well-known?? :?

And anyway eventhough you actually found a use for integrals :D, I think there are better methods...

Anyone a try? :)

How do u get the next limit...
B>1/15???
can show me???

Well I'll let everyone search a while first before I post the solution :D

[quote="Peter VDD"]You call this the "Wallis formula", is this well-known?? :?
[/quote]
yes, it is well-known if you learnt calculus.  :D

flanders MO is a high school contest, we don't have calculus in high school ;)

haku - the simple method show  by liyi:

A^2 = (1/2)(1/2) (3/4)(3/4) (5/6)(5/6) ..... (99/100)(99/100)
         >(1/2)(1/2) (2/3)(3/4) .(4/5)(5/6) ......(98/99)(99/100)
          > (1/200)
           >(1/15 )

Nothing special, the trick is   in middle term of A^2  each term of the form  (a/(a+1))(a/(a+1)   is between   ((a-1)/a))(a/(a+1) and (a/(a+1))((a+1)/(a+2))

Take either form and the  expression becomes simple. ..
(First form will give  A^2  < (1/101) and the next A^2  > (1/200)

Now a simple  extension of the problem and can be asked in middle/highschool
(Does NOT require calculus - iin fact it  is just too easy ..
)
[b]Prove the value is between 1/12 and 1/13.
[/b]
..

[quote="Gyan"]Now a simple  extension of the problem and can be asked in middle/highschool :
[b]Prove the value is between 1/12 and 1/13.[/b][/quote]

I don't think so... This question is from our national olympiad so from the top 60-70 players who were invited, not even all of them solved it! So what would a stricter version be for a school exam?? :D

On the contrary -   Once you have seen the solution (that is, if you understood the method given by Liyi - and repeated in my post - that it is between 1/10 and 1/15) the extension is laughably simple.
(A big hint!! that it is really very easy!!)

See if you can figure it out or let me know if you would like me to post this here...

P.S.

Yes, Wallis formula ((1/2)(3/4)(5/6) . ((2n-1)/2n)   = sqrt(1/(n*pi), as n tends to infinity),  is fairly well known. (I first saw it when I was in 6 or 7th grade). It can be proven even without calculus. (One can check any standard Math source) .  I have seen many bright High School students (even middle school students) who are fairly familiar  with it . (Some get fascinated by the way pi enters here).  Any way there is no harm to collect tools in your math tool box and learn something new.

Hope this helps.

[quote="Gyan"]Yes, Wallis formula ((1/2)(3/4)(5/6) . ((2n-1)/2n)   = sqrt(1/(n*pi), as n tends to infinity), ... [/quote]

and it's always between $\displaystyle \sqrt{\frac{1}{n\pi}}$ and $\displaystyle \sqrt{\frac{1}{(n+1)\pi}}$ ?? Amazing... :)

Yes, actually the other limit is little better, that is  it is between  sqrt(1/pi*n) and sqrt(1/(pi*(n+1/2)).

Now the spoiler for the  problem I asked: (Prove that  it is between 1/12 and 1/13)

[hide]Can you get  the upper (and lower) limit for the product if you  ignore the first few terms:  eg (7/8)(9/10)(11/12)(99/100)  ,  that is it is between  sqrt(7/101) and sqrt(6/100), multiply these by first few terms ( which is exactly = (1.3.5)/(2.4.6) = 5/16).  Rest is easy..  you can compare  the square terms are between 144 and 169)

[/hide]

Just read the number theory part in Engel, it's one of the problems there. Is engel printed before may 2001? Cuz if so, they just stole a problem for the national MO from engel, which would be kinda weak :?

[url=http://www.amazon.de/exec/obidos/ASIN/0387982191]Engel was printed in 1998.[/url] Actually, even the German BWM (Bundeswettbewerb Mathematik), where Arthur Engel is the head of the problem selection commitee, STILL has some problems from this book. For instance,

- problem 4 of the BWM 2003 second round coincides with a number theory problem in Engel's book (where it, however, was given without solution);
- problem 2 of the BWM 2003 second round is very similar and actually, I believe, analogous to a worked example in Engel's book.

  Darij

Yes, but this was just a problem given WITH solution, and the official solution for the problem was totally the same too (wordly), I guess they even could have copyright trouble with it if someone would be malicious...

Here is the solution of it's first part: $ \frac{1}{2}\cdot\frac{3}{4}\cdots\frac{99}{100}>\frac{1}{15}$.
LEMMA1:For any natural number $ n\geq 2$, $ \frac{n}{n\plus{}1}>\frac{n\minus{}1}{n}$. It comes from the fact, that $ \frac{n}{n\plus{}1}\minus{}\frac{n\minus{}1}{n}\equal{}\frac{1}{n(n\plus{}1)}> 0$.
Let, $ A \equal{}\frac{1}{2}\cdot\frac{3}{4}\cdots\frac{99}{100}$. Then, by LEMMA1 we get:
$ A^{2}>\frac{1}{200}$, or $ A^{2}>\frac{1}{225}$. So, $ A>\frac{1}{15}$, Q.E.D.


