From a strictly mathematical viewpoint:

What Makes 100%? What does it mean to give MORE than 100%? Ever wonder about those people who say they are giving more than 100%? We have all been to those meetings where someone wants you to give over 100%. How about achieving 103%? What makes up 100% in life?

Here's a little mathematical formula that might help you answer these questions:

If:
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z is represented as:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26.

Then:

H-A-R-D-W-O-R-K
8+1+18+4+23+15+18+11 = 98%

and

K-N-O-W-L-E-D-G-E
11+14+15+23+12+5+4+7+5 = 96%

But,

A-T-T-I-T-U-D-E
1+20+20+9+20+21+4+5 = 100%

And,

B-U-L-L-S-H-I-T
2+21+12+12+19+8+9+20 = 103%

AND, look how far ass kissing will take you.

A-S-S-K-I-S-S-I-N-G
1+19+19+11+9+19+19+9+14+7 = 118%

So, one can conclude with mathematical certainty that while Hard work and Knowledge will get you close, and Attitude will get you there, it's the Bullshit and Ass kissing that will put you over the top.  :D

Cool, and very true.
The - kissing equivalents (example: BROWNNOSING... heard it somewhere) will also take you over the top.

Hello,

[quote="eagle"]From a strictly mathematical viewpoint:
What Makes 100%? What does it mean to give MORE than 100%? Ever wonder about those people who say they are giving more than 100%? We have all been to those meetings where someone wants you to give over 100%. How about achieving 103%? What makes up 100% in life?
[/quote]

There is no strictly mathematical point of view in this. It is just like when someone says that "It is a mathematical fact" when they mean that it is quite obvius. (I haven't hear it in english, but in spanish, specially from old people).

Of course, everybody understand that when you give more than 100% in something, then you are giving more than what you would usually give, even in extreme situations.

[quote="eagle"]If:
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z is represented as:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26.

Then:

H-A-R-D-W-O-R-K
8+1+18+4+23+15+18+11 = 98%

K-N-O-W-L-E-D-G-E
11+14+15+23+12+5+4+7+5 = 96%
.
.
.
[/quote]

Yes, I remember that there was one of this kind of games, the first time that I saw it was in the mid 90's, that said gives Bill Gates = 666, and you would suppose what was the conclusion!!

Best,

HAHAHAHAHHAHAHAHAHA

Treething = 106%. Owned.

[quote="djimenez"][quote="yif man12"]Cmon, don't be a killjoy. It's F&G! This forum is meant for irrelevant stuff like this![/quote]

OK, OK, sorry, I was just making a comment, don't kill me for that, merci please!!!

But you are right, Next time I would try to make a less 'serius' comment.

Best,[/quote]

 ;)  :D

On a somewhat related note, if you add the values of a certain four letter word beginning with "f," their sum is 41.  That's where Sum 41 got there name from.  Nothing about meeting on the 41st day of the summer or anything.


I just put this here so that the mods wouldn't lock it or delete it. But this is a serious topic! [b]I am beginning to think that the mods are raptors themselves...[/b]

I have been cornered many times by raptors in my life, but I have been able to narrowly escape. In my most desparate attempt for survival, I was trapped in my house and hid within my big screen TV, but the raptors turned on the TV and saw me there! Wielding the cookie monster, I was able to smack the raptors away (This also explains the sudden disappearance of the cookie monster and the appearence of the cookie monster...Sorry, I had to survive!) for a while, 0.000000001 seconds to be exact, and I was able to drag myself out of my house with the two fingers I had left, as my legs and right arm had been chopped off. I have replaced these limbs with brooms and Buzz Lightyear action figure limbs, however, and am currently in peak physical condition.

The purpose of this thread is to describe your adventures with raptors and a list of tips on how to counter them. REMEMBER, THIS INFORMATION COULD SAVE LIVES! SPAM, AND KILL A HUMAN THAT IS FIGHTING TO SURVIVE.

My personal experience:

I used to live at a luxurious penthouse apartment with huge picturesque viewing windows. Since this was about forty stories in the air, I figured the place was safe from velociraptors. I was wrong! They came while I was getting up in the morning and dressed only in my PJs. The landed on the roof in a veloci-copter, broke the skylight, and dropped down on me while I was on the toilet. I was able to escape, but only after the velociraptors consumed my arms, legs and left ear. I used my mouth to flush myself down the toilet and came out in the bay, where I fortunately was washed ashore and found by a sympathetic doctor. This doctor (his name is Phil) arranged to have me patched up, and I am now in optimum condition even though I am mostly cyborg.

MORAL: Don't live in places with big windows, even if it is forty stories high in the air. Go through this thought process with every building you are considering living in:

[img]http://imgs.xkcd.com/comics/velociraptors.jpg[/img]


$ x,y,z \in \mathbb{R}_\plus{},0<x,y,z\leq \frac{4}{3}$ such that $ x\plus{}y\plus{}z\geq 1$

Prove that:

$ \frac{1}{1\plus{}x^2}\plus{}\frac{1}{1\plus{}y^2}\plus{}\frac{1}{1\plus{}z^2}\leq \frac{27}{10}$

[quote="tobeno_1"]$ x,y,z \in \mathbb{R}_ \plus{} ,0 < x,y,z\leq \frac {4}{3}$ such that $ x \plus{} y \plus{} z\geq 1$

Prove that:

$ \frac {1}{1 \plus{} x^2} \plus{} \frac {1}{1 \plus{} y^2} \plus{} \frac {1}{1 \plus{} z^2}\leq \frac {27}{10}$[/quote]
Standard multiplying gives $ 3 \le 7\sum_{cyc}x^2 \plus{} 17\sum_{cyc}x^2y^2 \plus{} 27x^2y^2z^2$
Let $ A(x,y,z) \equal{} 7\sum_{cyc}x^2 \plus{} 17\sum_{cyc}x^2y^2 \plus{} 27x^2y^2z^2$.
It is easy to see that $ A(tx,ty,tz) > A(x,y,z)$ if $ t>1$. Thus if we prove it for $ x\plus{}y\plus{}z \equal{} 1$, it obviously holds for $ x\plus{}y\plus{}z \ge 1$

$ 7\sum_{cyc}x^2 \plus{} 17\sum_{cyc}x^2y^2 \plus{} 27x^2y^2z^2 \ge 3 \equal{} 3(x\plus{}y\plus{}z)$. But since $ x\plus{}y\plus{}z \equal{} 1$. We just have to prove:
$ 7\sum_{cyc}x^2 * (x\plus{}y\plus{}z)^4 \plus{} 17\sum_{cyc}x^2y^2 * (x\plus{}y\plus{}z)^2 \plus{} 27x^2y^2z^2 \ge 3(x\plus{}y\plus{}z)^6$

Expanded and written in muirhead notation it looks like:
$ 12[6,0,0] \plus{} 60[5,1,0] \plus{} 126[4,2,0] \plus{} 90[3,3,0] \ge 18[4,1,1] \plus{} 204[3,2,1] \plus{} 66[2,2,2]$.
Which obviously is true...

It was a pretty ugly solution;)


modular arithmetic
number theory proposed
number theory

Solve $ \frac{1}{x}+\frac{1}{y}=\frac{2}{z} $ for positive integers $ x,y,z $.

$x={(x_0+y_0)kx_0 \over 2}$
$y={(x_0+y_0)ky_0 \over 2}$
$z=kx_0y_0$

where $gcd(x_0,y_0)=1$
and $k$ is positive and $k(x_0+y_0)=0 \pmod 2$

it is all solutions


[b]Solve this poblem: :P

$ \sqrt{5^x\minus{}2x}\minus{} \sqrt{2x\plus{}1}\equal{} 1\plus{}4x\plus{}4x.5^x\plus{}5^{2x}$[/b]

hello, your equation has exactly one solution, defining $ f(x)=\sqrt{5^x-2x}-\sqrt{2x+1}-1-4x-4x\cdot5^x-5^{2x}$,then it must be $ x\geq-\frac{1}{2}$. For $ x \geq 0$ it is $ f(x)<0$, so we need to  find all solutions for $ x<0$. We get $ f^'(x)=\frac{1}{2}\frac{5^x\ln(5)-2}{\sqrt{5^x-2x}}-\frac{1}{\sqrt{2x+1}}-4-4\cdot5^x-4x\cdot5^x\ln(5)-2\5^{2x}\ln(5)$,further we have $ f'(x)<0$ for all real $ x$,it is $ f(\frac{-1}{2})=\sqrt{\frac{1}{5}\sqrt{5}+1}+\frac{4}{5}+\frac{2}{5}\sqrt{5}>0$ and $ \lim_{x \to +\infty}f(x)=-\infty$,hence $ f(x)$ ist strictly monotonously decreasing and there is only one solution.
$ x \approx -.1850871205108$
Sonnhard.

[color=blue][b][quote="Dr Sonnhard Graubner"]hello, your equation has exactly one solution, defining $ f(x) = \sqrt {5^x - 2x} - \sqrt {2x + 1} - 1 - 4x - 4x\cdot5^x - 5^{2x}$,then it must be $ x\geq - \frac {1}{2}$. For $ x \geq 0$ it is $ f(x) < 0$, so we need to  find all solutions for $ x < 0$. We get $ f^'(x) = \frac {1}{2}\frac {5^x\ln(5) - 2}{\sqrt {5^x - 2x}} - \frac {1}{\sqrt {2x + 1}} - 4 - 4\cdot5^x - 4x\cdot5^x\ln(5) - 2\5^{2x}\ln(5)$,further we have $ f'(x) < 0$ for all real $ x$,it is $ f(\frac { - 1}{2}) = \sqrt {\frac {1}{5}\sqrt {5} + 1} + \frac {4}{5} + \frac {2}{5}\sqrt {5} > 0$ and $ \lim_{x \to + \infty}f(x) = - \infty$,hence $ f(x)$ ist strictly monotonously decreasing and there is only one solution.
$ x \approx - .1850871205108$
Sonnhard.[/quote]
How can you find $ x \approx - .1850871205108$[/b][/color]


heheh. this is a pretty idiotic problem. i'm solving a problem that asks for positive solutions and the answers are in terms of odd numbers and stuff. does odd integers only have 1,3,5,7,... or does it also have the negative ones -1,-3,-5,-7,... ? :D

I think integers are officially positive or negative (or zero!).  I suppose the question would say something otherwise.  I'm not 100% sure though.

Odd means not divisible by two, so negatives count too.

Integers are all -infinity to infinity. Counting 0. And within that limit. There are all whole numbers. No fractions! Or decimals!

Here's something to ponder about: The saying "To infinity, and beyond" Doesn't really make sense because if infinity is the most of something, then how is there a beyond?  :huh:

[quote="jonathanchou711"]There are all whole numbers. [/quote]

Note that whole numbers consists of only non negative integers.

[quote="jonathanchou711"]Really? Well... Unfractional numbas![/quote]

i.e. integers

[quote="crazypianist1116"]i.e. integers[/quote]

It would be rather awkward to define the word 'integer' as 'a number that is an integer', so a good definition of 'integer' would be 'a whole number multiplied by either 1 or -1'.

[quote="jonathanchou711"]Really? Well... Unfractional numbas![/quote]

What about irrational numbers? :P

Perhaps a string of digits without a decimal that is sometimes preceded by a negative sign.

um do you want the definition of an irrational number? :maybe:

A number that cannot be expressed in the form $ \frac{a}{b}$ where $ a$ and $ b$ are integers and $ b\neq0$. An example would be $ \sqrt{2}$.


Solve the Inequality SQRRT ( (x-2)/(1-2x) ) > -1

How to solve inequalities with SQR roots??? can some one teach me this?

[hide]since the LHS is a positive root, it is always >0>-1?[/hide]

[quote="t0rajir0u"]The problem as stated doesn't make sense.  Square roots are generally defined to be positive, as junggi says.  Then the problem is basically asking for $x$ such that the square root is defined.

That means the [b]real[/b] problem is

$\frac{x-2}{1-2x}\ge 0$[/quote]

[hide]LHS is positive if the num and the denom have the same signs.this happens in two cases:
Case 1: they are both positive
So $x-2>0\rightarrow x>2$ and
$1-2x>0\rightarrow 1>2x\rightarrow 1/2>x$
so no solutions there.
Case 2: they are both negative.
$x-2<0\rightarrow x<2$ and
$1-2x<0\rightarrow 1/2<x$
Case 3: the num is 0
$x=2$
So the solutions are
$1/2<x\leq2$[/hide]

[hide="Almost the same as the previous solution"]$\frac{x-2}{1-2x}\ge 0$
The terms inside the square root have to be positive (or 0):
CASE 1: $x-2\ge 0$, $1-2x>0$
CASE 2: $x-2\le 0$, $1-2x<0$

CASE 1:
The equations imply that $x\ge 2$, and $x<\frac{1}{2}$, which is clearly impossible.

CASE 2:
The equations imply that $1/2<x\leq2$[/hide]

Note: When doing problems like these, make sure the denominator is NEVER 0.

Thanks Everyone....this makes a lot of sense.... i think with all of you out here... i will master inequalities for sure!... :)


How do i find out how many cubic yards are in 2005 square feet?
I dont know where to begin...please help!! Please!!! :(

Please don't post the same question in two different forums. You already posted it [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106805]here.[/url]


It is easy to show that if $f : \mathbb{Q} \times \mathbb{Q} \to \mathbb{R}$ is any function in two variables, then there exists a function $g : \mathbb{Q} \to \mathbb{R}$ such that $f(x,y) < g(x)g(y)$ for all $x,y \in \mathbb{Q}$. Construct a function $f: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ for which this fails. That is,

[b]Problem. [/b] Construct a function $f : \mathbb{R}^2 \to \mathbb{R}$ such that, if $g : \mathbb{R} \to \mathbb{R}$ is any function, then there exists a pair $(x,y) \in \mathbb{R}^2$ for which $f(x,y)>g(x)g(y)$.

