AMC
AIME
USA(J)MO
USAMO
Putnam
AIME I
AIME II

So how to do you become certian you can write the CMO after you recieved your COMC results?

When do you get your official invitation?

official invitation comes very late... u just sorta have to know before hand

I just got the invitations today. they emailed my teacher, and my teacher told me.

:maybe: , my teacher hasn't told me yet.

If you have gone to Winter Camp, you go to CMO automatically. If you are about top 70 in Canada, you write CMO -- that is equivalent to getting 65+. If you won in your province, you also get invited.

Are you sure its not 66+ because 69 people got 66+ and then 84 ppl got 65+..because after you include the provincial winners that didnt get 65+ you have roughly 90 ppl writing it..Do they let that many people write it?

If you went to winter camp, you automatically write APMO as well.
But what about summer camp. I went to IMO camp in halifax in June 2006 as a local student. do they write cmo and apmo automatically?

@SM4RT: Maybe it's 66, I'm not sure.
@ssj4152: I am not sure, but I have never heard of summer camp ppl writing APMO...

o...im probably getting the invatation tmr...come to think about it...im just a sarter in math :P

Did anyone get the invitation yet? :maybe: I remember I got mine last year in December, and it is already the beginning of January...

ya, my teacher just gave me the problems book and the registration form today.

we just got the invitation......after the exam...
i believe they sent us the package on jan 19

i got the invitation letter today...theres another guy in my school who got invited too, but we got different books :lol: lol

i wonder when the apmo invitations would come...or would they come at all :rotfl: ...

[quote="rem"]OK.I guess I will get my invitation after the exams.[/quote]

What exams are you talking about???

I didn't get it yet???

End of semester exams (this event depends on your school's method of operation).

I got invited, so I know that the cutoff score is at most 68.

4 people from our school got invited.

[quote="1234567890"]I got invited!!!

Free book, too. :lol:[/quote]

Yeah, I think I know why it ended in 1993.

[quote="1234567890"]Shoot, this is not good.   :(

My school does not offer the AIME I (break), but it offers the AIME II. Unfortunately, the CMO is at the same time. What should I take?![/quote]
Um, take both. Ask your proctor to talk to the CMS or the MAA. You should be able to take one of them later in the day.

[quote="1234567890"]Shoot, this is not good.   :(

My school does not offer the AIME I (break), but it offers the AIME II. Unfortunately, the CMO is at the same time. What should I take?![/quote]

Oh, that does suck.

I just realized that too, I think, mate.

Edit: AIME II is 03/21/07
CMO is 03/28/07

Now you can just relax.

[quote="1234567890"]Really?

From the AMC website:
"Alternate AIME --
Wednesday, March 28, 2007"

Neil, you're taking AIME I, though, right?[/quote]

Yeah, if I qualify for AIME.

Because DK says March 21, 2007

Edit: You are right.

I guess you have to ask DK if I can take it at ICAE, or something like that, I don't know.

hey just wondering, Winter Camp allows writing USAMO without AIME right?

just making sure....cuz ya know... ya...CMO and AIME 2 are on the same day, and I might just skip AIME 2, if I don;t get a chance to write AIME 1

umm ya i think u can but then ur an unofficial contestant..i think..

Yes you write it unofficially, and it is not marked by AMC but by Canadian people.

PS I finally got my CMO invitation. :)

rem, do you know what you're going to do (about the AIME II/CMO conflict)?

I will try to do both.

By the way, you can still write AIME at your school during March Break. See here:
http://www.unl.edu/amc/e-exams/e7-aime/adminaime.shtml

Hi, rem
I know little about AIME. I'm wondering what are the requirements for the registration AIME? so that I can make sure if i'm eligble.
Thanks!

If your school has not yet registered for AMC, it must register as soon as possible, there's only one week left.

lol...got a call and got invited to apmo :lol: still have to contact a prof

[i]content deleted -- Admin[/i]

Once again, please do not comment about [b]any[/b] aspect of the AMC 10A/12A until Wednesday morning.  Thank you.

ooooo...im so sry...guess i got too excited :rotfl: good thing i didnt say much... :oops:

[quote="lowest_reynolds"]Yifan, you must write the AMC 12 and qualify with a score of >= 100.  The first set of exams was administered today, and the 12 B will be offered in about 2 weeks.

Are there even any schools in St. Catherines giving the 12 B?[/quote]
Thanks!
So A and B are actually two different but same-level exams?
I highly doubt there're any school in St.Catharines gving 12B, it's small town~
But I guess I'll ask my maths teacher tomorrow.

[quote="andy_tok"]and another question,  assume that I'm qualified, can I write AIME this year or next year?(i'm a senior :( ).
Thanks[/quote]
im not sure wut u mean...but if u qualify u take this years AIME...if thats what u mean

Dude seriously, why would you want to write the AIME next year.  Don't you have better things to do - like the Putnam?  :wink:

:oops: , maybe my question was too confusing.
Well, of course, I want to write it this year, what I was asking is that how long it will take for them to determine if someone is qualified?

They take the good-scoring AMC people from the same year.


limit
function
ratio
calculus
calculus computations

$ \lim_{x \to \plus{}\infty } (ch \sqrt{x\plus{}1} \minus{} ch \sqrt{x})^{\frac{1}{\sqrt{x}}}$

with $ ch(x)\equal{}\frac{e^x \plus{} e^{\minus{}x}}{2}$

hello, the searched limit is the Eulerian number $ e$, you can compute this with
the rule of L'Hospital.
Sonnhard.

The usual name for that function in English is $ \cosh$.

Wouldn't that be the symbol?  I call it hyperbolic cosine.

[hide] Let $ \sqrt {x} = z$

$ \therefore\lim_{x\to\infty}\left(\cosh\left(\sqrt {x + 1}\right) - \cosh\left(x)\right)^{\frac {1}{\sqrt {x}}}\right)\overbrace{\mapsto}^{\sqrt {x} = z}\lim_{z\to\infty}\left(\cosh\left(\sqrt {z^2 + 1\right) - \cosh(z)\right)^{\frac {1}{z}}}$

Now it is known by the relation of the Root and Ratio tests that

$ \lim_{n\to\infty}\sqrt [n]{f(n)} = \lim_{n\to\infty}\frac {f(n + 1)}{f(n)}$

$ \therefore\lim_{z\to\infty}\left(\cosh\left(\sqrt {z^2 + 1}\right) - \cosh(z)\right)^{\frac {1}{z}}$

$ = \lim_{z\to\infty}\frac {\cosh\left(\sqrt {(z + 1)^2 + 1}\right) - \cosh(z + 1)}{\cosh\left(\sqrt {z^2 + 1}\right) - \cosh(z)}$

$ = e$[/hide]

Another way to look at it:

By the Mean Value Theorem, $ \cosh(a\plus{}b)\minus{}\cosh a\equal{}b\sinh c$ for some $ c\in (a,a\plus{}b).$

$ \sqrt{x\plus{}1}\minus{}\sqrt{x}$ is asymptotic to $ \frac1{2\sqrt{x}}$ and $ \sinh x$ is asymptotic to $ \frac {e^x}2.$ Hence $ \cosh\sqrt{x\plus{}1}\minus{}\cosh\sqrt{x}$ is asymptotic to $ \frac{e^c}{2\sqrt{x}}$ for $ c$ between $ \sqrt{x}$ and $ \sqrt{x\plus{}1},$ and since $ \sqrt{x\plus{}1}\minus{}\sqrt{x}$ tends to zero, we can say that this is asymptotic to $ \frac{e^{\sqrt{x}}}{2\sqrt{x}}.$

Now raise that to the $ \frac{1}{\sqrt{x}}$ power. Since $ \left(2\sqrt{x}\right)^{1/\sqrt{x}}$ tends to $ 0,$ the limit is $ \left(e^{\sqrt{x}}\right)^{1/\sqrt{x}}\equal{}e.$


Solve  the  system  of  equation
$ x^3 \plus{} y^2 \minus{} 2 \equal{} 0 \\
x^2 \plus{} y^2 \plus{} xy \minus{} y \equal{} 0$

Please reply ASAP

First solv e the first equation for $ y^2$. In this way  we get  $ y^2 \equal{} 2 \minus{} x^3$,  and plugging in  this in the second equaton and solving for $ y$ we get $ y \equal{} \frac {x^3 \minus{} x^2 \minus{} 2}{x \minus{} 1}$, therefore by  inserting this in the second equation and after some simplification  we get to solve  $ x^6 \minus{} x^5 \minus{} x^4 \minus{} 3x^3 \plus{} 2x^2 \plus{} 4x \plus{} 2 \equal{} 0$.
But there exist only complex solutions for this polynomial equation.


Given that $ a, b, c$ are any real numbers, prove that;
\[ 9(a^6 \plus{} b^6 \plus{} c^6) \plus{} 6a^2b^2c^2 \geq 7(a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3) \plus{} 4abc(a^3 \plus{} b^3 \plus{} c^3)
\]

EDIT: sorry it was typo. I edited now. it is correct.

What about $ a \equal{} b \equal{} 0,c \equal{} \frac {1}{10}$?
Is it instead $ 9(a^6 \plus{} b^6 \plus{} c^6) \plus{} 6a^2b^2c^2 \geq 7(a^3 \plus{} b^3 \plus{} c^3)^2 \plus{} 4abc(a^3 \plus{} b^3 \plus{} c^3)$?

[hide="solution"]
From Schur and Muirhead
$ 2\sum a^6 \plus{} 6\sum a^2b^2c^2 \ge 2\sum a^4(b^2 \plus{} c^2) \ge 4\sum a^4bc$

So it suffices to prove

$ 7\sum a^6 \ge 7\sum a^3b^3$

Which is clearly true by muirhead Q.E.D :) [/hide]


The product of several distinct positive integers is divisible by ${2006}^{2}$. Determine the minimum value the sum of such numbers can take.

[hide]
$2006^{2}= 2^{2}\cdot 17^{2}\cdot 59^{2}$

By a little testing, we find the solution $59+59 \cdot 2+17+17 \cdot 2 = \boxed{228}$. Now, to prove that this is the minimum:
First, note that $17^{2}> 228$. So, this means that the two factors of $17$ must be kept in separate terms, as well as the two factors of $59$. Then, since the terms must be distinct, we must distribute the two $2$s, one to a factor of $59$, and the other to a factor of $17$. This gives the solution above. Now, if we were to have that the product is a larger multiple of $2006^{2}$, we would need either to multiply the extra factors into one of the terms in our sum, or add it at the end. Both would increase the sum; so, we have found our minimum. $\Box$ [/hide]


calculus
integration
function
calculus computations

If $ a > 0$, prove that
\[ \int_0^a y \int_{y^2/a}^{y} f(x,y) \textup{ d}x \textup{ d}y \equal{} \int_0^a\int_{x}^{\sqrt{ax}} f(x,y) \textup{ d}y \textup{ d}x.\]Deduce that
\[ \int_0^a y \int_{y^2/a}^{y} \frac{1}{(a\minus{}x)\sqrt{ax \minus{} y^2}} \textup{ d}x \textup{ d}y \equal{} \frac{\pi}{2}\]
Is the $ y$ in the middle of the 2 integral signs supposed to be there? If not would someone please explain - I drew diagrams but I can't see a reason for there to be an extra $ y$ there

[quote="aidan"]If $ a > 0$, prove that
\[ \int_0^a y \int_{y^2/a}^{y} f(x,y) \textup{ d}x \textup{ d}y \equal{} \int_0^a\int_{x}^{\sqrt {ax}} f(x,y) \textup{ d}y \textup{ d}x.\][/quote]

This is not true. Putting $ f(x,y)\equal{}1,$ gives $ \frac{a^3}{12}\equal{}\frac{a^2}{6}$.

Then I believe it's
\[ \int_{0}^{a}\int_{y^{2}/a}^{y}f(x,y)\textup{ d}x\textup{ d}y \equal{}\int_{0}^{a}\int_{x}^{\sqrt{ax}}f(x,y)\textup{ d}y\textup{ d}x.\]
and deduce that
\[ \int_{0}^{a}\int_{y^{2}/a}^{y}\frac{1}{(a\minus{}x)\sqrt{ax\minus{}y^{2}}}\textup{ d}x\textup{ d}y \equal{}\frac{\pi}{2}\]

Best way to see a change of order in a double integral is to draw the region. Then you can visually pick the limits off the graph for a simple one like this. You will notice that your $ x$ limits are

$ x \equal{} \frac {1}{a}y^2$
and $ x \equal{} y$

which intersect at $ y \equal{} a$, and $ y \equal{} 0$. So you are essectially integrating $ f(x,y)$over the region bounded by the curves given above. If you solve those for y as a function of $ x$, you get

$ y \equal{} \sqrt {ax}$
and $ y \equal{} x$

These intersect at $ x \equal{} a$, $ x \equal{} 0$, so your $ x$ limits become $ 0$ to $ a$, and since $ x \leq \sqrt {ax}$ on $ [0,a]$, your $ y$ limits become $ x$ to $ \sqrt {ax}$

So that gives the first result.


I have heard some of you guys say it. What does it mean?


search
inequalities
number theory
number theory unsolved

Find all natural $n$ such that [b]all[/b] numbers:$\binom{n}{1}, \binom{n}{2}, \cdots, \binom{n}{n}$ are odd

This is an immediate consequence of Lucas' theorem... use the search button for it

Pierre.

