With the two AIME exams coming up, there is bound to be a lot of pressure placed on the students who are hoping to make USAMO, but aren't the kids who get 14-15 easily on every practice AIME they take. Many students spend the entire year studying and preparing for these math contests, and for many high school students, the AIME is the most important test of the year. The pressure is even greater for the juniors and seniors, who are slowly aging and don't have many more high school years for math competitions ahead of them. Many are known to have horrific nightmares about performing poorly on the AIME.

So I was just wondering whether any of you have any tips for controlling emotional breakdowns that result either from too must pressure before the AIME or from the utter devastation that results from not making USAMO.

I know this may not sound like much. But relax! It's just a test. And the rest of your life doesn't hinge upon it's outcome. These competitions are meant to be fun. They aren't any indication of how good a mathematician you are, or can be, or how sucessful you will be in life. So just have fun! Do your best, but enjoy the fun of it and only compete against your own expectations.

get sick.  that's all I have to say.  or just prepare like hella and average a 12 or something.  that way you know you'll make it and there's no stress at all.

If you are actually stressed about AIME, then highlight the following: [hide]get a life![/hide]


AMC
USA(J)MO
USAMO
AIME
geometry
3D geometry
pyramid

How was the AHSME scored?  What score was needed to make USAMO?
Thanks, I just purchsed the AHSME's from 1950-1999, and wondering...

Scoring systems varied. In the later years, there were 30 problems scored at 5/2/0.

The USAMO didn't exist in the early years, and qualification has always been far more dependent on the AIME.

The other problem is that the standards twenty years ago are not the standards now.  People always get better and, as a general rule, it gets harder to qualify every year as more and more problems enter the training curriculum and problem-writers have to resort to more and more interesting and unexpected concepts.  In other words, qualifying on AHSMEs doesn't directly translate to qualifying now.

Hmph...

How much easier, do you think, are the old tests than the new ones?  For example, where would the "hardest" problem on an AHSME be placed in an AMC 12 today?

These still are good practice, right?

[quote="AIME15"]What score was needed to qualify for AIME, then?[/quote]
Well, when I took the AHSME (1969-70-71), there was no AIME and there was no USAMO. I was a 3-time school winner and it stopped with that. (And if you're wondering about the selection of the IMO team: the U.S. did not participate in the IMO, that was basically an Eastern European event.)

My scores were in the 70's and 80's, but that's with a scoring system under which negative scores were possible; they probably translate to scores in the 100-110 range under more recent systems.

Ahh...

I suppose that sending tests across the country to get graded and such wasn't very popular back then...

Do you guys think it's good practice though?

Also, what's diferent between the Contest Problem books and the CD's I got?  I got all the AHSMEs from 1950-1999 for 40 bucks.  Each contest book costs like 50 dollars...

[quote="AIME15"]I suppose that sending tests across the country to get graded and such wasn't very popular back then...[/quote]
You know, we had mail. And scantron forms. And even indoor plumbing. Yes, the AHSME forms were mailed off to be graded - to Lincoln, probably, although I don't know the full organizational history of the AMC.

The difference was the lack of a narrowing pyramid of tests, which puts a different sensibility on the whole thing. The AHSME was an end to itself and each school could be its own unit. My best bet - Steve Dunbar can correct me if I'm wrong on this - is that participation was higher than than it is now, because more schools took part.

The contest files on the AHSME disks have the scoring information in them, for example, from 1950 to 1954, the scoring is indicated in the instructions to each part.  Starting in 1956 there is a scoring table on page 2 of the contest pamphlet.

The difference between the CDs and the Contest Problem Books are:
-- The CDS contain the contests and solutions only
-- The books have history, commentary, and sometimes alternative solutions.
-- The books have indices, and tables of contests making it easier to find a particular problem or category of problem.
-- The books have been professionally typeset.
-- The books have tables of item difficulty (when available)
-- The books have all the portability and usability advantages of a book
-- The CDs have all the portability and usability advantages of a CD.

The history of the contests is documented, e.g. on our website
http://www.unl.edu/amc/e-exams/e6-amc12/amc12.shtml
as well as in this forum, the AoPS wiki and Wikipedia.
In addition, the most recent Contest Problem Book VIII and IX have thoughtful histories of the contests that explain the reasoning and philosophy behind the changes that the AMC has made over 60 years.

Steve Dunbar
MAA Director, American Mathematics Competitions

Thanks Mr Dunbar, but I was wondering what scores were needed to qualify for AIME back then?

The first AIME was in 1983, but the U.S sent their first IMO team to East Germany in 1974. How did people get from their local highschool to the IMO back then?

Didn't it go AHSME,  USAMO, IMO?

idk if there was mosp...


12. Al, Betty, and Clare split $ \$1000$ among them to be invested in different ways. Each begins with a different amount. At the end of one year they have a total of $ \$1500$. Betty and Clare have both doubled their money, whereas Al has managed to lose $ \$100$. What was Al's original portion?

$ (A) \$250 \qquad (B) \$350 \qquad (C) \$400 \qquad (D) \$450 \qquad (E) \$500$

I got $ \$400$.  Here's what I did: If we call $ x$, $ y$, and $ z$ the starting totals of Betty, Clare, and Al respectively.  Then we have $ 2x\plus{}2y\plus{}(z\minus{}100)\equal{}1500\Rightarrow 2x\plus{}2y\plus{}z\equal{}1600$.  We also know that their combined net gain is $ 600$.  In math, $ x\plus{}y\minus{}100\equal{}500\Rightarrow x\plus{}y\equal{}600$.  This is because both Betty and Clare double their initial totals and Al lost $ 100$ so I summed their three net outcomes.  Since we know $ x\plus{}y\equal{}600$, we can plug this into the initial equation and we get $ 2(600)\plus{}z\equal{}1600\Rightarrow z\equal{}400$.


geometry
symmetry
complex numbers
geometry proposed

Let $ a$, $ b$, $ c$ be three concurent lines passing through $ O$ and let $ d$ be another line meeting $ a$ at $ D$, $ b$ at $ E$ and $ c$ at $ F$. Let $ A$ and $ A'$ on $ a$ so that $ A$, $ A'$, $ D$ and $ O$ are in harmonic division. Define $ B$, $ B'$, $ C$ and $ C'$ similarly. Choose a point $ P$ on $ d$ and let $ PA'$ intersect $ BC$ at $ X$, $ PB'$ intersect $ CA$ at $ Y$ and $ PC'$ intersect $ AB$ at $ Z$. Prove that $ X$, $ Y$, $ Z$ are collinear.

Of course, we can apply a central projection which maps $ d$ onto the infinite line and then prove the property with the new figure. Indeed this can be easily done with complex numbers since $ O$ is the center of the symmetry which maps $ ABC$ onto $ A'B'C'$ and since the three lines $ PA'$, $ PB'$ and $ PC'$ are parallel.
But I'd like a direct proof using projective theorems and/or unicity of the cross ratio. Has anyone an idea ?


I'm assuming that at least some of the people here watch G4 TV.. what are your favorite shows?

reminds me of the good ole days of Tech TV.  then they merged and i didn't get G4 so i'm upset.  :(

the best shows i think are probably AOTS and X-Play..

[quote="xdl615x"]whatever you do, don't watch cheat.. The host is horrible.[/quote]

So true.  :D


The complex # $ z$ is $ 9\plus{}bi$, where $ b$ is a psotive real #. Given that the imaginary parts of $ z^2$ & $ z^3$ are equalm find $ b$.

This was 2007 AIME I Problem 3:
http://www.artofproblemsolving.com/Wiki/index.php/2007_AIME_I_Problems/Problem_3


Show that the equation $x^4 +y^4 +4z^4 =1$ has infinitely many rational solutions.

Let`s first we try to find such $ a,b,c,d\in N$, that $ a^4 \plus{} b^4 \plus{} 4c^4 \equal{} d^4$.
Let $ m,n,p\in N$ and $ a \equal{} 2^n$, $ b \equal{} 2^m \minus{} 1$, $ c \equal{} 2^p$, $ d \equal{} 2^m \plus{} 1$. Then we get:
\[ 2^{4n} \plus{} 2^{4p \plus{} 2} \equal{} (2^m \plus{} 1)^4 \minus{} (2^m \minus{} 1)^4
\]

\[ 2^{4n} \plus{} 2^{4p \plus{} 2} \equal{} 2^{m \plus{} 3}(2^{2m} \plus{} 1)
\]
Let $ p \geq n$, then:
\[ 2^{4n}(2^{4p \plus{} 2 \minus{} 4n} \plus{} 1) \equal{} 2^{m \plus{} 3}(2^{2m} \plus{} 1)
\]
Now we find such $ m \equal{} m(n)$ and $ p \equal{} p(n)$.
Let: $ 4n \equal{} m \plus{} 3$ and $ 4p \plus{} 2 \minus{} 4n \equal{} 2m$. From first we get: $ m \equal{} 4n \minus{} 3$. From second: $ p \equal{} 3n \minus{} 2$.

Finally for $ n\in N$:
\[ (a,b,c,d) \equal{} (2^n, 2^{4n \minus{} 3} \minus{} 1, 2^{3n \minus{} 2}, 2^{4n \minus{} 3} \plus{} 1)
\]
Return to our problem: Find $ x,y,z\in Q$, that $ x^4 \plus{} y^4 \plus{} 4z^4 \equal{} 1$.
If we put $ x \equal{} \frac {a}{d}, y \equal{} \frac {b}{d}, z \equal{} \frac {c}{d}$ we will get the final answer: for $ n\in N$
\[ (x,y,z) \equal{} (\frac {2^n}{2^{4n \minus{} 3} \plus{} 1}, \frac {2^{4n \minus{} 3} \minus{} 1}{2^{4n \minus{} 3} \plus{} 1}, \frac {2^{3n \minus{} 2}}{2^{4n \minus{} 3} \plus{} 1})
\]
In same way we can alsow prove, that equation $ x^4 \plus{} y^4 \plus{} 4^{2s \minus{} 1}z^4 \equal{} 1, s\in N$ has infinitely many solutions $ x,y,z\in Q$.
In this case the answer will be: For $ n\geq \frac{s \plus{} 1}{3}$
\[ (x,y,z) \equal{} (\frac {2^n}{2^{4n \minus{} 3} \plus{} 1}, \frac {2^{4n \minus{} 3} \minus{} 1}{2^{4n \minus{} 3} \plus{} 1}, \frac {2^{3n \minus{} s \minus{} 1}}{2^{4n \minus{} 3} \plus{} 1})
\]


geometry
inequalities
number theory
IMO Shortlist
graph theory
AMC

Hmm...
Does anyone know of any good books for preparing for
(a) Olympiad algebra
(b) Olympiad geometry
(c) Olympiad combinatorics (especially this one!!!)
(d) Olympiad number theory?

I have IMO Compendium and am considering getting Problem Solving Strategies and possibly Geometry Revisited, but I am not completely sure what to get. Does anyone have any suggestions?

Thanks!

My input:
Problem Solving Strategies is very good in the way that it has a lot of problems. However, most of the chapters do not have too much talking about the subject. Its mostly just problems.

Algebra: Im not really sure, but pythag011 suggests doing MOP problems(you can find some sets of homework problems on the internet)
Geometry: I learned from some chinese textbook that my grandparents sent me. Its really good, but im not sure if you would be able to get it :P.
Combo: I am not that good in this area, but i still use Intemediate Counting and Probability. Also, PSS has pretty nice combo chapters.
NT: 104 NT problems is very good. also, when you are really, PEN is also good.

I hope this helps!

Soooo many topics on the same subject...

Check out:

http://olympiads.win.tue.nl/imo/books.html

[quote="batteredbutnotdefeated"]I think ACoPS helps considerably...

This is the ultimate olympiad combo: ACoPS+Problem Solving Strategies=PWN(in that order)
the intermediates are good as well.[/quote]

@The last part: Books are helpful, but doing those alone probably isn't enough... do a lot of problems, try to do every problem in a book when you are working on it, don't give up on those problems too quickly...

For all of these, IMO Shortlist problems are helpful, but can be sort of...

Algebra: a) Inequalities: Hmm, looking through the inequality section on this website helped me a lot, Engel is also helpful for this, but the problems are not that hard. b) Other stuff: Engel is pretty good for this, after Engel do ISL probs...
Combinatorics: First of all, Engel is pretty good, then do Graph Theory by Diestel (I think I spelled the name correctly...) which is available free online, also you might want to do Enumerative Combinatorics by Richard Stanley and do Russian/ISL problems. (Russian combinatorics is pretty helpful...)
Geometry: Definitely Geometry Revisited for theory... also do all the problems from all the mop classes... I'm not very good at geometry, but doing the problems from the mop classes helped a lot...
Number Theory: 104 NT problems is pretty helpful, so is Engel and ACoPs. ISL is also pretty helpful for this (though some of the problems are not that helpful...)