Show that $ \forall x,y,z>0$ we have:

$ \sum \frac{x}{ \sqrt{(x^2+y^2)(x^2+z^2)} }  \leq \frac{1}{2} \frac{x^2+y^2+z^2}{xyz} $

cheers! :D :D :cool:

The sign is in fact <=. We have (x^2+y^2)(x^2+z^2)>=4x^2yz, so we have to prove that 1/2*sum 1/sqrt(yz)<=1/2*sum x/yz, or
  sum 1/sqrt(yz)<=sum x/yz. But  x/yz+y/zx>=2/z so sum x/yz>=sum 1/x>=sum 1/sqrt(yz).


A point inside an equilateral (actually the test said "equiliteral" dunno if that is a typo or not) triangle is chosen at random.  The perpendicular distances from the point to the sides are 1.25, .80, and .95 What is the area of the triangle?

[hide]If a point is chosen at random inside an equilateral triangle, the sum of the distances of the perpendicular segments to the sides of the triangle is the height.

In this case, $ h\equal{}1.25\plus{}0.80\plus{}0.95\equal{}3$.

$ A\equal{}\frac{3\cdot2\sqrt{3}}{2}\equal{}\boxed{3\sqrt{3}}$[/hide]

where did you get the \[ 2\sqrt{3}\] from?

In an equilateral triangle, $ h \equal{} \frac {s\sqrt {3}}{2}$. This can be shown with simple triangle relations.

For this particular problem, $ 3 \equal{} \frac {s\sqrt {3}}{2}\quad\Longrightarrow\quad s \equal{} 2\sqrt {3}$.


algebra
polynomial
number theory unsolved
number theory

Let $ f(x) \equal{} a_0 \plus{} a_1 x \plus{} ... \plus{} a_n x^n$ be a polynomial with integer coefficients where $ a_n > 0$ and $ n \ge 1$.  Prove that $ f(x)$ is composite for infinitely many integers x.

I can show that cases where $ a_0 \equal{} 0$ (clearly) and when $ |a_0| > 1$, but no the case where $ a_0 \equal{} \pm 1$

Thanks!

Show that a polynomial of the form $ f(x \minus{} a)$ has constant term not equal to $ \pm 1$ (which should be obvious), and then it reduces to the previous case.


parameterization
trigonometry
calculus
limit
calculus computations

for what values of parameter $x\in\mathbb{R}$ the following sequence is convergent?

$\Big(\cos\frac{1}{n}-\cos\frac{1}{n+1}\Big)x^{n}$

Might this be a bit too difficult for Intermediate?

[hide="In any case..."] Obviously it is bounded and monotonic for $|x| \le 1$.  For the other case we will use the cosine sum-to-product formula, which produces

$\cos \frac{1}{n}-\cos \frac{1}{n+1}=-2 \sin \frac{ \frac{1}{n}+\frac{1}{n+1}}{2}\sin \frac{ \frac{1}{n}-\frac{1}{n+1}}{2}$

After some simplification, our original expression can be written

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

The $\sin$ factors together grow $O(n^{-3})$ which is clearly not enough to overcome the $x^{n}$ term when $|x| > 1$, so it diverges in this case. [/hide]

it's exactly a test question with the possible answers:

A. convergent for $x=2$
B. convergent for $x\leqslant3$
C. divergent to $+\infty$ for $x=5$
D. divergent to $-\infty$ for $x>4$

Then are you sure you wrote it correctly?  Might that last expression be instead $n^{x}$?

t0rajir0u is right- there's definitely a mistake in either the original problem or your transcription, as none of the answers listed there are correct.

you're right, there should be:

$\Big(\cos\frac{1}{n}-\cos\frac{1}{n+1}\Big)n^{x}$

sorry, my fault :oops:

[color=darkblue]Prove easily that $b_{n}\equiv n^{3}\cdot\left(\cos\frac{1}{n+1}-\cos\frac{1}{n}\right)\rightarrow 1$ and $a_{n}(x)\equiv n^{x}\cdot \left(\cos\frac{1}{n}-\cos\frac{1}{n+1}\right)=-n^{x-3}\cdot b_{n}\ ,$ where $x\in \mathbb R\ .$
In conclusion, $\lim_{n\to\infty}a_{n}(x)=-\lim_{n\to\infty}n^{x-3}$ a.s.o.[/color]


[b]Let[/b] $ A \in M_{3} \mathbb{(R)}$ [b]such that :[/b] Tr(A)=1 , [b]I is identity 3x3 matrix.[/b]
[b]Prove that : [/b]
 $ 2.det(A + 2.I) \leq \int_{1}^{3} {det(A + xI).dx} \leq det(A + I) + det(A + 3.I)$
[b]"... To take from math Olympic exam .... "[/b]

[b]You can use Jensen inequality to Prove  !!![/b]
 :roll:


[hide="last warning SM4RT, hide ur answers"]$\frac{1}{\sqrt{5}}-\frac{1}{2\sqrt{5}}=\frac{1}{2\sqrt{5}}=\frac{\sqrt{5}}{10}$[/hide]


geometry
3D geometry
LaTeX
inequalities
logarithms
function

I do not know the answer to this problem; it seems like it may diverge, but I can't prove it either way.