Is Lucas theorem on the high school level or rather Colledge?

http://www.mathlinks.ro/Forum/viewtopic.php?t=54624
http://www.mathlinks.ro/Forum/viewtopic.php?t=54880
http://www.mathlinks.ro/Forum/viewtopic.php?t=5567
http://www.mathlinks.ro/Forum/viewtopic.php?t=30781

Enjoy! :)

Sorry, but you didn't answer my question?
I want to ask if it's Ok when I use this theorem on Polish Olympiad.
Anyway I can't even find polish name of this theorem and proof.

Well, if you use it in an olympiad, I'd say that, of course it's possible! :) (But, maybe you'll want to wait for an advanced user's advice ;))
One proof : http://planetmath.org/?op=getobj&from=objects&id=3917

Don't you have some more simple proof. I don't understand this one al all :(

Deleted!!!
See Megus post below for an explaination!

Thanks Zetax. but I ment more simple proof of Lucas Theorem :)

I was too late  :(  :mad:  :mad:  This one is from Polish Olympiad which is still in progress  :mad:

ZetaX, it's no problem  :) . The problem is what to make against posting by some users problems which should be done at home with no exterior help. This is not the first time as for example inequality from current Polish NO appeared at least 3-4 times I suppose. Fortunately the first round (doing problems at home) ends on 5th December...next year I'll try to deal with it better somehow...

Firstly this task isn't from Polish MO :!: . It from book Number Theory by Wacław Sierpiński. And secondly it's nothing wrong with asking about theorems like Lucas Theorem which is well-known. The task is very different from the one in Polish MO, so don't accuse me of posting problems from current Polish MO :!:

Firstly: nobody said asking for theorems is wrong

Secondly: I'm sorry [b]dondigo[/b] - I haven't seen this "all" word and I thought you posted the question about how many odd numbers among $\binom{n}{0}$...$\binom{n}{n}$ are there? (which is from Polish Olympiad). In this case, it's my mistake  :blush:


My only timepiece is a wall clock. One day I forgot to wind it and it stopped. I went to visit a friend whose watch is always correct, stayed awhile, and returned home. There I  made a simple calculation and set the clock right. How did I do this when I had no watch on me to tell how long it took me to return from my friend's house ?

[size=75]
from book "The Moscow puzzles" by Boris A. Kordemsky

Please hide Your solution . Write inside Hide ... /Hide
[/size]

[hide=" solution"]
You wound the clock before you left. You knew how long you spent at your friend's house. Subtract this and divide by $ 2$ to get the time to return home. Add this to the time you left your friend's house.
[/hide]


calculus
integration
trigonometry
algebra
binomial theorem
calculus computations

May be a silly question but is expanding $ (2x^{2} \minus{} 5)^{7}$ (by the binomial theorem) and then integrating term by term the only way of doing $ \int (2x^{2} \minus{} 5)^{7} dx$ or are there any neater ways?

Expansion is pretty much the most straightforward way.  You COULD do a trig substitution but I can't begin to imagine how nasty that will be.  That integral has 8 terms plus a constant so yeah, you might as well just expand it.

If you don't want to take up too much paper, you could use summation notation. Write $ (2x^2\minus{}5)^7$ as

\[ \displaystyle\sum_{i\equal{}0}^{7} {7\choose i}(\minus{}5)^{7\minus{}i} (2^i) (x^{2i})\]

Then,

\[ \int\displaystyle\sum_{i\equal{}0}^{7} {7\choose i}(\minus{}5)^{7\minus{}i} (2^i) (x^{2i})\, dx\]

is the same as

\[ \displaystyle\sum_{i\equal{}0}^{7} {7\choose i}(\minus{}5)^{7\minus{}i} (2^i)\int (x^{2i})\, dx\equal{}\sum_{i\equal{}0}^{7} {7\choose i}(\minus{}5)^{7\minus{}i} (2^i)\frac{x^{2i\plus{}1}}{2i\plus{}1}\]


Let be given triangle $ABC$. Let I be its incribed circle center.The line $AI,BI,CI$intersect $BC,CA,AB$  at  $D,E,F$ respectively.Let $r_{1}, r_{2}$ be the radius of the increbed circle of the triangles $IDE,IDF$ respectively. Prove that:$r_{1}= r_{2}$ if and only if the triangle ABC is an isosceles triangle.


How many units apart is any pair of parallel sides of a regular hexagon with a side of 8 units?  Express your answer is simplest radical form.

[hide]The answer is 8sqrt3. You can make a right triangle, and the hypotenuse is 16, and the length of one side is 8. [/hide]


Find the number of odd integers between 20,000 and 69,999 (inclusive) which have a 5 as the middle digit and which don't have any number more than once.

Here is what I did:

I wrote the number as: $ nk5ji$

Next, I split it into two cases:

Case I: $ i \equal{} 3$

Then we have $ nk(1)ji$ choices

Or since $ i \equal{} 3$ $ \Rightarrow nk(1)j(1)$ choices

Now $ n$ can't be $ 3$ or $ 5$ so it has $ 3$ choices $ (2, 4, 6)$

Analogously, $ k$ has $ 7$ choices and $ j$ has $ 6$ choices

Giving us: $ (3)(7)(1)(6)(1) \equal{} 126$

Case II: $ i \neq3$

We have $ 3$ choices for $ i$, $ 4$ for $ n$, $ 7$ for $ k$, and $ 6$ for $ j$ givings us: $ (4)(7)(1)(6)(3) \equal{} 504$

I then added the two together $ (126 \plus{} 504)$to get $ 630$

Is that correct?

Yes.

[hide="Another Solution"]Two cases:

Case 1: The first digit is even. Three choices for the first, then four choices for the second, then seven choices and six choices for the other two digits. $ 3 \times 4 \times 7 \times 6 \equal{} 504$.

Case 2: The first digits is odd. One choice for the first, then three choices for the third, then seven and six choices for the other two digits. $ 1 \times 3 \times 7 \times 6 \equal{} 126$.

$ 504 \plus{} 126 \equal{} \boxed{630}$.

[/hide]

[hide="And another solution:"]

Look at the number as $ \__ \__ \,5 \__ \__$, where the blanks are to be filled with the the digits 0-4 and 6-9 under the given restrictions. The first digit must be either 2, 3, 4, or 6; and the last digit must be either 1, 3, 7, or 9.

If the first digit is a 3, then there are only 3 possibilities for the last digit, followed by 7 possibilities for the second digit, followed by 6 possibilities for the fourth digit - for a total of $ (3)(7)(6) = 126$ possibilities.

If the first digit is a 2, 4, or 6 (3 possibilities), then the last digit has 4 possibilities, followed by 7 possibilities for the second digit, followed by 6 possibilities for the fourth digit - for a total for $ (3)(4)(7)(6) = 504$ possibilities.

So the total number of possible numbers is $ (3)(7)(6) + (3)(4)(7)(6) = (3)(7)(6) (1 + 4) = (3)(7)(6)(5) = 630$[/hide]

Use this first
_ _5_ _
its 20000 through 69000 so the only possibilitys for the first coloum are
2,3,4,5,and 6
teh 2nd and 4th coloumb have 10 possibilities (it dosent matter in the construction of an odd number)
teh 3rd has 1 possibility cuz it has to be 5
and the last coloum has 5 possobilities (odd numbers)
so
5*10*1*10*5=2500
so there are 2500 numbers that meet the criteria
(might be wrong)

[quote]Find the number of odd integers between 20,000 and 69,999 (inclusive) which have a 5 as the middle digit [b]and which don't have any number more than once[/b].[/quote]

That's the reason why this problem isn't trivial.


iTest results are out :D

Our NJ team got 88 pts (12th place) :)

[quote="azjps"]Hmm, didn't [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1255470#1255470]AAST[/url] post that they scored a 97?

Our team isn't up there, though we might bump you guys off from the state winners on the updated list on Friday, sorry.  :wink:[/quote]
Hm that's right... I'll wait for the updated list. That means AAST could be #1 (since the highest on that list is 95).

Ideally, the AAST team miscounted 17 questions and is the other team listed.  :maybe: Maybe we'll stand a chance then.

Either way though, NJ will probably have the highest-scoring team to not be a state winner.

Edit: Oh, and is WWP sending a PuMaC team?

Yay. :) We counted a $ 92$, though I'm not complaining.  :lol: And congratulations to you guys also.


AMC AIME algebra polynomial function inequalities probability \/closed

What sections of the AIME and AMC 12 do the problem series mainly work on?

From my experience with the AIME Problem Series:
There isn't really a section. The problems have varied from the very beginning to #15 on the AIME. Each week quickly covers a topic that is likely to appear on the AIME. This is probably why the instructors suggest taking a subject class over a problem series if time is not an issue.

Here is the way we designed the difficulty ranges for the problem series:

AMC 10 covers problems 8ish to 25 on the AMC 10 (6ish to 20ish on the AMC 12).

AMC 12 covers medium to hard AMC 12 problems (mostly hard) and a few easier AIME problems.

AIME Problem Series covers all levels of AIME problems.

All three of these problem series are broken into 12 lesson days.  During each day, we cover topics under a general theme.  For instance, here is the lesson plan for the AIME A Problem Series:

Lesson I: Equations
Lesson II: Complex Numbers and Polynomials
Lesson III: Functions
Lesson IV: Inequalities, Optimization Problems, Sequences and Series
Lesson V: Counting
Lesson VI: Probability
Lesson VII: Number Theory
Lesson VIII: Number Theory (Algebraic Methods)
Lesson IX: Trigonometry and Analytic Geometry
Lesson X: Euclidean Geometry
Lesson XI: Euclidean Geometry
Lesson XII: Extending Your Geometric Range

The goal is to cover nearly every general area that could be covered on the AIME.  That's hard in 12 weeks, so like towersfreak mentioned above, we do recommend subject courses for students who want to make a greater commitment to the material.  However, the problem series are great for shaking the rust off areas of problem solving you don't use as much but know and learn a few more concepts that relate to that particular level of problem solving.  It's also simply good problem practice at the level that a student is working on.

do you think you could post the 12 week lesson plan for the amc 12 problem series as well?  thanks.

AMC 12 Topics:

Lesson I: Arithmetic, Quadratics, Absolute Values
Lesson II: Algebraic Techniques
Lesson III: Number Theory
Lesson IV: Polynomials, Logarithms, and other Functions
Lesson V: Binomial Expansion, Inequalities, Optimization, Systems of Equations
Lesson VI: Counting and Probability
Lesson VII: Statistics, Sequences, and Series
Lesson VIII: Complex Numbers, Trigonometry, Law of Cosines
Lesson IX: Right Triangles, Quadrilaterals, Pick’s Theorem
Lesson X: Circles and Spheres
Leson XI: Triangles, Polygons, Polyhedra
Lesson XII: Coordinates, Graphs, and Geometric Inequalities

[quote="chess64"]What about AMC 10? :D Thanks. Why aren't these on the main class page? :?[/quote]

We may include them on the main class page at some point.

Here is AMC 10:

Lesson I: The Mechanics of Algebra
Lesson II: Word Problems
Lesson III: Patterns, Sequences and Series
Lesson IV: Functions and Graphs
Lesson V: Systems of Equations and Inequalities
Lesson VI: Number Theory
Lesson VII: Counting Methods
Lesson VIII: Probability
Lesson IX: The Mechanics of Geometry
Lesson X: Triangles
Lesson XI: Circles
Lesson XII: Polygons and Solids


Hi everyone,

I maintain a site called Cubefreak dedicated to speedcubing and blindfold cubing. I know there some cubers on AoPS, so I wanted to post the following message:

KS, a regular on my site's chatterbox, is selling cube shirts. Since
he's just starting out, he can only produce a new shirt design when 25
people are sure to buy it (he needs to buy screens for each new design).

I have an idea for a shirt design:
Front: "Schrödinger's Cube is scrambled."
Back: "Schrödinger's Cube is solved."

Black t-shirt, white font. Maybe you've seen the original:
http://www.thinkgeek.com/tshirts/science/6dff/

Since I want this shirt myself, I'm trying to find 24 other people who
want to buy one. The current price for the shirt is \$18. Please send an e - mail to smakisumi at gmail if you'd be willing to buy one, with your name and one of the following messages:

Subject: Schrödinger's Cube
"I'd buy the shirt for \$18."
"I'd buy the shirt only if the price becomes lower."

Thanks so much,
-macky

A request from KS: please also include your size (S, M, L, XL, XXL)
and address in the message, and send the e-mail to cubewear at gmail.
I'll forward the messages I've already received to that address.

Thanks,
-macky


can you get your username changed?

i was wondering because i know someone named indianamath but now his name is unimpossible...i dont see how that happened.

I think you have to contact the admins about it

They only let you change your name for special reasons though... is there a certain reason why you would want your name changed?

Indianamath is now [url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=22655]unimpossible[/url]


group theory
abstract algebra
superior algebra
superior algebra solved

Give me examples of 2 infinite groups G & H such that G is isomorphic to a subgroup of H and H is isomorphic to a subgroup of G. But G and H are not.
Hint: Use Infinite direct product of groups...........

Let $n<m$. Let $F_n$ and $F_m$ be free groups on $n$ and $m$ generators respectively.
It is known that there exist monomorphisms $F_n\to F_m$ and $F_m\to F_n$. But $F_n$ and $F_m$ are not isomorphic. For example their abelianizations are $\mathbb Z^n$ and $\mathbb Z^m$ and these are not isomorphic if $n\neq m$.