Enjoy! This is really beautiful!  ;)

--Vesselin

I think the first one has been posted (by me, If I'm not mistaken, but nobody answered :)).

For the second one (these are the [b]a)[/b] and [b]b)[/b] from an IMC problem, but I forgot the year), take $f(x,y)=\frac 1{|x-y|^2}$ whenever $x\ne y$ and $f(x,x)=0$, for example.

Actually, that problem from IMC (2003) was very similar, but it had $g(x)+g(y)$ instead of $g(x)g(y)$. There should be no problem, though. The idea is the same.

I knew this problem well before IMC 2003.  :) It's actually classical; too bad that IMC is using such well-known problems.  :?

Anyway, the function I had in mind was non-explicitly defined.

--Vesselin

[quote="grobber"]I think the first one has been posted (by me, If I'm not mistaken, but nobody answered :)).[/quote]
It is very strange, because I remember I solved it. It means I didn't see mentioned topic :(


combinatorics solved
combinatorics
games

A nice one :
Let n be a positive integer. Ann writes down n different positive integers. Then Ivo deletes some numbers (possibly none, but not all). He puts + or - signs in front of each of the remaining numbers and sums them up. If the result is divisible by 2003, Ivo wins. Otherwise Ann wins. For which values of n Ivo has a winning strategy ? For which values of n Ann has a winning strategy ?

It's interesting and not too complicated :)

Suppose n \geq 11. Let us consider all possible sums s1*a1+s2*a2+...+sn*an where a1,a2,...,an - Ann's numbers and s1,s2,...,sn \in {-1,1}. There are 2n>2003 such sums ==> there are two sums which identical modulo 2003 ==> (they difference)/2 is an answer for Ivo.

Suppose n \leq 10. Ann must write numbers 1,2,4,...,2n-1.


Hi everyone,

I'm a freshman currently taking pre-calculus. Which test, AMC 10 or AMC 12, will I have a better chance of making it to USAMO with?

Thanks.

i recommend seeing the topic "USMAO qualification".

If you can, try and get your school to participate on both the A and B days so you can do one of each.  That's what my school is doing.

However, if you can't do that, then you'll have to choose one or the other; thich choice depends on the sort of scores you expect.

I'll give an example: last year I got a 137 on the AMC10 and a 120.5 on the AMC12.  Last year the cutoff was either an index of 223 (calculated as AMC12 + 10*AIME) if you took the AMC12, or an 8 on the AIME if you took the AMC10.   So, according to the rules for USAMO qualification (which may have changed), I needed an 11 on the AIME if I took the AMC12, but only an 8 if I took the AMC10.

Suppose I had gotten a 150 on the AMC12.  Then I would have still needed an 8 on the AIME.

So what it boils down to is: do well on the AIME.  I personally recommend taking the AMC10, because then you'll only need a score of 8 or 9, as opposed to if you take the AMC12, where you will definitely need at least an 8, even if you ace the test.

but they may not do it this year. they were just being nice and overlooked it. be safe, take the amc10. but yet, if youre not able to get around a 120 on the amc 10, but you can get a 100 on the amc 12, take the amc 12

[quote]but didnt Simon say that they applied it to amc 12 takers in the 10th or below grade just because[/quote]

I'm a little confused, what was applied to AMC 12 takers in 10th grade or below?

the floor value of getting an 8 on the aime. but it may not always be applied

So no matter what I get on AMC 12 as long as I get an 8 or above on AIME, being a 9th grader, I will qualify for USAMO?

The floor might not be an 8, but if you're in 9th grade and you get at least the floor, you'll make USAMO.

[quote]one year, wasnt the floor as low as a 6? but yeah, its not guaranteed that the floor will be an 8, but theres a goode chance[/quote]

Eh, its going up every year for the past few years. You should be fairly safe with a 9 though.

really? i know the floor was 6 in 2000,2001,2002 since i have friends who got these scores and made it to usamo. only last year was the floor 8, probably due to how easy the first 3 problems were. i cannot believe that before 2000 the floor was anything less than 6, since that would make it too easy. so i think last year is just an exception to the usual floor.

[quote]i know the floor was 6 in 2000,2001,2002[/quote]

6 in 2002? Hmm, I think I heard it was 7 somewhere before. Could someone verify the cutoffs for previous years?

In 2000 and 2001 the qualifying structure was different.  In '00 they realized that they had to compensate in some way for the AMC 10 being easier, and so in addition to a selection process that didn't discriminate between 10 and 12 scores they invited all 12 takers who were grade 10 and below with at least an 8 or grade 9 and below with at least a 7.  I don't remember how the qualification worked on '01, but it was roughly similar.  (I only remember the '00 selection in such detail because I got in through the <=9th grade AMC 12 takers provision.)

If I remember correctly, there was no "floor" for AMC 10 takers in '01 (or '00).  The first selection of USAMO qualifiers was among all qualifiers based on grade and index alone (with no distinction between the tests) and then young AMC12 takers with AIME scores above a certain floor were added in to make things fair.  I know that in at least one of those years there were USAMO qualifiers who got in with a 150 AMC 10 and a 4 or 5 AIME because of that rule.

[quote="tetrahedr0n"]i guess i might be wrong about 2000, but from my state, anant gupta made usamo in 2001 with a 6 (and 150 on amc 10). jeff madsen made it in 2002 with a 6 again. this can be verified, try [url]http://www.calvin.edu/~lave/html/amc.html
[/url]. so there are two concrete examples that the floor is 6 in 2001 and 2002.[/quote]

soooo....for a freshman with no experience with [i dont know how to correctly word this], which test is better?

[quote="s0mp"]soooo....for a freshman with no experience with AMC things, which test is better?[/quote]
Umm... get some experience with AMC things before the test, regardless of the one you take.

So basically if you're confident you'll do good on AIME (or at least make it) then you should take the AMC 10 but if you're not (and you have a good chance at having a good score on the AMC 12) then you should take AMC 12?

Okay, I'm not sure I've ever weighed in on this type of question before, but I feel that 9th graders are probably better off taking the 10.  I think the number of people who are more likely to get a 100 on the 12 than a 120-or-so on the 10 is probably not very big.  I know I took the AMC 10 all 4 times I could.  I mean, it might be a point of pride to take the 12, but otherwise I think it's probably better to take the 10.


Are the sum of two twin primes >3 always divisible by 12? Why is this so?

Note that in order for twin primes to be >3,

1) they must be odd (even numbers are composite)
2) they must [i]not[/i] be a multiple of $ 3$

Note what this means: because every odd number is either a multiple of $ 3$ or a number right next to a multiple of $ 6$ (it should be simple to figure out why), and multiples of $ 3$ can't be prime, any set of twin primes must be consecutive with a multiple of $ 6$, and thus can be said to have the form $ x \minus{} 1$ and $ x \plus{} 1$, where $ x$ is a positive multiple of $ 6$. The sum of these two twin primes would thus be $ (x \minus{} 1) \plus{} (x \plus{} 1) \equal{} 2x$. So every sum is $ 2x$ where $ x$ is a multiple of $ 6$, and since every 2nd multiple of $ 6$ is a multiple of $ 12$, the sum of two twin primes must be a multiple of $ 12$. Sorry if this is confusing, but this is the clearest I can make it. If understanding issues occur, someone clarifying for me would be excellent.

EDIT: Thanks, math154, for providing an alternate solution.

[hide="More concise?"]
Since neither prime is a multiple of three, and they have a difference of two, they must be $ 3k\pm1$ for some integer $ k$, respectively. But if $ k$ is odd, then $ 3k\pm1$ is even, so they can't be prime, so $ k$ must be even. Thus, we can now express the primes as $ 3(2m)\pm1 \equal{} 6m\pm1$ for some integer $ m$, so their sum is $ 12m\equiv0\pmod{12}$, as desired.
[/hide]


Find all real functions f such that f(x+y)*f(f(x)+f(y))=(x+y) 2 .

For any commutative ring with unit element and without zero divisors : f(x) = u*x , where u is such that u3 = 1
 :D

If there is $t$ s.t. $f(t)=0$ then we put $x+y=t$ and we get $t=0$. Since we can take $x=y=0$, it means that there is $t$ s.t. $f(t)=0$, and we find $f(0)=0$. After $y=0$ we get $f(x)f(f(x))=x^2\ (1)$. Apply $(1)$ for $x\to x+y$ to find $f(x+y)f(f(x+y))=f(x+y)f(f(x)+f(y))=(x+y)^2$. If $x+y\ne 0\Rightarrow f(x+y)\ne 0\Rightarrow (x+y\ne 0\Rightarrow f(f(x+y))=f(f(x)+f(y)),\ \forall x,y\in\mathbb R\ (2)$.

$(*)\Rightarrow (f(x)=f(y)\Rightarrow x^2=y^2)\ (3)$. We apply $(3)$ to $(2)$ and we find $f^2(x+y)=[f(x)+f(y)]^2,\ \forall x,y\in \mathbb R,x+y\ne 0$.

Assume now there are $y\ne a$ s.t. $f(y)=f(a)$. Then $a=-y\ne 0$, and we have $f(y)=f(-y)$. Assume further that there is some $x\in \mathbb R$ s.t. $f(x+y)=-f(x)-f(y),f(x-y)=f(x)+f(-y)=f(x)+f(y)\Rightarrow f(2x)=0\Rightarrow x=0\Rightarrow y=0\Rightarrow$ contradiction, so for all $x\in \mathbb R\setminus\{\pm y\}$ we have $f(x+y)f(x-y)>0$. It means that $f(x+y)=f(x-y),\ \forall x\in \mathbb R\setminus\{\pm y\}$, and from $(3)$ we find $y=0$, which is a contradiction, so there is no $y\ne 0$ s.t. $f(y)=f(-y)$, so $f$ is injective, so $(2)$ actually translates to $f(x+y)=f(x)+f(y)$.

We now employ $(1)$; replace $x$ with $x+y$: $f(x)f(f(y))+f(y)f(f(x))=2xy,\ \forall x,y\in \mathbb R$. Take $y$ to be rational and non-zero (from the additivity it’s easy to show that $f(x)=x,\ \forall x\in \mathbb Q$). Then $f(f(y))+f(y)=2y$, and combining this with $(1)$ we get $f(y)=y,\ \forall y\in \mathbb R$.

Generalization: What is the sound of 2n + 1 hands clapping?

Similar to the sound of 2n hands when n is big enough. :D

why does everyone think one hand can't clap? :?


geometry
ratio
trigonometry
geometric transformation
rotation
homothety
complex numbers

Let $ \alpha \equal{} 1 \minus{} i$, where $ i \equal{} \sqrt { \minus{} 1}$. If $ z\in \mathbb{C}$, such that the numbers $ \alpha z, \alpha z^2$ and $ \alpha z^3$ are distinct vertices of an equilateral triangle, find $ \Im(\alpha z^2)$.

Note: If $ z \equal{} a \plus{} bi$, then $ \Im(z) \equal{} b$.

[hide="Answer"]
$ \Im(z) \equal{} \dfrac{1\pm\sqrt3}{2}$[/hide]

Edit: Since nobdy posted a solution, i think the statement was wrong. I did some changes and i think it is correct now.

Multiplication under a constant $ \alpha$, while not a rigid motion, does preserve length ratios, so if $ \alpha z, \alpha z^2$ and $ \alpha z^3$ are vertices of an equilateral triangle, than so are $ z$, $ z^2$, and $ z^3$. Therefore, the argument of z must be ${ \frac{\pi}{3}}, \frac{2\pi}{3}, \frac{4\pi}{3}$ or $ \frac{5\pi}{3}$, but we can't make any statements about it's modulus. As such, we also can't make any statement about the imaginary part of $ \alpha z^2$, so I think your problem statement still needs some fixing. We do know, however, that the argument of $ \alpha z^2$ must be one of $ \frac{2\pi}{3}$ or $ \frac{4\pi}{3}$ if I'm not mistaken.

You said that $ \alpha$ is constant?. $ \alpha\equal{}1\minus{}i$.

Sorry if i did not understood it correctly, my english is poor.

[hide="solution, i think, probably made mistakes"]
$ \alpha\equal{}\sqrt2\,\text{cis}\,\frac{\pi}{4}\\
z\equal{}r\,\text{cis}\,\theta$
Let $ s$ be the side length of the triangle. By Law of Cosines on the origin, $ \alpha z$, and $ alpha z^2$,
$ s^2\equal{}2r^2\plus{}2r^4\minus{}4r^3\cos\theta\equal{}r^2(2\plus{}2r^2\minus{}4r\cos\theta)$
On the origin, $ \alpha z^2$, and $ \alpha z^3$:
$ s^2\equal{}2r^4\plus{}2r^6\minus{}4r^5\cos\theta\equal{}r^4(2\plus{}2r^2\minus{}4r\cos\theta)$
Subtracting, $ 2r^2(r^2\minus{}1)(r^2\minus{}2r\cos\theta\plus{}1)\equal{}0$. If $ r^2\minus{}2r\cos\theta\plus{}1\equal{}0$, then the discriminant is only nonnegative (otherwise $ r$ is imaginary) if $ \theta\equal{}2\pi k$, and thus $ r\equal{}1$. However, then $ \alpha z$, $ \alpha z^2$ and $ \alpha z^3$ are equal and the points are nondistinct, contradicting the given. If $ r^2\equal{}0$, then all the points are actually equal to the origin, which again creates a contradiction. So $ r^2\minus{}1\equal{}0$ or $ r\equal{}1$. The $ 3$ points that constitute the triangle are now on a circle of radius $ \sqrt2$, so they must be $ \frac{2}{3}\pi$ apart in their arguments. Therefore, $ \theta\equal{}\frac{2}{3}\pi$ or $ \theta\equal{}\frac{4}{3}\pi$. Thus, the imaginary part of

$ \alpha z^2\equal{}\sqrt2\,\text{cis}\left(\frac{5\pi}{4}\pm\frac{\pi}{3}\right)$

is

$ \sin\left(\frac{15\pm4}{12}\pi\right)\equal{}\boxed{\frac{\mp(1\plus{}\sqrt3)}{2}}$[/hide]

[hide="Hint"]A theorem that will help you out:

http://www.cut-the-knot.org/Curriculum/Geometry/Connes.shtml

Also, you can scale the three points by any complex number, since multiplication by a complex number is essentially a rotation and a homothecy, and both preserve side length ratios.  So $ 1,z,z^{2}$ must also form an equilateral triangle.[/hide]

A more simple theorem is: $ x,y,z$ form an equilateral triangle in the complex plane if and only if $ x^2\plus{}y^2\plus{}z^2\equal{}xy\plus{}yz\plus{}zx$.