The square root of two is greater than the cube root of three which is greater than the fourth root of four.  If you continue this process, the billionth root of a billion is just more than one.

Find the sum of all outputs of f(x) if x starts at one and goes to infinity if f(x) is the number one less than the xth root of x.

In $\LaTeX$:

$\sum_{k=1}^{\infty} \left( \sqrt[k]{k} - 1 \right)$

Edit:  This problem seems a little difficult for intermediate.

[hide="Hint for the elementary solution"] Edit:  No, this inequality is not Bernoulli's inequality.  I mean the inequality $e^x \ge 1 + x$ for real $x$. [/hide]
[hide="Hint for the non-elementary solution"] $\sum_{k=1}^{\infty} a_k$ converges iff $\prod_{k=1}^{\infty} (1 + a_k)$ converges.  [/hide]

I am only a student that has completed half of Algebra 2, but based off what the first respondent said (about the products of 1+a[sub]k), my response is that there exists no solution (or, it is ∞).  Here is why:
Let us take the log of the product:
log(1th√1*2nd√2…).
Put that as an exponet of 10 so that the expression now = the original.
10^log(1th√1+…)
This =:
10^(1/1log1+1/2log2+1/3log3)
As the exponet is the sum of w+1/2x+1/3y+1/4z…, which is ∞ where all the variables are 1, and at a certain point (the 10th term), the variables are ≥1, the exponet must be infinity (this has already been shown in many basic books about math).  As 10^∞=∞, the answer is infinity.

This made me think, even at the beginning, about that infinite series of 1+1/2+1/3+1/4…, because this does not have a sum, even though it appears so.  I suspected that the same went for this series, and now it is known.

Nice :)  That's actually fairly impressive considering your relative lack of background.  Yes, that is the second solution I was hinting to.

[hide="Here is the first"]
$\sqrt[k]{k} = e^{ \frac{\ln k}{k} } \ge 1 + \frac{\ln k}{k}$

Hence $\sum_{k=1}^{\infty} \left( \sqrt[k]{k} - 1 \right) \ge \sum_{k=1}^{\infty} \frac{\ln k}{k}$, clearly divergent.  QED. [/hide]

Using the same method as in my previous post, the product of the n+kth root of n for n=1 to ∞ is 10^(1/(n+k)log(n)+1/(n+k+1)log(n+1)…), which is still 10^∞=∞ for all finite k.  This is because if one subtracts a finite value from an infinite one, the result is still ∞ (i.e. 1/1+1/2+1/3…=1/2+1/3…=(1/1+1/2+1/3…)-1/1).  Thus, something like (n+25th root of n)-1 or (n+68th root of n)-1 will never work.  By the way, I'm sorry that this isn't in LATEX; I have the software, but I have not yet read the documentation.

Recursive Acronym:  You don't need the software, it's part of the forum software.  All you need to do is enclose $\LaTeX$ within dollar signs, and more often than not you don't need any special tags, just regular mathematical text.

[quote="miyomiyo"]I thought that it diverged, like you said.  What about the ((n+1)th root of n)-1?  Or n+2?  Would any converge?[/quote]

Let's generalize to $\sqrt[f(k)]{k}$.

By either method the convergence depends on the convergence of $\sum_{k=1}^{\infty} \frac{ \ln k}{ f(k) }$.  We can therefore choose $f(k)$ any linear function, or any function growing slower than or up to $O(n \ln^2 n)$.

[quote="t0rajir0u"]By either method the convergence depends on the convergence of $\sum_{k=1}^{\infty} \frac{ \ln k}{ f(k) }$.[\quote]

Yes.

[quote="t0rajir0u"]We can therefore choose $f(k)$ any linear function, or any function growing slower than or up to $O(n \ln^2 n)$.[/quote][/quote]

Nope.  For example, if we have $f(n)=n \log^2n\log\log n$, then $\sum_{k\geq 3} \frac{\log k}{f(k)}=\infty$

Here is another simple proof of the original result.

It is well-known that $\left(1+\frac{1}{n}\right)^n < n$ for $n\geq 3$.  Hence $n^{\frac 1n}-1>\frac{1}{n}$ for $n\geq 3$, and this is enough to show divergence.