[quote="amfulger"]It is known that there exist monomorphisms $F_n\to F_m$ and $F_m\to F_n$. [/quote]

Can you give me explicit examples,I know nothing about it.Can one find an example using infinite direct product of groups.

For the free groups an example for $n=2, m=3$ (it easy generalizes for all $m,n >1$):
let $F_m : = < x,y,z> , F_n : = <u,v>$.
$s: F_n \to F_m$ is given by $s(u) : = x , s(v): =y$.
$t: F_m \to F_n$ is given by $t(x): =u^2 , t(y): =v^2 , t(z) : = uv$.

A previous thread on this subject, with an abelian example: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=17801]Here[/url].

You've asked this question before, and been given the same example and link.

O sorry I forgot :blush:. I will try to be careful in future.


Find all functions $f: \mathbb{Z}\to \mathbb{Z}$ such that for all $m,n\in \mathbb{Z}$: \[f(m+f(n)) = f(m)+n.\]

$ m \equal{} n \equal{} 0$ gives $ f(f(0)) \equal{} f(0)$. $ n \equal{} 0$ gives $ f(m\plus{}f(0)) \equal{} f(m)$, but $ n \equal{} f(0)$ also gives $ f(m\plus{}f(0)) \equal{} f(m)\plus{}f(0)$. From the previous two statements we conclude that $ f(0) \equal{} 0$. Then $ m \equal{} 0$ gives $ f(f(n)) \equal{} n$.

Suppose $ f(k) \equal{} 0$ for some $ k$. Then $ n \equal{} k$ gives $ f(m) \equal{} f(m)\plus{}k$, so $ k \equal{} 0$. Now put $ m \equal{} f(1), n \equal{}\minus{}1$ in to get $ f(f(1)\plus{}f(\minus{}1)) \equal{} 0$, so $ f(1) \equal{}\minus{}f(\minus{}1)$.

We now prove by induction that $ f(k) \equal{} kf(1)$ for all positive integers $ k$. Base case $ k \equal{} 1$ is trivial. Suppose it is true for $ k$. Then $ m \equal{} k, n \equal{} f(1)$ gives $ f(k\plus{}1) \equal{} (k\plus{}1)f(1)$, completing the induction.

We also prove by induction that $ f(\minus{}k) \equal{}\minus{}kf(1)$ for all positive integers $ k$. Base case $ k \equal{} 1$ has already been established. Suppose it is true for $ k$. Then $ m \equal{}\minus{}k, n \equal{} f(\minus{}1)$ gives $ f(\minus{}k\minus{}1) \equal{}\minus{}kf(1)\plus{}f(\minus{}1) \equal{} (\minus{}k\minus{}1)f(1)$, completing the induction.

So we have $ f(k) \equal{} kf(1)$ for all integers $ k$. Let $ k \equal{} f(1)$, and we get $ f(1)^{2}\equal{} f(f(1)) \equal{} 1$. So $ f(1) \equal{} 1$ or $ f(1) \equal{}\minus{}1$. We see that $ f(k) \equal{} k$ and $ f(k) \equal{}\minus{}k$ both work, and they are the only solutions.

$n=-f(m)$ gives $f(m+f(-f(m)))=0$. Substituting $n$ by $m+f(-f(m))$ gives $f(m)=f(m)+m+f(-f(m))$, so $f(-f(m))=-m$. Substitute $n$ by $-f(n)$, we get $f(m-n)=f(m)-f(n)$. Letting $m-n=x,n=y$ gives Cauchy equation.

Let $P(m,n)$ be $f(m+f(n))=f(m)+n$.It is easy to show that $f$ is bijective.

$P(0,0)\Rightarrow f(f(0))=f(0)$.From injectivity,$f(0)=0$.

$P(0,n)\Rightarrow f(f(n))=n$.

Then $f(m+f(n))=f(m)+f(f(n))$.From surjectivity, $f(m+n)=f(m)+f(n)(\forall m,n\in \mathbb{Z})$.This is the [b]Cauchy equation[/b].So ∃$c\in \mathbb{Z}$ s.t. $f(m)=cm$.Substituting this into $P(m,n)$, we get $c^2n=n$.Then $c=\pm 1$, so $\boxed{f(m)=m, f(m)=-m}$.It is easy to check that both solutions satisfy the condition.$\blacksquare$

First by fixing m and moving n we get that f is a covering function .
If f(a)=0 then n=a , we get that f(0)=0 . Now we get f(f(n))=n (1).
If f(b)=1 then n=b , we get that f(m+1)=f(m)+b , so f(m)=mb (2).
By using (1),(2) we understand b=1 or -1 , From this we get f(m)=m , f(m)=-m

Let $P(m,n)$ denote the given assertion.

$P(m,0)$ forces $f(0)=0.$ Then $P(0,n)$ gives $f$ is an involution. Now $P(m,f(n))$ gives $f$ is additive. Hence $f(x)=kx.$ And we get $f(x)=x$ and $f(x)=-x$ only, which work.


[quote="Calculus"]I have already done past aime tests, challenging problems in geometry, and HMMT problems. I am still struggling with the mid-hard aime geometry problems. My goal is to be able to solve every aime geometry problem in a reasonable time. Where can I find problems at that level? I was thinking about taking the olympiad geometry class but it says its offered in the winter and I don't think I will have time then.[/quote]

You've been asking this same type of question in counting, conics, and bunch of other areas of math contest. My best advice is that if you went through past AIME tests, challenging problems in geometry, and HMMT problems, try to go attack some Olympiad short-answer problems for your question. The problems are found in AoPS or can be Googled.

Since this issue was asked by other people in the past as well, search the forum and the thread is locked.


abstract algebra
vector
algebra
polynomial
Ring Theory
superior algebra
superior algebra unsolved

Prove or disprove that the tensor product $ I \otimes_R J$ of two injective $ R$-modules $ I$ and $ J$ is again injective.

* The dual statement for projectives is true: one can use the adjoint isomorphism theorem for this. And if $ R$ is a P.I.D. then the above result is positive, because then a module is divisible if and only it is injective, and that makes the situation a lot easier. I don't know how to approach the general case though.

http://mathnet.kaist.ac.kr/mathnet/thesis_content.php?no=342205 could help.

  darij

For an example, take k to be a field, R to be k[x,y]/(x,y)^2, and I=J=E(k), the injective envelope of the simple module.  E(k) has k-dimension 3 and every injective is a direct sum of copies of E(k).  E(k) ⊗R E(k) has k-dimension 4, which is not a multiple of 3, so is not injective.

More specifically: Let I be the k[x,y] module (x,y)/(x^2,y^2) with k-basis {x,y,xy}.  Then check that (x,y)^2 * I = 0, so I is an R-module.  Then apply Baer's criterion to check injectivity: R has only one non-zero, proper ideal, (x,y).  Hom((x,y),I) has k-dimension 2, with basis elements sending x or y to xy (and the other to 0).  Hom(R,I) contains extensions of those homomorphisms, namely the maps sending 1 to y or x (and so x,y to xy,0 or 0,xy).  Hence I is injective.

(I ⊗R I)/(x,y) ≅ I/(x,y) ⊗R I/(x,y) = I/(x,y) ⊗k I/(x,y) has dimension 2*2 since I/(x,y) is a k-module with basis {x,y}.  Hence I ⊗R I has k-dimension at least 4.  On the other hand I ⊗k I has dimension 9 with basis vectors { x⊗x, x⊗y, x⊗xy, y⊗x, y⊗y, y⊗xy, xy⊗x, xy⊗y, xy⊗xy }.  Now in I ⊗R I, y⊗xy = yy⊗x = 0, and similarly for any of the basis vectors involving xy.  Hence I ⊗R I has k-dimension at most 4.  Hence I ⊗R I has k-dimension exactly 4.

This construction is called Northcott's inverse polynomials, and is described in textbook form as example 3.69 on page 94 of Lam's Lectures on Modules and Rings (actually it is described 5 or 6 times in that chapter, but this particular setup is worked out there).

The fundamental duality of flat and injective is initially described in:

Ishikawa, Takeshi. "On injective modules and flat modules."
J. Math. Soc. Japan 17 (1965) 291-296.
http://www.ams.org/mathscinet-getitem?mr=188272
http://dx.doi.org/10.2969/jmsj/01730291

The more complete version for commutative noetherian rings is described in:

Enochs, Edgar E.; Jenda, Overtoun M. G.
"Tensor and torsion products of injective modules."
J. Pure Appl. Algebra 76 (1991), no. 2, 143-149.
http://www.ams.org/mathscinet-getitem?mr=1145862
http://dx.doi.org/10.1016/0022-4049(91)90056-8

Both of these papers take as given the beautiful theory of injectives worked out by Macaulay, Northcott, and Matlis and described in chapter 3 of Lam's text.

The simple summary of the result is that tensor products of injectives are injective (over an associative, unital, commutative, noetherian ring) if and only if the injective hull of the ring is flat.  Rings with this property have been studied in a number of papers.

Ishikawa's work was extended in other ways as well, and a summary of many of those extensions for the non-commutative case is contained in Enochs&Jenda's textbook on Relative Homological Algebra.


inequalities
function
inequalities proposed

I think we can solve arqady's in the same manner as we did with the following nice one. \[\frac1{1+a^{2}}+\frac1{1+b^{2}}\geq\frac2{1+ab}.\]

I am sorry. I have corrected a mistake:
Let $ab\geq1.$
Prove that $\frac{1}{(a+2)^{2}}+\frac{1}{(b+2)^{2}}\geq\frac{2}{3(ab+2)}.$

[quote="Albanian Eagle"]arqady maybe you meant $ab\le 1$ :wink:[/quote]
No, $ab\geq1.$
By the way, if $ab\leq1$ then the inequality is wrong. :wink:

[quote="Albanian Eagle"]if you put $a=b$ you get $a^{2}\le 1$ :wink:[/quote]
Oh! You are right. Thank you. I have again corrected a mistake:
Let $ab\geq1.$
Prove that $\frac{1}{(a+2)^{2}}+\frac{1}{(b+2)^{2}}\geq\frac{2}{3(ab+2)}.$

yes now it's correct!
I solved it like this:
the requested inequality is equivalent to $ab(a-b)^{2}+2(ab-1)(a^{2}+b^{2})+4(ab-2)(a+b-2)\geq 0$ so it is obvious we should prove it only for the case $ab=1$ in this case it reduces to $\frac{(a+1)^{2}}{a^{2}}\geq \frac{4}{a}$ so the inequality is proved!  :)

[quote="Albanian Eagle"]
the requested inequality is equivalent to $ab(a-b)^{2}+2(ab-1)(a^{2}+b^{2})+4(ab-2)(a+b-2)\geq 0$ so it is obvious we should prove it only for the case $ab=1$ in this case it reduces to $\frac{(a+1)^{2}}{a^{2}}\geq \frac{4}{a}$ so the inequality is proved!  :)[/quote]
I don't understand your proof. Why we should prove it only for the case $ab=1$$?$ Thank you.

let me try to explain please,
as we see in the last inequality function everything is ok (ok means not rational), hence if we prove the inequality for its minimal value (of ab) so that means we are done for all ab>1.

davron

sure, but I did't say that, ok i will explain it later, thanks

Sincerely
davron

well ,
arqady you may ask what you didn't understand,

sincerely
davron

[quote="arqady"]My proof:
$\frac{1}{(a+2)^{2}}+\frac{1}{(b+2)^{2}}\geq\frac{2}{3(ab+2)}\Leftrightarrow$
$\Leftrightarrow(ab-1)(3(a-b)^{2}+4(\sqrt{a}-\sqrt{b})^{2})+$
$+(a+b-2)^{2}+(\sqrt{ab}-1)^{2}(ab+4\sqrt{ab}+3)\geq0.$ :)[/quote]

arqady , i think it is

$4(\sqrt{ab}-1)^{2}(ab+4\sqrt{ab}+3)$ and it is nice

 :)

[quote="Lovasz"]I had a solution, based on function's properties.
Tomorrow...[/quote]
What about your solution Lovasz ? Can you show ?

[quote="shyong"]
arqady , i think it is
$4(\sqrt{ab}-1)^{2}(ab+4\sqrt{ab}+3)$ [/quote]
Yes, you are right, shyong. Thank you.

What a pity, my solution is lost. The papers is now away, and I don't know where it is.
But the following is the main idea:
  - In this problem, I think we can normalize, i.e. let $a+b=C$, a large enough constant.
  - Change LHS to the form which contain $a+b$ and $ab$ only,...
  - Consider function $f(ab)$.

Is it possible?

Set $s=a+b\ge2, t=ab\ge1.$ Then we get
$\frac{1}{(a+2)^{2}}+\frac{1}{(b+2)^{2}}\ge\frac{2}{3(ab+2)}\Leftrightarrow$
$\Leftrightarrow(s-2)^{2}+(t-1)(3(s^{2}-4t)+4(s-2)+4(t-1))\ge0$
which is close to arqady's solution :)

[color=red]Generalization (see[/color] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=100817][color=purple]a and b only - 2[/color][/url][color=red]):[/color]
For every $0\le c\le2$ and for every positive $a,b$ such that $ab\ge1$ the inequality $\frac{1}{(a+c)^{2}}+\frac{1}{(b+c)^{2}}\ge\frac{2}{(c+1)(ab+c)}$ holds.


inequalities
logarithms
calculus
algebra unsolved
algebra

$ 12^x\plus{}15^x\plus{}16^x>25^x$

Greetings  :D

[quote="TALENTO"]$ 12^x \plus{} 15^x \plus{} 16^x > 25^x$

Greetings  :D[/quote]

[hide="My solution"]

$ 12^x \plus{} 15^x \plus{} 16^x > 25^x\Longleftrightarrow \left(\frac {12}{25}\right)^x \plus{} \left(\frac {15}{25}\right)^x \plus{} \left(\frac {16}{25}\right)^x \minus{} 1 > 0$

Let $ f(x) \equal{} \left(\frac {12}{25}\right)^x \plus{} \left(\frac {15}{25}\right)^x \plus{} \left(\frac {16}{25}\right)^x \minus{} 1$

$ \implies f'(x) \equal{} \left(\frac {12}{25}\right)^x\ln\left(\frac {12}{25}\right) \plus{} \left(\frac {15}{25}\right)^x\ln\left(\frac {15}{25}\right) \plus{} \left(\frac {16}{25}\right)^x\ln\left(\frac {16}{25}\right)$

As $ \ \ln x < 0\ \forall\ x\ \in (0,1),\implies f'(x) < 0\implies f(x)$ is a decreasing function.