(the theorem that they give is actually two equations, one for each nonreal 3rd root of unity, one for each orientation, etc, this just combines them).


[quote="markcampion"]Howdy
after spending well over an hour typing in source code for a differentiation question sheet for my students i saved the file. When i quickly got out of my chair for something i pulled the cable on my mac and, because my battery is dead, my laptop shut down.
When i logged back in and typeset my file it had converted into gobbledygook some of which is shown below.
Any ideas on how i can retrieve the original?
thanks
campo
[hide="gook"]Ø?”@ÿña\|VoÒ??v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"?ä?w¸„¶V¶H?v"[/hide][/quote]

Isn't that just a sequence of "?ä?w¸„¶V¶H?v" 's?

This has nothing to do with $\LaTeX$; you could have gotten a corrupted file for any program.
Recovering something might be possible, but it's probably more hassle than redoing it even if it is possible. An hour or two isn't that bad.


calculus
integration
Ring Theory
algebra
polynomial
abstract algebra
function

if A is a subring of B, and $b \in B$ is integral over A means that it is a root a monic polynomial with coefficients in A
that means b^n + a_1 b^{n-1} + .... + a_n = 0 , $a_{i}\in A$
one mostly chooses monic polynomials to have a unique minimal polynomial, but can it happen that there exists a polynomial (not monic) such that
a_0 b^n + a_1 b^{n-1} + .... + a_n = 0 , $a_{i}\in A$
and a_0 is not a unit. If so, why don't we allow such polynomials?

Different point of view: one of the most important properties of integral extensions is the relationship to finite extensions. (A ring extension $A\to B$ is called finite if $B$ is a finitely generated $A$-module.) This fails if you allow arbitrary coefficients: $\mathbb Z[\tfrac12]=\mathbb Z[X]/(2X-1)$ is not a f.g. $\mathbb Z$-module, but every element $x=n/2^{k}$ satisfies an equation of the type $2^{k}x-n=0$. owk

[quote="jmerry"]Of course there are such polynomials- but they're not irreducible.

Exercise: Consider a polynomial $p(x)=a_{0}+a_{1}x+\cdots+a_{n}x^{n}$ with coefficients in a PID $A$ such that $a_{0},a_{1},\dots,a_{n}$ are relatively prime and $a_{n}$ is not a unit. Let $K$ be a field that $p$ splits in. Let $F$ be the smallest field containing $A$ in $K$, and let $R$ be the smallest ring containing $A$ and all roots of $p$ in $K$.
Show that $F\cap R\neq A$; it's strictly larger.
(I may have assumed more than necessary about $A$; don't worry about it.)

That's the reason; "rational" elements force you to extend the base ring within its quotient field.[/quote]

Indeed, the idea works much more generally and is at the  heart of the notion of an algebraic integer. If a domain R contains all the roots of a monic polynomial f then R also contains all the coefficients of f (by the Factor Theorem and Vieta's formula for the coef's in terms of the roots). Thus if f has some coef c being a [i]proper[/i] fraction  over the base domain A, then c is in the root ring R but not in A, i.e. the contraction of R to the fraction field of A contains a proper fraction over A, so it's larger than A. This is an essential part of the [u]motivation for the definition of an [i]algebraic [/i]integer[/u], e.g. see my post below which explains this motivation at length to someone who mistakenly thinks he has devised a better notion of an algebraic integer:

sci.math, Aug 12 2003, How I know, linchpin of my FLT proof
http://google.com/groups?selm=y8zllty36hr.fsf%40nestle.ai.mit.edu
http://groups.google.com/group/sci.physics/msg/2125df8b858cda73

[quote="jmerry"]Of course there are such polynomials- but they're not irreducible.[/quote]

Not irreducible???


if $a,b,c>0$ with $abc=1$ show that

\[ \frac{a}{b}+\frac{b}{c}+\frac{1}{a}\geq a+b+1  \]

 :)


geometry inequalities circumcircle AMC USA(J)MO USAMO conics

You may choose from one of the following categories; as an am. oddsmaker I will set the lines

Counting 3:1
System of Equations 10:1
Sequences and Series 6:1
Circle Geometry 3:1
Triangle Geometry 3:1
Divisibility-related Question 10:1
Inequality 6:1
Analysis 4:1 (see definition - I don't mean "analysis" in the usual way)
Conics/Loci 10:1
Other 7:1
---

(loose) definitions
Counting - if it asks you "how many ways"
System of equations - has to have 2 or more equations requiring solution / analysis
Sequences and Series - has a sequence involved/defined
Circle Geom - if it has 1 circle, its circle geom, except if this is the circumcircle of a triangle, and no other circles appear.
Triangle Geom - dealing with triangles
Divisibility - if it asks to show that some fraction is an integer, or deals with modular arithmetic
Inequality - asks you to prove an inequality
Analysis - Euclid 2005 last Q,  CMO "lollipop" question, USAMO last Q :  new situation that requires thorough analysis
Conics/Loci - dealing with the loci of points, or ellipses/hyperbola
Other - not specific to any of the categories  (ie., CMO '82 #3)

I sure hope it's either Counting or Systems of Equations.  I pwn at those.

I hope its geometry, or equations. [Please don't let it be conics!!!] (Haven't got a clue what those are)

[quote="startogalaxy"]Wice, I would say it is either sequence or circle geometry. But I wouldn't bet with you. I always lost in Texas hold'em; luck is just not with me. (Of course, texas hold'em is not just luck... And you're pro in poker. I don't have a chance of winning.) Anyway, wish you all luck tomorrow.[/quote]

Ha, at least u were half right.


geometry
circumcircle
geometric transformation
reflection
geometry unsolved

A triangle $ ABC$ has $ \angle ACB > \angle ABC$.

The internal bisector of $ \angle BAC$ meets $ BC$ at $ D$.

The point $ E$ on $ AB$ is such that $ \angle EDB \equal{} 90^{\circ}$.

The point $ F$ on $ AC$ is such that $ \angle BED \equal{} \angle DEF$.

Show that $ \angle BAD  \equal{} \angle FDC$.

Let $ K$ be the intersection of $ (ADF)$ with $ AB$.It's easy to see $ DF \equal{} DK$ and K on segment $ AB$.Applying the law of sines.We have $ \dfrac{DF}{sinDEF} \equal{} \dfrac{ED}{sinDFE} \equal{} \dfrac {DK}{sinDEK} \equal{} \dfrac {ED}{sinDKE}$ => $ sinDFE \equal{} sinDKE$. Morever $ K$ on segment $ AB$.We get $ \angle DFE \equal{} \angle DKE$ (QED)
 Sorry I have solution in my mind so It's can be wrong.I'll edit later :)

Draw the circle passing through $A$ and $D$ and tangent to $BC$, see that it is the homothetic of $\odot (ABC)$ at $A$, hence it has it’s circumcenter on $DE$ and intersects $AB$ and $AC$ at 2 points lying on a parallel to $BC$, consequently, the reflection of $AC$ in $DE$ will pass through the intersection point of this circle with $AC$.

Best regards,
sunken rock


given n distinct points are on a circe of radius 1.Let q be the number of the line segments whose length is greater than  :sqrt: 2 and whose endpoints lie in the given points .Prove that 3q <=  n 2
I trie but I failured.Can you help me?.Thanks you very much.

It's been posted before, and it was solved with the aid of [url=http://planetmath.org/encyclopedia/TuransTheorem.html]Turan's Theorem[/url].

Consider the graph having the points as vertices, and in which two points are connected iff the distance between them is $>\sqrt 2$. This graph cannot have any $K_4$'s as subgraphs (complete graphsd with $4$ vertices), so, by Turan's Theorem, it has at most $\frac{n^2}3$ edges.

Grobber,thanks you.Can you give me proof of Turan theorem?

I fixed the link in my first post. You can probably find it if you click that.


All of the faces of a convex polyhedron are squares or regular pentagons.  How many faces of each type does it have?  That is, how many different such polyhedra are there?

I'm not really sure about this, but let's try:

[hide]
Since all angles of our faces are at least 90°, it's impossible that more than three faces share a vertex. Less than three is of course also impossible, so we can state that every vertex is shared by exactly three faces.

If x is the number of square faces and y the number of pentagon faces, we get for Euler's formula ($ v\minus{}e\plus{}f\equal{}2$):
$ \frac{4x\plus{}5y}{3} \minus{} \frac{4x\plus{}5y}{2} \plus{} x\plus{}y \equal{} 2$
$ 8x\plus{}10y\minus{}12x\minus{}15y\plus{}6x\plus{}6y\equal{}12$
$ 2x\plus{}y\equal{}12$

x and y are nonnegative integers, so the possible values for (x,y) are:
(0,12)
(1,10)
(2,8)
(3,6)
(4,4)
(5,2)
(6,0)
But here's the flaw. Neither can I prove, nor do I believe they're all possible polyhedrons, the only ones that I think are possible are:
(0,12): a dodecahedron.
(2,8): with the two squares on the opposite sides (and 45° rotated to eachother).
(5,2): two parallel pentagons, connected by five squares.
(6,0): a cube

The reason I think (3,6) and (4,4) are impossible is that you will have two squares sharing a vertex, so you'll need two parallel pentagons and you'll end up with the (5,2). The only reason I think (1,10) is impossible is that I can't imagine a polyhedron which faces are one square and ten regular pentagons.

[/hide]

That's an excellent start, and it took me a while after getting that far to eliminate the other cases geometrically (since I think the Polyhedral Formula basically exhausts the combinatorial/algebraic approach).  I think the key observation is that [hide]if $ P$ and $ Q$ are two regular polygons in space that share consecutive vertices $ B$ and $ C$, so that $ ABCD$ are consecutive vertices of $ P$ and $ A'BCD'$ are consecutive vertices of $ Q$, then $ m\angle ABA' \equal{} m\angle DCD'$.[/hide]  (Actually, as a result of this observation we see that one of the polyhedra you've claimed can't be realized with all-regular faces.)

[quote] (Actually, as a result of this observation we see that one of the polyhedra you've claimed can't be realized with all-regular faces.) [/quote]
Well, it looked quite possible on my drawing, but I have to admit you're right.


trigonometry
inequalities unsolved
inequalities

Prove that:
$\tan x > x^3 ,\forall x \in \left( {0,\frac{\pi }{2}} \right)$

and
$x\cos x < \frac{{\sqrt 3 }}{3},\forall x \in \left( {0,\frac{\pi }{2}} \right)$

Just a kamikaze thought: do you think derivates would help? :D :D


How would I extend DeMoivre's to fractions?

If I raise $z$ to a fractional power, $\frac{x}{y}$, there would be y values satisfying it right? So if I raise these new values to $\frac{y}{x}$ power, instead of getting just plain old $z$, I get bombarded with extraneous values. I basically end up getting $x*y$ values.

Someone please disabuse my confusion and or reasoning.

If this help you: http://math.fullerton.edu/mathews/c2003/demoivre/DeMoivreBib/Links/DeMoivreBib_lnk_1.html

Okay, so I think you do end up with xy values, but the primary one is equal to original z.

i think u have to raise the new values to x/y to get it back to z
because
if u do
$z^{\frac{x} {y}}$ then $z = (rcis{\theta})^{\frac{y} {x}}$ so if u raise the new value to ${\frac{y} {x}}$ then ull get $z = [(rcis{\theta})^{\frac{y} {x}}]^{\frac{y} {x}}$ which gives us $z = (rcis{\theta})^{\frac{y^2} {x^2}}$
but if u raise to ${\frac{x} {y}}$ then ull get back to z
I THINK! :P

actually, I think you end of with $x$ values.  Think of it this way: first time you get $y$ value that are all the $y$th root of $z^x$, so raising them all to the $y$th power again gives you $z^x$.  now get the $x$th root of this number and you basically get $x$ roots of the number $z^x$.

I think breez actually got his own question right...

Because when you raise to $\frac{x}{y}$, you get $y$ values, then you can think of it as raising each of these to $\frac{y}{x}$, which will give $x$ values for each of the $y$ values we had before, meaning $\boxed{xy}$ total values...

And the principal value will obviously be $z$, because $((rcis{\theta})^{\frac{x}{y}})^{\frac{y}{x}}=(rcis{\theta})^{\frac{xy}{yx}}=(rcis{\theta})^1=z$.


Solve in integers: $x(x-y)+y(y-z)+z(z-x)=1$

[quote="shadysaysurspammed"]All i can get to is

$(x-y)^2+(y-z)^2+(z-x)^2$=2[/quote]

Each of the terms are $\geq 0$, and $x,y,z$ are integers, so...

so generally speaking the ordered pair of $x\geq y\geq z$ is $(x,y,z)=(k+1,k,k)$ and $(k,k,k-1)$ for integer $k$ .