[quote="miyomiyo"]So what would be an f(x) that converges?  What would id converge to?[/quote]

For example, $f(k) = k^2$ will converge.  But the values to which they will converge are not going to be very neat...

blahblahblah, I had not considered $f(k) = n \log^2 n \log \log n$, what a nice counterexample to my claim :P  I think it is not easy in general to characterize $g(k)$ such that $\sum_{k=1}^{\infty} g(k)$ converges...

[quote="t0rajir0u"]I think it is not easy in general to characterize $g(k)$ such that $\sum_{k=1}^{\infty} g(k)$ converges...[/quote]

The brief answer is that there isn't a complete characterization.


[quote=hophinhan]Let $ a,b,c >0$ . Prove that :  (Pham Kim Hung)

\[ \frac{ab}{a^2\plus{}b^2\plus{}3c^2}\plus{}\frac{bc}{b^2\plus{}c^2\plus{}3a^2}\plus{}\frac{ca}{c^2\plus{}a^2\plus{}3b^2} \le \frac{3}{5}\][/quote]
See [url=https://artofproblemsolving.com/community/c6h402647p2243406]here.[/url]

And for this exist one nice generalization:

\[\sum_{cyc} \frac{k(k-3)c^2+2(k-1)ab}{a^2+b^2+kc^2}\le \frac{3(k+1)(k-2)}{k+2}\]

for any $ a,b,c>0 $ and $k>1 $


I took these examples out of the chapter i am working on in my book. To be quite honest, im lost on it.

Thx for the help in advance guys....

1.  Simplify and evaluate:
(42x3y-2z-3/x-5y8z-9)2

2.  Convert to scientific notation (here you may use x as multiplication):
0.00083

3.  For f(x) = 6x – 7, find f(2).

4.  Perform the indicated operations:
(-x2 – 6x + 3) – (-9x2 + x – 1) – (x2 + 5x – 2)

5.  Use FOIL to multiply:
(8x – 1)(2x + 5)

6.  Multiply:
(x4 – 4x2 – 5)(x2 – x + 2)

[quote="theflash0216"]1.  Simplify and evaluate:
(42x3y-2z-3/x-5y8z-9)2
[/quote]

[hide]
$ \frac{(42x3y-2x-3)}{(x-5y8z-9)}\times 2$

$ \frac{2(42x3y-2x-3)}{(x-5y8z-9)}$

$ \frac{84x6y-4x-6}{x-5y8z-9}$
[/hide]

[quote="theflash0216"]2.  Convert to scientific notation (here you may use x as multiplication):
0.00083
[/quote]

[hide]
$ 0.00083 = 8.3 \times 10^{-4}$
[/hide]

[quote="theflash0216"]3.  For f(x) = 6x – 7, find f(2).
[/quote]

[hide]
$ f(2) = 6(2)-7$

$ f(2) = 5$
[/hide]

[quote="theflash0216"]4. Perform the indicated operations:
(-x2 – 6x + 3) – (-9x2 + x – 1) – (x2 + 5x – 2)
[/quote]

[hide]
$ (-x^{2}-6x+3)-(-9x^{2}+x-1)-(x^{2}+5x-2)$

$ (-x^{2}-6x+3)+9x^{2}-x+1-x^{2}-5x+2$

$ 7x^{2}-12x+6$
[/hide]

[quote="theflash0216"]5.  Use FOIL to multiply:
(8x – 1)(2x + 5)
[/quote]

[hide]
\begin{eqnarray*}(8x-1)(2x+5)&=&16x^{2}+40x-2x-5\\ &=& 16x^{2}+38x-5 \end{eqnarray*}
[/hide]

These are all the ones I can do I have to go now but I will finish the rest later.


Let $ ABCD$ be a convex quadrilateral with $ AB\equal{}AD$ and $ BC\equal{}CD$. On the sides $ AB,BC,CD,DA$ we consider points $ K,L,L_1,K_1$  such that quadrilateral $ KLL_1K_1$ is rectangle. Then consider rectangles $ MNPQ$ inscribed in the triangle $ BLK$, where $ M\in KB,N\in BL,P,Q\in LK$  and $ M_1N_1P_1Q_1$  inscribed in triangle $ DK_1L_1$  where $ P_1$ and $ Q_1$  are situated on the $ L_1K_1$, $ M$ on the $ DK_1$ and $ N_1$ on the $ DL_1$. Let $ S,S_1,S_2,S_3$ be the areas of the $ ABCD,KLL_1K_1,MNPQ,M_1N_1P_1Q_1$ respectively. Find the maximum possible value of the expression:
$ \frac{S_1\plus{}S_2\plus{}S_3}{S}$

I have an ugly solution (and I don't know whether it is right or not ) here is my result  :blush:   : [hide]2/3[/hide]
And my solution:[VERY SORRY TO SEPARATE IT INTO FOUR PARTS, because I am not good at editing images on computer ...REALLY SORRY!! :blush:  :blush: ]

This question is another version of an old Putnam problem: https://artofproblemsolving.com/community/c7h1889190p12883634.