As $ \ f(2) \equal{} 0$

$ \implies f(x) > 0\ \forall\ x\in( \minus{} \infty ,\ 2)$

$ \implies12^x \plus{} 15^x \plus{} 16^x > 25^x\ \forall\ x\in( \minus{} \infty ,\ 2)$
[/hide]

[quote="TALENTO"]$ 12^x \plus{} 15^x \plus{} 16^x > 25^x$

Greetings  :D[/quote]

[hide="My alternate Solution without calculus"]

$ 12^{x}\plus{}15^{x}\plus{}16^{x}> 25^{x}\Longleftrightarrow\left(\frac{12}{25}\right)^{x}\plus{}\left(\frac{15}{25}\right)^{x}\plus{}\left(\frac{16}{25}\right)^{x} > 1$

as we know that for $ \ x\in (0,1),\  x^i<x^j\Longleftrightarrow i>j$

$ \left(\frac{12}{25}\right)^{x}>\left(\frac{12}{25}\right)^2\equal{}\frac {144}{625}\Longleftrightarrow x<2$

$ \left(\frac{15}{25}\right)^{x}>\left(\frac{15}{25}\right)^2\equal{}\frac {225}{625}\Longleftrightarrow x<2$

and $ \left(\frac{16}{25}\right)^{x}>\left(\frac{16}{25}\right)^2\equal{}\frac {256}{625}\Longleftrightarrow x<2$

$ \implies\left(\frac{12}{25}\right)^{x}\plus{}\left(\frac{15}{25}\right)^{x}\plus{}\left(\frac{16}{25}\right)^{x}>\frac {144}{625}\plus{}\frac {225}{625}\plus{}\frac {256}{625}\equal{}1 \Longleftrightarrow x<2$
[/hide]

To be honest, my solution is exactly same as yours, makar.

For example, so ''find solve $ 3^x\plus{}4^x\equal{}5^x$'' did, didn't you? :)

[quote="kunny"]To be honest, my solution is exactly same as yours, makar.

For example, so ''find solve $ 3^x \plus{} 4^x \equal{} 5^x$'' did, didn't you? :)[/quote]

yes i do agree with you it is a trivial problem :lol:


algebra
polynomial
inequalities
calculus
integration
inequalities unsolved

5. Let $P(x)=1+a_1 x+\cdots+a_{n-1} x^{n-1}+x^m$ be a polynomial with complex coefficient. Suppose the roots of $P(x)$ are $\alpha_1, \cdots$ with
$|\alpha_1|>1$ for i=1 to j
 and
$\alpha_i| \leq 1$ for i=j+1 to n.

Prove that $\prod{\alpha_i| \leq \sqrt{|a_0|^2+|a_1|^2+\cdots}}$

Look in the algebra proposed problems for a topic of vess called " Inequalities for polynomials". I gave there a solutions using integrals and vess gave nice elementary solution. It is a really hard and nice problem.


topology
real analysis
real analysis unsolved

$ E$ is a compact set, $ \{E_\gamma\}$ is a series of compact set which cover the $ E$.Can we choose a Finite covering in
covering  $ \{E_\gamma\}$ to cover  $ E$? I think it's true ,but i can't prove it!

Take $ E\equal{}[0,1]$ and $ \{E_\gamma \}\equal{}\{ \{\gamma\} : \gamma \in [0,1] \}$ in the metric space $ (\mathbb{R},d)$ where $ d$ is the usual Euclidean metric.

yannick, thank you very much!
But i have another question .
If the index set $ \{\gamma\}\equal{}N$,can we get the conclusion?

thanks again!

Nope, just take $ (E_n)_{n \in \mathbb{N}}$ as follows: $ E_0 = \{1\}$ and $ E_n = [0,1 - 1/n]$.


trigonometry
analytic geometry
Euler
geometry
3D geometry
quadratics
algebra

I'm not sure how to do this problem, but it looked pretty easy in the beginning.
\[x^{6}=1\]
The obvious answers are 1 and -1 but there are more. What are those solutions?

[quote="Quevvy"]I'm not sure how to do this problem, but it looked pretty easy in the beginning.
\[x^{6}=1 \]
The obvious answers are 1 and -1 but there are more. What are those solutions?[/quote]
[hide]
Well you could take the sixth root of both sides leaving

$x=\sqrt[6]{1}$

[/hide]

[quote="7h3.D3m0n.117"][quote="Quevvy"]I'm not sure how to do this problem, but it looked pretty easy in the beginning.
\[x^{6}=1 \]
The obvious answers are 1 and -1 but there are more. What are those solutions?[/quote]
[hide]
Well you could take the sixth root of both sides leaving

$x=\sqrt[6]{1}$[/hide][/quote]
But then  $x=\sqrt[6]{1}=\pm1$ and that's what I already stated as an obvious answer but there are supposed to be 4 more answers besides 1 and -1.

let $a=x^{2}\Rightarrow{x^{6}=a^{3}}$
so..
$a^{3}=1$
$a^{3}-1=0$
$(a-1)(a^{2}+a+1)=0$
$a=1\Rightarrow{x^{2}=1\Rightarrow{x=\pm{1}}}$ or
$a=\frac{1\pm\sqrt{-3}}{2}=x^{2}$

if i didnt scrwe up then the other solutions are complex??? :maybe:

[quote="robinhe"][hide]The roots are $x=\cos\left(\frac{k\pi}{3}\right)+i\sin\left(\frac{k\pi}{3}\right)$ for $k=0,1,2,3,4,5$.[/hide][/quote]
k=0 doesn't work, because then x=0 and $x^{6}=0$

[quote="SM4RT"]it does..

$x=\cos(0)+i\sin(0)=1+0=1$
$x=\cos(\pi)+i\sin(\pi)=-1+0=-1$[/quote]

Since this is the Basic forum i think everybody would appreciate it if you didn't use trig

[quote="robinhe"]That's the only/simplest way to answer the problem: trig and complex numbers.[/quote]

lol then perhaps move it?

well how would u get other solutions? oh wait...give me an example of a solution other than $-1$ and $1$...? because didn't the original poster say something about other solutions??

Yeah, there are 6 solutions and I was only able to find two; $\pm1$.

..the trig solution by robinhe shows them all...

The roots to this are known as the sixth roots of unity.

Robinhe's solution is completely correct. Please don't doubt it. Later when you study further, you will see that it does not obstruct at all  :)

yes, there are only 2 [color=red][i]real[/i][/color] sixth roots of 1, [hide="but"] imagine a regular n-gon inscribed in a unit circle in the coordinate plane with one vertex at $(1,0)$. someone (de moivre?) proved that if an angle has its vertex at the origin, one ray on the positive x-axis, and another extending through one vertex of the n-gon, and the measure of that angle is $\theta$, then $(\cos\theta+i\sin\theta)^{n}=1$. This can be very helpful in complex number problems.[/hide]

The n-th roots of 1 are cis(2kpi/n) because if we rewrite in exponential form, we get $e^{2ik\pi/n}$, then clearly if we raise it to the nth power we get 1 (again by Euler's Formula). And this generates n solutions, so these are the only solutions.

2 real roots (1 and -1) but using demoives you can get the other 4 imaginary ones. Also you can use eulers but I still prefer demovies because i'm more familiar with it  :D

this definetly should not be in the hsb forum if we are using trig.

[quote="Quevvy"]I'm not sure how to do this problem, but it looked pretty easy in the beginning.
$x^{6}=1$
The obvious answers are 1 and -1 but there are more. What are those solutions?[/quote]
$\begin{equation*}\begin{aligned}x^{6}-1 &= 0 \hfill \\ (x^{2}-1)(x^{4}+x^{2}+1) &= 0 \hfill \\ (x+1)(x-1)(x^{4}+x^{2}+1) &= 0 \hfill \\ \end{aligned}\end{equation*}$

The roots that you're lookin' for are solving the equation $(x^{4}+x^{2}+1)=0$, but $\Delta < 0$, so they're complex roots.

Just wondering, how did you find the discriminant? The discriminant of a quartic is uber complicated, did you like set $a=x^{2}$ or something?

Not exactly.

Just:

$\begin{equation*}\begin{aligned}x^{4}+x^{2}+1 &= 0 \hfill \\ (x^{2})^{2}+x^{2}+1 &= 0 \hfill \\ x^{2}&= \frac{{-1 \pm i\sqrt 3 }}{2}\hfill \\ \end{aligned}\end{equation*}$

But at the end, is the same thing  :ninja:

Sorry that I didn't notice this earlier so I couldn't move it, but roots of unity are beyond HSB so this is moved.

More generally, the nth roots of unity can be expressed as

$w$, $w^{2}$, $w^{3}$, $w^{4}\cdots$ $w^{n}$, where $ w=e^{\frac{2\imath\pi}n}$.

[quote="Quevvy"][quote="boxedexe"]$x^{6}=e^{i2k\pi}\implies x=e^{i\frac{k\pi}{3}},\ k=\overline{0,5}$.[/quote]
Does $k=\overline{0,5}$ mean $k=0,1,2,3,4,5$?[/quote]

yes

We have:
$x^{6}=1\implies x^{6}-1=0$

Difference of Squares:
$\implies (x^{3}+1)(x^{3}-1)=0$

Difference and sum of cubes factorizations:
$(x+1)(x^{2}-x+1)(x-1)(x^{2}+x+1)$

The first and third give $1$ and $-1$, respectively.

Solving the two quadratics with the quadratic formula, you get:
$x=\frac{1\pm i\sqrt{3}}{2}$ and $x=\frac{-1 \pm i\sqrt{3}}{2}$

So the solutions to $x^{6}=1$ are:

$x=\pm 1,\;\;x=\frac{\pm 1 \pm i\sqrt{3}}{2}$


LaTeX
geometry
inequalities
inequalities unsolved

with a,b,c are three sides of a triangle:
prove that:
(p-b)(p-c)/(2p-a)+(p-a)(p-b)/(2p-c)+(p-a)(p-c)/(2p-b)<=p/4
wiht p:=a+b+c/2

sorry because i don't know how to use latex :blush:  :(  :(  :(

No problem, I will rewrite your question so or so since I use different notations:

[color=blue][b]Problem:[/b] Let a, b, c be the three sidelengths of a triangle ABC, and let $s=\frac{a+b+c}{2}$ be the semiperimeter of triangle ABC. Prove that

$\frac{\left(s-b\right)\left(s-c\right)}{2s-a}+\frac{\left(s-c\right)\left(s-a\right)}{2s-b}+\frac{\left(s-a\right)\left(s-b\right)}{2s-c}\leq\frac{s}{4}$.[/color]

And here the [i]solution[/i]:

Denote x = s - a, y = s - b and z = s - c. From the fundamentals of triangle geometry, it is well-known that these numbers x, y, z are positive. Now, we have 2s = a + b + c (since $s=\frac{a+b+c}{2}$), and thus

y + z + 2x = (s - b) + (s - c) + 2 (s - a) = 4s - (b + c + 2a) = 2s + 2s - (b + c + 2a) = 2s + (a + b + c) - (b + c + 2a) = 2s - a,

and similarly, z + x + 2y = 2s - b and x + y + 2z = 2s - c, and finally,

x + y + z = (s - a) + (s - b) + (s - c) = 3s - (a + b + c) = 3s - 2s = s.

Thus, the inequality in question,

$\frac{\left(s-b\right)\left(s-c\right)}{2s-a}+\frac{\left(s-c\right)\left(s-a\right)}{2s-b}+\frac{\left(s-a\right)\left(s-b\right)}{2s-c}\leq\frac{s}{4}$,

rewrites as

$\frac{yz}{y+z+2x}+\frac{zx}{z+x+2y}+\frac{xy}{x+y+2z}\leq\frac{x+y+z}{4}$.

But this inequality must be known to you! For the proof, I again cite http://www.mathlinks.ro/Forum/viewtopic.php?t=14289 , posts #2-#4.

  darij


[quote="sharkman"]The layout of a Little League playing field is a square, 60 feet on a side.  Overlay a rectangular coordinate system on a Little League baseball diamond so that the origin is at home plate, the positive x-axis lies in the direction from home plate to first base, and the positive y-axis lies in the direction from home plate to third base.

What are the coordinates of first base, second base, and third base?  Use feet as the unit of measurement.[/quote]
dude, analytic, dude, dude

I think you mean to say that the bases are 60 feet, not the entire field (this is how the real Little League field is layed out).[hide] If home plate is the origin (0,0) then first base is (60,0).  See what you can do from there.[/hide]

[quote="sharkman"]No help at all.