[quote="warut_suk"]so two of the terms must be $1$ and one must be $0$, right?[/quote]

Thats what i thought in the first place,how can u get numerical solutions,there'll be tons of them.

Did you miss the condition that $x,y,z$ are integers?

yields the ordered pair: $(-1, -1, 0)$ in any order

It doesn't matter that x,y,z are integers.  There are that infinitely many ordered pairs that satisfy it.

You mean there are infinitely many ordered pairs of integers $(x,y,z)$ satisfying
$(x-y)^2+(y-z)^2+(z-x)^2=2$? :o

$(x-y)^2 + (y-x)^2 + (z-x)^2 = 2$
since we are dealing with integers, and squares (x-y)^2, etc must be greater that one, and 3 of them must equal 2, since 0 and 1 are the only perfect squares $0 \ge x \ge 2$, we must have (0,1,1) for each of the terms being (x-y)^2, etc
so without loss of generality, we assign the 0 to the first part, so $(x-y)^2=0$
x=y, substituting and comparing the next term to 1 (since the 2 others have to sum to 2), we get (y-y)^2 =1, and we get 0=1, which is a contradiction to the fact that we assumed that there were solutions, therefore, there are
[b]no solutions[/b]

[quote="Altheman"]$(x-y)^2 + (y-x)^2 + (z-x)^2 = 2$

so without loss of generality, we assign the 0 to the first part, so $(x-y)^2=0$
x=y, substituting and comparing the next term to 1 (since the 2 others have to sum to 2), we get (y-y)^2 =1, and we get 0=1, which is a contradiction to the fact that we assumed that there were solutions, therefore, there are
[b]no solutions[/b][/quote]

You'll get $(x-z)^2+(z-x)^2 = 2$...

[quote="fourierseries"]There ARE infinite number of solutions!

Check that (k+1, k, k) works for any integer k.[/quote]

Right!  :)

Yeahhh!!!I got it right first!!!!!Even i said there are infinite solutions :lol:  :lol:

i see my problem...copied the expression wrong, i had (x-y)^2 2 times, just i reversed one of them... :wallbash:
$(x-y)^2 + (x-z)^2 + (y-z)^2 = 2$
$(x-y)^2=0$
$x=y$,
$(x-z)^2 + (x-z)^2 = 2$
$(x-z)^2 = 1$
$x = z\pm 1$
so we get $(x,x,x\pm1)$


Let $f(n)$ be a function  $f:\mathbb{N}-\{0\} \to \mathbb{N}-\{0\}$ such that

$ f(1)=1 $ and for all $n  \in \mathbb{N}-\{0\}$;
$ f(f(n))\cdot f(n+2)+1 = f(f(n+1))\cdot f(n+1)$.
 Prove that  $f(n)=n$.

[Admin Edit: Made Latex Compliant].

Hey, guys, there are still some great problems unsolved in this section? This one is one of them:
  I will prove that for any n we have $ f(f(n))<f(n+1)$. Then, by a known (IMO) property, f will be the identity function. The property will be proved by induction. For n=1, it's trivial (just take n=1 in the initial relation!). Suppose it is true for n and also suppose it's not true for n+1. Thus, $ f(n+2)\geq f(f(n+1))$. But $ f(f(n+1))=\frac{1+f(n+2)f(f(n))}{f(n+1)}\leq\frac{ f(n+2)(f(n+1)-1)+1}{f(n+1)} $ and this combined with the inductive step gives us the only possibility $ f(f(n))=f(n+1)-1$ and $ f(n+2)=1 $ . Thus, $ f(f(n+1))=1 $. Now, put n+1 instead of n in the initial relation. You will find $ f(n+3)=0$ contradiction!


Q)
    a, b, c are the real number which are larger than 0.
    The sum of a, b, c is 3.
    Prove this inequality.
      (1/a^2)+(1/b^2)+(1/c^2) ≥ a^2+b^2+c^2

Please help me proving inequality!!

[quote="sigma phoenix"]Q)
    a, b, c are the real number which are larger than 0.
    The sum of a, b, c is 3.
    Prove this inequality.
     $ \frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}\geq a^2\plus{}b^2\plus{}c^2$

Please help me proving inequality!![/quote]

Let $ s\equal{}a\plus{}b\plus{}c$  , $ p\equal{}ab\plus{}bc\plus{}ca$  , $ q\equal{}abc$

 We have $ p^2\geq3sq\equal{}9q$ or  $ p\geq3\sqrt{q}$(***)

 Inequality is equivalent with $ \frac{p^2\minus{}6q}{q^2}\geq9\minus{}2p$ or $ p^2\plus{}2pq^2\geq9q^2\plus{}6q$

 Follows (***) is suffices to prove that $ 9q\plus{}6q^2\sqrt{q}\geq9q^2\plus{}6q$ wich is obvious because $ (\sqrt{q}\minus{}1)^2(2q\plus{}1)\geq0$

[quote="sigma phoenix"]Q)
    a, b, c are the real number which are larger than 0.
    The sum of a, b, c is 3.
    Prove this inequality.
     $ \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\geq a^2 \plus{} b^2 \plus{} c^2$

Please help me proving inequality!![/quote]

By AM-GM we have $ \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\ge \frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca}$
We only need to prove $ \frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca}\ge a^2 \plus{} b^2 \plus{} c^2$
$ \leftrightarrow 3\ge abc(a^2 \plus{} b^2 \plus{} c^2)$ which true by AM-GM
$ 3abc(a\plus{}b\plus{}c)(a^2 \plus{} b^2 \plus{} c^2)\le (ab\plus{}bc\plus{}ca)^2(a^2 \plus{} b^2 \plus{} c^2)\le(\frac{2(ab\plus{}bc\plus{}ca)\plus{}(a^2 \plus{} b^2 \plus{} c^2)}{3})^3\equal{}27$
so we are done!

setting $ F(a,b,c) \equal{} \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2} \minus{} a^2 \minus{} b^2 \minus{} c^2$

$ F(a,b,c) \minus{} F(a,(b \plus{} c)/2,(b \plus{} c)/2) \equal{} (b \minus{} c)^2[\frac {b^2 \plus{} c^2 \plus{} 4bc}{b^2c^2(b \plus{} c)^2} \minus{} \frac {1}{2}]$

we need prove that $ \frac {b^2 \plus{} c^2 \plus{} 4bc}{b^2c^2(b \plus{} c)^2} \minus{} \frac {1}{2} \ge0$

$ \longleftrightarrow \frac {1}{b^2c^2} \plus{} \frac {2}{bc(b \plus{} c)^2} \ge \frac {1}{2}$ it is right  :)

by Mixing Variable Theorem we need prove that $ F(3 \minus{} 2t,t,t)\ge 0$ (easy...)


function
integration
vector
inequalities
real analysis
real analysis solved

Let C be a positive number,E the space of continuous functions defined on [0,1], with real values and let F be a subspace such that for all f in F we have                 $ (\sup_{0\leq x\leq 1}{|f(x)|})^2\leq C^2\cdot\int_{0}^{1}{f^2(x)dx}$. Then $dim F$ is at most $C^2$.

$F=\{f\in C^{0}([0,1],R); ||f||_{\infty}^2\leq C^2||f||_{2}^2  \}$

The question is about $dim \; F$, we should understand that $F$ is sub-vector space
of $C^{0}([0,1],R)$, I don't see why if $f,g \in F \Rightarrow af+bg \in F$ where a,b are reals.

No, Moubi, F is just an arbitrary subspace of the space of continuous functions with that magnificient property. It is not the space of all the functions f that verify the given inequality, which in fact is not a subspace. F is just a subset of the set of the functions that verify the inequality.

Ok, just few remarks

$||f||_2\leq ||f||_{\infty}$ so in $F$,  $||..||_{\infty}, ||..||_2$ are equivalent norm.

$(C^{0}([0,1],R), ||..||_{\infty})$ is a complete space

$F$ is closed

...

Let $f_i$ ($i=\overline{1,n}$) be orthonormal basis in $F$.
We will prove that there exist $a_1,...,a_n\in\mathbb{R}$ s.t. $a_1^2+...+a_n^2=1$ and $n\leq \max_{t\in[0,1]}|a_1f_1(t)+...+a_nf_n(t)|^2$. Note that in this case
\[n \leq \max_{t\in[0,1]}|a_1f_1(t)+...+a_nf_n(t)|^2 \leq C^2\int_0^1(a_1f_1(t)+...+a_nf_n(t))^2dt=C^2.\]
Suppose contrary. Let $t\in [0,1]$. Then for all $a\in B_1(0)\subset\mathbb{R}^n$ we have $|a_1f_1(t)+...+a_nf_n(t)|<\sqrt{n}$. Take $a_i=f_i(t)/S$, $S=\sqrt{f_1^2(t)+...+f_n^2(t)}$. Then $\sqrt{f_1^2(t)+...+f_n^2(t)}< \sqrt{n}$ for all $t\in [0,1]$. It implies that
\[n=\int_0^1(f_1^2(t)+...+f_n^2(t))dt<n.\]
Contradiction.


Hi, can anyone help me with this?

Prove that:

  a                 +             1/(b(a-b))               >=                3.

for a>0 and b>0. sorry about that.:D

Your problem:
$ a \plus{} \frac {1}{b(a \minus{} b)}\geq 3$.
It is wrong if $ a < 0,b > 0$.
If $ a < b < 0$.Then take $ a,b\rightarrow\infty$.
If $ b < a < 0$.The problem is incorrect too.
If $ a > 0 > b$.It is wrong too.For example $ b\rightarrow 0_{ \minus{} }$.
But if $ a\geq b\geq 0$.We make a substitution:
$ (a \minus{} b) \equal{} x\geq 0$,$ b \equal{} y$,hence
$ a \equal{} x \plus{} y$.And your problem is:
$ x \plus{} y \plus{} \frac {1}{xy}\geq 3$.
Which is AM-GM.

[quote="Jakey"]to Erken, how do you do it by cauchy schwartz inequality? Sorry, I am just starting to review myself on inequalities. Does it hold true for a>0, b>0, the specified condition?[/quote]
This inequality holds if and only if $ a\geq b\geq 0$.
I didn't use Cauchy Schwarz,i used AM-GM,in such way:(sorry,it was my typo)
$ x\plus{}y\plus{}\frac{1}{xy}\geq 3\sqrt[3]{x\cdot y\cdot\frac{1}{xy}}$

well, if i present the proof, it's ok right if i say assume a>=b aside from a>0, b>0... Is it right erken? thanks for your help!:D

anyone has another solution? I would just like to see how beautiful and creative inequalities are!:D

as the equation isn't symmetric you can't say $ a\ge b$; you can however say $ a>b$ OR $ a\equal{}b$ OR $ a<b$ and look at each case separately.

[quote="bertram"]as the equation isn't symmetric you can't say $ a\ge b$; you can however say $ a > b$ OR $ a \equal{} b$ OR $ a < b$ and look at each case separately.[/quote]
I mean that his conditions to this problem are not complete,he forgot to write that $ a\geq b$.


Is this general term formula for #4 correct?

$T_{n}= \frac{\sqrt{3}}{6}\left[ (2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}\right]$

Could someone prove this for me?

(I already proved it, I just want to see if there is a more elegant proof)

This is wrong...
Just trying the n=1,2 it fails...

The correct general formula is $T_{n}=\frac{(2+\sqrt{3})^{n}-(2-\sqrt{3})^{n}}{2\sqrt{3}}$.
To prove this, consider the characteristic polynomial $x^{2}-4x+1$.

[quote="samath"]Umm ...
$\frac{\sqrt{3}}{6}[(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}]=\frac{(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}}{2\sqrt{3}}$. You guys have the same formula.[/quote]Not quite.  Their formulas have a sign difference.


Is it possible place the numbers: $1,2,3,...,16$, in a square board with side $4$, such that the difference of two neighbours is at most $3$(Two numbers are neighbours if they are in two squares with a commom side).

$come$ $on$ $boys.$ $I$ $am$ $waiting$ $for$ $a$ $great$ $solution.$

No, it is NOT possible.

Note that numbers 1 and 2 can have as neighbors only the following three numbers: 3, 4, and 5 (even worse: 5 cannot be neighbor of 1 but we ignore this fact). Consider the following cases:
1) Both 1 and 2 are at corners. Then each has two neighbors and these neighbors are all different. This case is impossible.
2) At least one of 1 and 2 is not at a corner and they are not neighbors. Then 1 and 2 have at least 2+3=5 neighbors one of which may be shared. So they have at least 4 different neighbors, which is impossible.
3) Both 1 and 2 are not at corners. Then they have at least 3+3=6 neighbors, two of which they may form by themselves (if 1 and 2 are neighbors). Anyway they have at least 4 different neighbors, which is impossible.

Therefore, one of 1 and 2 must be at a corner, and the other must at neighboring square. In such position, 1 and 2 have three other neighbors that are 3,4, and 5. So the position looks like:
x x y z
y y z *
z z * *
* * * *
where each x is either 1 or 2, and each y is either 3,4, or 5.
Take a look at z's. They must be 6,7, or 8. But there are four z's! So this is also impossible.

In fact it`s an old and very well known problem (which probably is in the forum millions of times) that if you arrange the numbers $1,2,..,n^{2}$ in a $n\times n$ board, then there are two numbers $a,b$ that share a side such that $|a-b|\geq n$.