[quote="limac"]Sorry to revive this old post...but where can I find the russian combinatorics problems?[/quote]
http://www.artofproblemsolving.com/Forum/resources.php?c=143


Hiya, just wondering, what are some things you do for christmas?  Especially in warmer countries without snow.

Our family always goes sledding Christmas Eve on this awesome hill half an hour away, then we come back and have hot chocolate and dinner.

Yay I love Christmas!!

We go quail or dove hunting at Granddad's house.

We always have fondu on Christmas Eve night. Beef and Chicken. Very yummy........ :lol:

I usually go to a restraunt for Christmas.  But i'm not Christian.

To Round Table...

and we don't have any special Christmas traditions. Hmm...


can anyone tell me how to figure this out (without just plugging in the values experimentally)?

If x and y are positive integers such that the greatest common factor of x^2y^2 and xy^3 is 45, then which of the following could y equal?
a. 45
b. 15
c. 9
d. 5
e. 3
Please and Thank you :)

The greatest common factor of $ x^2 y^2$ and $ xy^3$ is $ xy^2 \gcd(x, y)$; we have that this is equal to 45.

Since $ y^2 | 45$, it follows that $ y$ cannot have a factor of 5, and the exponent of 3 in $ y$ cannot be greater than 2, so $ y \equal{} 1$ or $ y \equal{} 3$. If $ y \equal{} 3$, when $ x \equal{} 5$, we get that $ xy^2 \gcd(x,y) \equal{} 45$, so e is an answer, and it is the only answer (1 isn't on the list :P.)


Find the last three digits of

$ 2008^{2007^{2006^{2005}}} + 2009^{2008^{2007^{2006}}}$

[hide="After you solve it"]Is there a way to solve it without the use of CRT, ${ a^{\varphi (n)} \equiv 1 \pmod n}$, etc?[/hide]

What would a solution without the use of the totient theorem (in some form or another) look like?  I'm not sure what you're expecting.  There's literally nothing else to do; even if you wanted to use binary exponentiation, that doesn't work on towers unless you know what mod to take the exponents - which precludes knowledge of the totient theorem.

Look at the topis:
http://www.mathlinks.ro/viewtopic.php?t=202883
There is a general method of my own idea how to "cut off" digits when dealing with problems involving form $ 9^{2k}$. So the second one will go with only elementary congruence skills with not much problem (the source I gave you contains mistakes, but they are easily corrected).
About the fist form with $ 8$ I don't have any immediate idea. But still what [b]facis[/b] presented is the easiest way to approach a problem like that. :wink:

The solution facis gave uses a nice method, it's somewhat similar to my solution.
I split the problem into finding out the value of the power tower $ \pmod{8}$ and $ \pmod{125}$, and combined that with the Chinese Remainder Theorem to find the value $ \pmod{125\cdot 8 \equal{} 1000}$. However, I suppose that's pretty much unnecessary with the Carmichael function (I didn't know about it, and was using Euler's totient).

That's how Carmichael's theorem is proven.


You are confronted with the following formula:

A * [(B + C)^(D - E) - F^(G*H) ] / J = 10

Knowing that each variable is a unique, single-digit, nonzero number, and that C - B = 1, and H - G = 3, what is the number ABCDEFGHJ, where each letter is a digit? For example, if A = 4, B = 2, and C = 7, ABC would equal 427.

Can you solve it?

Btw sorry all but apparently the (D - E) and (G*H) didn't want to go up, but (D - E) is a power to (B + C) and (G*H) is a power to -F, so I guess to your question on the post below there is a negative :/

I have a small query on the formula.

Say, B = 2, C = 3 and G = 4, H = 7 and let A = 9, D = 8, E = 6, F = 1, and J = 5.

Does this A * [(B + C)(D - E) - F(G*H) ] / J
= 9 * [(5)(2) - 1(28) ] / 5
= 9 * [10 - 28] / 5     OR    = 9 * [52 - 128] / 5

I think it is the former. Please, confirm.

Shamik

Wasn't it A * [(B + C)^(D - E) - F^(G*H)] / J = 10?

I guess there were exponents!

Yah sorry about that, no wonder u guys couldn't get it.

I  will just write the formula in $\LaTeX$ so it is easier to read:
$\frac{A\times ((B+C)^{D-E}-F^{G\times H})}{J}=10$

can u tell us what some of the numbers r?


Let $ R$ be the real numbers and $ S$ the set of real numbers excluding $ 0$ and $ 1$. Find all functions $ f : S\rightarrow R$ such that $ f(x) \plus{} \frac{1}{2x} f( \frac{1}{1\minus{}x}) \equal{} 1$ for all $ x$.

[hide="Hint"]Call $ g(x)\equal{}\frac{1}{1\minus{}x}$ then $ g(g(g(x)))\equal{}x$[/hide]


Find all odd positive integer n bigger than 1 that fullfills that for any couple of prime factors a and b of n, a+b-1 also divides n.

Thankyou

Are you sure you don't mean relatively prime, that $ (a,b)\equal{}1$ and $ a,b|n$, and not only prime factors?

According to my exam it only says prime.

   Do you have a solution with relative primes??

ty, please help


An astronaut lands on the equator of a spherical asteroid. He travels due north 100 km, without reaching the pole, then east 100 km, then south 100 km. He does not pass the same point more than once, and finds out that he is due east of his original starting point by 200 km. How many kilometers more would he now need to travel by continuing in an easterly direction in order to reach his original starting point?

Consider the cross sections parallel to the plane of the equator circle; he traveled 100 km north to a circle that had half the circumference than the equator circle. Now take a cross section of a plane perpendicular to the equator plane. You have a circle,

[code]

        ____
      /   x
    /--------
   /               arc=100km
   |--------o center
         2x[/code]

You have to determine the amount of the arc that the 100 km subtends; it is trivial that we have pi/3 radians. Hence $ d=r\theta\implies r=\frac{300}{\pi}$. The circumference of the equator circle is then $ 600$, he has traveled 200, he must go another $ 600-200=\boxed{400}$.


function
analytic geometry
integration
calculus
calculus computations

A function may be defined by

$\frac{dy}{dx}=x(1-x^{2})e^{-x^{2}}$
and passes through $(0,0)$

Find the coordinates of the turning points of this function.

A similar function is defined by $\frac{dy}{dx}=x(1-x^{2})e^{-x^{3}}$ and has a turning point at $(1,k)$ where $k\in\Re$.

Show that $k>\frac12e^{-1}$

[hide="My Solution"]
The turning points of the function occur when $x=1,-1,0$
To find $y$ coordinates we must integrate

$\int{x(1-x^{2})e^{-x^{2}}}\,dx=\int{xe^{-x^{2}}}\,dx-\int{x^{3}e^{-x^{2}}}\,dx$

$=\frac{x^{2}e^{-x^{2}}}{2}+constant$

The curve passes through the origin hence $constant=0$

Thus we have $y=\frac{x^{2}e^{-x^{2}}}{2}$

This allows us to easily show that the turning points are $(0,0), (1,\frac12e^{-1}), (-1, \frac12e^{-1})$

When considering the second function, it is clear that if its gradient is greater than that of the first function at a point in the range $x\in(0,1]$ then its turning point will occur at a greater value of $y$ than that of the first function, which was $\frac12e^{-1}$(this is because both functions have a turning point at $x=1$ and are increasing in this range).

Hence, let $x=\frac12$

First function $\frac{dy}{dx}=\frac{15}{4}e^{-0.25}$

Second function $\frac{dy}{dx}=\frac{15}{4}e^{-0.125}$

Thus, as $e^{-0.125}>e^{-0.25}$ we can deduce that $k>\frac12e^{-1}$[/hide]


This is a good bread-and-butter problem where you just need to follow through on all the steps.  [hide]Factoring the bottom, finding a common denominator, taking the difference, and cancelling gives C.[/hide]


One taker so far... thanks to Max Schindler for submitting a very tough team round (found at the bottom of this page if you are interested:  [url]http://www.agmath.com/169763.html[/url]... one solution is not given and seems brutal).    Thanks to those who IM'd me with leads to more student-created practice sets.

Here's a collection of mock tests that I've accumulated during MATHCOUNTS...(all have been posted on AoPS before so there shouldn't be permission issues if you just provide a link to this thread).

Of course, not all of them have answers, but they're still good practice...

Wow.  This is incredibly useful.  Mock MC Tests are awesome!  I downloaded all of them.  Is there a way I can create my own Mock MC Test?

Wow, thanks for sharing this! This should be useful for people (like me) who have done and redone all of the real mathcounts tests, and want more things to practice from.

Thank you very much math154... I will see if I can track down permission, but I suspect you are right that few will mind having their mock competitions shared.  I will try to post ones with solutions and create missing keys when I have time.  Some of these are great!

If you authored one of the above, please let me know if I can post it on agmath.com (or let me know if you would prefer I didn't).

Mine is under the name "Victor's Mock Team" or something like that; feel free to post it, but a lot of the questions are too bashy and in my opinion don't particularly resemble MATHCOUNTS problems...

I was about to PM you ... I was pretty sure that was who you were.  Thanks.

[quote="jbatterson"]Some of you may have visited my site, AGMath.com.  Last summer I put together some files I created for teaching in my middle school classroom in NC so that others could borrow/steal my stuff to practice math that was otherwise unavailable to them for free.

Since then, I have built the site to its now somewhat respectable form and I have reached a point where I need help continuing to grow a site that I hope supplements AoPS, especially for younger students who are not quite at the level of the typical AoPS regular.

I know many of you, and I know that you all write problems ... lots of problems.  If you have something you would like to share with the greater math community ... preferably something in Word or PDF form, I would love to post it on a new page of my site, appropriately called "Submit Content" [url]http://www.agmath.com/169763.html[/url] (I'll think of something better once I actually get some content submitted... any ideas?

If you have not seen my site already, have a look ... there is a lot there including a book I just completed on math competition preparation.

If you would like to contribute, I would love your submission!  Instructions are included on the linked page above and there is a template (relatively challenging) Warm-Up that I wrote this morning for anyone interested in solving some problems.

Edit:  I think the Word template is working now.[/quote]

Do you accept rounds made by 2 people?

^ Absolutely... any materials that would be helpful to others studying for competitions up to MC Nats and AMC10+ level.

If it looks good, I will try to post it... thanks for everything that people have contributed, I am a little overwhelmed today sifting through what can/should be posted and what may not be ready yet.

What level are these Mathcounts Competitions supposed to simulate:
school, chapter, state, or nationals?

Most of what I have received has been pretty hard ... but don't worry too much about the level of difficulty, most sets will have some problems that you can do and others that are too difficult (for most of us, myself included).  I haven't had time to look to carefully, but at first glance most appear to be state-level.

I am one of those who are still using agmath.com to choose problems :) Does anyone have an answer key (especially for the geometry section). There are tons of problem but it will take me a long time to solve them all. @jbatterson . I have found the PDFs to be immensely useful and thanks for having them in the open.


How can you tell when to count the dice rolls twice?

So 2 dice is pretty simple, but what about $n$ dice?

[hide] So $6^n$ total ways to roll the dice. We proceed with complementary counting. So $n$ choices for which die is a non-composite. Then the rest are all one's. There are 4 choices for this number. So we have $4n.$ But subtract 1 because we overcounted when the non-composite is also equal to $1.$ So the ways to get a prime is $4n-1$ and composite is $37-4n.$ So our probability is $\frac{37-4n}{6^n}.$ [/hide]

[quote=Maxima]We count the number of ways that we can roll a product that yields a number that is not composite; the utilization of complementary counting makes this much easier.

Basically, for this to happen, we have $ (1, n)$ where $ n$ is a prime number or $ 1$.

So, we can have $ (1, 1)$, $ (1, 2)$, $ (1, 3)$, $ (1, 5)$, $ (2, 1)$, $ (3, 1)$, and $ (5, 1)$ for a total of $ 7$ ways to get what we don't want.  Since there are $ 6^2 \equal{} 36$ ways in total, the desired probability is $ \frac{36\minus{}7}{36} \equal{} \boxed{\frac{29}{36}}$.[/quote]

[b]Please hide your solution  :help: [/b]

[quote=Bob.The.Builder]Wouldn't that probability be $\frac{5}{6}$ because (1,1) is neither prime nor composite?[/quote]

Well since it's NOT composite, it's still 29/36 I think.