Can someone show me how to answer this question?[/quote]
Dude, what are you talking about? Analytic. That's the key to solving this problem.

From what I understand that this questions is asking me,

The home plate is 0,0
The 1st base is at 60,0
The 2nd base is at 60,60
The 3rd base is at 0,60

I don't think this question requires any anylytic.. unless I misunderstood it.

[quote="sharkman"]Are all the answers on the xy-plane?[/quote]
What else would they be on?
The problem tells you to turn the field into a rectangular coordinate system (that means xy) and to find the coordinates of the points.  Draw it out if you get confused.

And bpms and ckck, stop being mean :)  :mad: .

1st base: (60, 0)
2nd base: (60, 60)
3rd base: (0, 60)

this is rather easy. i think it should have been posted in classroom math.

u(sharkman) r aware that aops is NOT a homework or a "tutoring" site, aren't u? because, no offense, u don't exactly enhance the aops community, but instead post incredibly similar problems over and over again. you've yet to learn latex, so even reading your posts becomes a hassle. I understand that I don't have to read them, and that it's really not my business, but I feel that you should take others into consideration when posting. This is not an attack;it is a suggestion.

if its not an attack
why does it sound so....
mean?

yes the problem is easy
but people can get just confused
so dont be so critical

Maybe the reason is that ALL of his posts have been homework problems. He barely shows any work at all. AoPS is not for posting homework problems;I think we do those enough at school. It's about posting challenging problems that actually require you to think and not just apply some formula or something.


Solve the equation in complex numbers:

$ (z\plus{}1)^n\minus{}(z\minus{}1)^n\equal{}0$

[quote="monkeyman2"]Solve the equation in complex numbers:

$ (z \plus{} 1)^n \minus{} (z \minus{} 1)^n \equal{} 0$[/quote]

Since $ z\equal{}1$ is not a solution, we can divide by $ (z\minus{}1)^n$ and we get $ (\frac{z\plus{}1}{z\minus{}1})^n\equal{}1$ $ \iff$ $ \frac{z\plus{}1}{z\minus{}1}\equal{}e^{\frac{2k\pi}ni}$

And so $ z\equal{}\frac{e^{\frac{2k\pi}ni}\plus{}1}{e^{\frac{2k\pi}ni}\minus{}1}$ for $ k\equal{}1,2,...,n\minus{}1$


[color=purple]Em thay' trong muc duoi' avatar ko' kai' rate no' la` ji` vay a[/color]

Đấy là thông số đánh giá chất lượng các post của các thành viên, như vậy đa số các post của chú là ở mức kém, còn anh thì đa số là trung bình.

Cái này linh tinh quá, trẻ con hay bầu nhầm lắm :D

[quote="evarist"][color=purple] Gia' tri ? Gia' tri ji` a gia su em la` 2.7 la` the' nao` a[/color][/quote]
Anh vừa bị ông gemath chơi thì phải? :D Hắn ấy bầu mình toàn là Poor không hà :P Làm bây giờ dưới cả trung bình rồi, nhưng hắn ấy bầu đa số đều đúng  :rotfl:

Em rate cho anh Tuân là excellent rồi đó  :D

Của anh bị lão ấy chơi ác quá, lập nick khác để vote mãi mà cũng ko được trung bình :D ( Đương nhiên anh chọn các bài tử tế của anh để vote :) )  . Bây giờ đi vote 5* cho mọi người tham gia topic này nhé! Mà dạo này box Vietnam vui lên rồi đấy, các bác các chú tích cực post bài vào, vượt các box khác đê :P

[quote="tunganh"]Hình như cái này chỉ có từng người xem đc đúng ko ạ :P
Em chả thấy của người khác như thế nào :D[/quote]
Của tớ bây giờ là 1,3  :rotfl:

He` kho than bac'   :lol:  nhung em ko`n kho hon chua dc vote cho ai kung nhu chua dc ai vote :P

Mấy cái này chỉ để cho vui thôi chứ chả ai thấy rate của ai, mà cũng chẳng mấy ai bầu chọn một cách đàng hoàng. Có lẽ admin dùng nó để nhắc nhở các members!

[quote="The soul of rock"]Mấy cái này chỉ để cho vui thôi chứ chả ai thấy rate của ai, [b]mà cũng chẳng mấy ai bầu chọn một cách đàng hoàng[/b]. Có lẽ admin [b]dùng nó để nhắc nhở các members[/b]![/quote]
Đen 1: Em nhầm, trong các box Toán ĐH họ bầu cực kì nghiêm túc và rất nhiều người bầu.
Đen 2: Nói chung là đúng, nó dùng để cho chính thành viên đó xem các bài của mình có chất lượng hay không? Từ đó tự điều chỉnh và post cho đoàng hoàng.

P/S: ý tưởng thì hay, nhưng có 1 vài người dùng cái này để bầu xấu nhau thì không nên, không ai biết rate của người khác cả, việc chơi xấu là vô nghĩa. Vậy hãy bầu nghiêm túc vào :)

Liệu có cách nào biết thằng nào rate mình không anh ? Biết tên tuổi để đễ bề thanh tóan   :starwars:
Nếu trong ngày mai chưa thấy hồi âm là em cho anh POOR !   :lol:

Âý! Đừng làm vậy chứ? :P Anh không biết, chắc chỉ Admin mới biết ai bầu cho mình thôi :D Các Mod thì biết thông số rate của tất cả các thành viên. Mà chú đừng bầu Poor cho anh, anh đang bị dưới trung bình đấy  :rotfl:

Nói vậy thôi chứ làm người ai làm thế nhỉ  :P ! Excellent

[quote="N.T.TUAN"]Chỉ Admin mới biết ai bầu cho mình thôi :D [/quote]

Chết em rùi !  :P ! Sao anh không nói sớm một chút , làm em cho VV poor không  :D !

Em vừa được rate 1 lần,Chả biết có được Excellent không!

em chợt nảy ra 1 ý định đen tối thế này, mấy bác nghe coi  :D
Mọi người cứ bầu Excellent cho nhau vậy là ai cũng vui cửa vui nhà :P
Em mới được có 4,7 thui à  :rotfl: Mấy bác thử làm coi nhá, ai đồng em ý bầu cho còn ai trái lệnh em cho poor liền à :D Danger  :P

Không biết chú nào vote cho hungkhtn ở topic này đây http://www.mathlinks.ro/Forum/viewtopic.php?t=148547
Các chú quý nhau thì quý cho nó đúng chỗ, kẻo thành viên khác họ cười cho  :rotfl:
P/S: zai này, anh ko đồng ý và cũng không hưởng ứng, chú đừng bầu cho anh nhá! Xin chú đấy! :P

[quote="N.T.TUAN"]Sự đo lường về giá trị của member :D[/quote]
Ủa em mới pot có vài bài mà giờ đã rate 5.5 rồi
Thế là sao ạ

Có nghĩa là có người vote cho chú 6* rồi có người lại vote thấp hơn, thế thôi, lúc này đang có ít người vote cho chú. Nhưng chú cứ tham gia nhiều mà xem, số đó sẽ giảm đi nhiều đấy! :D

Vậy điểm Rate trung bình là bao nhiu nhỉ?(5 ạ?)


I don't know...I didn't even make Nats....can someone give me a sample of some of the target problems?

8. To win a coin flipping game, you must flip $ n \plus{} 1$ heads in a row, where $ n$ is the number of tails flipped so far. For example, flipping heads automatically wins, since 1 head in a row is equal to 0 + 1 (0 tails flipped), and flipping THH wins as well. What is the probability that you will win the game by flipping four heads in a row?

The wording of the last sentence (quoted verbatim except for the second sentence - that was just added for clarity) made me do infinite series junk :mad: and as a result I got an answer of $ \frac 3{256}$

Of course, the actual answer was

[hide]$ \frac {21}{1024}$[/hide]

Which was sooooo easy once I figured out how to do it. Just list them out :no:

lol that problem must sound really hard for me :mad:  :mad:  :mad:

According to the statistics, four people got a perfect on the target, which must mean at least those four people got that question right. :)

[quote="Yongyi781"]8. To win a coin flipping game, you must flip $ n \plus{} 1$ heads in a row, where $ n$ is the number of tails flipped so far. For example, flipping heads automatically wins, since 1 head in a row is equal to 0 + 1 (0 tails flipped), and flipping THH wins as well. What is the probability that you will win the game by flipping four heads in a row?

The wording of the last sentence (quoted verbatim except for the second sentence - that was just added for clarity) made me do infinite series junk :mad: and as a result I got an answer of $ \frac 3{256}$

Of course, the actual answer was

[hide]$ \frac {21}{1024}$[/hide]
how the hell did you go to infinite series? :?:  :!:   :P
Which was sooooo easy once I figured out how to do it. Just list them out :no:[/quote]

I thought the question asked that [b]given you win[/b], what is the probability that you flipped four heads in a row to win?

I thought #8 too, but when I did it after the competition, it seemed really easy to me.

I didn't like this year's target in general, because there was too much counting and probability.  :mad:

The question was pretty ambiguous...but I've seen [b]much[/b] worse as in like 4 or 5 possible cases... :wink:

I miscounted and got an answer of $ \frac{9}{512}$  :(  :(

i failed target with a score of $ 6$!!!!!!!!out of $ 16$, btw.

total score at mc nats: $ 27$!!!!! :rotfl:

epic fail.

How many did you get on sprint and target?

yongyi can u email me the nats tests?

[quote="Yongyi781"]Ummm.... and your e-mail address is...? :maybe:[/quote]
I emailed the request to ya :D

Someone voted the Triangleland problem as the hardest. Could he/she explain why he/she thought it was hard?

[quote="Bachukas"]Target 8 solve:

heads - H
Tails - T

Different situations:

TTTHHHH - $ 1/{2^7}$
THTTHHHH - $ 1/{2^8}$
TTHTHHHH - $ 1/{2^8}$
THTHTHHHH - $ 1/{2^9}$
TTHHTHHHH - $ 1/{2^9}$
THTHHTHHHH - $ 1/{2^{10}}$

$ P(A) \equal{} \frac {1}{2^7} \plus{} \frac {2}{2^8} \plus{} \frac {2}{2^9} \plus{} \frac {1}{2^{10}} \equal{} \frac {8 \plus{} 8 \plus{} 4 \plus{} 1}{2^{10}} \equal{} \frac {21}{1024}$[/quote]
I found that easy too! YAY!!!!!!!! NATS POTENT! :D

[quote="Yongyi781"]Yeah, it looked so easy afterwards.

Justin Ahmann (last year's written winner) mentioned that the "easy" way was just $ \frac 12 \cdot \frac 34 \cdot \frac 78 \cdot \frac 1{16}$, and I still haven't found out why that was so. :maybe:[/quote]

I think that $ \frac 12$ - probability that first throw is tails
$ \frac 34$ - probability that tails will show in 2 or less than 2 throws (otherwise the game would due)

$ \frac 78$ - probability that tails will show in 3 or less than 3 throws (otherwise the game would due)

$ \frac 1{16}$ - probability for four heads in a row

Yay! I thought #3 was easiest too because it was geometry-and geometry is my best subject! :P

[quote="Alysero"]Where do the 1/2^n come from. Where do they come out?[/quote]

Each throw has probability 1/2 , so if there are n throws, then probability for particular situation is (1/2 * 1/2 * 1/2 ... 1/2) - multiplying n times.
We get that probability is $ 1/2^n$

[quote="isiahbunshart10"][quote="dynamo729"]Definitely number 8. **scowls at counting and probability....**[/quote]

i thought number 8 was hardest also. i got 13/64 i think. what did u place at nationals[/quote]

11/6= 17= 169th. ... I dont like these people who think 10 points higher than me is an epic fail :P

[quote="Yongyi781"]So did I. Same score - 27 :(
Did you get 74th?[/quote]

@Yongyi781:
I must have done worse than you... I got 88th or something.

[quote="Quickster94"]According to the nats stats, the target problems in order of easiest to hardest are:
3,4,5,2,1,7,6,8
25 ppl got number 8 right[/quote]

Like everyone I talked to at nats said they missed number 1. What a surprise, numbers 1,7,6,8 were all Counting/Probability. :o  :o


invariant
ratio
analytic geometry
geometry theorems
geometry

Sorry, can somebody explain about Affine transformations:
when can we used them?
What do they do?
What changes and what remains invariant?
What figures can we obtain from others via these transformations?

Thanks

We can use them whenever the properties we want to prove depend only on incidence between lines and points and ratios in which points divide segments. They preserve parallelism, incidence relations and ratios in which a point divides a segment.

If we denote a point of the plane by its homogenous coordinates $(x,y,z)$, the transformation has the form $\left (\begin {array}{c}x'\\y'\\z'\end{array}\right )=\left (\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\0&0&1\end{array}\right)\cdot \left (\begin{array}{c}x\\y\\z\end{array}\right )$. This is necessary because we want the line at infinity to be invariant, i.e. the points having their $z$-coordinate equal to $0$ have to turn into points with the same property.

In fact, this is why the transformation preserves parallelism: two parallel lines concur on the line at infinity, so the images of the two lines must concur on the image of the line at infinity, which is also the line at infinity, so the images of two parallel lines must also be parallel.


Three married couples arrange a party. They arrive at the party one at a time, the couples not necessarily arriving together. They all, upon arriving, shake the hand of everyone already there, except their own spouse. When everyone has arrived, someone asks all the others how many hands they shook upon arriving, and gets five different answers. How many hands did he himself shake upon arriving?