[quote="mr. math"]A bag contains eight yellow marbles, seven red marbles, and five black marbles. Without looking in the bag, Igor removes N marbles all at once. If he is to be sure that, no matter which choice of N marbles he removes, there are at least four marbles of one color and at least three marbles of another color left in the bag, what is the maximum possible value of N?[/quote]

[hide]Ok, to get four marbles the same colour we have to think of the worst case, not semi-worst.  :rotfl: So he'll have to pick 3 of each colour before getting the fourth of one of them. Then, there must be 3 marbles remaining of a different colour, if black was four then we have 5 y and 3 r, if red was four, then 5 y and 2 r, and if y was 4 we have 4 y still remaining anyway. I think it's done?   :D [/hide]

Igor cannot afford to have two colors end up short (with only 2 or less marbles left in the bag), and that would happen if he removed 5 red marbles and 3 black ones.
So N<8 (Igor cannot afford to remove 8 marbles).

Would it be safe to remove seven marbles?
[hide]If Igor removes 7 marbles, at most one color could end up short (less than 3 marbles of that color left in the bag), but it can be proven that there will be at least four marbles of a second color and at least three marbles of a third color left in the bag. So N=7 is the solution.[/hide]
Proving the value for N stated above:
[hide]If the color left short is black: Two or less black marbles left in the bag means that 3 or more black marbles were removed. Then, 4 or less non-black marbles must have been removed, leaving at least 4 yellow marbles and at least 3 red ones in the bag.  [b]Igor wins![/b]
If the color left short is not black: There are at least 3 black marbles left in the bag, and Igor only needs to find 4 marbles of another color left in the bag. Since 2 or less marbles of the short color (yellow or red) are left in the bag, at least 5 marbles of that color have been removed (5 red or 6 yellow), along with at most 2 marbles of the other (non-black, non-short) color. That leaves at least 5 red marbles (if yellow is the short color), or at least 6 yellow marbles (if red is the short color).  [b]Igor wins by a wider margin.[/b][/hide]

Same solution as KMST (but worded differently):

[hide]There are 20 marbles in the bag to begin with.  The highest number of marbles that can be in the bag without satisfying the required conditions (there must be at least four marbles of one color and at least three marbles of another color) is 12: 8 yellow, 2 red, and 2 black.  Therefore, Igor cannot remove 8 marbles (20-12).  The correct answer is 7.[/hide]


inequalities
triangle inequality
inequalities proposed

Prove that the inequality

$ \sqrt {a^{2} \plus{} ab \plus{} b^{2}} \plus{} \sqrt {c^{2} \plus{} cd \plus{} d^{2}} \geq \sqrt {(a \plus{} c)^{2} \plus{} (a \plus{} c)(b \plus{} d) \plus{} (b \plus{} d)^{2}}$

holds for any for any $ (a,b,c,d)\in \mathbb{R}^{4}$.

[b]Salutations.[/b]
--
P. S. Wonder whether there is a proof that doesn't require any expanding.

Triangle inequality:
$ \omega\equal{}\frac{\minus{}1\minus{}\sqrt{3}i}{2}$
$ \sqrt {a^{2} \plus{} ab \plus{} b^{2}} \equal{} |a \minus{} \omega b|$
$ \sqrt {c^{2} \plus{} cd \plus{} d^{2}} \equal{} |c \minus{} \omega d|$
$ \sqrt {(a \plus{} c)^{2} \plus{} (a \plus{} c)(b \plus{} d) \plus{} (b \plus{} d)^{2}} \equal{} |a \plus{} c \minus{} \omega(b \plus{} d)| \equal{} |a \minus{} \omega b \plus{} c \minus{} \omega d|$

Equivalently:
[hide]Can also be written as
$ \sqrt {(a \plus{} \frac {b}{2})^2 \plus{} (\frac {\sqrt {3}}{2}b)^2} \plus{} \sqrt {(c \plus{} \frac {d}{2})^2 \plus{} (\frac {\sqrt {3}}{2}d)^2} \ge \sqrt {(a \plus{} c \plus{} \frac {b \plus{} d}{2})^2 \plus{} (\frac {\sqrt {3}}{2}(b \plus{} d))^2}$
Which is just a private case of Minkowski Inequality (which, in turn, is the triangle inequality, again...)[/hide]

I like the first alternative better... How did you think of that $ \omega$, [i]man[/i]?

Regards.