[quote=Bob.The.Builder]Wouldn't that probability be $\frac{5}{6}$ because (1,1) is neither prime nor composite?[/quote]

[hide]it is still 29/36 by complementary[/hide]

[quote=sleepypuppy][quote=GameBot]Two fair six-sided dice are rolled.  What is the probability that the product
of the two numbers is a composite number?  Express your answer as a common fraction.[/quote]

[hide=Sketch]\frac{29}{56} for composite. Others would be different.[/hide][/quote]

no it is 29/36

[quote=sleepypuppy][quote=GameBot]Two fair six-sided dice are rolled.  What is the probability that the product
of the two numbers is a composite number?  Express your answer as a common fraction.[/quote]

[hide=Sketch]\frac{29}{56} for composite. Others would be different.[/hide][/quote]

lol. [hide=quick sol without proof, i can post that later if necessary] using complementary counting, there are $7$ ways to form a non composite number.  So the answer is $\frac{36 - 7}{36} = \frac{29}{36}$ as said above.[/hide]

[quote=IbrahimNadeem][hide="complementary counting"]
Note that the product is non-composite only if one roll is 1 and the other prime, or both of them are 1. You can choose one of the two rolls to be 1, then the other to be 1, 2, 3, or 5. Subtract 1 since we overcounted where both rolled 1.

This gives $\frac{2\cdot 4 - 1}{36} = \frac{7}{36}$ as the probability to subtract; $p = 1 - \frac{7}{36} = \boxed{\frac{29}{36}}$.[/quote]

well, you provided that proof for me, thank you XD

[quote=sleepypuppy][quote=falco_sparverius][quote=sleepypuppy]complementary counting is overkill[/quote]

tell me how[/quote]

too hard for comp[/quote]

wdym comp
if you mean competition how is it hard
it's literally a standard probability technique

[quote=Bumfuzzle][quote=programmeruser][quote=sleepypuppy]
too hard for comp[/quote]

wdym comp
if you mean competition how is it hard
it's literally a standard probability technique[/quote]

I think sleepypuppy was referring to comprehend.[/quote]

Yes I was. :D

[quote=sleepypuppy][quote=falco_sparverius][quote=sleepypuppy]complementary counting is overkill[/quote]

tell me how[/quote]

too hard for comp[/quote]

if u think it's too hard to comprehend, u just need to relearn it then

[quote=sleepypuppy][quote=falco_sparverius][quote=sleepypuppy]complementary counting is overkill[/quote]

tell me how[/quote]

too hard for comp[/quote]

only for you

[quote=peelybonehead]
only for you[/quote]

Ah come on man, let's be nice lol.

That statement is also false, [hide=Because][size=200]It'S HARD FOR ME TOO![/size][/hide]

[hide=sol]It's honestly easier if we just count the cases in which we can generate prime numbers as products or 1. Assuming the two die are distinguishable, that gives $7$ different combinations (1,1 1,2 1,3 1,5 2,1 3,1 5,1) out of $6^2$ or 36 combinations, and subtracting it from 1, we have our answer as $\boxed{\dfrac{29}{36}}$[/hide]

this is my 500th (counted posts) post on AoPS

[hide=Sol with Cases]I did it the longer way of cases.
Case 1: First is not 1 and the second is not 1. Likelihood is $\frac{5}{6}\cdot\frac56 \rightarrow \frac{25}{36}$
Case 2: First die is 1, and second die is 4 or 6. We get $\frac16\cdot\frac13 \rightarrow \frac{1}{18}.$
Case 3: First die is 4 or 6, and second die is 1. We get $\frac16\cdot\frac13 \rightarrow \frac{1}{18}.$
Now adding them all we get $$\frac{1}{18}+\frac{1}{18}+\frac{25}{36} \rightarrow \boxed{\frac{29}{36}}$$[/hide]


[hide]$ S\equal{}\sum_{n\equal{}1}^9\sin^2\frac{n\pi}{18}$

$ \Rightarrow9\minus{}2S\equal{}\sum_{n\equal{}1}^91\minus{}2\sin^2\frac{n\pi}{18}$

$ \Rightarrow9\minus{}2S\equal{}\sum_{n\equal{}1}^9\cos\frac{2n\pi}{18}$

$ \Rightarrow9\minus{}2S\equal{}\sum_{n\equal{}1}^9\cos\frac{n\pi}{9}$

Since $ \cos\frac{n\pi}{9}\plus{}\cos\frac{(9\minus{}n)\pi}{9}\equal{}0$ we have

$ 9\minus{}2S\equal{}\cos\frac{9\pi}{9}\equal{}\minus{}1$

$ \Rightarrow S\equal{}5$
[/hide]

[quote="cosinator"][hide]$ S \equal{} \sum_{n \equal{} 1}^9\sin^2\frac {n\pi}{18}$

$ \Rightarrow9 \minus{} 2S \equal{} \sum_{n \equal{} 1}^91 \minus{} 2\sin^2\frac {n\pi}{18}$

$ \Rightarrow9 \minus{} 2S \equal{} \sum_{n \equal{} 1}^9\cos\frac {2n\pi}{18}$

$ \Rightarrow9 \minus{} 2S \equal{} \sum_{n \equal{} 1}^9\cos\frac {n\pi}{9}$

Since $ \cos\frac {n\pi}{9} \plus{} \cos\frac {(9 \minus{} n)\pi}{9} \equal{} 0$ we have

$ 9 \minus{} 2S \equal{} \cos\frac {9\pi}{9} \equal{} \minus{} 1$

$ \Rightarrow S \equal{} 5$
[/hide][/quote]m feeling stupid, but how do you get $ \cos\frac {n\pi}{9} \plus{} \cos\frac {(9 \minus{} n)\pi}{9} \equal{} 0$

$ \cos\frac {(9 - n)\pi}{9} = \cos(\pi - \frac {n\pi}{9}) = \cos\pi\cos\frac {n\pi}{9} + \sin\pi\sin\frac {\n\pi}{9} = - \cos\frac {n\pi}{9}$

Alternately one can multiply both sides of $ 9 \minus{} 2S \equal{} \sum_{n \equal{} 1}^9\cos\frac {n\pi}{9}$ by $ \sin\frac{\pi}9$ and then convert the product of sine and cosine to the sum of sine and cosine to obtain a telescoping series.


abstract algebra
group theory
analytic geometry
vector
superior algebra
superior algebra unsolved

Classify the quotient group (Z_4 cross Z_4 cross Z_8)/<(1,2,4)> according to the fundamental theorem of finitely generated abelian groups.

Yes:
Try to show that $ \mod (1,2,4)$, every element of $ \mathbb Z_4 \times \mathbb Z_4 \times \mathbb Z_8$ corresponds to exactly one element of type $ (0,x,y)$. Thus $ (\mathbb Z_4 \times \mathbb Z_4 \times \mathbb Z_8)/(1,2,4)$ corresponds to the subgroup of elements of that type, being $ \mathbb Z_4 \times \mathbb Z_8$.

So, you are saying apply the homomorphism phi: Z_4 cross Z_4 cross Z_8 -> Z_4 cross Z_4 cross Z_8 mod (1,2,4). It seems like the image of that homomorphism should be Z_2 cross Z_4. I know thats wrong, but I can't figure out why.

I still don't see why the following reasoning fails:

Define: phi: Z_4 cross Z_4 cross Z_8 -> ? by phi(x,y,z) = (x,y,z) mod (1,2,4)

Since ker(phi) is clearly <(1,2,4)>, we just need to figure out what the question mark is.  Clearly the first Z_4 does no matter since the first coordinate of the image is always 0, so we ignore it. So, we really just need to find Z_4 cross Z_8 mod (2,4). But that is clearly Z_2 cross Z_4! But that doesn't make sense because the group has to have order 32...

Grr. This is really annoying me.

You are reasoning as if the modulo group was the product of 3 cyclic groups, one in each component of the numerator. This is not true. This vector $ (1,2,4)$ does not lie parallell to an axis so it is a completely different situation.

This statement:
" Clearly the first Z_4 does no matter since the first coordinate of the image is always 0, so we ignore it."
also makes no sense to me

I am using (x,y,z) mod (1,2,4) to mean (x mod 1, y mod 2, z mod 4), what does it mean to you?

EDIT: wait, if I use that definition then (1,0,4) is also in the kernel of phi, so that's definitely not what it should mean...

I said that the first Z_4 does not matter because everything mod 1 is 0.

Kalle, I want to find I way to do this problem that will be applicable to other similar problems as well

EDIT: I might as well list the possibilities:

(Z_2)^5
Z_32
(Z_2)^3 cross Z_4
Z_16 cross Z_2
Z_8 cross Z_2 cross Z_2
Z_2 cross Z_4 cross Z_4
Z_8 cross Z_4

[quote="FibonacciFan"]
Kalle, I want to find I way to do this problem that will be applicable to other similar problems as well
[/quote]
It's not that easy. Here is a thread for a similar problem in the general case:
http://www.mathlinks.ro/Forum/viewtopic.php?t=112335

Finding a general method to solve these problems mechanically will require competence. I would reccomend not trying to do that, and instead focusing on trying to solve these particular case. ZetaX "saw" an isomorphism, geometrically or otherwise, while I just played around with some crap. It's nicer to see the isomorphism, but all else fail you will probably not get too large groups to just try some stuff

OK. Fair enough.

You said "In the quotient group, all elements have order less than 8 obviously."

Why is that obvious?

I listed two groups that have elements of order greater than 8.

In the "numerator-group" every element has order $ \leq 8$. This carries over to the quotient group

[quote="Kalle"]With no intuition or understanding:
It is an abelian group of order $ 2^5$. The theorem in question gives all possible such groups. In the quotient group, all elements have order less than $ 8$ obviously. But $ (0,0,1) \plus{} \left < \left(1,2,4\right)\right >$ has order 8. So the group is either $ \mathbb{Z}_4 \times \mathbb{Z}_8$ or $ \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_8$. But by writing down the elements in $ \left < \left(1,2,4\right)\right >$ you can see that there 3 elements of order 2 so the group is $ \mathbb{Z}_4 \times \mathbb{Z}_8$.

edit: ok fixed it now[/quote]

I think that there are 3 elements of order 2 in either group.

EDIT: never mind


analytic geometry
geometry
similar triangles
power of a point

Alright, I guess I have to keep continuing my daily routine  :D . Maybe I will change the topic later.

In semicircle diamater AB of radius 3 and center C, the perpendicular to AB through C meets the semicircle at D. Let E and F be the trisection points of DC. Let the perpendiculars to DC though E and F meet arc DB at G and H and arc AD at I and J, respectively. Let GH and IJ meet at K. Find KD.

[hide="coordinate system sol"]

wlog the circle is centered at $ (0,0)$ and we have $ x^2 \plus{} y^2 \equal{} 9$

For $ y \equal{} 1, y \equal{} 2$ we have the coordinates $ (2\sqrt {2},1)$ and $ (\sqrt {5},2)$

The intersection of the two lines is the y-axis intercept of the line passing through these points. $ KD$ is the $ y$ intercept minus three, hence;

$ \left(2 \minus{} \sqrt {5}\cdot \frac {1}{2\sqrt {2} \minus{} 5}\right) \minus{} 3 \equal{} \frac {2(\sqrt {10} \plus{} 1)}{3}$[/hide]

[hide="similar triangles"]
basicly the same as above but we use triangles to show that

$ \frac {2\sqrt {2}}{KD \plus{} 2} \equal{} \frac {\sqrt {5}}{KD \plus{} 1} \Rightarrow KD \equal{} \frac {2(\sqrt {10} \plus{} 1)}{3}$[/hide]

[hide="Other Way"]Use Power of a Point to find $ FH\equal{}\sqrt{5}, EG\equal{}\sqrt{8}$. Then letting $ DF\equal{}h$, we have $ \frac{h}{\sqrt{5}}\equal{}\frac{h\plus{}1}{\sqrt{8}} \iff h\equal{}\frac{\sqrt{5}}{\sqrt{8}\minus{}\sqrt{5}}\equal{}\frac{\sqrt{5}(\sqrt{8}\plus{}\sqrt{5})}{3}\equal{}\frac{2\sqrt{10}\plus{}5}{3}$. Then our answer is this subtracted by $ 1$, or $ \frac{2\sqrt{10}\plus{}2}{3}$.[/hide]

That's right, I used power of a point. :)


Find all triples of functions $f,g,h$ such that the identity $f(h(g(x))+y)+g(z+f(y))=h(y)+g(y+f(z))+x$ holds for every reals $x,y,z.$

See all the problems from Vol. 12(2006), #1-3 at [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=465391#p465391]Ukrainian journal "In the World of Mathematics" problems contest[/url]

[quote="rogue"]Find all triples of functions $f,g,h$ such that the identity $f(h(g(x))+y)+g(z+f(y))=h(y)+g(y+f(z))+x$ holds for every reals $x,y,z.$

See all the problems from Vol. 12(2006), #1-3 at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=465391#p465391]Ukrainian journal "In the World of Mathematics" problems contest[/url][/quote]


geometry
perimeter
rectangle
geometry proposed

Let $P$ be a point in the plane of triangle ABC such that the triangles $PAB$, $PBC$ and$PAC$ have the sane perimeter and the same area,
(1)Show that the triangle $ABC$ is equilateral if and only if $P$ lies inside the triange $ABC$.
(2)Show that the quadrilateral $ABCP$ is a rectangle if $P$ lies outside the triangle $ABC$. Does the converse necessarily hold true?

Posted before as [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=55650]Beautiful geometry problem[/url].