A. 0
B. 1
C. 2
D. 3
E. 4

[hide="Answer"]
C
If the people are $ A_1,A_2,B_1,B_2,C_1,$ and $ C_2$, where two people with the same letter are married, then the first 3 to arrive must be all from different couples. (Simple casework proves this.) Then the order of the rest does not matter. An example order is $ A_1,B_1,C_1,A_2,B_2,C_2$, who, on arrival, shake 0, 1, 2, 2, 3, and 4 hands, respectively. Thus the person has shaken 2 hands, which gives answer C.
[/hide]


function
probability
calculus
derivative
algebra
domain
expected value

Suppose $ r\equal{}4$ animals are captured and tagged then released into the population of unknown $ N$ animals. A sample of $ n\equal{}3$ is then selected at random from the same population.  Let $ Y$ be the number of tagged animals among the $ n$.
Find
1) $ P(Y\equal{}1)$ as a function of $ N$
2) What value of N will maximize $ P(Y\equal{}1)$?

My answer
1) I used the probability distribution to get $ P(Y\equal{}1) \equal{} \frac{12(N\minus{}4)(N\minus{}5)}{N(N\minus{}1)(N\minus{}2)}$

2) I was thinking of differentiating my answer 1 and then set it to zero to isolate $ N$ but I it's tedious and I got stuck with a large polynomial. Can anybody help with this?

One caveat on the use of the derivative: this is discrete function, a function whose domain is some subset of the integers. Technically, it doesn't have a derivative.

OK, that's an overly pedantic objection - but you don't want the real number $ N$ that maximizes that, not unless it's an integer.

Getting practical: For what value of $ N$ would the expected value of $ Y$ be $ 1?$ Since $ E(Y)\equal{}\frac{12}{N},$ this happens when $ N\equal{}12.$ That's a first guess as to where the maximum might be. Now, just evaluate this probability for $ N\equal{}12$ and work backwards and forwards from there: $ N\equal{}11, N\equal{}10, N\equal{}13, \dots$ until the picture is clear. It's a small number of specific evaluations. We see that we get a tie: both $ N\equal{}11$ and $ N\equal{}12$ give $ P(Y\equal{}1)\equal{}\frac{28}{55},$ which is maximal.

(The real function you were trying to maximize probably has a maximum somewhere in the interval $ (11,12),$ which you don't care about.)


induction
combinatorics solved
combinatorics

An $n \times n$ table is filled with real numbers in such a way that no two rows are identical (viewed as vectors). Prove that it is possible to subtract a column of the table in such a way that  the resulting $n \times (n-1)$ table  also has pairwise non-identical rows.

Yes, it's not Catalan conjecture. But if I remember well, this problem is really well-known. It was given in the Tournament of the Towns as well as in the Kurschak Competition. Anyway, it's a very nice problem.

Yes, a nice problem with a graph-theoretic approach.
You may find a solution, in french, in the course on graphs available at :
http://www.animath.fr

Pierre.

It was also in Hungary and Australia somewhere in 70-80s.
Suppose it is not true. Then for any i<=n we can find at least two rows in which only ith elements are different. For any i fix define such a pair. Make a graph in which vertexes are the rows and edges join defined pairs. Then it has n vertexes and n edges, so it contains a cycle a1,a2,...,ak. This leads to a contradiction (a1 and ak differ in ith element, all other aj coincide in ith element).

there is also another proof.
Call a column bad if there are two rows coinciding after deletion of this column.
We claim for any table of n columns and m rows there are at most m bad columns by induction.
Basis is obvious.
Now take m+1. Suppose the contrary.
Write the first m+1 pairs of form (R1,R2,C1) that mean R1 R2 coincide after deleting colum C1.
By induction assumption the rows entering this pairs are all m+1. Moreover if we consider this as graph and edges the graph is conex (otherwise split into conex components and use induction)
Then as the graph is conex the last n-m-1 columns coincide at all rows so the induction step is proven


linear algebra
matrix
absolute value
linear algebra unsolved

Let $a,b,c$ be distinct real numbers such that $a+b+c>0$. Let $M$ be the set of $3\times 3$ matrices with the property that each line and each column contain all given numbers $a,b,c$. Find $\{\max \{ \det A \mid A \in M \}$ and the number of matrices which realise the maximum value.

[i]Mircea Becheanu[/i]

The maximum value is $a^{3}+b^{3}+c^{3}-3abc$.
The number of such matrices is $6$.

Here is my solution.We know that the absolute value of a determinant does not change with a permutation of the rows and columns.Hence assume WLOG that the first row is is given by $a,b,c$.Also fix the first element of the second row which is eithe $b$ or $c$ as $b$.This forces the determinant to be:
$a b c$
$b c a$
$c a b$
which is $3abc-(a^{3}+b^{3}+c^{3})$.As $a+b+c>0$,the result follows.
Also it s easy to see that the first row has 6 possible arrangements of $a,b,c$.Also,each of these arrangements gives exactly one determinant with the needed value.

Cheers
Bhargav


Well, AoPS has obviously changed over the last two weeks. One thing in particular that I'm wondering about...what are these bots that show up in the members list? What do they do? And is there any way to revert back to the old forum layout?

The bots sort of just crawl around the web site building up an index so that when you search on their respective search engine you'll find AoPS/Mathlinks in the search results.

[quote="filletwho"] And is there any way to revert back to the old forum layout?[/quote]

The old forum layout is history.  The new layout and server provides significant advantages in performance and programming.


trigonometry
function
ratio
analytic geometry
graphing lines
slope

Solve for x if $ \arctan \frac{x}{2} \plus{} \arctan \frac{2x}{3} \equal{} \frac{\pi}{4}$

(the a in arctan is supposed to be capatilized.. I'm not sure how to do that in latex.)

[hide]We begin by taking the tangent of both sides, that way we are not dealing with inverse tangent functions. That gives us $ \tan(\arctan \frac {x}{2} \plus{} \arctan \frac {2x}{3}) \equal{} \tan \frac {\pi}{4}.$ Note that $ \tan \frac {\pi}{4} \equal{} 1.$ Using the sum of tangents, we have
\[ \dfrac{\tan(\arctan \frac {x}{2}) \plus{} \tan(\arctan\frac {2x}{3})}{1 \minus{} \tan(\arctan\frac {x}{2})\tan(\arctan\frac {2x}{3})} \equal{} 1.\]
Since $ \tan(\arctan y) \equal{} y,$ for any number $ y,$ we have
\[ \dfrac{\frac {x}{2} \plus{} \frac {2x}{3}} {1 \minus{} \left(\frac {x}{2}\right)\left(\frac {2x}{3}\right)} \equal{} 1.\]
Simplifying this, we get $ 2x^2 \plus{} 7x \minus{} 6 \equal{} 0,$ which should give you the solution. [/hide]

fixed, thank you JBL.

[quote="limac"]Since $ \tan(\arctan y) \equal{} y,$ for any angle $ y$[/quote]
The arctangent function takes pure numbers (ratios of lengths) as its input and outputs angles; it doesn't take angles as input  :wink:  (Otherwise, looks fine.)

See this post:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1734694#1734694

for a nice way to do these problems (anything involving arctan addition/subtraction) with complex numbers.

Which is actually exactly the same as the method in this post, just cast in different language ;).  Indeed, we can see the tangent angle summation formula in the final expression in your signature.  (Note that this is not a criticism.)

Edit:
Hmm, I don't know if signatures are preserved for all eternity; the preceding poster's signature reads
[quote]$ \arctan\alpha \plus{} \arctan\beta \equal{} \text{arg}\{(1\plus{}\alpha i)(1\plus{}\beta i)\} \equal{} \arctan\left( \frac{\alpha\plus{}\beta}{1\minus{}\alpha\beta} \right)$[/quote]

Why does $ \arctan\frac{b}{a}\equal{}\text{arg}\{a\plus{}bi\}$?

[hide="Why is this incorrect?"]

So the LHS tells us that we have two angles which sum to 90 degrees. Further, we also have one angle whose tangent is x/2 and another whose tangent is 2x/3.

call the angles a and b, then we know that a+b = 90, which means b= 90 -a

There is an identity which states that:
$ \tan{a} \equal{} \frac {1}{\tan90^{\circ} \minus{} a}$

Do the substitution and I get tan a = 1/tan b, which means that x/2 = 3/2x, so $ x \equal{} \sqrt {3}$
[/hide]

EDIT: Please excuse my stupidity  :oops:

[quote="mathemagician1729"]So the LHS tells us that we have two angles which sum to 90 degrees.[/quote]
How do you get that? $ \frac{\pi}{4}\equal{}45^\circ$. :wink:

[quote="JBL"]Draw the picture: those are two different ways of refering to the slope of the same line.[/quote]

Ok, I think I understand it intuitively now.  Is there a way to prove this rigorously or is it mostly the same kind of argument?

I'm curious what rigor you think is lacking: for $ (a, b) \neq (0, 0)$, $ \arg(a \plus{} bi)$ is defined to be a particular angle, and the tangent of this angle is defined to be $ \frac{b}{a}$.  Actually, I come across this idea that simple proofs are non-rigorous reasonably often on the forum.  Sometimes mathematical facts are simple.  This particular fact is hardly anything more than a definition -- what should "rigor" mean in this case that is more important than clarity?

hello, the solution is
$ x\equal{}\frac{1}{4}\left(\minus{}7\plus{}\sqrt{97}\right)$
one need the addition formulas for the arctan function.
Sonnhard.

[quote="JBL"]I'm curious what rigor you think is lacking: for $ (a, b) \neq (0, 0)$, $ \arg(a \plus{} bi)$ is defined to be a particular angle, and the tangent of this angle is defined to be $ \frac {b}{a}$.  Actually, I come across this idea that simple proofs are non-rigorous reasonably often on the forum.  Sometimes mathematical facts are simple.  This particular fact is hardly anything more than a definition -- what should "rigor" mean in this case that is more important than clarity?[/quote]

Ah, I was just confused on if there was a proof for a distinct angle to correspond to every distinct slope or if it was defined to be so.  I mean I do think it's pretty obvious that this is so.  Also I think your post just clarified the $ \tan \frac {b}{a}$ part.

Agh I'm getting confused on what I was confused on now.

Um in conclusion, thanks JBL  :D

Also does anyone think it would be messy to use the argument method because instead of computing arctans we're solving for x which is in the arctan, or is it just me?


I am learning C++ (I already know java), and I was wondering if someone could tell me how to create multiple methods in one class in C++?

You can define as many methods as you want for each class in C; not sure what you're asking for there. There's an example of C++ multi-method definition here:

http://www.horstmann.com/ccj2/ccjapp3.html

okay, so its essentially the same as java, but you have to declare the methods at the top?

You can put them anywhere within the function, just like java
The only difference between c++ and java methods is with the access types and some other inheritance issues


This problem is from the Rose-Hulman 2005 math contest.  It is somewhat interesting, and I would sure appreciate seeing a slick solution.  :)  (I have found two solutions so far: one based on approximations and one that is rather labor-intensive.)  Here's the problem:

A list of consecutive integers beginning with $ 1$ is written on the chalkboard.  One of the numbers is erased.  If the average of the numbers that remain is $ 35\frac{6}{17}$, then find the deleted number.

The sum of the numbers from $ 1$-$ n$ is known to be $ \frac {n(n \plus{} 1)}{2}$. Now I'll find the least possible values and highest possible for each $ n$ and use it to bound where $ n$ is. The highest $ n$ can be is when the subtracted number is one. By using this, we get the upper bound average as $ \frac {\frac {n^{2} \plus{} n \minus{} 2}{2}}{n \minus{} 1}$. The lowest $ n$ can be is when the subtracted number is $ n$. By using this, we get the lower bound average as $ \frac {\frac {n^{2} \minus{} n}{2}}{n \minus{} 1}$. These two terms can be simplified to $ \frac {n \plus{} 2}{2}$ and $ \frac {n}{2}$. So we have $ \frac {n \plus{} 2}{2}$ ≥ $ \frac {601}{17}$ ≥ $ \frac {n}{2}$.
Multiplying it all by $ 34$, we get $ 17n \plus{} 34$ ≥ $ 1202$ ≥ $ 17n$ or $ 68\frac {12}{17}$ ≤ $ n$ ≤ $ 70\frac {12}{17}$. So we now know $ n$ must be either $ 69$ or $ 70$. Because the denominator of the average is $ 17$, that means that $ n$ is one more than a multiple of $ 17$, so $ n \equal{} 69$. Now, we just use simple math to find the answer. The sum of the numbers from $ 1$ - $ 69 \equal{} \frac {70*69}{2} \equal{} 2415$. Multiplying $ 35\frac {6}{17} \equal{} \frac {601}{17}$ by 68 gives $ 2404$. Because the difference between these two numbers is $ 11$, the number deleted was $ 11$.

Actually, I think that you've found the number of numbers in the original sum, not the deleted number.  Good approach, though!  I think that it is rather easy to proceed from here, although the arithmetic is fairly messy.  Any ways to save some labor?

Oops  :blush:

I'll edit it to include the answer.

As for any way to make the arithmetic easier, specifically which part do you find hard?

Hmmm... on second thought, that arithmetic was not messy at all.  It seems that I was overcomplicating things when I did the problem.  Thanks for an elegant solution.  :)


summer program
MathZoom
function
MATHCOUNTS
AMC 8

So what kind of things will we do the last week of mathzoom?