[quote="shfdfzhjj"]I think it is hard.[/quote]
It's just a well-known
\[ \sum\frac{\sqrt{ab\plus{}4bc\plus{}4ac}}{a\plus{}b} \ge \frac{9}{2}
\]
for $ a,b,c \ge 0$. :)

[quote="can_hang2007"][quote="shfdfzhjj"]I think it is hard.[/quote]
It's just a well-known
\[ \sum\frac {\sqrt {ab + 4bc + 4ac}}{a + b} \ge \frac {9}{2}
\]
for $ a,b,c \ge 0$. :)[/quote]
To prove this inequality, squaring it
\[ \sum \frac{bc+4ac+4ab}{(b+c)^2} +2\sum \frac{\sqrt{(4ab+c(4a+b))(4ab+c(a+4b)}}{(a+c)(b+c)} \ge \frac{81}{4}
\]
Using the Cauchy Schwarz,
\[{ \sqrt{(4ab+c(4a+b))(4ab+c(a+4b)}} \ge 4ab+c\sqrt{(4a+b)(a+4b)}
\]
But it still have root, (note that the key for us to solve root inequality always is "damage" the root), so we will find a lemma that help us to do this and the lemma is
\[ \sqrt{(4x+y)(x+4y)} \ge \frac{2(x^2+3xy+y^2)}{x+y}
\]
:)
Using this lemma and the $ p,q,r$ techniques, we can prove the orginal inequality.

Question: How to figure out the lemma? It's interesting, isn't it? :)

[quote="can_hang2007"]

Question: How to figure out the lemma?  :)[/quote]

Great mental ability :idea:

can_hang2007 you're right it's really interesting :lol:

To prove the lemma, just notice that

$ \sqrt{(4x\plus{}y)(x\plus{}4y)}\minus{}2(x\plus{}y)\equal{} \frac{9xy}{\sqrt{(4x\plus{}y)(x\plus{}4y)}\plus{}2(x\plus{}y)}\\
\ge \frac{9xy}{\frac{(4x\plus{}y)\plus{}(x\plus{}4y)}{2}\plus{}2(x\plus{}y)}\equal{}\frac{2xy}{x\plus{}y}$

[quote="shalex"]can_hang2007 you're right it's really interesting :lol:

To prove the lemma, just notice that

$ \sqrt {(4x \plus{} y)(x \plus{} 4y)} \minus{} 2(x \plus{} y) \equal{} \frac {9xy}{\sqrt {(4x \plus{} y)(x \plus{} 4y)} \plus{} 2(x \plus{} y)} \\
\ge \frac {9xy}{\frac {(4x \plus{} y) \plus{} (x \plus{} 4y)}{2} \plus{} 2(x \plus{} y)} \equal{} \frac {2xy}{x \plus{} y}$[/quote]
Yes, you're right Shalex. That's what I want to talk about. It's really nice. :)

[quote="can_hang2007"]

......so we will find a lemma that help us to do this and the lemma is
\[ \sqrt {(4x \plus{} y)(x \plus{} 4y)} \ge \frac {2(x^2 \plus{} 3xy \plus{} y^2)}{x \plus{} y}
\]
:)
Using this lemma and the $ p,q,r$ techniques, we can prove the orginal inequality.

[/quote]

You need not p,q,r terchnique, as using your lemma and famous Iran 1996

$ \sum \frac{ab\plus{}bc\plus{}ca}{(a\plus{}b)^2} \geq \frac{9}{4}$

the inequality reduces to

$ 3\sum \frac{a}{b\plus{}c} \plus{} 8\sum \frac{ab}{(b\plus{}c)(a\plus{}c)} \plus{} 4\sum \frac{c(a^2\plus{}3ab \plus{} b^2)}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \geq 18$

which is surprisingly just Shur  :)

Very very nice prove can_hang2007.  :wink:

Congratulations!!!


quadratics
calculus
integration
algebra
partial fractions
real analysis
real analysis theorems

The first one, I can tell you from experience, gets nasty.  You first have to split it up into two irreducible quadratic polynomials.  Then use partial fractions to get two smaller integrals.  Then you split those two up again into four smaller integrals.  Those integrals then require double substitutions.  It's nasty, but the calculations are not very difficult.  Just for simplicity, I would stick with
$ \int\frac{dx}{x^{4}\plus{}1}$

Why?  As written, $ x^4 \plus{} 4 \equal{} (x^2 \minus{} 2x \plus{} 2)(x^2 \plus{} 2x \plus{} 2)$, which has roots $ \pm 1 \pm i$.  It's nice to have integer coefficients, and indeed the coefficients of the answer turn out much nicer this way (they're rational instead of involving surds).

I'm not sure what "surds" are but I guess this case might actually be easier since there are no $ \sqrt{2}$'s in the factorization.

Quick note: Surds are irrational numbers written in the form of (square) roots. In other words, it is a square root not reducable to an integer.


has anyone taken the SSAT? what's it like?

I don't know anything about it but their website ([url=http://www.ssat.org/]www.ssat.org/[/url]) seems halpful--- though you probably already saw it!