If M is a point on the circle and AB is its diameter than prove :

1)$ MA \plus{} MB\le 2 \sqrt {2} r$

2)$ MA*MB\le 2r^{2}$

Is $ r$ radius? If it is:
[hide="Solution"]
If we take that $ \angle MAB\equal{}\alpha$, then:
$ MA\plus{}MB\equal{}2r(sin\alpha\plus{}cos\alpha)$
$ MA\cdot MB\equal{}4r^2(sin\alpha\cdot cos\alpha)$
From $ (sin\alpha\minus{}cos\alpha)^2\ge 0$ it follows that $ sin\alpha\plus{}cos\alpha\le \sqrt{2(sin^2\alpha\plus{}cos^2\alpha)}\equal{}\sqrt{2}$
and $ sin\alpha\cdot cos\alpha\le \frac{1}{2}$.
I didn't explain every step, you should know that. [/hide]


G is the centroid of triangle ABC.  E is on AB and F is on AC.  If E, G and F are colinear, prove that EG   <= 2GF.

There's a simple proof with vectors.  Let $\bf{a}$ and $\bf{b}$ be position vectors, and let $\bf{b}$ be parallel to the $x$-axis.  Let $\bf{e}=$ $m\bf{a}$, and it's well-known that $\bf{OG}=\frac {a+b}{3}$.  From here, we can solve for $\bf{EG}$.  We know that $\bf{GF}$ is parallel to $\bf{EG}$, and using the fact that $\bf{OG}+\bf{EG}$ lies on the $x$-axis, we get that:

$f(m)=\frac {\vert\bf{EG}\vert}{\vert\bf{GF}\vert}=\frac {1}{3m-1}$.  Since $m\in [\frac {1}{2},1]$, it is clear that $\frac {1}{2}\leq f(m)\leq 2$.  This also proves that the centroid divides medians in a $2:1$ ratio.


A few recent developments (taken from Mathworld News):

April 12, 2004 - Arbitrarily Long Progressions of Primes
A new preprint by Green and Tao appears to establish the existence of prime arithmetic progressions of any length k, a result that has been conjectured for three quarters of a century but until now has defied proof.
http://mathworld.wolfram.com/news/2004-04-12/primeprogressions/

June 1, 2004 - 41st Mersenne Prime Announced
Josh Findley, a participant in the Great Internet Mersenne Prime Search, identified the 41st known Mersenne Prime on the morning of May 15. The discovery was confirmed on May 29, making 2^24036583 - 1 the largest known Mersenne prime, as well as the large prime number known.
http://mathworld.wolfram.com/news/2004-06-01/mersenne/

June 9, 2004 - Twin Prime Proof Proffered
A May 26 preprint by Vanderbilt University mathematician R. F. Arenstorf appears to come close to settling the longstanding question of the infinitude of twin primes. While a hole has recently been found in the proof, mathematicians remain hopeful that the proof can be fixed up as needed.
http://mathworld.wolfram.com/news/2004-06-09/twinprimes/

Riemann Hypothesis "Proof" Much Ado About Nothing
A June 8 Purdue University [url=http://news.uns.purdue.edu/UNS/html4ever/2004/040608.DeBranges.Riemann.html]news release[/url] reports a proof of the Riemann Hypothesis by L. de Branges. However, both the 23-page preprint cited in the release (which is actually from 2003) and a longer preprint from 2004 on de Branges's home page seem to lack an actual proof. Furthermore, a counterexample to de Branges's approach due to [url=http://arxiv.org/abs/math.NT/9812166]Conrey and Li[/url] has been known since 1998. The media coverage therefore appears to be much ado about nothing.

A few of my own comments:

It should be noted that [url=http://www.math.ucla.edu/~tao/]Terence Tao[/url] (involved in the proof that there are arbitrarily long arithmetic progressions of primes) was the youngest person to ever earn a Gold Medal at the IMO. He did so in 1988 at the age of (just barely) 13. Previous to that, he had earned a Bronze Medal in 1986 and a Silver Medal in 1987. He went on to get his PhD from Princeton University in 1996 at the age of 20. He is now a professor at UCLA.

The Mersenne Prime that was just found has 7,235,733 digits. There's a $100,000 [url=http://www.mersenne.org/prize.htm]prize[/url] to the first person who finds a prime number with 10 million or more digits.

L. de Branges has made claims before of proving the Riemann Hypothesis that have not panned out - so a lot of people are skeptical of any claims that he makes to proving it now. However, he did a similar thing with the long outstanding [url=http://mathworld.wolfram.com/BieberbachConjecture.html]Bieberbach Conjecture[/url], and he turned out to be right! So his claims are at least always given scrutiny. The fact that one of his former graduate students (Xian-Jin Li) has provided a counterexample to his approach though, is pretty bad news.


For $ x,y,z,a,b,c$ positive reals and $ k$ positive integer prove

$ \sum \sqrt[k\plus{}1]{\frac{2x}{x\plus{}y}a^kb} \leq \sum a$

It is here:
http://ddbdt.tk/forum/ca-c-cua-c-thi-trasn-vimf/136-ae-thi-cha-nh-tha-c-cua-c-thi-vicfj.html

[quote="manlio"]For $ x,y,z,a,b,c$ positive reals and $ k$ positive integer prove

$ \sum \sqrt [k \plus{} 1]{\frac {2x}{x \plus{} y}a^kb} \leq \sum a$

It is here:
http://ddbdt.tk/forum/ca-c-cua-c-thi-trasn-vimf/136-ae-thi-cha-nh-tha-c-cua-c-thi-vicfj.html[/quote]
Using the AM-GM Inequality, we have
$ 2\sqrt{\frac{2x}{x\plus{}y}ab}\plus{}(k\minus{}1)a \ge (k\plus{}1) \sqrt [k \plus{} 1]{\frac {2x}{x \plus{} y}a^kb} .$
It follows that
$ 2\sum \sqrt{\frac{2x}{x\plus{}y}ab} \plus{}(k\minus{}1)\sum a \ge (k\plus{}1) \sum  \sqrt [k \plus{} 1]{\frac {2x}{x \plus{} y}a^kb} .$
Thus, it is enough to prove that
$ \sqrt{\frac{2x}{x\plus{}y}ab}\plus{}\sqrt{\frac{2y}{y\plus{}z}bc}\plus{}\sqrt{\frac{2z}{z\plus{}x}ca} \le a\plus{}b\plus{}c.$
This is a known inequality.

Nice proof, my dear friend. :)

Sorry, but I don't remember how to prove

$ \sqrt {\frac {2x}{x \plus{} y}ab} \plus{} \sqrt {\frac {2y}{y \plus{} z}bc} \plus{} \sqrt {\frac {2z}{z \plus{} x}ca} \le a \plus{} b \plus{} c.$


How many zeros appear at the end of the whole-number representation of $\displaystyle \frac{20!}{10^4}$?

MCrawford wrote:How many zeros appear at the end of the whole-number representation of ?

[hide]

None.

Due to the factors {5, 10, 15, 20}, 20! ends in exactly four zeros.

Dividing by 10,000 will eliminate them.[/hide]

How many zeros appear at the end of the whole-number representation of 20!/10^4?

[hide]20! will end with 4 zeros because of the multiples of 5, but 10^4 cancles them out.

The anser is non![/hide]

[hide]we first have to find how many zeroes 20! ends with and then subtract four from the number of zeroes.
So:
In order to determine the number of zeroes n! ends in, we see how many multiple os 5, 25, 125, etc. are in n. so applying this fact we get there will be 4 zeroes at the end of 20!. Now we subtrct our four to get: 4-4 = [b]0[/b][/hide]

[hide]There should be no zeros at the end of this.  Any set of numbers that can make it, 10,20,5,15,two even numbers that pairs with 5 and 15 all get cancelled when divided by 10^4.[/hide]


Gauss MATHCOUNTS blogs ARML AMC USA(J)MO USAMO

:(  :(  :(

Gaussy and ThinkFlow are sent on a mission to murder a massacre of bunnies that are invading guineapiggies's house. Gaussy can murder 20 bunnies per hour, and ThinkFlow can murder 15 bunnies per hour. If an initial horde of 1337 bunnies invades guineapiggies's house, how long would it take for Gaussy and ThinkFlow to murder the entire horde of bunnies if the following things happen:

After 5 hours, the bunny king shoots and kills Gaussy, therefore leaving ThinkFlow to confront the bunnies alone, until after 15 more hours, pythag011 uses his magical IMO powers to revive Gaussy, therefore letting her continue in the bunny rampage.

huh? lol I didn't go anywhere.

In the first 5 hours they killed 35 * 5 = 175 bunnies.  In the next 15 hours thinkflow murders 15 * 15 = 225 (yay a square number! :D).

So they have 1337 - (400) = 937 bunnies left to exterminate.

937/35 = 26.77

20 + 26.77 = [b]46.77 hours[/b].

P.S.  I recruited someone from IL to start posting on the IL forum.  :ninja:

Yayness! lol

Mieh, next stuff: ILLINOIS TRIVIA. lol.
See how many you can answer! (no cheating pplz!)

1. Which Illinois school had the most state championship victories?

[hide]Kennedy Junior High School[/hide]

2. How many years has gauss1181 been to state mathcounts?

[hide]2 years.[/hide]

3. How many years has suma_milli done mathcounts?

[hide]1 year.[/hide]

4. Which Illinois user's blog has the most stories about school life and life in general/love rather than math?

[hide]It's a tie between Stargirl240 and guineapiggies.[/hide]

5. Of the four Mathletes on this year's Nats team for Illinois, who is the tallest?

[hide]David Wang.[/hide]

6. Which two users in the Illinois forum seem to have such a humorous, silly friendship?

[hide]suma_milli and gauss1181.[/hide]

7. Which user particularly stated he wore a striped polo shirt at this year's state contest and would be eying Ravi?

[hide]helagha.[/hide]

8. Which user has the longest username?

[hide]ManHatTaNaTiVe94[/hide]

9. Including gauss1181, how many users in Illinois are from gauss1181's chapter?

[hide]Three.[/hide]

10. Who is the best Illinois middle schooler at countdown?

[hide]jgjgjmath.[/hide]

11. Which two Illinois middle schoolers originally intended to go to Chicago ARML tryouts but were turned down by their parents?

[hide]suma_milli and gauss1181.[/hide]

12. Which Illinois user  :rotfl: ed the most during state mathcounts?

[hide]suma_milli.[/hide]

13. Which Illinois user, in particular, made USAMO in eighth grade and Red MOP in ninth grade?

[hide]Aryth.[/hide]

14. (don't be offended by this) Two Illinois users did not make the top 25% of individuals at this year's state MATHCOUNTS. Who were they?

[hide]suma_milli and RussianRocket.[/hide]

15. Which Illinois user has the most views and entries on his/her blog?

[hide]gauss1181.[/hide]

OH AND THE HARDEST QUESTIONS OF ALL TIME:

16. Why do so many people laugh at the pictures of gauss1181 at states?

[hide]Because she was wearing green and white which are the exact two colors of  :D . Also, she looked like a boy wearing a female wig. :rotfl: [/hide]

17. What's so funny about suma_milli?

[hide]She laughs a lot, and is a really nice and entertaining person. Also, she likes playing antics with her teammates on the elevator at states, and sometimes frightens gauss1181 when she scares her by creeping behind her and tapping her on the shoulder. :rotfl: [/hide]

18. Which Illinois user is never annoyed by excessive smiley spam?

[hide]Stargirl240.[/hide]

Ok my guesses before I click on them:

1> KJHS
2> 2
3> 1
4> guineapiggies
5> Patrick?
6> suma_milli & gauss1181
7> helagha
8> guineapiggies
9> 2??!?
10> Ravi Jagadeesan
11> suma_milli & gauss1181
12> gauss1181
13> Aryth
14> suma_milli & ?
15> gauss1181
16> ...... ok seriously..... the answer to this question is : a certain someone looks like they are wearing a gaussy costume
17> ???
18> suma_milli??

Edit: I did pretty well lol.  I think only we would actually know this stuff.  :|

LOL I WOULD REALLY BE OFFENDED IF SOMEONE WENT TO NATS WEARING A DISGUISE OF ME........................................................................!?!?!?!!!!

EDIT: gaussy costume= :rotfl:  :rotfl:  :rotfl:

LOL.   :rotfl:  :rotfl:  :rotfl:

Oh I'm supposed to be doing my science project that's due tommorow right now..

*sigh* We have nothing to talk about on the IL forum now,  since we didn't go to ARML tryouts.

You and cheesymath could talk about nats though :D

By posting, I am currently adding to the boredness of this forum.  :maybe:

lol go stargirl and me!!! randomest blogs about life!!!! lol i think gaussy is trying to mathify my blog though xD


April 1, 2002 is a Monday. What day is April 23, 2002?

If April 1st is a Monday, so is April 8th, 15th, and 22nd,  Because each adds 7 days, or 1 week.  Thus, April 23rd is a tuesday.