During the last week, the curriculum involves in depth topics (recursions, functional equations, generating functions, etc., depending on the group).  For the Red Group, we review the strategies for AMC 8 and Mathcounts, and some easy AMC 10/12 materials.  Last year the Red Group had a bridge building project during the last week, and we may do something similar this year.  There will be several contests: each group has its own contest, and then the M.C. Escher Math Olympiad, and the MathZoom Olympiad Team Challenge at the closing ceremony.

After class, there are group activities, and evening seminars.

There will be a field trip to Disneyland during one of the weekends.


I've got a problem, perhaps more than 1! :-) Would someone please work this and post the answer? It's a very straightforward multi-variable calculus problem. I just can't get what the book gets.

Find the unit normal vector $ N(t)$ to the curve $ r(t)\equal{}\left<t,2t,t^2\right>$ when $ t\equal{}1$.

Thanks,
Tom

[hide="Using the gradient which has nothing to do with colors  :) "]Well, we have $ r(t) \equal{} t \hat{i} \plus{} 2t \hat{j} \plus{} t^2 \hat{k}.$ So that means that $ \nabla{r} \equal{} \hat{i} \plus{} 2\hat{j} \plus{} 2t\hat{k},$ the gradient of the given function. Substituting $ t\equal{}1,$ we get $ \nabla{r(1)} \equal{} \hat{i} \plus{} 2\hat{j} \plus{} 2\hat{k}.$ So the unit normal is

\[ \frac{\nabla{r(1)}}{\|\nabla{r(1)}\|} \equal{} \frac{1}{3}\hat{i} \plus{} \frac{2}{3}\hat{j} \plus{} \frac{2}{3}\hat{k}
\]

$ \blacksquare$[/hide]

I have three stupid questions.

1.  Does "normal" not mean "perpendicular" in this context?  I note that the vector Karthik gave is tangent to the curve in question, not perpendicular to it.  If so, why is it not called the tangent vector?

2.  What does gradient have to do with colors?

3.  Why do you put hats on the $ i$s and whatnot?  It seems pretty cumbersome.

Q1: Isn't the gradient perpendicular to the given curve? I thought that gradients are always perpendicular (though it is true that you do take derivatives)...

Q2: If you go to PowerPoint/Photoshop/Flash or something, you can fill in shapes and stuff with backgrounds of gradients... this is when one color slowly blends into another (similar to a rainbow)

Q3: I usually put hats over my unit vectors just as a form of notation. I don't write $ i$ as a unit vector just to ensure that it isn't accidentally confused with $ i \equal{} \sqrt{\minus{}1}.$ Many people use this notation.

Hm, well, I like no hats, becuase I think that $ i$ [i]should[/i] be confused with the complex number.  I guess this might raise issues if you were in a complex vector space, though.

Hm, well, I'm on sort of shaky footing, but a gradient seems to be a vector which points in the direction of the greatest rate of change of some (real?) quantity, in this case, $ t$, which seems to me like it would be anything but perpendicular.  I am not on shaky footing when I say that the derivative of the curve (with respect to $ t$) is $ \langle 1,2, 2t \rangle$, and anyhow, the thing Karthik computed looks like a scalar multiple of that.

Anyhow, thanks for answering my questions, Karthik.

But to return to the conversation, if normal=perpendicular, how do we get a unique unit normal curve?  The only way I can think of involves finding a convenient plane to which we must restrict this vector.  Perhaps we could define a sort of unique "tangent plane" in the same way we define a tangent line, by taking the limit of the planes passing through our target point and two nearby points, as these two nearby points tend toward the target point.  I'm not sure if this would always work, but it doesn't seem too unreasonable.  You'd probably want nonzero curvature, though.

But anyhow, here such a limiting process is superfluous, as the curve conveniently lies in the plane $ 2x\equal{}y$.  The unit vectors perpendicular to the derivative $ \langle 1,2,2 \rangle$ which fit into this plane are $ \pm \frac{1}{3\sqrt{5}} \langle 2,4,\minus{}5 \rangle$.

(Note :  It is possible that I just spewed useless nonsense.)

[quote="Boy Soprano II"]Hm, well, I like no hats, becuase I think that $ i$ [i]should[/i] be confused with the complex number.  I guess this might raise issues if you were in a complex vector space, though.

Hm, well, I'm on sort of shaky footing, but a gradient seems to be a vector which points in the direction of the greatest rate of change of some (real?) quantity, in this case, $ t$, which seems to me like it would be anything but perpendicular.  I am not on shaky footing when I say that the derivative of the curve (with respect to $ t$) is $ \langle 1,2, 2t \rangle$, and anyhow, the thing Karthik computed looks like a scalar multiple of that.

Anyhow, thanks for answering my questions, Karthik.

But to return to the conversation, if normal=perpendicular, how do we get a unique unit normal curve?  The only way I can think of involves finding a convenient plane to which we must restrict this vector.  Perhaps we could define a sort of unique "tangent plane" in the same way we define a tangent line, by taking the limit of the planes passing through our target point and two nearby points, as these two nearby points tend toward the target point.  I'm not sure if this would always work, but it doesn't seem too unreasonable.  You'd probably want nonzero curvature, though.

But anyhow, here such a limiting process is superfluous, as the curve conveniently lies in the plane $ 2x \equal{} y$.  The unit vectors perpendicular to the derivative $ \langle 1,2,2 \rangle$ which fit into this plane are $ \pm \frac {1}{3\sqrt {5}} \langle 2,4, \minus{} 5 \rangle$.

(Note :  It is possible that I just spewed useless nonsense.)[/quote]

It is true that there are multiple (actually infinite) "normals" to the curve. But the procedure about finding this normal is to find the plane which is "tangent" to the curve (also known as the "kissing plane") and fine the normal to this plane. I think there's a cross product involved somewhere which ensures one direction... it's probably a torque-like cross product like $ \hat{r} \times z$ or something where $ \hat{r}$ is the radial vector pointing away from the center of curvature and $ z$ is the instantaneous velocity of the curve. This vector would make the normal unique.

EDIT: I just realized, that's what you just stated above, more or less.  :P


Galois Theory
superior algebra
superior algebra unsolved

Let p(x) = 5x^3 + x+1, and let E be a splitting field of p(x) over the rationals Q. Determine all elements of Gal (E:Q). Further, determine [E:Q].

$ P\equal{}x^3\plus{}px\plus{}q$ with $ p\equal{}1/5,q\equal{}1/5$. $ P$ has not any rational root then is irreducible. $ D\equal{}\minus{}4p^3\minus{}27q^2<0$ ; thus $ D$ is not the square of a rational number and the Galois group of $ P$ is $ S_3$.
Let $ Q\equal{}x^2\plus{}27qx\minus{}27p^3$ and $ w$ be one of its roots. Let $ u$ be s.t. $ u^3\equal{}w$ and $ v\equal{}\minus{}3p/u$. Then the roots of $ P$ are $ a\equal{}(u\plus{}v)/3,b\equal{}(uj^2\plus{}vj)/3,c\equal{}(uj\plus{}vj^2)/3$ where $ j\equal{}e^{2i\pi/3}$.
The $ 6$ elements $ \sigma_i$ of Galois group are defined as follows (with the associated permutation) $ \sigma_i(j)\equal{}j$ and
$ \sigma_1(u)\equal{}u$ , $ id$
$ \sigma_2(u)\equal{}uj$ , $ (acb)$
$ \sigma_3(u)\equal{}uj^2$ , $ (bca)$
$ \sigma_4(u)\equal{}v$ , $ (bc)$
$ \sigma_5(u)\equal{}vj$ , $ (ab)$
$ \sigma_6(u)\equal{}vj^2$ , $ (ac)$


Let f(x)=ax^2+bx+c.
Find f_n(x)=fofo...of(x)(n times)

I am sure you ment some thing else. This cannot be called a problem.

Yes, thta Is a problem, but who can solve it?

Fermat great theorem is also a problem, but who can solve it? Of course, Wiles is an exception!

What do you want to say, amfulger? This is not a solution.I think the proposer wanted an expression for f_n(x) depending only on x.

I didn't say I posted a solution.
I just gave a recursive (better that the first) method for computing f_n. If you think it can be done better, be my guest.


1. What is the probability that a woman with three children has more than 1 girl given that she has at least 1 girl?

2. A woman gave birth to a quintuplet. The probability of the woman giving birth to a boy is 0.5. What is the probability that the woman will give birth to 2 boys and 3 girls given that one of her child is a boy?

[hide="Solution 1"]
Given that she already has one girl, the only way she could NOT have more than 1 girl is if the other two children are boys. This occurs with probability (1/2)*(1/2). To obtain the desired probability, we now take the complement of this for 1-(1/4) or [b](3/4)[/b]
[/hide]

[hide="Solution 2"]
If she already has one boy, there must be ONLY one boy amongst the remaining four children to meet the restriction of 2 boys and 3 girls. One such combination is BGGG out of a total of four similar. So, we have four successful outcomes out of a total of 16 for 4/16 or [b]1/4[/b].
[/hide]

"At least one of several children is male" and "the first of several children is male" are not the same statement: if you randomly select a large group of people, the probability that the first one will be male is $ \frac{1}{2}$ no matter how large the group is, but the probability that at least one member of the group is male gets very close to 1 very quickly.

Similarly, it's twice as likely that flipping a fair coin twice gives you one head and one tail as it is that you'll get two heads.

These problems are simple enough that I would do them by direct counting of possible outcomes.

the answers were 4/7 and 10/31. Any ideas how?

Hmm...

For number 1, how can it be 4/7? Since the woman must have at least 1 girl, our only possibilities are

Girl, girl, girl
Girl, boy, girl
Girl, boy, boy

So isn't the probability 2/3?

@The Archer, I think in your case, the compliment should have been $ \dfrac{\dfrac{1}{4}}{\dfrac{3}{4}},$ the denominator can't be 1 because only 3 out of the 4 possibilities have one girl. So, that would have given us 1/3, and subtracting that from 1 gives 2/3.

1. A woman can have children in the following ways:

3 boys, no girls (BBB)
2 boys, 1 girl (GBB, BGB, BBG)
1 boy, 2 girls (BGG, GBG, GGB)
no boys, 3 girls (GGG)

for a total of total $ 8$ ways.
However, the woman has at least one girl, so we throw out the case of BBB, leaving $ 7$ total ways.
Since there are $ 4$ ways for the woman to have more than one girl (GGG, BGG, GBG, GGB), the probability that she has more than one girl given that she has at least one girl is $ 4/7$.

Problem 2 can be done in a similar manner.

perpetualmotion, aren't BGG, GBG, and GGB the same thing? They represent having two girls and a boy. In the context of this problem though, shouldn't the order not matter?

[quote="limac"]In the context of this problem though, shouldn't the order not matter?[/quote]

They have the same end result insofar as that the family ends up with two girls and one boy, but that does not mean that it is equally likely for a family to have two girls and one boy versus having three girls.


AMC AIME probability counting distinguishability geometry

In these problems, I am hoping for someone to show me what i did wrong, so thanx in advance .

First....

Nine tiles labeled 1,2,3...9 are randomly divided among three players, 3 tiles each. Find the probability that the sum of each player's tiles is odd.

[hide]the answer for this was 3/14, but i got 3/7. I did

(5C3)(4c2)(2c1) as the total count of getting this odd feature. This stems from the first guy getting 3 odds out of 5, the second getting 2 evens out of the 4 possible, and then he gets 1 odd out of the 2 remaining. The last guy gets whatever is left. Then I divided this by (9C3)(6C3)/3! as the total count. I think that i possibly should have divided the count of the odd feature by 2 because i could be counting when the first guy gets AB and the second guy gets CD and when the first guy gets CD and the second gets AB, which are no different.[/hide]

And...

Find the number of ordered 4-tuples (a,b,c,d) of odd pos ints with sum 98.

[hide]We have that a = 2g+1 ... b = 2h+1 ... c = 2j+1... d = 2k+1,

and plugging these into a+b+c+d = 98, we get that h+j+k+g = 47. These 4 ints can be any pos int, so the answer is just 46C3.

But the answer given is about 4000 greater. I dont see what i did wrong on this problem.[/hide]

For the second one..

[hide]

Where did you get 46C3? Is that the number of ways of selecting the first three positive integers? that won't work, because, what say the first three are 30, 29, and 28? That shouldn't count as a possibility.

And don't forget the question asks for ordered.

[/hide]

TripleM -- [hide]he was using a divider method.  That is, there are 46 gaps between the 47 objects and he needs to pick 3 of them to make the division into 4 sets.  The real problem is that the numbers could be 0, as well -- you need to use (2n - 1) for n :ge: 1 in order to generate positive numbers.  (2n + 1) gives you n :ge: 0[/hide]

For the first one, [hide]you are correct about why you got it wrong -- you need to divide by 2 because you are treating the people as indistinguishable (which is also why you divided by 3!).

If you were treating the people as distinguishable, you would not divide by 3! in the denominator, but would multiply the numerator by 3 in order to account for the 3 choices of guy who can get odd-odd-odd.[/hide]

Does that make sense?

response:

[hide]we have that a+b+c+d = 98 and they all have to be odd pos ints, so they all equal 2j +1 for some j for each int. After we substitute this in for the corresponding int, we get that these new values, j_1, j_2, j_3, and j_4 sum to 47. The thing is that these can be any pos ints such that they sum to 47. And what do those numbers 30, 29, 28 ur saying correspond to? I dont understand where they come from...