To Cuenca.....¿Te conozco?


geometry
ratio
circumcircle
trigonometry
geometric transformation
dilation
angle bisector

Let an acute triangle $ABC$ , it’s orthocenter$H$  and the midpoint  $M$  of side $BC$. The bisector of angle  $A$ of triangle $ABC$  meets $HM$  at the point  $Z$.
If $ZE \perp AC$ , where $E$ is a point on side $AC$ , prove that
     $HE \perp AZ$.

[u]  babis[/u]

We may have an easy proof of this nice problem, considering the follows alternative configuration.

     Through the orthocenter H of a given triangle ABC, we draw a line perpendicular to the line AA', angle bisector  through vertex  A, which intersects the sidelines AC, AB, at points E, E', respectively.

     If now, we denote as Z, the intersection point of the lines through the points E, E' and perpendicular to AC, AB respectively, it is enough to prove that the point Z, which lies on the line AA' ( it is clear ), lies also on HM, where M is the midpoint of BC.

     But we can easy to prove this result, based on the follows simple idea:

     Lemma. If the orthogonal projections of three given points, on two lines intersect each other, form segments with the same ratio, then these points are collinear with the same ratio. ( Easy to prove by Thales theorem ).

     We denote as D, D', the traces of altitudes of ABC, through vertices B, C, on AC, AB, respectively. Also we denote as F, F', the orthogonal projections of M, on AC, AB, respectively.

     So, from the similar right triangles HDE, HD'E', we have that (HD)/(HD') = (DE)/(D'E'), (1).

     From the similar right triangles HDC, HD'B, we have that (HD)/(HD') = (DC)/(D'B), (2).

     (1), (2), => (DE)/(D'E') = (DC)/(D'B), =>  (DE)/(DC) = (D'E')/(D'B),  (3).

     From (3) =>  (DE)/(DF) = (D'E')/(D'F'),  (4)  ( because of F, F', are the midpoints of DC, D'B ).

     From (4), based on the Lemma, we conclude that the points H, Z, M, are collinear and the proof is completed.

     Some further notes about the Lemma, have been posted in this forum:

     http://www.mathlinks.ro/Forum/viewtopic.php?t=80163

     Kostas Vittas.

[color=blue]To Albanian Eagle[/color]

  Yes , point $E$ is on side $AC$ .I have written this in the wording now(edited).

 [u] Babis[/u]

[quote="stergiu"]Let an acute triangle $ABC$ , it’s orthocenter$H$  and the midpoint  $M$  of side $BC$.
The bisector of angle  $A$ of triangle $ABC$  meets $HM$  at the point  $Z$.
If $ZE \perp AC$ , where $E$ is a point on side $AC$ , prove that $HE \perp AZ$.[/quote]
[color=darkblue][b]A metrical proof.[/b] Denote : the second intersection $A'$ of the line $AZ$ with the circumcircle $w=C(O,R)$ of the triangle $ABC$ ;
the projection $E_{1}$ of the point $Z$ to the line $AC$ ; the point $E_{2}\in AC$ for which $HE_{2}\perp AZ\ .$
It is well-known that if the triangle $ABC$ is acute, then $AH=2R\cos A$and$MA'=R(1-\cos A)\ .$
Generally, the length of the $A$- angle bisector is $l_{a}=\frac{2bc\cos\frac{A}{2}}{b+c}$and$AA'=\frac{bc}{l_{a}}$, i.e.$AA'=\frac{b+c}{2\cos\frac{A}{2}}\ .$ Therefore,

$1.\blacktriangleright\ AH\parallel MA'$ $\Longrightarrow$ $\frac{ZA}{AH}=\frac{ZA'}{MA'}=\frac{AA'}{AH+MA'}$ $\Longrightarrow$ $\frac{ZA}{2\cos A}=\frac{ZA'}{1-\cos A}=\frac{AA'}{1+\cos A}$ $\Longrightarrow$
$ZA=\frac{2\cos A}{1+\cos A}\cdot AA'=\frac{2\cos A}{1+\cos A}\cdot \frac{b+c}{2\cos\frac{A}{2}}$ and $AE_{1}=ZA\cos\frac{A}{2}$ $\Longrightarrow$ $\boxed{\ AE_{1}=\frac{(b+c)\cos A}{1+\cos A}\ }\ \ (1)\ .$

$2.\blacktriangleright$ The [i]Sinus' theorem[/i] in the triangle $AHE_{2}$ $\Longrightarrow$ $\frac{AH}{\cos\frac{A}{2}}=\frac{AE_{2}}{\sin \left(C+\frac{A}{2}\right)}$ $\Longrightarrow$ $AE_{2}\cos \frac{A}{2}=2R\cos A\cdot \sin\left(C+\frac{A}{2}\right)\ .$ Multiply with $2\cos\frac{A}{2}$ :
$AE_{2}\cdot 2\cos^{2}\frac{A}{2}=2R\cos A\cdot 2\cos\frac{A}{2}\sin\left(C+\frac{A}{2}\right)$ $\Longrightarrow$ $(1+\cos A)\cdot AE_{2}=2R\cos A(\sin B+\sin C)$ $\Longrightarrow$ $\boxed{\ AE_{2}=\frac{(b+c)\cos A}{1+\cos A}\ }\ \ (2)\ .$
[b][i]Otherwise ![/i][/b] Denote $R\in HE_{2}\cap AZ\ .$ Thus, $AE_{2}=\frac{AR}{\cos\frac{A}{2}}=$ $\frac{AH\cos\frac{B-C}{2}}{\cos \frac{A}{2}}=$ $\frac{AH\cdot 2\cos\frac{A}{2}\cos\frac{B-C}{2}}{2\cos^{2}\frac{A}{2}}=$ $\frac{2R\cos A(\sin B+\sin C)}{1+\cos A}=$ $\frac{(b+c)\cos A}{1+\cos A}\ .$

From the relation $(1)$ and $(2)$ obtain $AE_{1}=AE_{2}$, i.e. $E_{1}\equiv E_{2}=E$ and $\boxed{\ AE=\frac{(b+c)\cos A}{1+\cos A}\ }\ .$[/color]

[quote="stergiu"]Let an acute triangle $ ABC$ , it’s orthocenter$ H$  and the midpoint  $ M$  of side $ BC$. The bisector of angle  $ A$ of triangle $ ABC$  meets $ HM$  at the point  $ Z$.
If $ ZE\perp AC$ , where $ E$ is a point on side $ AC$ , prove that
     $ HE\perp AZ$.

[u]  babis[/u][/quote]

 Vittasko told me that this problem apeared recently as proposal in Hyakinthos group. Probably , new members could try again for a new short and nice proof .

[u] babis[/u]

Very nice problem!  I still have to think about it, but it appears something more general is true (I have Sketchpad)

Let $ H$ be the orthocenter of $ \triangle{ABC}$.  Let $ P$ be a point on the circumcircle of $ \triangle{BCH}$.  Now let $ Q$ be the isogonal conjugate of $ P$, and let $ R$ be the image of $ Q$ after a dilation by a factor of $ 2$ from $ A$.  Let the bisector of $ \angle{A}$ hit $ PR$ at a point $ Z$.  Let $ E$, $ F$ be the feet of the perpendiculars from $ Z$ to $ AB$, $ AC$.  Prove that $ P$ lies on $ EF$. (The proposed problem reduces to the above in the case $ P \equal{} H$)

EDIT: Corrected (Thanks vittasko!)

This problem is very similar to that one of the SL 2005
http://www.mathlinks.ro/viewtopic.php?p=519896#519896

There exists a very nice solution using similar triangles and bisector theorem .

About conjugate ratio,see [url]http://webs.uvigo.es/adg2006/ [/url].


algebra
polynomial
advanced fields
advanced fields unsolved

Suppose that $ f \in k[x_1,\dots,x_n]$ is a homogeneous polynomial of degree $ d$. That is, degree of each term of  $ f$ is equal to $ d$. And let $ k$ be an algebraically closed field. Prove or disprove that $ f$ can be written as multiplication of $ d$ linear polynomials, i.e, homogeneous polynomials of degree $ 1$.

[hide]My professor told me that it is true (without any proof). I neither can prove it nor give a counterexample.[/hide]

Let $ x^2\plus{}yz \equal{} (x\plus{}by\plus{}cz)(x\minus{}b'y\minus{}c'z)$. Look at coefficient of $ xy$, we see that $ b \equal{} b'$. Look at coefficient of $ xz$, we see that $ c \equal{} c'$. Then $ x^2\plus{}yz \equal{} x^2 \minus{} (by\plus{}cz)^2$. Look at coefficient of $ y^2$ and $ z^2$, $ b \equal{} c \equal{} 0$, contradiction.

Are you sure that you considered the complex possibility for that coefficients?
I don't know how to calculate them but I think there are appropriate complex coefficients for such a factorization.

No, Soarer's computation is valid.  The statement is true for $ n \equal{} 1$ trivially and for $ n \equal{} 2$ by the fundamental theorem of algebra but false thereafter.  Your professor is wrong; the problem statement is equivalent to the claim that every projective variety is a union of hyperplanes, which is patently false.

The mentioned counterexample, along others, is easier seen by Eisensteins theorem.

I am completely mixed up.
Could you explain the relation between Eisenstein's theorem and this problem?

furthermore, I cannot precise the prove for case $ n\equal{}2$. The problem is in the linearity of $ x_i$s. When one write the polynomial as:
$ (a_1 y \minus{} a'_1 x^{i_1})...(a_d y \minus{} a'_d x^{i_d})$, why do you claim that each $ i_j\equal{}1$?

$ y$ is prime in $ k[y,z]$, thus $ x^2\minus{}yz$ satisfies Eisensteins criterion.

[quote="fesghelbahal"]When one write the polynomial as:
$ (a_1 y \minus{} a'_1 x^{i_1})...(a_d y \minus{} a'_d x^{i_d})$, why do you claim that each $ i_j \equal{} 1$?[/quote]  It's a simple consequence of homogeneity: write $ f(x, y) \equal{} x^d \cdot f(1, y/x)$, then apply the Fundamental Theorem of Algebra to the right-hand side in the variable $ y/x$.

It is the thing that came to my mind, but what about the points $ x\equal{}0$?
for changing variables in such manner, wee need to add a nonzero condition to $ x$. Am I wrong?

No; everything's formal until you write down the factorization, and once the factorization's written down everything makes sense.  You shouldn't think of $ x$ as necessarily taking values at all; that is, temporarily you perform a computation over $ k(x, y)$, the field of fractions.

Now, everything's all clear. Thank you all. :roll:


inequalities
LaTeX
inequalities unsolved

I didn't see this inequailties very well. But, I'm also from korea..

Nice to meet you..

[quote="Inequalities Master"]Find the greatest real p>1,s.t. if a,b,c are greater or equal than 1/p and smaller or equal than p, the following inequality holds:
 9(ab+bc+ca)(a^2+b^2+c^2) is greater or equal than (a+b+c)^4.[/quote]

LaTeX:

Find the greatest real $ p>1$ such that, for all $ \frac{1}{p}\le a,b,c\le p$, the following inequality holds:

\[ 9(ab\plus{}bc\plus{}ca)(a^2\plus{}b^2\plus{}c^2)\ge (a\plus{}b\plus{}c)^4\]

Aargh, I thought I solved it.But there are some huge mistakes.Isn't there any fixed idea for this?

Maybe the answer is $ p\equal{}\sqrt2$,because is easy to prove that the inequaity holds if $ a,b,c$ are lengths of sides of triangle


A flight from San Francisco to New York takes five hours, but the return flight takes six hours.  When it is 4 p.m. in New York, it is only 1 p.m. in San Francisco.  The flight leaves San Francisco at noon and arrives in New York at $ x$ o'clock, and the return flight the next day leaves New York at noon and arrives in San Francisco at $ y$ o'clock.  What is the positive difference of $ x$ and $ y$?

[hide]
When going from San Fran to NY you add 3 hours. And when going from NY to San Fran you subtract 3 hours. This statement is now:
\[ (5\plus{}3)\minus{}(6\minus{}3)\equal{}\boxed{3}
\]
[/hide]

[quote="LiBoy"]Um... The answer on FTW is 5, not 3.
However, I don't know why.
Can someone tell me?[/quote]

[quote="hunter34"]
When going from San Fran to NY you add 3 hours. And when going from NY to San Fran you subtract 3 hours. This statement is now:
$(5+3)-(6-3)=\boxed{3}$.
[/quote]

hunter simply messed up when subtracting.
$(5+3)-(6-3)=8-3=\boxed{5}$  :lol: .


Determine all positive integers $n > 2$ , such that \[ \frac{1}{2} \varphi(n) \equiv 1 ( \bmod 6)  \]

Well, this follows straightforward from multiplicativity:
You get that there is only one odd prime divisor or none if the number is divisible by $4$ (so $=4$ itself).
Now just consider the cases of different residues $\mod 6$ that occure and you are done.

[hide=Solution]
This is equivalent to
\begin{align*}
\varphi(n) &\equiv 2 \pmod{4} \\
\varphi(n) &\equiv 2 \pmod{3}
\end{align*}
We can easily find \(n = 4m\) where \(m\) is odd. Furthermore, \(\varphi(m)\) must be odd, implying \(m=1\). Thus, \(n=4\) is the only solution.
[/hide]

@above What about $n=3$?
[hide=Writing out post 2's sol]As mentioned in the post, $n=4k$ implies $n=4$. Otherwise, as mentioned there, $n=p^k$ or $n=2p^k$. In either case, $\phi(n)=(p-1)p^{k-1}$, which has to be $2\pmod{12}$. This means that either $p=3$ (meaning $k=1$) or $p\equiv 2\pmod{3}$ (meaning $k$ is even). As $p\equiv 3\pmod{4}$, we finally have the solutions $\boxed{n=3,4,6,p^{2k},2p^{2k} \text{ for } k\equiv 11\pmod{12}}$. [/hide]


Solution to 2.