There are 22 days from April 1st to 23rd.
22/7=3 R1
You count one day from monday, giving us $ \boxed{\text{Tuesday}}$


Let $a,b,c$ be positve real numbers such that $a+b+c=1$, prove or disprove that \[\frac{ab}{\sqrt{\frac1c-1}}+\frac{ca}{\sqrt{\frac1b-1}}+\frac{bc}{\sqrt{\frac1a-1}}\leq\frac{\sqrt2}6.\]

The fact is that it's very easy. By C-S ineq \[(\sum\frac{ab\sqrt{c}}{\sqrt{1-c}})^{2}\le (\sum\frac{ab}{a+b})(3abc).\] The rest it's not problem. I think if you think a little instead of only "creating and leaving, non-thinking", this problem is not hard to you, PVThuan.

I can't understand why:
$(\sum\frac{ab}{a+b})(3abc)\leq \frac{\sqrt2}6?$

It's certainly, \[\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le \frac{1}{2}(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\le \frac{1}{2}\sqrt{3(ab+bc+ca)}\] and moreover, \[ab+bc+ca\le 1/3, abc\le 1/27\]


What do you guyz know about the UCSD Math tourney...?

Yikes.. this is a long-shot...

Thanks,
IW.OAV

If you're talking about the High School Honors Mathematics Contest, it is held in late April/early May.  Part I is 25 multiple choice questions, while Part II consists of 4 harder questions (explanations/proofs required).  Each is worth a maximum of 100 points and is 90 minutes long.  The highest Part I scores on each team (top 5?) counts for the team score, while Part I and Part II are both factored into the individual rankings.

-Timothy


I am asking because I was told by the editors that before sending the paper, it should not contain macros.

What is a macro? Could you give me an example please?

Thanks
Jean

[url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=lshort-english]The Not So Short Introduction to LaTeX2e[/url] says:[quote]LaTeX is a macro package that enables authors to typeset and print their work at the highest typographical quality, using a predefined, professional layout.[/quote]so if you took the instructions literally then you couldn't use LaTeX but would need Plain TeX, which is a bit unlikely.

When TeX first came out it was quite difficult to use, so macros were added which put together TeX commands to make them easier to use. Any command that isn't a command in TeX is a macro. So \newcommand, \section, \frac ..., you name it, is a macro. Since style files contain macros you couldn't use \usepackage either.

So I think you need to get back to the journal and ask exactly what they require. Do they provide, as many journals do, a style or class file for you to use? It would be helpful to you if they could supply a sample tex file which accords with their requirements, then you could see what is allowed and what isn't.

Yes, you are very right. The full paper was that I sent to a journal. This one, a shorter version, is to be published in the Proceedings of a conference, locally.
I guess they reffered to the [quote] \newtheorem{theorem}{Theorem}
\newtheorem{proposition}[theorem]{Proposition}
[/quote]

They also added "use Latex 2e standard". I used the packages amssymb, amsmath and amsthm. And I used the newtheorem, indeed.

Questions: 1) Are those packages contained in "Latex 2e standard" or I have to remove them?

2) If I don't use newtheorem, I cannot use begin{proposition} and end{proposition} for propositions. Is there any away to solve this problem or I have to make the propositions manually, in italics?

Thanks,
Jean

Certainly those packages are standard and recommended for LaTeX2e use and are used by most mathematicians. I would personally describe them as standard but only the editors can tell you if they think the same, so you will need to ask them.


limit
trigonometry
calculus
calculus computations

Let $ p \equal{} \frac {\pi}{2^x} \iff 2^{x \minus{} 1} \equal{} \frac {\pi}{2p}$

$ \lim_{x\to\infty} 2^{x \minus{} 1}\left(\sin{\frac {\pi}{2^x}} \plus{} \tan{\frac {\pi}{2^x}}\right) \equal{} \lim_{p \to 0} \frac {\pi}{2} \left( \frac {\sin p}{p} \plus{} \frac {\tan p}{p} \right) \equal{} \frac {\pi}{2} (1 \plus{} 1) \equal{} \pi$


inequalities
geometry
inradius
circumcircle
trigonometry
geometry proposed

In an [u]acute[/u] triangle $ ABC$ denote $ 2s \equal{} a \plus{} b \plus{} c$ , the length $ r$ of  inradius , the length $ R$ of circumradius, the length $ h_a$ of $ A$-altitude

and the length $ m_a$ of $ A$-median.  Prove that there the inequality $ \boxed {2(R \minus{} 2r)\ \ge\ m_a \minus{} h_a \plus{} \frac {r}{2a(s \minus{} a)}\cdot (b \plus{} c \minus{} 2a)^2}$ .

[u]Consequence[/u].  $ \boxed {2(R \minus{} 2r)\ \ge\ m_a \minus{} h_a}$ ([u][b]L.Panaitopol's [/b]inequality[/u]).

This inequality and the inequality from [url=http://www.mathlinks.ro/viewtopic.php?p=1686316#1686316][color=red][b]here[/b][/color][/url] will appear in an article from the future number of the review [url=http://www.arhimede.ro/][color=darkred][b]"Arhimede", Romania.[/b][/color][/url]

Firstly we will show that, in a triangle $ABC$ satisfying $\angle A\ \le\ 90^{\circ}$ the following inequality holds : $\boxed{\ m_a-h_a\ \le\ 2R\sin^2\frac {B-C}{2}\ }\ \ (\ast)$ . [hide="Proof"]

Since $\angle A\ \le\ 90^{\circ}\ \implies\ m_a\ \le\ R(1+\cos A)$ . Thus, $m_a-h_a\ \le\ R(1+\cos A) - R\left(\cos A+\cos (B-C)\right)=2R\sin^2\frac {B-C}2$ .

I used the relation (easy to prove) : $\boxed{\ h_a=R\left(\cos A+\cos (B-C)\right)\ }$ .

[/hide]

Now we'll use the identities : $ \boxed{\ \begin{array}{cccc}
1\blacktriangleright & R-2r=R\left(1-4\sin\frac{A}{2}\cos\frac{B-C}{2}+4\sin^{2}\frac{A}{2}\right)\\ \\
2\blacktriangleright &\frac{r}{2a(s-a)}\cdot(b+c-2a)^{2}=2R\left(\cos\frac{B-C}{2}-2\sin\frac{A}{2}\right)^{2}\ \end{array}} $ and the upper inequality :

Thus, $m_a-h_a+\frac{r}{2a(s-a)}\cdot (b+c-2a)^2\ \stackrel{(\ast)\wedge (2)}{\le}\ 2R\sin^2\frac {B-C}2$ $+2R\left(\cos\frac {B-C}2-2\sin\frac A2\right)^2\ \iff$

$\iff\ m_a-h_a+\frac{r}{2a(s-a)}\cdot (b+c-2a)^2\ \le\ 2R$ $\left(1-4\cos\frac{B-C}2\sin\frac A2+4\sin^2\frac A2\right)\stackrel{(1)}{=}2(R-2r)$ .

[u]Thank you[/u], [b]C. Mateescu[/b]. Another proof ?


inequalities
logarithms
triangle inequality

Prove for positive reals x,y,z

$xyz \geq (x+y-z)(y+z-x)(z+x-y)$.

No schur allowed!

[quote="Phelpedo"]Prove for positive reals x,y,z

$xyz \leq (x+y-z)(y+z-x)(z+x-y)$.

No schur allowed![/quote]
This doesn't work for $x = 1,\ y = 2,\ z = 4.$ [b]amcavoy[/b], your solution is very cryptic -- at least kindly provide some explanation  for your steps. Anyways, the problem is incorrect in its current form.

The inequality doesn't hold: Choose $x,y$ arbitrarily and put $z=x+y$. Then LHS is positive while RHS is zero.

The inequality sign is just reversed. It expands out to Schur, which can be proved by ordering $x \ge y \ge z$.

[quote="ch1n353ch3s54a1l"][quote="FieryHydra"][quote="Phelpedo"]Prove for positive reals x,y,z

$xyz \leq (x+y-z)(y+z-x)(z+x-y)$.

No schur allowed![/quote]
The correct problem has the inequality sign reversed:
Prove for positive reals x,y,z

$xyz \geq (x+y-z)(y+z-x)(z+x-y)$.[/quote]
[hide]
$x=a+b$
$y=b+c$
$z=a+c$
$(a+b)(b+c)(c+a)\ge (a+b+b+c-a-c)(b+c+c+a-b-a)(a+c+a+b-b-c)$
$(a+b)(b+c)(c+a)\ge (2a)(2b)(2c)$
$a^{2}b+ab^{2}+b^{2}c+bc^{2}+ca^{2}+c^{2}a\ge 6abc$ which is true by AM-GM[/hide][/quote]
The substitutions made for $x,\ y,\ z$ are valid only if $x,\ y,\ z$ are the lengths of the sides of a triangle. It is not true if $x,\ y,\ z$ are any positive numbers. My guess is that you learnt this technique from Hojoo Lee's online book on Inequalities. Lee calls this the "Ravi" substitution -- perhaps named after Ravi Vakil though I am not very sure if that's true.

For those who may not know this, the "Ravi" substitution (according to Hojoo Lee) basically means if $x,\ y,\ z$ are the sides of a triangle then we can find positive $a,\ b,\ c$ such that $x=a+b,\ y = b+c,\ z = c+a.$ This is not hard to see for yourself. This special technique is sometimes very useful when solving inequality problems involving lengths of sides of a triangle.

Darn, another incorrect problem posted. It's this silly latex. You're right, the sign was reversed. Also, biffandoc is right- the substitution x = a+b, y = b+c, z = a+c imposes the triangle inequality on the variables.

That hole is rather easily filled:  if they are not the sides of a triangle, exactly one of $x+y-z, x-y+z$ and $-x+y+z$ is negative.

[quote="amcavoy"][hide]$\left(x+y-z\right)\left(y+z-x\right)\left(z+x-y\right)\geq\left(\sqrt{xyz}+\sqrt{xyz}-\sqrt{xyz}\right)^{2}=xyz$[/hide]
Edit:  Cauchy doesn't extend to cases like this...[/quote]
wait a sec how is that cauchy?
anyway $x,y,z$ majorizes $x+y-z,...$
is there a general theorem that if $a_{1},a_{2},...a_{i}$ majorizes $b_{1},b_{2},...b_{i}$ then \[\prod a_{i}\geq \prod b_{i}\large \large ?\]

See

http://www.mathlinks.ro/Forum/viewtopic.php?p=354109

conceive

[quote="me@home"]anyway $x,y,z$ majorizes $x+y-z,...$ [/quote]

In fact, it doesn't.  $max(x, y, z) \le max(x+y-z, y+z-x, z+x-y)$.  WLOG $max(x, y, z) = z$ and so either one of $z+(x-y), z-(x-y)$ is greater than $z$ or both are equal to $z$.

Edit:  I actually suspect the reverse might be true.  In any case, the proof by Karamata's is still not valid.

[quote="me@home"]
is there a general theorem that if $a_{1},a_{2},...a_{i}$ majorizes $b_{1},b_{2},...b_{i}$ then \[\prod a_{i}\geq \prod b_{i}\large \large ?\] [/quote]

There is quite a general theorem, but your statements are wrong.  For example, $(1, 4) \succ (2, 3)$ but $1 \cdot 4 < 2 \cdot 3$.

The correct statement:  $f(x) = \ln x$ is concave, so by Karamata's Inequality we have

$\sum f(a_{i}) \le \sum f(b_{i})$

Notice that this proof is valid only when $a_{i}, b_{i}$ are positive (I do not think the inequality is true in the negative case), which cannot be guaranteed in this problem.

[quote="Phelpedo"]Prove for positive reals x,y,z

$ xyz \geq (x + y - z)(y + z - x)(z + x - y)$.

No schur allowed![/quote]
[hide]
\[ \begin{aligned*}(x+y-z)(y+z-x)=y^2-(x-z)^2\leq y^2;\\
(y+z-x)(x+z-y)=z^2-(x-y)^2\leq z^2;\\
(x+y-z)(z+x-y)=x^2-(y-z)^2\leq x^2;\end{aligned}\]
Multiply these and we are done; Equality holds iff $ x=y=z.\Box$[/hide]
:)

Denote $ x \plus{} y \plus{} z \equal{} s$ . Then exists and is uniquely a triangle $ ABC$ so that $ \left\|\begin{array}{c} x \equal{} s \minus{} a \\
\ y \equal{} s \minus{} b \\
\ z \equal{} s \minus{} c\end{array}\right\|$ , where $ s \equal{} a \plus{} b \plus{} c$ . Our inequality

becomes $ (2a \minus{} s)(2b \minus{} s)(2c \minus{} s)\le (s \minus{} a)(s \minus{} b)(s \minus{} c)$ which is equivalently with $ 16Rr\le s^2 \plus{} 5r^2$ what is well-known.