O, and, the number of ordered arrangements of a set of n pos ints such that a_1 + a_2 + ... a_n = P is (P-1)C(n-1). Thats where the 46C3 comes from.

[/hide]

jelyman -- [hide]TripleM thought you were doing something like picking 3 numbers 46 or less (46C3) and then saying "the other one is the left-over part."  He didn't realize you were using dividers.

Did you see my earlier comment explaining what you did wrong?[/hide]

Cinjon, for #2, instead of using a, b, c, and d, use 2w + 1, 2x + 1, 2y + 1, and 2z + 1 instead of a, b, c, and d where w, x, y, and z are all greater or equal to 0.  And then add them up and you'll see you'll have 2(w+x+y+z) + 4 = 98.  w+x+y+z = 47, so then there are 50 C 3 ways, not 47 C 3 or whatever you had.   From there on it shouldn't be too hard.  Joel is right, for what you did you'd need it to be in the form 2n - 1 though, but this method could still work, you just need to account for 0.

jese.. yannik, at least read the post...
And yes, JBL, i see what i did wrong... should have noticed that. I dont think its so bad though, just took this aime tonight and in 2 hrs (i had gone to a math comp earlier today and was just about dead, so i didnt do a third hour), i got a 10 minus these 2. So, this is very promising.

i remember looking at this aime, it was pretty easy if i recall. i liked the two mathematicians question.

any suggestions for learning a bunch of similar triangle, geometry like concepts. usually i can get some using what i've learned from mu alpha theta, but there's still a lot that i don't know. i'm talking about things i hear like ceva's theorem, stewart's theorem, and others. you get the idea.

should the art of problem solving volume one be enough?

thanks
-ben

For the second one, I have a different way to do that. a+b is an even number, and c+d is an even number. (2, 96), (4, 94), (6, 92)......(96,2). There is one possibilily for 2, and 48 possibilities for 96.  There is 2 possibilities for 4, 47 posibilities for 94.....  So the total equals  :Sigma: n(49-n) where 0<n<49.  :Sigma: n(49-n)= :Sigma: 49n-n :^2: = :Sigma: 49n- :Sigma: n :^2: =49(49)(48)/2-(48)(49)(97)/6=48*49*50/6=19600

Again, for the second problem...

The number of ways that you can sum four odd pos integers to 98 is equal to the number or ways to sum four pos even integers to 102, which is equal to the number of ways to sum four pos integers to 51. This is simply 50C3.

[/hide]

We have already been through that throughout the whole thread.

"
w+x+y+z = 47, so then there are 50 C 3 ways

"

could someone please explain this?

Read the first 4 to 5 posts of the thread -- together, I think they explain it reasonably well.


Can anyone explain what it is and what can you use it for?

If you have, in triangle $ABC$, three concurrent segments, $AD$, $BE$, and $CF$ where $D$ likes on $BC$, $E$ on $AC$ and $F$ on $AB$, then
$AF*BD*CE=AE*BF*CD$
There are places where this can be used to prove equalities, etc.

I must say Trig Ceva's is also useful.

Given K81o7's triangle, $\frac{\sin{ABE}}{\sin{CBE}}\cdot\frac{\sin{BCF}}{\sin{ACF}}\cdot\frac{\sin{CAD}}{\sin{BAD}} = 1$

In my proof of Ceva, use the fact that $S=\frac{1}{2} ab \sin{C}$, and we get the trig form.


In the circle with center $ Q$, radii $ AQ$ and $ BQ$ form a right angle. The two smaller regions are tangent semicircles, as shown. The radius of the circle with center $ Q$ is 14 inches. What is the radius of the smaller semicircle? Express your answer as a common fraction.
[asy]draw((1,0)..(0,1)..(-1,0)..(0,-1)..cycle);
draw((-1,0)--(0,0)--(0,-1));
draw((0,0)..(-.5,-.5)..(-1,0));
draw((0,-1)..(-1/3,-2/3)..(0,-1/3));
label("$Q$",(0,0),NE);
label("$A$",(-1,0),W);
label("$B$",(0,-1),S);[/asy]

Refer to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1388227#1388227]this[/url]; the answer is $ \boxed{\frac{14}3}$.


How many quarter-inch cubes would be needed to form a solid one-inch cube?

It is easy to see that this is $ 4^3\equal{}\boxed{64}$.


The set {1,2..,3n} is partitioned into three sets A,B and C with each set containing n numbers. Determine with proof if it's always possible to choose one number out of each set so that one of these numbers is the sum of the other two.

Indeed, it is beautiful question, but it appeared many times here... ;)  (so that I'm wondering if there would be no solution using Beatty's sequences :D )

I think that last time is :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=schweitzer&t=32567

You will find a link to the solution and to a generalization.

Pierre.

Supposing that {1,2,...,3n} is partitioned into three sets A,B and C, each set containing n numbers.
Take that (a,b,c) as good if a in A, b in B, C in C and one of the numbers a,b,c is the sum of the remaining two.
Assume that 1 in A and, if k is the smallest number not in A, that k in B. Assuming there're no good triples, we claim if x in C then x-1 in A then it contradicts.
Clearly, if C={c1,...,cn} then A contains the numbers c_1 -1, c_2-1,...,c_n-1, (all num.>1 since 2 isn't in C).
But 1 in A, so A have at least n+1 elements.
*Now,let's prove the claim.
i) Assume it's true.Then there is a number x in C such that x-1 isn't in A so x-1 isn't in B (otherwise (1,x-1,x) will be good).
ii) For x in C and x-1 in C, proving x-k is in C and x-1-k in C:
......If x-k is in A then (x-k,k,x) is good, if x-k is in B then (k-1,x-k,x-1) is good.(similiarly we can repeat this argument and ) x-ik, x-ik-1 are in C for i=1,2,...: But x-ik will be one of 1,2,3,...,k for some i.
Hence it will be en element of A or B,contradiction. So x-1 is in A and it will be proved.

[quote="matematikator"]pbornsztein! Thanks for your links...each link was very usefull...  :?[/quote]

turkce konush kardesh

tut turki ili je ruskie ya ne ponal?


Is there somewone able to prove this?
If a,b,c are positive reals prove that:

     8*(a+b+c)[ab*sqrt(ab)+bc*sqrt(bc)+ca*sqrt(ca)]>=9*(a+b)*(b+c)*(c+a)

this ineq is not correct because degree of each part of the sum on the left is equal to 4 and on the right it is equal to 3


socom 2 comes out tomorrow, november 4, 2003, and i was just wondering who else was going to buy it and if they do, care to join my to be clan?

Bleh...might as well wait til PS3 comes out...I dun wanna spend money on that...

you dont want to spend 50 dollars?

ps3 comes out in 2006 most likely buddy


function
advanced fields
advanced fields unsolved

This is very very cute: suppose that $A$ is a complex Banach algebra such that $\parallel x^{2}\parallel =\parallel x\parallel ^{2}$ for all $x\in A$. Then $A$ is commutative.

can you show us the solution? I tried to work out the gelfand-transform $ A \to C(Spec A)$ in the non-commutative case, which is isometrically in our case, but this failed.

There are some tricky steps: first, observe that by the spectral radius theorem, $ \parallel x\parallel $ will equal the spectral radius of $ x$. But it is very easy tp see that $ xy$ and $ yx$ have the same spectral radius, so $ \parallel xy\parallel =\parallel yx\parallel $. Now, for fixed $ x,y$ look at the function $ exp(zx)yexp(-zx)$ and apply to it Liouville's theorem to deduce that it is constant. From here everything is obvious.

$ \lVert xy\rVert=\lVert yx\rVert$ also follows from $ \lVert xy\rVert^{2^{n}}=\lVert x(yx)^{2^{n}-1}y\rVert\leq\lVert x\rVert\,\lVert yx\rVert^{2^{n}-1}\,\lVert y\rVert$. owk


Let $ w\equal{}a/b$ etc. with $ wxyz\equal{}1$. Then
\[ 7w^{10}\plus{}x^{10}\plus{}y^{10}\plus{}z^{10}\geq 10w^6\]
\[ 3z^6\plus{}2y^6\plus{}x^6\geq 6z^3y^2x\]
But $ z^3y^2x\equal{}\frac{bcd}{a^3}$. So summing both inequalities cyclically yields the result.


Does anyone know how to prove that the number of solutions to $x^{2}+y^{2}=m$ is not bounded for choices of $m$.

I have seen this statement many times without proof, but I don't know how to prove it myself. Thanks

The number of representations gets big when the number of prime divisors $\equiv 1 \mod 4$ of $m$ is big.
Let $n=a^{2}+b^{2}=c^{2}+d^{2}$ and $q=x^{2}+y^{2}=x^{2}+(-y)^{2}$ be two different representations of $n,q$ with $q$ prime and $q \nmid n$ and all numbers $\neq 0$ (thus $n$ no perfect square since only then this can happen). Then we get four representations of $nq$ by:
a) $(ax-by)^{2}+(ay+bx)^{2}$
b) $(ax+by)^{2}+(-ay+bx)^{2}$
c) $(cx-dy)^{2}+(cy+dx)^{2}$
d) $(cx+dy)^{2}+(-cy+dx)^{2}$
Lets show they are all different:

Compare a) to b):
$ax-by=ax+by , ay+bx=-ay+bx \implies by=0=ay \implies$ contradicting that all are $\neq 0$.

Compare a) to c):
$ax-by=cx-dy , ay+bx=cy+dx$
$\implies x(a-c)=y(b-d) , y(a-c)=-x(b-d)$.
Write $s=a-c, t=b-d$ (they can't be both $0$ cause of distinctness of the representations of $n$), then it is $sx=ty , sy=-tx$. Multiply them to get $xys^{2}=-xyt^{2}\implies xy(s^{2}+t^{2})=0$, which is only possible when $x=0, y=0$ or $s=0=t$, contradiction.

Compare a) to d):
Thats the ugliest case, I will let it to you ;)
It's the one and only case you will need Number Theory for.

All others follow from these by changing the names of the variables.
So we know that $nq$ has twice that much representations than $n$.

So take $k$ different primes $\equiv 1 \mod 4$, by the above their product will have at least $2^{k}$ different representations.

Well, all this would be so easy in Gaussian integers:
the product of different sets of Gausian primes (seen as equal when equal up to units) is always different by unique factorisation. To get different numbers with same modulus^2=norm, just take $\pi_{1}$ into one and $\pi_{2}$ into the other set where $\pi_{1}\pi_{2}=p \equiv 1 \mod 4$.

With gaussian integers and primes one can show more:
[quote="Formulary"][b]2)[/b] Sum of two squares:
A positive integer $n$ can be represented as sum of two perfect squares iff all prime factors $p \equiv 3 \mod 4$ of $n$ occur an even number of times in the factorisation of $n$. $n$ can be written as sum of squares $\neq 0$ iff the previous condition holds and it has at least one prime factor $\equiv 1 \mod 4$ or $v_{2}(n)$ is odd.
There are exactly \[\mathrm{r}_{2}(n) = 4 \cdot \sum_{\substack{ d \in \mathbb{N}\\ d|n \\ d \equiv 1 \mod 2}}(-1)^{\frac{d-1}{2}}= 4 \cdot \prod_{\substack{p \in \mathbb{P}\\ p\equiv 1 \mod 4}}(v_{p}(n)+1)\] different solutions $(a,b) \in \mathbb{Z}^{2}$ to $n= a^{2}+b^{2}$.[/quote]

PS: As you may have noticed, your title is not considered to be the best ;)
Please choose a better one.

[hide="Index"]Representation as sums of squares[/hide]


The following problem is a generalization of an old problem of mine, to which Treegoner gave a super-nice solution with his geometric trick. Here is the problem:

   For any positive real numbers $ x_1,x_2,,x_n $ such that $\sum_{k=1}^n\frac{1}{x_k}=1 $ we have $\prod_{i=1}^n\sum_{j<>i} x_j^2 \geq \frac{(n-1)^n}{n^{n-1}}\cdot (\sum_{i=1}^n x_i^2)^{n-1} $.

Is
$\sum_{k=1}^{n}\frac{1}{x_k}=1$
or
$\sum_{k=1}^{n}\frac{1}{x_k}=n$  ?
In the last case, it is not true for $n=2$.

Oh, sorry, it's the last case, but here n>2.

I just wanted to remember you that there are still some hard algebraic inequalities that deserve to thought at.

up up up ,is some one have a solution or just a hint ?


Show that an angle of measure 45 degrees can be trisected using only a compass and straightedge. Hint:15 degrees = 60 degrees - 45 degrees.

Very easy
first, draw an equilateral triangle so we got a 60-degee-anlge.
THen use the compasses to divide it into 4 smaller angles and we got it.


combinatorics solved
combinatorics
permutations

How many permutations pi = {a_1,a_2,...,a_n} of natural numbers 1, 2, ...,n are there such that for each i = 2, ...,n the following condition is satisfied:

a_i = k => there is a index j with 1 \leq j < i and a_j \in {k-1,k+1}.

Example: (4,3,5,2,6,1), (3,4,2,1,5,6), (4,3,1,5,6,2)

The first two permuatations are valid but not the third one (the third digit must not be a 1, before that digit neither 0 nor 2).

It is clear any such permutation is a merging of two sequences l,l-1,...,1 and l+1,l+2,...,n (l=1,2,...,n) with the first member l ==> for numbers l-1,l-2,...,1 there are C(n-1,l-1) placements ==> number of permutation is \sum C(n-1,l-1) = 2n-1

[quote=Quaratinium]What does leq mean?[/quote]

\leq