$ \mathbf{r}\equal{} r\mathbf{e_{r}}$ and $ \frac{d\theta}{dt}\equal{}\omega$

$ \mathbf{v}\equal{}\frac{d\mathbf{r}}{dt}\equal{} r\frac{d\mathbf{e_{r}}}{d\theta}\cdot\frac{d\theta}{dt}\equal{} r\omega\mathbf{e_{\theta}}$

$ \mathbf{a}\equal{}\frac{d\mathbf{v}}{dt}\equal{} r\omega\frac{d\mathbf{e_{\theta}}}{d\theta}\cdot\frac{d\theta}{dt}\equal{}\minus{}r{\omega}^{2}\mathbf{e_{r}}\equal{}\minus{}{\omega}^{2}\mathbf{r}$

Hence the magnitude of the acceleration is $ {\omega}^{2}r$ where $ r$ is the radius of the circle. It acts in the opposite direction to the position vector of a point on the circumference, so is directed towards the centre.

Solution to 3.

Let X be the displacement from M. Right will be considered positive, and left will be considered negative.

$ r \equal{}\frac{b}{\sin{\theta}}$

$ \Rightarrow PM \equal{} r\cos{\theta}\equal{} b\frac{\cos{\theta}}{\sin{\theta}}\equal{} b\cot{\theta}$

$ \Rightarrow X \equal{}\minus{}b\cot{\theta}$

$ \Rightarrow\frac{dX}{dt}\equal{} b (\csc^{2}{\theta})\frac{d\theta}{dt}$

$ \Rightarrow u \equal{} b (\csc^{2}{\theta})\omega$

$ \Rightarrow\omega \equal{}\frac{u}{b}\sin^{2}{\theta}$

But $ \sin{\theta}\equal{}\frac{b}{r}$

$ \Rightarrow\omega \equal{}\frac{u}{b}\frac{b^{2}}{r^{2}}\equal{}\frac{ub}{r^{2}}$

Solution to 7.

I assume we are finding the velocity of a point above the point at which the wheel makes contact to with the track.

$ \omega$ is contant at all parts on the wheel.

Lets conisider travelling at the same velocity of the train justnow, and consider the relative velocity of the point. The tangential velocity of the circumference must be equal to the linear velocity of the train. What we actually see is a circle with fixed centre in space, the relative velocity of the point is directed in the opposite direction to the train's motion. We take the direction of the train's velocity to be positive.

$ \Rightarrow\frac{v_{1}}{r_{1}}\equal{}\frac{v_{2}}{r_{2}}$

$ \Rightarrow v_{1}\equal{} v_{2}\frac{r_{1}}{r_{2}}$

Since $ \frac{r_{2}}{r_{1}}$ is a dimensionless ratio, there  is no need to convert everything to the same units.

$ \Rightarrow v \equal{}\minus{}60\cdot\frac{0.2}{0.4}\equal{}\minus{}60\cdot\frac{1}{2}\equal{}\minus{}30\,km\,h^{\minus{}1}$

Hence the actual velocity of the point is $ v \equal{} 60\minus{}30 \equal{} 30\,km\,h^{\minus{}1}$ in the same direction to the train.


Dear Members,
                      Its been 13 days that i failed everytime i wanted to register in our "AoPs" blog forum.I haven't sign
anything yet.But a notice always appear saying that my request
is already send waitng to be appove in 3 days. I tried again.But
there was nothin to sign for me! Tell me:Just how will i get into registering my own "AoPs Blog.?"

                                               Sincerly,
                                              Mikhail S.

I'm not sure what you mean by this.
Possibility 1: You don't already have a blog and want to make one here and don't know how to (unlikely).
Answer: Go to "Profile", then click on "Create a new blog"
Possibility 2: You already have a blog and want to register it (unlikely).
Answer: Go to the Blogs page.  There should be instructions for registering.
All other possibilities: You are having problems with either of those.    Go to the help forum and ask this question.

Either way, this belongs on the help forum, not the Graphical Design forum.

You seem to already have a (albeit blank) blog...

http://www.artofproblemsolving.com/Forum/weblog.php?w=538

Post there. Then you have your blog.

Just read their terms and conditions first and then post it. It will be sure live if you done all the formalities. Thanks you!!!

[url=http://ioffersearch.com]Ioffersearch.com Blogs - Just another Ioffersearch.com weblog[/url]


superior algebra
superior algebra unsolved

What is the cardinality of the set of automorphisms of the complex field $ \bf{C}$?

It's as large as possible- $ 2^c$. Given an uncountable set of algebraically independent transcendental numbers, we can permute them.

Why does an infinite set $ S$ have $ 2^{|S|}$ permutations?
This took me quite a while to figure out.
(the rest is easy using a [url=http://en.wikipedia.org/wiki/Transcendence_basis]transcendence basis[/url] of $ \mathbb{C}$ over $ \mathbb{Q}$
and extending automorphisms to algebraic closures.)

Every subset $ T\subseteq S$ gives a different bijection
$ \sigma_T: S\times\{0,1\}\to S\times\{0,1\}$
$ (s,0)\mapsto\{
\begin{array}{cl}
(s,0)&\text{for }s\in T\\
(s,1)&\text{otherwise}\end{array}$
$ (s,1)\mapsto\{
\begin{array}{cl}
(s,1)&\text{for }s\in T\\
(s,0)&\text{otherwise}\end{array}$.

So $ S\times\{0,1\}$ has at least $ |\mathcal{P}(S)|=2^{|S|}$ permutations (even involutionary)
and so does $ S$, as $ |S|=|S\times \{0,1\}|$.


An $ r$-partition of $ n$ is a collection, without regard to order, of $ r$ integers $ (\ge 1)$ whose sum is $ n$. Galileo was given, and solved, the following problem. There are six 3-partitions, which use no integers above $ 6$, of both $ 10$ and $ 12$, yet when a $ 3$ six-sided fair dice are thrown simultaneously the total $ 10$ appears more frequently than $ 12$. Why?

[hide="Answer."]
=====
$ \{3,3,4\}$ is a single $ 3$-partition of $ 10$, yet $ (3,3,4),(3,4,3),(4,3,3)$ are all distinct rolls of three dice, so you can't count the outcomes of the dice based on the $ 3$-partitions. In essence, the fact that order isn't regarded is what ruins it.
=====
[/hide]


[size=150][b]Two ships leave port at noon. One's ship's speed is 15 km/h, and its heading is 75 degrees. The other ship's speed is 24 km/h, and its heading is 155 degrees. How far apart are the ships at 3 P.M. and what is the bearing of each from the other?[/b][/size]

The distance after three hours (3 PM) is $\sqrt(7209-6480cos(80))$ which is approximately....78km. The bearings from each other do not change. fro ship 1 to ship two, it is 80 degrees conterclockwise from ship 1 and from ship 2 to ship 1 it is 80 degrees clockwise from ship 2.


I found a really easy way of multiplying!

Sya you have 3x51.

$ 3\times{51}$

Circle the last number and write it down:

$ 3\times{5\boxed{1}}\equal{}1$

DO the same with second-to last

$ 3\times{\boxed{5}1}\equal{}15$

And first number:
$ \boxed{3}\times{51}\equal{}153$

That's a cool trick that works, even with random numbers like 8x86=688!

$ 6 \times 68 \neq 866$

Sorry for popping your bubble.

It doesn't work if you have to carry numbers. :wink:

[quote="mewto55555"]I found a really easy way of multiplying!

Sya you have 3x51.

$ 3\times{51}$

Circle the last number and write it down:

$ 3\times{5\boxed{1}} \equal{} 1$

DO the same with second-to last

$ 3\times{\boxed{5}1} \equal{} 15$

And first number:
$ \boxed{3}\times{51} \equal{} 153$

That's a cool trick that works, even with random numbers like 8x86=688![/quote]

...rrofl?

51*3 :D

1*11?  how did you come up with this "genius" way, btw?

if you use mews method, don't mulitply by one :P

In general, just don't use mewto's method. :harhar:


An 8-foot board is cut into two pieces whose lengths are in the ratio of 7 to 5. What is the number of inches in the length of the longer piece?

This is my method for solving ratio problems.

Since two numbers are in a ratio, they are both multiplied by some factor, which we can call $ x$, and when added, give us the total number.

So in this example, the two numbers are in a ratio of 7 to 5, so we have an equation that looks like this:

$ 7x \plus{} 5x \equal{} 8$

$ 12x \equal{} 8$

$ x \equal{} \frac{2}{3}$

We wish to find the longer piece, so $ \frac{2}{3} \times 7$ yields $ \frac{14}{3}$ feet. However, we want this in inches, so $ \boxed{56}$.

Or simply  , the longer piece would be (7/(7+5)) *8*12 ( 12 inches per foot) = 7*8 = 56


parameterization
vector
analytic geometry
trigonometry
real analysis
real analysis unsolved

A dog buries a bone at $(0,1)$. He is at the end of a 1-unit long leash, and his master walks down the positive $x$-axis, draggin the dog along. Since the dog wants to get back to the bone, he pulls the leash taut. Give parametric equations for the dog's path in terms of
$a)$  $\theta$
$b)$  $t$

where $\theta$ is the angle made by the positive $x$-axis and the vector connecting the position of the master to the position of the dog, and $t$ is the $x$-coordinate of the master.

[quote="drunner2007"]anybody know of a way?[/quote]This is known as Huygens' tractrix. First show that $\frac{dy}{dx}+\frac{y}{\sqrt{1-y^{2}}}=0$ from the definition of the curve. Since $y = \sin \theta$, this should allow you to get the second parametrization rather quickly. For the other parametrization, set $z = y^{-1}$ and play around a little :).


function
analytic geometry
graphing lines
slope
algebra
domain
calculus

I'm having a bit of trouble understanding what uniformly continuous exactly is.  I can prove whether a funciton is uniformly continuous or not without too much difficulty.  But I'm having trouble understanding exactly what it is.

For example, a way to think of continuous (in a non-rigorous mathimatical way) is if you can draw the function without lifting up your pencil.  The way to think of differentiable is if the function is smooth.  Is there such a way that I can think about uniformly continous like this?

The reason why I ask is that on tests, my professor normally asks for examples of functions which have specific properties.  For example:

Two uniformly continuous functions f,g: E->R such that the product fg is not uniformly continuous on E.

And I'm just not sure how to have an educated guess at whether or not a function is uniformly continous.

My analysis professor used this explanation:

Imagine a huge guy bursting into the room and whipping out an $\epsilon$ and an $x$. If you produce a $\delta$ so that the continuity condition is satisfied for the function $f$ at $x$, then he can't do anything, and you win. But if you can't, then he'll smack you around with the $\epsilon$ and $x$. That's continuity -- he doesn't do anything sneaky to overcome you.

But with uniform continuity, when he bursts in, he only pulls out the $\epsilon$. You need to find your $\delta$ to fend him off knowing that he might pull out an $x$ but not knowing what it is. You can only see his $\epsilon$.

This is sort of silly but it illustrates the difference in the definitions pretty well, I think.

To see what uniform continuity means, just consider some examples of functions that are continuous but not uniformly so. Take $f(x) = \frac{1}{x}$ on $(0, \infty)$. If I give you an $x$ and an $\epsilon$, you're home free. But if you can only see the $\epsilon$, after you choose I can pick an $x$ close enough to 0 that no matter what $\delta$ you chose is not a winning $\delta$.

My example illustrates one important difference -- a uniformly continuous function is bounded on any finite interval, while a regular continuous function is bounded only on compact (that is, closed and finite) intervals. In fact, a regular continous function on a compact set is uniformly continuous on that set. It's with open sets that continous functions can have wild behavior that uniformly continuous functions don't.

[quote="Ricky"]That helps quite a lot, thanks.  Would it be safe to say that anything with an unbounded slope is not uniformly continuous?[/quote]
No, it would not be safe to say that. The function $f(x)=\sqrt{x}$ on the interval $[0,1]$ has an unbounded slope, but (like any continous functions on a closed bounded inteval) it is uniformly continuous.

Definition: $f$ satisfies a Lipschitz condition if $\exists\,M$ such that for all $x,y$ in the domain of $f,\,|f(x)-f(y)|\le M|x-y|.$

When someone talks about a function "having a bounded slope" I assume they mean it satisfies a Lipschitz condition.

Here are three implications. In each case, the converse is false:

1. If $f$ has a bounded derivative, then $f$ satisfies a Lipschitz condition.

2. If $f$ satisfies a Lipschitz condition, then it is uniformly continuous.

3. If $f$ is uniformly continuous, then it is continuous.

Here are easy to remember counterexamples to each of the converses:

1. $f(x)=|x|$ on $\mathbb{R}.$

2. $f(x)=\sqrt{x}$ on $[0,1].$

3. $f(x)=\sin\left(\frac1x\right)$ on $(0,1].$

[quote]Two uniformly continuous functions f,g: E->R such that the product fg is not uniformly continuous on E. [/quote]
If $f$ is uniformly continuous on some domain, then it can be uniquely extended to a continuous function on the closure of that domain.

Suppose $f$ and $g$ are both uniformly continuous on a bounded domain. Then each can be extended to be continuous on the closure of that domain, which is closed and bounded, hence compact. Then $fg$ can be considered to be a continuous function on that compact set, hence uniformly continuous.

What does that tell you? It tells you that if you want an example of two uniformly continuous functions whose product is not uniformly continuous, then you must be looking on an unbounded domain and concerning yourself with the behavior of the function as $x\to\infty.$

Once you see that, it's easy to find an example.