Wow so many posts for one silly inequality (of course, the sign is reversed).

JBL was the only one right: if x,y,z aren't sides of a triangle, then LHS is positive, and RHS is negative so it's done. If they are sides of a triangle, make Ravi's substitution, or AM-GM on each 2 factors on the RHS.

[quote="xyy"][quote="Potla"][quote="Phelpedo"]Prove for positive reals x,y,z

$ xyz \geq (x + y - z)(y + z - x)(z + x - y)$.

No schur allowed![/quote]
[hide]
\[ \begin{aligned*}(x + y - z)(y + z - x) = y^2 - (x - z)^2\leq y^2; \\
(y + z - x)(x + z - y) = z^2 - (x - y)^2\leq z^2; \\
(x + y - z)(z + x - y) = x^2 - (y - z)^2\leq x^2;\end{aligned}\]
Multiply these and we are done; Equality holds iff $ x = y = z.\Box$[/hide]
:)[/quote]

$ (x + y - z)(y + z - x),(y + z - x)(x + z - y),(x + y - z)(z + x - y)$. What happen when one of them is negative???[/quote]
Multiplying these gives us
\[ x^2y^2z^2\geq \left[(x+y-z)(y+z-x)(z+x-y)\right]^2\]
So that
\[ xyz\geq |(x+y-z)(y+z-x)(z+x-y)|\geq (x+y-z)(y+z-x)(z+x-y).\]
:)

[quote="Potla"][/quote][quote="xyy"][quote="Potla"][quote="Phelpedo"]Prove for positive reals x,y,z

$ xyz \geq (x + y - z)(y + z - x)(z + x - y)$.

No schur allowed![/quote]
[hide]
\[ \begin{aligned*}(x + y - z)(y + z - x) = y^2 - (x - z)^2\leq y^2; \\
(y + z - x)(x + z - y) = z^2 - (x - y)^2\leq z^2; \\
(x + y - z)(z + x - y) = x^2 - (y - z)^2\leq x^2;\end{aligned}\]
Multiply these and we are done; Equality holds iff $ x = y = z.\Box$[/hide]
:)[/quote]

$ (x + y - z)(y + z - x),(y + z - x)(x + z - y),(x + y - z)(z + x - y)$. What happen when one of them is negative???[/quote][quote="Potla"]
Multiplying these gives us
\[ x^2y^2z^2\geq \left[(x + y - z)(y + z - x)(z + x - y)\right]^2\]
So that
\[ xyz\geq |(x + y - z)(y + z - x)(z + x - y)|\geq (x + y - z)(y + z - x)(z + x - y).\]
:)[/quote]

no if $ (x + y - z)(y + z - x)>0,(y + z - x)(x + z - y)>0,(x + y - z)(z + x - y)<0$ or etc...
when we Multiply them the sign of ineq is contrary

[quote="Phelpedo"]Prove for positive reals x,y,z

$ xyz \geq (x \plus{} y \minus{} z)(y \plus{} z \minus{} x)(z \plus{} x \minus{} y)$.

No schur allowed![/quote]
The following inequality is also true.
Let $ x,$ $ y$ and $ z$ are positive numbers. Prove that:
\[ \frac {xyz(xy \plus{} xz \plus{} yz)^2}{(x^2 \plus{} y^2 \plus{} z^2)^2} \geq (x \plus{} y \minus{} z)(y \plus{} z \minus{} x)(z \plus{} x \minus{} y)\]
[hide="PS"] $ \frac {xyz(xy \plus{} xz \plus{} yz)^3}{(x^2 \plus{} y^2 \plus{} z^2)^3} \geq (x \plus{} y \minus{} z)(y \plus{} z \minus{} x)(z \plus{} x \minus{} y)$ is wrong already. :wink: [/hide]


Hi.

It looks like a simple straightforward calculation, doesn't it? But don't you mean [tex](n+2)^{n-1}[/tex] on the RHS?

We have to prove that for all [tex]n\in{\mathbb N}[/tex] the following equality holds:
[tex]\prod_{k=0}^{n-2} (C_k+2C_{k+1}+C_{k+2})=\frac{(n+2)^{n-1}}{(n+1)!}\prod_{k=1}^{n} (C_{k-1}+C_{k})[/tex]
Since [tex]C_{k-1}+C_{k}={n\choose k-1}+{n\choose k}={n+1\choose k}[/tex] this rewrites as
[tex]\Leftrightarrow \prod_{k=0}^{n-2} {n+2\choose k+2} = \frac{(n+2)^{n-1}}{(n+1)!}\prod_{k=1}^{n} {n+1\choose k}[/tex]
[tex]\Leftrightarrow \prod_{k=2}^{n} \frac{(n+2)!}{k!(n+2-k)!}= \frac{(n+2)^{n-1}}{(n+1)!}\prod_{k=1}^{n} \frac{(n+1)!}{k!(n+1-k)!}[/tex]
[tex]\Leftrightarrow\frac{(n+2)!^{n-1}}{(2!\cdot 3!\cdots n!)^2}= \frac{(n+2)^{n-1}}{(n+1)!}\frac{(n+1)!^{n}}{(1!\cdot 2!\cdots n!)^2}[/tex],

but this is obviously correct.

Hanno


function
geometry
geometric transformation
search
complex analysis
complex analysis unsolved

Suppose $f(z)$ is an entire function (that is not a translation); does it then follow that $f(f(z))=z$ has a solution? The source of this problem is a great book called "problems for mathematicians young and old" by Halmos that I found in my schools library.

http://www.mathlinks.ro/viewtopic.php?search_id=1826678109&t=85368  By the way, please post complex analysis problems in appropriate forum. Thanks.


[quote="myc"]Can someone get me started on this problem?  not sure where to start...
[url]http://www.nspsmo.org/documents/2001trig-03.pdf[/url]

thanks so much![/quote]
[hide]Since we know the angles, $ \angle BEC$ and $ \angle CBE$, we can find $ \angle BCE$. So using sine rule, we have, $ \frac{BE}{\sin \angle BCE} = \frac{CE}{\sin \angle CBE}$. Now we know the length $ BE$. So we can use cosine rule for $ \triangle ABE$ to find the length $ AB$ as,
\begin{eqnarray*}
AB^2 & = & BE^2 + AE^2 - 2BE \cdot AE \cdot \cos \angle AEB
\end{eqnarray*}

You can find $ BC$ using sine rule as $ \frac{BC}{\sin \angle BEC} = \frac{CE}{\sin \angle CBE}$.

In the same way, using sine rule, you can find $ BD$ as, $ \frac{BD}{\sin \angle BED} = \frac{BE}{\sin \angle BDE}$.

Remember that you need to use minutes and seconds when computing your results, not just degrees.[/hide]

[quote="Patterns_34"][quote="mthiyaga"]Remember that you need to use minutes and seconds when computing your results, not just degrees.[/quote]
What's the significance of the time variables next to the angles?[/quote]
[hide]They're degrees, minutes and seconds. 1 deg = 60 minutes and 1 min = 60 seconds.[/hide]


Hi, I was wondering how the difficulty of AIME has changed since 1983, through until now. How difficult, relative to the other years, are, for example, the 80's, early 90's, late 90's, early 00's, and late 00's tests? Thanks in advance.

Geez...

Do the problems, you will know  :wink:

Generally, competitions tend to get harder as years pass...

yeah, thats true. there was some poll on what aimes people thought were ahrdest. i think majority said 91,94,95. However, that doesnt necissarily 92 and 93 were that difficult


If a half of a number, a third of the number, a fourth of the number and two-sevenths of the number are added together, the sum is 575. What is the number?

hello, we have $ \frac{1}{2}x\plus{}\frac{1}{3}x\plus{}\frac{1}{4}x\plus{}\frac{2}{7}x\equal{}575$
solving for $ x$ we get $ x\equal{}420$.
Sonnhard.


sin va cos e ie adade mokhtaleto( :!: ) che joori hesab mikonan?

Take a look at http://en.wikipedia.org/wiki/Trigonometric_function .


Which is a better Lewis acid and why: $ \ce{AlCl3}$ or $ \ce{FeCl3}$?

[quote="JRav"]It's the one that's a better donor/recipient of electrons.  I just can't remember which is the acid and which is the base definition.[/quote]
its the one that is a better recipient of electrons. But the question is which between the two is better

$ \ce{AlCl3} \ \plus{} \ : \ce{Cl\minus{}} \ \ce{\minus{}>} \ \ce{[AlCl4]\minus{}} \ \ \ \text{tetrachloro aluminate ion}$
$ \ce{FeCl3} \ \plus{} \ : \ce{Cl\minus{}} \ \ce{\minus{}>} \ \ce{[FeCl4]\minus{}} \ \ \ \text{tetrachloro ferrate (III) ion}$

Thanks Kouichi Nakagawa, but how does that answer the question?

I think $ \ce{FeCl3}$ is the better lewis acid...

Since you've given the answer I'll atleast try to explain it. :D

AlCl3 because the d-orbitals are empty and thereby it can accomodate the e- well.

[quote="Cyber-Shreyas"]AlCl3 because the d-orbitals are empty and thereby it can accomodate the e- well.[/quote]
Even $ \ce{FeCl_3}$ has got it.
But probably the extra gained electrons go to p-obitals in case of $ \ce{AlCl_3}$ rather than d-orbitals of Fe so that's why it is a better lewis acid than $ \ce{FeCl_3}$

i am not contradicting the reason what u said Cyber......

The reasson that chemrock told is not complete

[quote="Cyber-Shreyas"][quote="valeriummaximum"]Lewis acid property is only accepting electrons and not to where iot gets filled into???

CAn u eluycidate on ur answer???[/quote]

Well it can't accept electrons and just keep it in its pocket?? It needs some empty orbitals to keep it. Al has 3d orbital fully empty.[/quote]
Excuse me shre but Fe has also fully empty 4p and 4d(and 4f ofc) so?? :huh:  :huh:  :huh:


Find the component of the force $ \bold{F}\equal{}(3\bold{i}\plus{}2\bold{j})N$ in the direction of the vector $ 2\bold{i}\minus{}\bold{j}$.

Thanks. Please don't use super-advanced methods, and explain :)

Let $ \overrightarrow{F}\equal{}(3,\ 2),\ \overrightarrow {u}\equal{}(2,\minus{}1) ,$ we have the desired vector is $ \frac{\overrightarrow{F}\cdot \overrightarrow{u}}{|\overrightarrow{u}|^2}\overrightarrow{u}\equal{}\frac{8}{5}\overrightarrow{i}\minus{}\frac{4}{5}\overrightarrow{j}$.


Is (a,b,c) and (c,b,a) considered the same set?  Like if those were the two possible sets, and you were asked a question that asked for the number of possible set, would it be two or one?  I don't know much about sets, so anything else would be helpful.

Yes those are considered the same, look [url=http://en.wikipedia.org/wiki/Set_(mathematics)#Describing_sets]here[/url]. If those were two sets to choose from, then you would count them as one.

If the question says:

"how many pairs of triplets $ (x,y,z)$ exist such that..........."
Then ORDER DOES MATTER!!!! In other words, (a,b,c) and (c,b,a) ARE different sets. (well, as long as a,b,and c are different).
Say the question was how many pairs of triplets $ (x,y,z)$ exist such that $ x\plus{}y\plus{}z\equal{}3$. Well, in this case $ (2,1,0)$ and $ (1,2,0)$ are different, because they correspond to different values of x,y, and z.

And someone correct me if im wrong, but if the question just asks  "how many sets are there such that.......", then order does NOT matter, eg {0,1,2} is the same as {2,1,0}

[quote="Thunder365"]how many pairs of triplets  exist such that...[/quote]pairs of triplets means sets of two triplets

@Ganesh: If {a,b,c} and {c,b,a} are possible sets, then so are {a,c,b}, {b,a,c}, {b,c,a}, and {c,a,b}. These are all the same set. However, (a,b,c) and (c,b,a) are ordered triples, which are not the same as sets. For this, these would be two different triples.

if a, b and c are distinct then they are different. This can be seen when you plot them on a 3-D graph


AMC
AIME
USA(J)MO
USAMO
ratio
AMC 12
AIME I

For people who took the AIME II on Wednesday March 28, 2007.  Post what you got on it on the poll.

I DON'T KNOW YET!! YAY!! (I'll post after the Math Jam, I guess)

If my test was not the AIME, but rather, not to make careless mistakes, I literally would have failed (55% solved correctly). yuck.

I don't quite know but i'm pretty sure i got a 4. i'm just really happy i reached my goal :D . i attempted 8 of them (made a careless mistake on #2). oh, well. i'm not complaining. i'm in eighth grade and it's my first year doing it.

It's really frustrating to spend like 4 hours a day studying for the two weeks before the test and then make 4 stupid mistakes. Oh well. If the floor's a 6 AND I don't have to worry about tiebreaks I qualify. Yeah, right...

Eh, oh well. I did better than last year (3).