The SSAT is used by many prep schools in the admission process, at least in the Northeast anyways. There's 3 sections: Math, Reading Comprehension/verbal, and Writing. (sounds familiar?) Each section goes from 200 to 350 (I think).

I'm pretty sure that it's only used for prep schools and not colleges, but I can't confirm that. Maybe colleges do look at your SSAT scores...

I took SSATs this past Janurary to apply to Andover and Exeter (applied for 11th grade).

The first thing they do is give you an 20ish minute essay responding to some random thoughtful/philosophical topic (mine was: tell us a moment when you had failed, and tell us how you overcame your failure). The essay isn't graded. It's just sent to the schools you're applying to so the admin counselors can do what they want with it.
Hint on essay section: WRITE SMALL!!! You only got 1 page to write your essay. I ran out of room. (So I squeezed in on the last line: I have no more room to write a conclusion.)

Math section was pretty straightfoward.

Verbal section is ALL vocab. There's 30 questions on synonyms (e.g. COLD:: A.HOT, B.BURNING C.ORDER D. SUBTOTAL E. FREEZING) , and 30 analogies.

Reading was basiclaly reading comprehension - analysis, deriving the meaning of words from context, interpreting an author's logic, attitude and tone, etc.  Has  maybe 4-6 page-long passages for you to read and analyse. I thought it was the toughest section for me.

If you're taking SSATs and if you haven't already done so, then I STRONGLY recommend getting their SSAT preperation book. I didn't do a single practice test,  and I kinda ran out of time on reading section cuz i didn't know what the pace is.

Anyways, good luck on SSATs the if you take them! If you have any questions, feelf ree to ask me.

BTW, what schools are you thinking about applying?

[quote="h_s_potter2002"]I'm almost 100% sure that the SSAT is only used for prep schools.  Colleges have the SAT and ACT already.[/quote]

I'm sure that's correct. The SSAT probably plays no role in college admission whatever. (Indirectly, where you go to high school determines how credible your grades and high school activities are regarded.)

I am assuming that it is sort of like ISEE. They are both for prep school admission only. I took one when I applied for a prep school in Seattle for 5th grade.

to tekno10m:
Phillips Academy Andover
 :D
how hard is it compare to the SAT?

Nice nice, PA is an awesome school.  :D Good luck!!!

For my SSAT (Grade 10), i would say the SSAT is about 60% as difficult as the SATs. The biggest difference is that SATs are much more "advanced" than SSATs. The vocab words are bigger in SATs, the math section had much more advanced materials, and the reading comprehension sections in SATs are much harder to understand. The reading passages on SSATs are like passages from 5th grade story books. So if you can do well on SATs, I would say you will have very little trouble with SSATs, although I would still recommenend doing a couple of sample SSAT tests.

where can i get the practice tests? or do i have to buy that book?

There are third-party books of made-up practice tests for the SSAT, but my advice on test prep is always to get REAL actual previous tests from the test publisher, and not to bother with third-party imitations.

There are many 3rd party SSAT/ISEE books  from Kaplin, Princeton Review, Petersons, Arco, Barron's, etc. You can get the third party books online - just search "SSAT" at Amazon.com or barnesandnoble.com.

You can also get the SSAT's own practice book at SSAT.org for $22.

tim whats up.  it's steve, you got AIM?  my sn is garycoleman996, hit me up some time.

i think i have the peterson's and arco SSAT prep books, and they both suck pretty hard; they dont even come close to the real questions.  stick with the real book, and maybe a more well-known brand like princeton review if you think u need more practice.  if you study well for the test now (for the verbal part at least), you'll make your life a lot easier when it comes time to prepare for the SAT I.  a lot of the vocab i memorized for SSAT i found i still remembered when i studied for the SAT I, since theres a lot of overlap with the words they tend to use.

which voc books have you used so far? Thanks for your recommendation.

i brought the SSAT's own book
pretty good

i think use SAT books to study for vocabulary.  the one book that helped me the most was 'up your score' because they had helpful mnemonics.  also i tried doing http://www.frevocabulary.com   , i only got to the end of the a's, but i think that helped a good amount.  Knowing the words is important to an extent but there will always be words that you havent come across on the test, and so long as you're confident and take time to reason it through you should have a good shot at getting it right.

so how much high are the good scores in each part and/or average?

thanks!

ok i think it's the percentile that matters not necessarily the score, so a good 'score' is gonna vary from test to test.  my brother went to andover and his impression was that people didnt necessarily have good SSAT scores, so i dont think its that big of a factor in admissions, and that's something important to remember.  going by my brothers scores i think he was in like 70th percentile for reading comprehension, and other scores were ~80/90th percentile, so nothing remarkable.  other things like essays would seem to matter more.  but there's people here who go to exeter so maybe they can help you more.


Find all positive integers $ a$ and $ b$ such that

$ \frac{4}{ab} \plus{} \frac{2}{a} \plus{} \frac{1}{b} \equal{} 1$.