You can include diagrams in your solutions file (as opposed to in a separate file) by creating them in JPG or PDF form (the former is much simpler, but you could use Go2PDF to convert to PDF) and using the command

\includegraphics{filename}

within the TeX file, where you want the picture to go, and where filename.JPG, filename.PDF, etc. is the name of your diagram.  You need only save the file in the same directory as solutions.tex.

More information is available on the [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_PictHow.php]LaTeX --> Pictures[/url] section of this site.

[quote="vinny919"]But see, it would be a lot harder for me if I created a separate file for each picture.  I have many diagrams.  I could easily make them in Word and just label them and refer to them in the solution.  Would this be better?[/quote]
Only if you mailed the solutions in, or compiled them into one PDF document.  If you sent in two documents via e-mail and said "refer to the other document" in a problem, it's likely your grader would be confused, as he would not have received both documents, but just the problem solution that you submitted.  So, either you break up the images and incorporate them into your LaTeX document, or you find some way to compile the images and laTeX together without splitting up your images.

[quote="vinny919"]One of my solutions to one of the problems has a lot of diagrams.  Can I use paint or Word to make the diagrams and label them, save them as PDFs using Go2PDF, and then just send that as a separate file along with the solutions file, and then just refer to the diagrams within the solution??  If I can't do this, then what way should I include the diagrams??[/quote]

I assume you're using TeX from your other posts - you can make them all one PDF, include that one PDF in your TeX document, and refer to them in your solution.


LaTeX
inequalities proposed
inequalities

hello, i think in your case exists no maximum.
Sonnhard.

[quote="Victory.US"]Let $ x,y,z$ such that : $ x\gey \ge z \ge 1$; $ 2y + 3z\ge 6$ and $ 11x + 27z\ge 54$
Find the maximum value of $ P = \frac {1}{x^2} + \frac {2008}{y^2} + \frac {2009}{z^2}$
 :arrow:[/quote]

$ x=z=2,y->0$

[quote="Victory.US"]Let $ x,y,z$ such that : $ x\gey \ge z \ge 1$; $ 2y + 3z\ge 6$ and $ 11x + 27z\ge 54$
Find the maximum value of $ P = \frac {1}{x^2} + \frac {2008}{y^2} + \frac {2009}{z^2}$
 :arrow:[/quote]

it must be Let $ x,y,z$ such that : $ x\ge y \ge z \ge 1$; $ 2y + 3z\ge 6$ and $ 11x + 27z\ge 54$
Find the maximum value of $ P = \frac {1}{x^2} + \frac {2008}{y^2} + \frac {2009}{z^2}$

[hide]mistake in latex[/hide]


Kim Hung proposed the following inequality
Prove that if $a,b,c>0$ then
\[\frac{a^2+bc}{(b+c)^2}+\frac{b^2+ca}{(c+a)^2}+\frac{c^2+ab}{(a+b)^2}\geq\frac32.\]
Here is another proof I have just found. Sorry, I have made silly mistakes.

I have serious doubts on how you managed to have the inequalities $\frac{a^2+bc}{\left(b+c\right)^2}\geq\frac{a}{b+c}$, $\frac{b^2+ca}{\left(c+a\right)^2}\geq\frac{b}{c+a}$ and $\frac{c^2+ab}{\left(a+b\right)^2}\geq\frac{c}{a+b}$ simultaneously hold. However, your idea would work if you would apply Vornicu-Schur. See the thread http://www.mathlinks.ro/Forum/viewtopic.php?t=24555 for a number of different proofs of this inequality.

  darij

It's so easy, I think.

By Vornicu Schur

\[\sum \frac{a^{2}+bc}{(b+c)^{2}}-\sum \frac{a}{b+c}=\sum \frac{(a-b)(a-c)}{(a+b)^{2}}\ge 0.\]

What happens here, Arqady?  :lol: You and me posted at the same minute.


search
number theory unsolved
number theory

Prove: if $ \alpha \mid n$ $ \forall \alpha \epsilon \mathbb{Z}$ such that $ 0 < \alpha \le \sqrt_{n}$ then $ n \mid 24$

I have proved this, but my proof is not very elegant and involves a small amount of exhaustive verification.  I would really like a proof using minimal modern (last 100 years) results in Prime Number Theory.

EDIT: sorry this may belong in Proposed & Own

$ lcm(1,2,\dots,x)|n$ for $ x\le \sqrt{n}$
$ b^2\le n < (b\plus{}1)^2$

If you use the result of
[url]http://www.mathlinks.ro/viewtopic.php?search_id=345139475&t=292849[/url]

$ lcm(1,2,\dots,b)\ge 2^{b\minus{}1} \ge (b\plus{}1)^2$ for $ b\ge 7$
So you only have to check $ n\le 48$

Suppose $ k,n,a \epsilon \mathbb{Z}$, $ k \ge 5$, $ k^2 \le n < (k + 1)^2$ and $ a \mid n$ $ \forall a \le \sqrt_{n}$
then $ k \mid n$ and $ k - 1 \mid n$ since $ k$ and $ k - 1 \le \sqrt_{n}$
$ gcd(k,k - 1) = 1$ because they are consecutive so $ k(k - 1) \mid n$

$ k(k - 1) \le k^2 \le n$ so $ n \neq k(k - 1)$ and since $ n \epsilon \mathbb{Z}$ we know $ n \ge 2k(k - 1)$
so $ n \ge 2k^2 - 2k > (k + 1)^2$ $ \forall k \ge 5$ which is assumed. This contradicts the hypothesis that $ n < (k + 1)^2$

Now check $ n \le 24$

and we are done :)


counting
distinguishability
geometry
geometric transformation
rotation
reflection
AMC

There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible to place one on the other, using translations, rotations, and/or reflections, so that their corresponding small triangles are of the same color.

Given that there are six different colors of triangles from which to choose, how many distinguishable large equilateral triangles may be formed?

[hide]This comes down to case analysis:

Start by ignoring the middle triangle, because you can make it whatever you want and it will be distinct. If the three colors from the outer triangles are the same you can always translate it to another version so that it will not be distinct with that version unless the colors are different. Thus there are 3 cases:

Case 1: 3 different colors for outer triangles
Number: (6 3)

Case 2: 2 different colors for outer triangles
Number: (6 2) (2 1)

Case 3: All colors of outer triangle are the same
Number: (6 1)

(6 3) + (6 2) (2 1) + (6 1) =
20 + 30 + 6 = 56

Now multiply be 6 for inner triangle:

56 * 6 = 336.[/hide]

[hide]
Notice that all 3!=6 colorings of triangles with a fixed center triangle are indistinguishable. For every fixed center, we need to color the 3 outer ones with 6 colors, so we need the number of solutions to the equation a+b+c+d+e+f=3 in nonnegative integers, which is just $\binom{8}{3}=56$. Since there are 6 possible colorings of the center triangle, our answers is $6\cdot 56 = \boxed{336}$.
[/hide]

Can someone go through this casework solution and tell me where the error is/where it breaks down? Thanks in advance.

PS: I understand the above solutions, just didn't think of them while doing the test. So I want to know why mine doesn't work.

[hide="My (Flawed) Solution"]
We will break this problem down into four cases, based on how many colors are being used.

Case 1: All four triangles are the same color.

For this case, we simply choose which color of the six we will be using in $\binom61 = 6$ ways.

Case 2: Three of the triangles are one color, and the other triangle is a different color.

First, we choose the two colors we are using in $\binom62$ ways. Then we choose which will be used three times and which will be used once in $2!$ ways. Then, there are exactly two possible triangles that can be formed--one where the first color is in the center, and another where the other color is in the center. This makes a total of $\binom62 \cdot 2! \cdot 2 = 15 \cdot 2 \cdot 2 = 60$ ways for this case.

Case 3: Two of the triangles are one color, and the other two are distinct colors.

First, we choose the colors we are using in $\binom63$ ways. Then, we choose which two colors of the three will only be used once in $\binom32$ ways. Now, there are exactly three possible triangles that can be formed, one for each of the three colors being the center triangle. This makes a total of $\binom63 \cdot \binom32 \cdot 3 = 20 \cdot 3 \cdot 3 = 180$ ways for this case.

Case 4: All four of the triangles are different colors.

First, we choose the colors we are using in $\binom64$ ways. There are exactly four triangles that can be formed now, one for each of the differently colored triangles being the center one. So for this case, there are $\binom64 \cdot 4 = 60$ ways.

Finally, the total number of colorings is the sum of the number of ways for each of the cases, and thus our answer is $6+60+180+60 = \boxed{306}$.
[/hide]

I'm pretty sure you're missing the "2 triangles one color, 2 triangles another color" case. That'd bring you to the 336 you wanted.

Yeah wow that was stupid. Thanks a lot. Now, here's a full solution for those needing it and are not clever enough to come up with JSteinhardt/chess64's.

[hide="My Correct Solution"]
We will break this problem down into five cases, based on how many colors are being used and how many of each color.

Case 1: All four triangles are the same color.

For this case, we simply choose which color of the six we will be using in $\binom61 = 6$ ways.

Case 2: Three of the triangles are one color, and the other triangle is a different color.

First, we choose the two colors we are using in $\binom62$ ways. Then we choose which will be used three times and which will be used once in $2!$ ways. Then, there are exactly two possible triangles that can be formed--one where the first color is in the center, and another where the other color is in the center. This makes a total of $\binom62 \cdot 2! \cdot 2 = 15 \cdot 2 \cdot 2 = 60$ ways for this case.

Case 3: Two triangles are one color, and the other two are both another color.

First, we choose which two colors we will be using in $\binom62$ ways. Now there are two possible triangle colorings based on which color is in the middle. This means there are $\binom62 \cdot 2 = 15 \cdot 2 = 30$ ways for this case.

Case 4: Two of the triangles are one color, and the other two are distinct colors.

First, we choose the colors we are using in $\binom63$ ways. Then, we choose which two colors of the three will only be used once in $\binom32$ ways. Now, there are exactly three possible triangles that can be formed, one for each of the three colors being the center triangle. This makes a total of $\binom63 \cdot \binom32 \cdot 3 = 20 \cdot 3 \cdot 3 = 180$ ways for this case.

Case 5: All four of the triangles are different colors.

First, we choose the colors we are using in $\binom64$ ways. There are exactly four triangles that can be formed now, one for each of the differently colored triangles being the center one. So for this case, there are $\binom64 \cdot 4 = 60$ ways.

Finally, the total number of colorings is the sum of the number of ways for each of the cases, and thus our answer is $6+60+30+180+60 = \boxed{336}$.
[/hide]

[hide="Burnside's Lemma"]
There are 6 transformations of an equilateral triangle with rotations and reflections.
Identity: $6^4$ colorings
Rotation (x2): $6^2$ colorings
Reflection (x3): $6^3$ colorings
$\frac{6^4 + 2\cdot6^2 + 3\cdot6^3}{6}=\boxed{336}$
[/hide]

Hmm for the above solution how do you get the the (x2) and (x3)? Shouldn't it be x3 for rotation and x2 for reflection?

You can rotate the triangle by 120 or 240 degrees (0 degree rotation counted as identity).  The lines that the triangle can reflect over go through each of the 3 vertices (and bisect the opposite side), hence 3 reflections.

I noticed the answer is $\dbinom{8}{3}$ x 6, and that using the info from the cases I think I found a combo identity. The identity is Vandermonde's and the proof is just explaining casework(trivial proof).

Well you didn't come up with an identity if you can't prove it.

[b]Solution:[/b] Notice that the center triangle can be any of the $6$ colors, so we will multiply our result by $6$ at the end. For the three surrounding triangles, first consider the case where they are all distinct. In that case, there are $\frac{6 \cdot 5 \cdot 4}{3!} = 20$ distinct possibilities (as we have to divide by $3!$ to preserve distinction even in the case of rotations/reflections). Next, suppose that two of the exterior triangles are the same color, then we have $6 \cdot 5 = 30$ possibilities in that case. Of course, if they are all the same color, we have $6$ distinct arrangments. Hence our answer is $6(20 + 30 + 6) = \boxed{336}$.


[quote="rnwang2"]Err...how is this fun? It's pretty pointless. :maybe:

EDIT: Here is a fun problem

Compute $4\tan^{-1}\left(\frac{1}{5}\right)-\tan^{-1}\left(\frac{1}{239}\right)$.[/quote]
I think that was sarcasm
[hide]
$\sqrt{\frac{529}{10}+\frac{529}{36}}$
$\sqrt{\frac{(529)(46)}{360}}$
Now just take out the squares[/hide]

O I deleted that post and I just posted that problem in another thread so we wont get confused.

[quote="rnwang2"][quote="mets501"]=======Post removed.  The one below is much better :blush: ======[/quote]
Uhh...yeah that was not a random problem.  :wink:[/quote]
Yeah, I realized that afterwards.

 :rotfl:


I hear a lot about quants and quantitative finance in this forum, but how is it possible to actually have a genuine definable advantage when everything is constantly changing in the market?  And I see forums where quantitative finance people argue about strategies, but if they were discussing those strategies publicly then they would become useless, right?  Or is it possible to use those strategies to consistently beat the market?  It seems like once an idea becomes public it should be no more successful than random stock picking.

Also I do not really see what traders do, if you have quants who develop strategies which do have a definite advantage, why do you need traders, it seems like everything could be automated after that, which would reduce human errors.

Just to clarify: a quant trys to apply what they know about probability, statistics, game theory ect. to the financial market, and does simulations of their strategies to see if they have identified a genuine inefficiency, which can be exploited, is that correct?