Edit: I got 3-5,8,11-12.

If the floor is lower than a 9 and there are no tiebrakers then I'll qualify. If there are tiebrakers, I'm dead  :(  Do the tiebrakers apply to all <10 graders or only those at the floor?

[quote="mkkool64"]i'm a little sad about messing up number 1  :(  but happy overall.  It seemed comparable to other recent years, in my opinion, with some later questions easier and some earlier ones harder.[/quote]

ya, i missed #1 twice. like i was working on #9 and i just realized "whoa, number one is wrong. let me go change it." when i redid it, i got it wrong, because i forgot something while doing the casework, when i did it the third time, i got it right.

BTW, i got 336 for my first answer...apparently a lot of ppl got that answer...

Man I am jealous of you guys for having a much better test :(.

Eh, I got a 6. Only attempted 9 of them though. I'm kinda angry because if I was only able to read my own - handwriting, I would've for sure gotten a 7. Another one was in my reach too, if it wasn't for a silly arithmetic mistake. Unfortunately, 6 is probably not going to be the floor and, even if it is, I have a slim chance if tiebreakers count. Oh man oh man...

Got a 7- I think I'll make it, but there's just as much chance I won't.
Missed number 2 because I carried a 2 instead of a 1 when adding the total.. I got 300.

7...kinda disappointed how do tiebreakers work?

8.

Made four dumb mistakes, oh well. Qualifying is really all that matters...

Ugh, just looked at the AIME II. SUCH a better test. I'm jealous. Grr.
Oh well, there's always next year, and hopefully the AIME I next year won't be such crap.

[quote="Danbert"]The AIME II was horrible.  I really need to work on not making dumb mistakes - one misreading error cost me a perfect AMC12, and 6 stupid mistakes got me a 5 on AIME.  I solved 11 problems, checked ten of those, and got a 5.  After a 144 AMC I was sure I was going to qualify for USAMO, but noooo.....  This is crushing for me: USAMO was one of my goals for high school math, then I f*** it up.  Sorry.  I'm upset.  A 194 index likely won't make it.

[EDIT]
I just checked the official answers.  I got one more wrong than I thought.  A 4.  (collapses)
184?  Forget it. :wallbash_red:
 :censored:
[/EDIT][/quote]

Was that your last chance, i.e. are you in 12th grade? If so, man, that sucks...damn.

[quote="Danbert"]The AIME II was horrible.  I really need to work on not making dumb mistakes - one misreading error cost me a perfect AMC12, and 6 stupid mistakes got me a 5 on AIME.  I solved 11 problems, checked ten of those, and got a 5.  After a 144 AMC I was sure I was going to qualify for USAMO, but noooo.....  This is crushing for me: USAMO was one of my goals for high school math, then I f*** it up.  Sorry.  I'm upset.  A 194 index likely won't make it.

[EDIT]
I just checked the official answers.  I got one more wrong than I thought.  A 4.  (collapses)
184?  Forget it. :wallbash_red:
 :censored:
[/EDIT][/quote]

You double checked 10 problems and still made 7 errors?

Maybe you need to evaulate the way you are double checking, are you simply glancing over your work? I usually don't even look at my past work, approach the problem in a different way, and if I get the same answer move on, if I get a different answer I review my work/methods. This helped me minimize my careless errors from originally four to two (although I only checked three of those four problems)

[quote="Danbert"

[EDIT]
I just checked the official answers.  I got one more wrong than I thought.  A 4.  (collapses)
184?  Forget it. :wallbash_red:
 :censored:
[/EDIT][/quote]

hey! a four isn't [i]that[/i] bad... :( . the sad thing is that i attempted 8 problems. i made like a zillion really dumb mistakes. on #2, i considered not counting 0's, but only for like a second and for a totally wrong reason.... i got 1, 3, 5, and 9 right.

[quote="Phelpedo"]The sad thing is I solved 13 problems...must have lost focus in the second hour because it was like

right right right rihgt right riight wrong wonrg wrong wrong wrong right rightr[/quote]

I see the loss of focus, but not the second hour part.

I worked the diagonal question 6 times until I added correctly. Mistakes are eliminated with practice, at least in my opinion. But as mentioned before, one of my goals in hs was making MOP and I've missed it every year (it's very unlikely that I'll make it this year).

I think because the problems were like in the middle it was 2nd hour....

[quote="ra5249"]Hey DPopov, how can you not make MOP with your scores?
You are so gonna make mop.[/quote]

Well, he [i]is[/i] a senior. So he needs to make Black MOP, which is a nontrivial task.

I was referring to more spelling wise >_> <_<

[quote="Phelpedo"]The sad thing is I solved 13 problems...must have lost focus in the second hour because it was like

right right right rihgt right riight wrong wonrg wrong wrong wrong right rightr[/quote]

[quote="Phelpedo"]Yeah the number of rights and wrongs might not have been totally accurate but you get the picture.[/quote]

Oh, I was going to say that I got exactly the same as you.  (I got 8: #1-#6, #12, #13)    :ninja:

[quote="davidyko
Was that your last chance, i.e. are you in 12th grade? If so, man, that sucks...damn.[/quote]

Yeah, I'm a senior.  Otherwise I wouldn't be upset about it.

@Ignite - true, my checking method isn't great.  I'll try to fix it in time for ARML...

Omg I got a 6!

Haha, just got the official word from my math teacher -- I actually had decided not to check my answers for fear I would get a 0 and feel really bad.

Sorry for resurrecting the thread, but I had to post this somewhere!

Oh yea! Woo hoo!

(PS: and yeah, I'm a senior... but this makes up for not qualifying the last 3 years, heh)

It's actually kinda funny.. the one year I don't study AT ALL (I haven't looked at AOPS books or any other contest math practice stuff since I worked through them almost 2 years ago) I do the best.. *sigh* At least I proved myself that I can do it, hehe!


Does anyone know how long it usually takes for Mathcamp to get back to applicants about admission? Thanks.

in http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12190&postdays=0&postorder=asc&start=0

caroline said 1 or 2 weeks.  I remember in other threads people saying they found out ~1st week of may last year.

Hmm, well has anyone here found out yet?

i doubt it, else people would have posted about it.  1 week sounds too quick though, I imagine there's a lot of stuff to go through.

[quote="cats..."]caroline said 1 or 2 weeks.  I remember in other threads people saying they found out ~1st week of may last year.[/quote]

I think that this year will be slower because so many of the invitations to apply were mailed out so late this year. MathCamp will decide as rapidly as it can, but it will take extra care to look at applications that arrrived late this year, I venture to guess.

We've just finished making the first round of admissions decisions, and emails should go out in the next 24 hours.   (If you don't want to wait to find out, email admissions@mathcamp.org to inquire about your status.)

At this point we're admitting enough people to (almost) exactly fill all the spots at the camp, which means there will <i>definitely</i> be more people accepted later, because not everyone we accept now will end up coming to camp.   The next round of acceptances will most likely be in a week, after the deadline for alumni to make final decisions about whether they'll attend.

Do they give a lot of scholarships?  Because I really need them.

i dont know, but from the website it sounds like you can get a LOT out of the fundraising , so you may not actually not need the scholarships.

I also got the email today.
I can't wait to go!
Interestingly enough, I know someone who  found out that he got in last year on May 10th as well.

Got my email of acceptance last night 10 PM here.

Got email a couple days ago.  oh yeah, i got in.


If
  $ a,b,c > 0$
then
$ 2\frac {a}{b} \plus{} 3\sqrt {\frac {b}{c}} \plus{} 4\sqrt [3]{\frac {c}{a}}\geq\frac {15}{2}$

it's wrong, take $ a\equal{}b\equal{}c\equal{}1$

[quote="greenvert"]it's wrong, take $ a \equal{} b \equal{} c \equal{} 1$[/quote]

Sorry!!!!!

$ 2\frac {a}{b} \plus{} 3\sqrt {\frac {b}{c}} \plus{} 4\sqrt [3]{\frac {c}{a}}\geq\frac {15}{2}$


geometry
geometric transformation
reflection
perimeter

Hey I was wondering how to do this problem:

Let ABC be an acute angled triangle

D is on the interior of AB.

Locate E on AC and F on CB such that the inscribed triangle DEF will have minimum perimeter.

It hints to me to reflect D in AC (call it D') and then again in CB (call it D'') and to look at that line segment (D'D''). For the life of me, I cannot see how to use this. I tried making it into a right triangle DD'D'' but still, I just don't know what to do.

Thanks for your help.

The perimeter of triangle DEF is exactly the length of the "broken" line D'E - EF - ED''. Since D' and D'' are fixed this sum is minumum when D',E,F,D" are aligned. I'm sure you now know how to build E and F [hide](it's the intersection of D'D" with AC and BC)[/hide].


Determine (with proof) whether there is a subset $X$ of the integers with the following property: for any integer $n$ there is exactly one solution of $a + 2b = n$ with $a,b \in X$.

I posted (or meant to post) essentially the same problem a few weeks ago, here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=354033.

I will try to put up a solution soon.

Let $ f(x) \equal{} \sum_{a \in X} x^a$; the given condition is equivalent to $ f(x) f(x^2) \equal{} \frac {1}{1 \minus{} x}$, which immediately gives it away:  recall that unique binary expansion implies the identity

$ \frac {1}{1 \minus{} x} \equal{} (1 \plus{} x)(1 \plus{} x^2)(1 \plus{} x^4)(1 \plus{} x^8)...$

so take $ f(x) \equal{} (1 \plus{} x)(1 \plus{} x^4)(1 \plus{} x^{16}) ...$.  In other words, $ X$ consists of the set of positive integers whose binary expansions in base $ 4$ contain only $ 0$s and $ 1$s.  (The bijective proof is straightforward:  consider the base-$ 4$ expansion of $ n$ and isolate its digits of $ 2$ and $ 3$, etc.)  Agrippina's problem is similar:  the identity is $ f(x) f(x^2) f(x^4) \equal{} \frac {1}{1 \minus{} x}$ and we can take $ f(x) \equal{} (1 \plus{} x)(1 \plus{} x^8)(1 \plus{} x^{64})...$.

Well, first of all this problem's been posted before (although without the source) and this solution given by several MOPpers, so it wasn't hard to remember.  It was also given as a very nice list of examples [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1247963905&t=46819]in this thread[/url].  Generally, it is very natural to analyze solutions to equations like $ a \plus{} b \equal{} n, a \in A, b \in B$ by studying the generating functions of $ A$ and $ B$ because $ AB$ gives all the possible sums at once.  For example, the following can be solved by similar means.

[b]Putnam 2003 A6:[/b]  For a set $ S$ of non-negative integers let $ r_S(n)$ denote the number of ordered pairs $ (s_1, s_2) \in S^2, s_1 \neq s_2$ such that $ s_1 \plus{} s_2 \equal{} n$.  Is it possible to partition the non-negative integers into disjoint sets $ A$ and $ B$ such that $ r_A(n) \equal{} r_B(n)$ for all $ n$?

The important step is to realize that if $ S(x) \equal{} \sum_{s \in S} x^s$, then the set of all sums of [i]distinct[/i] elements of $ S$ is given by $ S(x)^2 \minus{} S(x^2)$ (why?).  The rest is computation, and once you've figured out what the answer should be it is not hard to give a direct proof.

I'll note that even if you didn't think of binary expansion, repeated application of the problem condition allows you to perform the following calculation:

$ f(x) \equal{} \frac {1}{(1 \minus{} x) f(x^2)} \equal{} \frac { f(x^4) (1 \minus{} x^2)}{1 \minus{} x} \equal{} \frac {(1 \minus{} x^2)}{(1 \minus{} x)(1 \minus{} x^4) f(x^4)} \equal{} \frac {(1 \minus{} x^2)(1 \minus{} x^{8}) f(x^{16})}{(1 \minus{} x)(1 \minus{} x^4)} \equal{} ...$

Even if this is not rigorous, it tells you what the answer [i]should[/i] look like, but even more it strongly suggests that this is the [i]only[/i] answer, not just an answer that works.

My apologies; I misread "integer" as "non-negative integer," and the solution I gave doesn't work as is.  I'll keep thinking.

Use induction: Suppose we have integers $ x_1...x_n$ such that every $ x_i \plus{} 2x_j$ is distinct. Suppose this set misses the value $ n$. Then add to the set $ k, n\minus{}2k$, so now we have $ n$ included as a sum. It is clear that if you take $ k$ arbitrarily large it wont overlap with any previous sums. So such a set exists.