[hide="Solution"]
$ \frac{4}{ab}\plus{}\frac{2}{a}\plus{}\frac{1}{b}\equal{} 1$

$ \frac{4}{ab}\plus{}\frac{2}{a}\plus{}\frac{1}{b}\equal{} 1$

$ 4\plus{}2b\plus{}a\equal{}ab$

$ a\equal{}\frac{2b\plus{}4}{b\minus{}1}\equal{}\frac{2(b\plus{}2)}{b\minus{}1}\equal{}2\plus{}\frac{6}{b\minus{}1}$

Therefore, it is sufficient for $ b\minus{}1$ to divide $ 6$. The factors of $ 6$ are $ 6,1,2,3$.
The corresponding values of $ a$ and $ b$ are:
$ b\equal{}7$ and $ a\equal{}3$
$ b\equal{}2$ and $ a\equal{}8$
$ b\equal{}3$ and $ a\equal{}5$
$ b\equal{}4$ and $ a\equal{}4$[/hide]

Dont you also need the negative divisors of 6? that will give you 4 more solutions...

[quote="AndrewTom"]Find [b]all positive integers[/b] $ a$ and $ b$ such that

$ \frac {4}{ab} \plus{} \frac {2}{a} \plus{} \frac {1}{b} \equal{} 1$.[/quote]

 :)


I want to know something about notation: is $2\geq 2$ correct??? All right, I know that there is a better notation, that is $2=2$, but is $2\geq 2$ correct, although is not the best option??????

Well, in words, it means 2 is greater than OR equal to, so I belieev it is correct.

Yeah, I believe it is correct.  I once had to show that $\geq$ is not an equivalence relation, and it [b]did[/b] satisfy the symmetric property $a\geq a$.  So yes, $2\geq 2$ although you may just as well say $2=2$.

of course it is true...if you say $x\ge2$, then we consider $x=2$, which is what you are saying, but i suppose that you want to be as specific as possible, you most likely would not write that on a test, maybe if you were showing an inequality was true, but again it would be better to be specific


For what value of $ x$ does $ \frac{20}{50} \equal{} \sqrt{\frac{20}x}$?

Square both sides to get $ \frac{400}{2500} \equal{} \frac{20}{x}$. Cross-multiplying yields $ 400x \equal{} 20 \cdot 2500 \implies x \equal{} \boxed{180}$.

Umm, apparently there's a computation mistake here?

20 times 2500 is 50,000. 50,000 divided by 400 is 500 divided by 4 which is $ \boxed{125}$, not 180. :wink:

hello, we have $ \frac {4}{25} \equal{} \frac {20}{x}$ and from here we get $ x \equal{} 125$
Sonnhard.

I believe it should be

$ \frac{4}{25} \equal{} \frac{20}{x}$

maybe some typo error i guess.. :)

[quote="darthfader"]I believe it should be

$ \frac {4}{25} \equal{} \frac {20}{x}$

maybe some typo error i guess.. :)[/quote]
Yeah, that was probably what Sonnhard meant. Actually, the arithmetic would be much simpler if you cancelled out the zeros in the original fraction so that it's simplified to $ \frac{2}{5}$ before squaring the equation.


Find the sum of all solutions of the equation $ |2\minus{}x|\equal{} 3$.

The solutions are $ 5$ and $ \minus{}1$.  Summing, we get $ \boxed{4}$.

How to find these solutions:

Because we have absolute value, to solve this equation we have to write this two ways.

Way 1: $ 2 \minus{} x \equal{} 3$

Clearly, $ x \equal{} \minus{}1$.

Way 2: $ 2 \minus{} x \equal{} \minus{}3$

$ x \equal{} 5$.

Thus, we add these two numbers together to get $ \boxed{4}$.

[quote="isabella2296"]How to find these solutions:

Because we have absolute value, to solve this equation we have to write this two ways.

Way 1: $ 2 \minus{} x \equal{} 3$

Clearly, $ x \equal{} \minus{} 1$.

Way 2: $ 2 \minus{} x \equal{} \minus{} 3$

$ x \equal{} 5$.

Thus, we add these two numbers together to get $ \boxed{4}$.[/quote]

Well, that may have been written if somebody says that he/she doesn't understand how AIME15 got the solution.
Until then there's no need.
And, http://www.mathlinks.ro/viewtopic.php?t=234171 is really good, but when person who writes that post doesn't stick to that rule then... (Sorry if judging too harsh, but I don't know many children who want to help so much, so that's the reason why I'm so suspicious(sorry for bad spelling)).

Bugi, AIME15 did not elaborate how to arrive at those solutions. If the person who asked the question knew how to find solutions to an absolute value equation, I'm very sure he/she is capable of adding two numbers together.

The solution was very unclear. I clarified. That does not violate the guidelines I came up. If you wish to continue this discussion, please private message me.