Unfortunately, there are no superior and enduring strategies. Typically, you make money by exploiting mis-pricing effects, arbitrage opportunities, e.g., between professional and retail market, and, although in its strict legal sense forbidden, leveraging insider know-how

The common misconception is that financial companies make a lot of money by finding some inefficiency and exploiting it for a long time. This isn't so. Nowadays, arbitrage opportunities are very short-term and difficult to identify. So the change is what makes this kind of stuff possible, and the people exploiting it keep driving the change. (There are other factors, too, but not so relevant to what you asked.)

You will never see a quant publicly discuss in detail any strategy that actually makes his firm money. If people are discussing strategies, it's either too vague to actually harm anything (e.g., "fundamental investing" is a strategy), or they're specific, but they're really old strategies that don't work anymore.

Generally purely automated trading would be VERY risky. If your program decides to do something stupid, it could lose you a LOT of money before you noticed. "Traders" in quantitative firms might not actually do all of the trades, but they will instruct the computer to do specific trades, supervise everything, etc. They know what's going on in the markets and decide what new approaches might prove fruitful. Also, at some firms, quants pretty much work in the dark and only do mathematical analysis, and the traders are the ones who apply their math to the market. So the quants will write a program that prices a security, and the traders will be on the floor and they'll see a security and put the parameters into the program and if the "right" price is higher than the offer the trader sees then the trader will make the trade. So traders are still useful, even at firms that mostly trade on quantitative strategies.

Not nearly all firms use this kind of strategy anyway. So all of these things only apply to the quant groups at banks or to the firms that specifically trade according to quant-style strategies.

And yes, basically, what you said about what quants do is accurate.

There are "stock picking" quant groups and some have had prolonged success but many suffered really quite badly in Q4 2007. But in general I agree with the others, quants & traders develop the frameworks for strategies, but the "tactics" require constant updating, and have to be thrown away with depressing frequency.

Some of the techniques are trying to spot behaviour patterns like groups of buyers or sellers who trade in particular circumstances. Catching those foolish enough to buy into technical analysis forms part of that, but one looks also for people who are trying to lose inventory, or those who by the rules under which they operate, must buy/sell.

A big, but less glamorous part of the work is optimising the trades so that you don't move the market against yourself, but at the same time making sure you make enough of them.
Also, a good place to start understanding the structure of this line of work is the Kelly Criterion. Won't give you any trading ideas, but will help you manage the risks involved.

[quote="Dominic Connor"]A big, but less glamorous part of the work is optimising the trades so that you don't move the market against yourself, but at the same time making sure you make enough of them. [/quote]
Actually this part seems really interesting to me!

[quote]Actually this part seems really interesting to me![/quote]

I can see that :)
One can feel a sense of accomplishment in making something work that one understands. Many high frequency trading strategies don't seem to have much sense of "mechanism" at all. In some ways they resemble primitive peoples who have worked out that the rains come every so many moons, but have no idea what the moon is, or that their planet is going around the Sun.

Hard for someone trained in science to really enjoy working in that mode.


Find $ x,y \in \mathbb{R}\ ,\ x\ge 3$ if $ x^2\plus{}y^2\minus{}xy\minus{}3x\minus{}y\plus{}4\equal{}0$

[hide]
Factorise to get
$ (x\minus{}y)^2 \plus{} (x\minus{}1)(y\minus{}3) \plus{} 1 \equal{} 0 \Leftrightarrow (x\minus{}y)^2 \plus{} 1 \equal{} (x\minus{}1)(3\minus{}y)$

Since the $ LHS$ is positive and $ x\ge 3$ we have $ y<3$. Also if $ y\le 0$ then substituting $ y'\equal{}\minus{}y$ gives

$ (x\plus{}y)^2 \plus{} 1 \equal{} (x\minus{}1)(y\plus{}3)$ which is clearly impossible because $ x\plus{}y\ge 3\plus{}y$ and $ x\plus{}y > x\minus{}1$

So we have $ x\ge 3, 0\le y < 3$ So substitute $ a\equal{}x\minus{}3 \ge 0$ and $ b\equal{}3\minus{}y > 0$

This gives $ (a\plus{}b)^2 \plus{} 1 \equal{} b(a\plus{}2)$

$ \therefore b(a\plus{}2) \equal{} (a\plus{}b)^2 \plus{} 1 \ge 2(a\plus{}b) \Longrightarrow b\ge 2$

Since $ b\ge 2$

$ b(a\plus{}2) \equal{} (a\plus{}b)^2 \plus{} 1 \ge (a\plus{}2)^2 \plus{} 1 \Longrightarrow b \ge a\plus{}2 \plus{} \frac{1}{a\plus{}2} \ge 5\plus{}\frac{1}{5}$

Thus $ b>5$, but $ 3\minus{}y\equal{}b>5 \Rightarrow y <\minus{}2$ which contradicts $ 0\le y < 3$

Hence there are no solutions  :) [/hide]


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ G$ be a non-abelian group with commutator subgroup $ G'$. Prove that $ G/G'$ cannot be the union of an ascending chain $ (g_1) \subset (g_2) \subset \ldots (g_n) \subset \ldots$ fo cyclic subgroups.

[quote="owk"]So you mean "an infinite, strictly ascending chain"? owk[/quote]

Sorry, actually that was not necessary strictly ascending,  we might have some equal subgroups in between. But it could be infinite or finite, we don't specify this point.

$ G\equal{}S_3$ if you don't require it to be infinite and strict.

Ok, what if it is infinite, but still not strict ?

[quote="ZetaX"]Then you can take a constant chain and the same group.[/quote]

Well, i mean non-trivial case. Let's say "Is it possible to find such a non-constant infinite chain ?".

Take the first groups to be the trivial group, then switch at some point to $ \mathbb Z / 2 \mathbb Z \cong S_3/ A_3$ (note that $ A_n \equal{} (S_n)^\prime$).
You need that there are infinitely many "jumps", but then you can already assume it not to be eventually constant or to be strict.

Eugene, did u intend to refer to locally cyclic factor commutator group ?


I know that preparing for the AMC is extremely fun, but we at the AMC board want to take extremely fun to the extreme!

That said, I have decided that the AMC board will host its first mock AMC competition.

When will this take place?  Well, in order to make this fair, I'm announcing that problems will be posted tomorrow at 3:00 PM PST.  (I hope I don't mess up and forget...:D)

I'll be taking the problems from various sources, but I won't disclose them until [i]after[/i] the contest, to guard against cheats.  (Although I don't think anyone would...still...this has to be as realistic as possible).

Oh yeah, what do you win?  Well, you get bragging rights, and after a while, your AMC score will probably improve...there...

- mathfanatic

P.S. See the Notes on Scoring/Scoreboard post for information on how this thing will be scored, how many questions there will be, etc.

P.P.S. In compliance with Mr. Rusczyk, I have de-announcemented this post in order to free up the space at the top of the AMC board.


$n$ is a positive integer, $f(n)=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$.

(1) When $n>1$, show that $f(2^{n}) > \frac{1}{2}(n+2)$.

(2) When $n>1$, define $A_{n}= f(1)+f(2)+\cdots+f(n)$ and $B_{n}= n [ f(n)-1 ]$.  Find which one of $A_{n}$ and $B_{n}$ is larger.


A bicycle is traveling at 20 feet per minute. What is the bicycle's speed expressed in inches per second?

First, we want to convert our units.  $ 20 \text{ feet}\equal{}20\cdot12\equal{}240 \text{ inches}$.  $ 1 \text{ minute}\equal{}60\text{ seconds}$.  So, inches per second would be expressed as $ \frac{240}{60}\equal{}\boxed{4}$.


[color=red][b]ΑΣΚΗΣΗ 1 [/b][/color]

Έστω $ ABC$ ένα ισοσκελές τρίγωνο με $ AB \equal{} AC$ και $ A \equal{} 20$. Θεωρούμε σημείο $ D$ πάνω στο τμήμα $ AB$ τέτοιο ώστε $ AD \equal{} BC$. Να αποδείξετε ότι $ ACD \equal{} 10$ :)

 [color=red]Edited : $ \angle ACD \equal{} 10$[/color]

[quote="maniopas"]Έστω ότι $ D:$ εσωτερικό του $ AB$ τότε

$ \widehat{ACD} \equal{} 20^o \Leftrightarrow$
$ \widehat{DCB} \equal{} 60^o$
και
$ CD \equal{} AD \equal{} BC \Leftrightarrow \widehat{DBC} \equal{} \widehat{CBD} \equal{} 80^o$

άρα το άθροισμα των γωνιών του τριγώνου $ BDC$ βγαίνει $ 220^o \Leftrightarrow$ άτοπο

Έστω ότι $ D:$ εξωτερικό του $ AB$ τότε πρέπει (επειδή τα τρίγωνα ABC και ACD ίσα)
$ CD \equal{} AC \Leftrightarrow$ άτοπο

Μήπως έχεις κάνει κάποιο λάθος; Μήπως κάνω κάπου εγώ λάθος;[/quote]

Εαν οπως είπες έχουμε $ ACD\equal{}20$ και $ DCB\equal{}60$ τότε ανaγκαστικά πρέπει και $ BDC\equal{}40$ εφόσον $ DBC\equal{}80$

Ακριβώς αυτό λέω:
η $ \widehat{BDC}$ πρέπει να είναι $ 40^o(I)$
επειδή είναι DC = AD = BC πρέπει $ 80^o \equal{} \widehat{CBD} \equal{} \widehat{CDB}(II)$
αυτά τα δύο δίνουν άτοπο

[quote="Γιώργος"]

[color=red][b]ΑΣΚΗΣΗ 1 [/b][/color]
Έστω $ ABC$ ένα ισοσκελές τρίγωνο με $ AB \equal{} AC$ και $ A \equal{} 20$. Θεωρούμε σημείο $ D$ πάνω στο τμήμα $ AB$ τέτοιο ώστε $ AD \equal{} BC$. Να αποδείξετε ότι $ \boxed{ACD \equal{} 10}$ :)[/quote]

[color=blue][b]ΛΥΣΗ[/b][/color]

Και να η λύση μου:
Θεωρώ μια ευθεία $ CD'$ τέτοια ώστε $ \angle{ACD'} \equal{} 10$ και θα αποδείξω ότι $ AD' \equal{} BC$ με αποτέλεσμα το $ D'$ να ταυτίζεται με το $ D$.

Πράγματι θα έχουμε απο θεώρημα ημιτόνων στα τρίγωνα $ AD'C, BD'C$:

$ \frac {D'C}{AD'} \equal{} \frac {\sin{20}}{\sin{10}} \equal{} 2\cos10$  $ (1)$ και

$ \frac {D'C}{BC} \equal{} \frac {\sin{80}}{\sin{30}} \equal{} 2\cos10$ $ (2)$  αφού $ \sin{80} \equal{} cos{10}$)=

Από $ (1)$, $ (2)$

$ \frac {D'C}{AD'} \equal{} \frac {D'C}{BC}\Longleftrightarrow AD' \equal{} BC$

[quote="Γιώργος"]Έστω $ ABC$ ένα ισοσκελές τρίγωνο με $ AB \equal{} AC$ και $ A \equal{} 20$. Θεωρούμε σημείο $ D$ πάνω στο τμήμα $ AB$ τέτοιο ώστε $ AD \equal{} BC$. Να αποδείξετε ότι $ ACD \equal{} 10$ :)

 [color=red]Edited : $ \angle ACD \equal{} 10$[/color][/quote]

Γιώργο ,
 πολύ καλή άσκηση ! Κάθε μαθητής της Γ΄Γυμνασίου που θέλει να ασχοληθεί  με ολυμπιάδες , πρέπει να συναντήσει αυτό το πρόβλημα για να δει τις κλίσεις του !!!
 Βάβαια , υπάρχουν πάνω από πέντε λύσεις. Η τριγωνομετρική που έστειλε ο Δημήτρης είναι απλή και καλή.

 Μπάμπης

Αυτό ακριβώς και συγνώμη για την ταλαιπωρία, αλλά το παραδέχομαι πως είμαι απρόσεχτος :blush: ..  Για να σας αποζημιώσω όμως να μία άλλη άσκηση( σωστή αυτή τη φορά)....

[color=red][b]ΑΣΚΗΣΗ 2 [/b][/color]

Έστω $ ABCD$ τετράγωνο πλευράς $ a$ και $ M,N$ τα μέσα των πλευρών $ AB$ και $ CD$ αντίστοιχα. Τα τμήματα $ AN$ και $ AC$ τέμνουν τη $ DM$ στα $ E$ και $ Z$ αντίστοιχα. Τα τμήματα $ CM$ και $ AC$ τέμνουν τη $ BD$ στα $ T$ και $ H$ αντίστοιχα. Να αποδείξετε ότι η ακτίνα του εγγεγραμμένου κύκλου στο τετράπλευρο $ MTHZ$ είναι διπλάσια από την ακτίνα του εγγεγραμμένου κύκλου στο $ AZE$.


Find all f:R+->R+. such that:
f(x+y)+f(x)f(y)=f(x)+f(y)+f(xy).

I don't find this problem very hard, but also not at all trivial.
  We obvioulsy have f(2)=2 (take x=y=2). Put also x=y=1 and find that f(1) is 1 or 2. Suppose it is 2. Then take y=1 and find that f(x)=2 for any x>1. Now take any x>0 and take any y>1 with x+y>1 and xy>1. It follows that f(x)=2. Thus in this case f is constant.
   Suppose now that f(1)=1. Take y=1 to find that f(x+1)=1+f(x). Take y+1 instead of y and use the initial relation and the one obtained after the substitution (and of course the fact that f(1+x)=1+f(x)) to find that f(x)+f(xy)=f(x+xy). Thus f is aditive. And thus f is also multiplicative from the initial relation. f must be strictly increasing from the aditivity and also f(x)=x for any rational x. Thus f(x)=x for any x.