Since nobody has come up with a (complete) solution, let me post mine (inspired by Kalva):

First, let us verify that a "base -4" can work.  That is, every number can be expressed uniquely as $a_1-4a_2+16a_3-64a_4+\ldots$ where $a_i\in\{0, 1, 2, 3\}$.  Uniqueness is evident: if we have $a_1-4a_2+\ldots=b_1-4b_2+\ldots$ then suppose $k$ is the first natural number such that $a_k\neq b_k$.  Then

$(-4)^{k-1}a_k+(-4)^ka=(-4)^{k-1}b_k+(-4)^kb$

for the rest of the integer $a$ and $b$.  Then that means that, dividing by $(-4)^{k-1}$, we have $a_k-b_k=4(a-b)$, so $a_k-b_k$ is divisible by 4, which is impossible because $a_k$ and $b_k$ are not equal and between 0 and 3.  So the uniqueness is proved.

Now take the $4^k$ numbers $a_1-4a_2+16a_3-\ldots+(-4)^{k-1}a_k$ where $a_i\in\{0, 1, 2, 3\}$.  The minimal number possible is $3(-4-64-\ldots)$ and the maximal number is $3(1+16+256+\ldots)$, which cover a range of

$3(1+16+256+\ldots)-3(-4-64-\ldots)+1=3(1+4+16+\ldots+4^{k-1})+1=4^k$

numbers.  Since there are $4^k$ numbers in the possible range and all numbers are expressed at most once, all numbers must be expressed exactly once.  Thus base -4 exists and we can work with it like any normal base.

So now as before, we place all numbers with only 0's and 1's in their base -4 expansions in the set.  Then for any integer $n$, take its base -4 expansion $n_1-4n_2+16n_3-\ldots$.  If $n_i=0$, let $a_i=b_i=0$; if $n_i=1$, let $a_i=1$ and $b_i=0$; if $n_i=2$, let $a_i=0$ and $b_i=1$; and if $n_i=3$, let $a_i=b_i=1$, so that in any case, $a_i+2b_i=n_i$.  Then let $a=a_1-4a_2+16a_3-\ldots\in X$ and $b=b_1-4b_2+16b_3-\ldots\in X$ also.  Then clearly $a+2b=n$ by construction.

We must show uniqueness.  For any $a, b\in X$, we have

$a-b=(a_1-b_1)-4(a_2-b_2)+16(a_3-b_3)-\ldots$.

The first nonzero $a_i-b_i$ is either $1$ or $-1$.  Thus $a-b=\pm(-4)^{k-1}+(-4)^kx$ where $x$ is an integer.  Hence $|a-b|$ is divisible by $4^{k-1}$ and not $2\cdot4^{k-1}$, so the highest power of two that divides $a-b$ is also a power of 4.

Now if $a+2b=c+2d$ for $a, b, c, d\in X$, then $a-c=2(d-b)$.  If $a\neq c$ then let us look at the highest power of 2 that divides this common difference.  It must be a power of 4 that divides $a-c$, but also is a power of 4 that divides $d-b$ and so is twice a power of 4 that divides $2(d-b)$.  No number is both a power of 4 and twice a power of 4, so this contradiction shows uniqueness and we're done.

Using generating functions since they generalize to alot of counting problems, (though this uses neg exponents, but it still is legit)

Let $f(x)=(1+x^{1})(1+x^{-4})(1+x^{16})(1+x^{-64})*...(1+x^{(-4)^{a}})$ (the same base -4 thing)
Then
$f(x)f(x^{2})=\frac{(1+x^{1})(1+x^{2})(1+x^{4})(1+x^{8})....(1+x^{2^{n}})}{x^{r}}$, where $r$ is a pretty huge number,

which means at the limiting case, as $a$ tends to infinity,
$f(x)f(x^{2})=...+1/x^{3}+1/x^{2}+1/x^{1}+1+x^{1}+x^{2}+x^{3}...$ which solves the problem.

No that's not legit.  It's only legit if the generating function converges for some $x$, which this one does not.  The generating function approach does not prove the answer, it just points you in the direction.  You need another approach to show that it actually works.  In the limit, what is $r$?  How do you know that the generating function doesn't just become $\ldots+\frac{1}{x^4}+\frac{1}{x^3}+\frac{1}{x^2}+\frac{1}{x}+1$ and stop there?

Sorry, I was in a rush, and put only the general idea down. It doesn't HAVE to converge for some X, because of the fact that all it's doing is manipulating the exponents of the terms, as a representation of a sumset problem. I'm not ever actually going to evaluate the function $f(x)$. I know it makes absolutely no sense if you considered $x$ as an actual number, but this wasn't the point here.

And sorry, I can't calculate right now, but essentially it's pretty easy to show through calculation that the most negative exponent and most positive exponents are increasing arbitrarily large as $a$ approaches infinity.

My point is that you cannot just rush into these calculations.  You need the theory of formal power series or (in this case) Laurent series.  The calculations you do, taking products of sums and expanding them, is only justified when the power (Laurent) series converges.  That's why everyone was up in arms when Euler calculated $\sum_{i=1}^{\infty}\frac{1}{i^2}=\frac{\pi^2}{6}$ by setting $\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}-\ldots=x(x^2-\pi^2)(x^2-4\pi^2)\ldots$; he had no rigorous justification for saying two power series were equal just because they had exactly the same roots.  And this is a bit like what you're doing; you're saying that the power series $\frac{(1+x)(1+x^2)(1+x^4)\ldots}{x^r}=\ldots+\frac{1}{x^3}+\frac{1}{x^2}+\frac{1}{x}+1+x+x^2+x^3+\ldots$, whatever $r$ means in the limiting case.

Now I'm not saying I don't believe you, because I just gave the same solution set as you.  I'm just pointing out that your taking the limit of $f(x)f(x^2)$ as $a$ or $n$ goes to infinity and saying that it equals the product of $\lim_{n\rightarrow\infty}f(x)$ and $\lim_{m\rightarrow\infty}f(x^2)$ is unjustified.  It would be justified if it converged for some number $x$, but it doesn't.

Sorry if I was unclear with my words, by limit, I didn't actually mean that the numerical value for some particular $x$, of $f(x)*f(x^2)$; by limit, I just meant that the larger integers (in absolute value) could be represented uniquely once we increased the size of our sequence. I'm not ever using $f(x)*f(x^2)$ as any evaluation for a numerical answer, I'm just using $f(x)*f(x^2)$ conveniently because it just simplified the concept of "looking at all sums".
Your example indeed uses these generating functions as actual polynomials, which evaluate numerical answers, which isn't the same as what I'm doing.

I can easily just as well put this into an argument without generating functions, but still using the same concept of "the terms acting like a binary string". It's just that the algebraic manipulations are more representative/clear of what the concept is.

zero.destroyer's argument is correct. You don't need convergence for formal power series at all, but it is indeed necessary to be careful with Laurent series' that are infinite in both directions; but this is mostly due to the fact that these do not form a ring, not even a module over the power series' (but over polynomials or finite Laurent sums).

The argument here never compares terms in a non-formal nature, unlike Euler, who, as said above, plugged in real numbers to to speak of "roots". It is enough to show that additional factors in the product do not contribute to those summands of exponent $s$, where $|s|$ is bounded by some growing bound dependent on the number of factors. This is the case here.

If one wants to do it very formally, an algebraic version would be to state that the module of Laurent series is the projective limit over $n$ of Laurents sums whose exponent's modulus is bounded by $n$. This is just a less comprehensible way to state what I said in the previous paragraph, though.

Also, note this "fact": $\sum_{n \in \mathbb Z} x^ n = \sum_{n=1}^ \infty x^{-n} + \sum_{n=0}^ \infty x^n = \frac{x^{-1}}{1-x^ {-1}} + \frac{1}{1-x} = \frac{1}{x-1} + \frac{1}{1-x} = 0$. The error here is that to apply geometric series, you would need it to be a module over power series' (i.e. multiplying any Laurent series with a power series would need to make sense; try to multiply the above one by $\sum_{n=0}^ \infty x^n $ to see the problem). But the problem is not that we lack a common are of convergence for those sums.
Actually, the fractions are the meromorphic continuations of the sums (which in turn are the Laurent expansions around $\infty$ and $0$) and as an identity of meromorphic functions, this is completely correct!

Not clear or not if $a$ and $b$ are distinct.

Let $f(X) = \sum_{a \in X} X^a$. The necessary and sufficient condition for $X$ to satisfy the condition is for $$f(X)f(X^2) = \frac 1{1-X}$$ for all $x \neq 1$. One can notice that we have $$f(X^4) = \frac 1{(1-X^2)f(X^2)} = \frac 1{(1-X^2) \cdot \frac 1{(1-X)f(X)}} = \frac{f(X)(1-X)}{1-X^2} = \frac{f(X)}{1+X},$$ so the infinite product $$\frac{f(X)}{(1+X)(1+X^4)(1+X^{16}) \cdots} = f(X^{2^n}) \to f(0) = 1,$$ so the function $$f(X) = (1+X)(1+X^4)(1+X^{16}) \cdots$$ can be checked to work.

In more concrete terms, we may pick $X$ to be the set of numbers that can be represented as the sum of some distinct nonnegative powers of 4.

[hide=solution]
It suffices to find an infinite set of numbers $X = \{a_j\}$ such that $$\left(\sum_{a_i\subset X}x^{a_i}\right)\left(\sum_{a_i\subset X}x^{2a_i}\right) = 1+x+x^2+\dots = \frac{1}{1-x}.$$ Note that $1 = (1-x)\prod_{k\geq0}(1+x^{2^{k}}).$ So, we let $\sum_{a_i\subset X}x^{a_i} = \prod_{k\geq0}(1+x^{2^{2k}})$ and $\sum_{a_i\subset X}x^{2a_i} = \prod_{k\geq0}(1+x^{2^{2k+1}}),$ meaning for all $n\subset \mathbb{Z},$ there is exactly one solution to $a+2b = n,$ with $a,b\subset X.$
[/hide]

We interpret this using generating functions. Defining $A(x) = \sum_{k \in \mathcal{X}} x^k$, our condition requires
\[A(x) A(x^2) = \ldots + x^{-2} + x^{-1} + x^0 + x^1 + x^2 + \ldots.\]

to cover all integers exactly once. From here, we note that the functions
\begin{align*}
A(x) &= \prod \left(1+x^{(-4)^i}\right) = (1+x^1)(1+x^{-4})(1+x^{16})(1+x^{-64}) \ldots \\
A(x^2) &= \prod \left(1+x^{2 \cdot (-4)^i}\right) = (1+x^2)(1+x^{-8})(1+x^{32})(1+x^{-128}) \ldots
\end{align*}

indeed have the desired product, so our construction for $\mathcal{X}$ is simply
\[\boxed{\{\mathcal{X}\} = \text{Integers with only 0 and 1 as digits in base -4}}. \quad \blacksquare\]


function
limit
algebra
domain
algebra proposed

Does there exist a function $f: R\to R$ that fulfils all the following conditions:
.$f(1)=1$;
.there exists $M>0$ such that $-M<f(x)<M$ for all real x;
.$f(x+\frac{1}{x^2})=f(x)+(f(\frac{1}{x}))^2$ for all real x?

Edit:  Forgot a possible case :oops_sign:

For any $a = x + \frac{1}{x^2}$ we have $a > x$ and $f(a) > f(x)$ as long as $f(\frac{1}{x})^2 \neq 0$.  The sequence $a_1 = a, a_n = a_{n-1} + \frac{1}{a_{n-1}^2}$ is strictly increasing, and it can also be shown (I think) $\lim_{n \to \infty} a_n = \infty$ (I will post a proof of this later).  Take the sequence $b_n = f(a_n)$ - it is also strictly increasing, and because it grows by the same rule, $\lim_{x \to \infty} f(x) = \infty$ [i]unless $f( \frac{1}{a_n} )^2 = 0$ for every value of $a_n$ [/i].  Because $a_1$ can be any value in the domain, $a_n$ can be any value in the domain, and so we have $\boxed{f(x) = 0}$ for every value in the domain.  Q.E.D.

Actually, there is not any function satisfying the above conditions..
You can prove it by contradiction ;)

You forgot that $f(1)=1$ so there isn´t a function that satisfies all the conditions.


If the code is too long for the popup screen, it just continues below the bottom line, there doesn't appear a scrollbar in the window. Since asymptote codes can be quite lengthy sometimes, it would be helpful for some people to have a scrollbar there. Just a suggestion.

I would like to add to this post the comment that apparently some people can still scroll through the asymptote code by using page up/down or home/end.  This does not however work on any computer that I have tried.  What causes this?


[i]Prove that with all positive integers $ n$ , $ n \geq 2$ , we always have :

      $ \varphi(n) \ \geq \ \frac{ \pi(n)}{2}$[/i]

You can prove that exist a $ v \in \mathbb{N}_0$ such that $ \varphi(n) > \pi(n)$ for every $ n > v$.  :roll:
(More interesting is making clever lower bound on $ v$..)


if $ {x},{y}> 0$ ,find the min value of the following:

$ {x}+{y}^{2}+\frac{1}{xy}$

it's a Very nice problem,trust me! (given to me by a friend...may has a stronger source! )

oh yes,thanks arqady,your nice solution is similar to mine...
any other (even non-elementary) solutions?


Do you offer a proof with that statement?