hi i have here a problem solving equations. i havent slept because of this. can u guys pls help me? pls....

1.) Olan is 3 years older than his sister rose.  one-half of olan's age is thirteen years more than one-forth his sister's age one year ago.  find their ages now?

2.)Tess is 4 years older than her sister ana.  in four years, she will be 37 years more than one-forth as old as ana will then be.  how old are the girls now?

3.) a mother is 20 years older than her daughter.  five years from now, she will be thrice as old as her daughter will then be.  How old is the daughter?

4.) the sum of the ages of bobby and mercy is 60. nine years ago, bobby was three years more than one half as old as mercy was then.  find their ages now?

5.) when ramon was 12 years old, his father was 40.  how many years must pass for ramon to be 8 yyears older than one-forth as old as his father?

Um...Ok i dont know why you would lose sleep over these problems but i can help.

#1
[hide][quote="eeyahn23"]1.) Olan is 3 years older than his sister rose. one-half of olan's age is thirteen years more than one-forth his sister's age one year ago. find their ages now?[/quote]

if we say that Olans sisters age is x than Olans Age is x + 3.  therefore we know that $ \frac{x \plus{} 3}{2}$ = 13 + $ \frac{x}{4}$ multiplying everything by 4 yeilds 2x + 6 = 52 + x.  Solve this like a regular equation, we have x = 46.  So $ \boxed{\text{Olan is 49 and his sister is 46.}}$[/hide]

#2
[hide][quote="eeyahn23"]2.)Tess is 4 years older than her sister ana. in four years, she will be 37 years more than one-forth as old as ana will then be. how old are the girls now?[/quote]

so if ana is x than tess is x + 4, in four years ana will be x + 4 and Tess will be x + 8.  Than we can make an equation from this x + 8 - 37 = $ \frac{1}{4}$x + 1 solving this equation yeilds x - 30 = .25x this yeilds .75x=30 and x=40 so the ages are $ \boxed{\text{Tess \equal{} 44 and Ana \equal{} 40}}$[/hide]

#3
[hide][quote="eeyahn23"]3.) a mother is 20 years older than her daughter. five years from now, she will be $ \boxed{\text{thrice}}$ as old as her daughter will then be. How old is the daughter?[/quote]

Daughter = x Mother = 20 + x in five years mother = 25 + x and daughter = 5 + x and becaue she she i one third of her mothers age we multiply her age by 3 to create equality.  we have x + 25 = 15 + 3x, there fore x = 5 so $ \boxed{\text{mother is 25 and daughter is 5}}$[/hide]

#4
[hide][quote="eeyahn23"]4.) the sum of the ages of bobby and mercy is 60. nine years ago, bobby was three years more than one half as old as mercy was then. find their ages now?[/quote]

bobby = x and mercy = 2x - 3 but this was nine years ago so we add nine to each value giving us  x + 9 and 2x + 6 these add to 60 so we have 3x + 15 = 60 therefore 3x = 45 and x = 15 but this was nine years ago so nine years ago we had Bobby = 15 and mercy = 27 so add nine and we have $ \boxed{\text{Bobby \equal{} 24 and mercy \equal{} 36}}$[/hide]

#5
[hide][quote="eeyahn23"]5.) when ramon was 12 years old, his father was 40. how many years must pass for ramon to be 8 yyears older than one-forth as old as his father?[/quote]

so if we say that the years past = x then we know that 12 + x = 1/4(40 + x) + 8 which yeilds 12 + x = 18 + .25x which is 6 = .75 x so x = 8 so the number of years that will pass untill ramon is 8 years older than on fourth the age of his father is $ \boxed{8}$[/hide]

It seems like you just are too lazy to do your homework, or that you don't know how to do them. In any case, i will tell you HOW to solve that and i will start you off on #1.
First, you have to set easy variable for the ages such as O for Olan and R for Rose.
Then, you have to set up the equations. while in this process it is important to know that no matter how many years pass, the difference in 2 ppl's ages never gets bigger or smaller. Also dont get deterred by confusing phrases. tackle the phrases one or 2 words at a time if they look too confusing to deal with as a whole.

So for number 1, i will use O and R like i said before.

The first equations tells us that Olan is 3 years older that Rose
O=3+R
Now for the toughie. One half of Olan's age is 13 more than one fourth of Rose's age one year ago.
What is important here is that one quantity is 13 more than another quantity.
I will start with the first quantity: "one half of Olan's age"
that represents (1/2)O
the second quantity is "one fourth of Rose's age one year ago"
(1/4)(R-1)
Remember that this is not (1/4)R-1, the subtract 1 is first.

so now the first equations is         O=3+R
and the second equation is    (1/2)O=13+(1/4)(R-1)
then we can substitute 3+R for the O in the second equation and you get an equation in one variable. then plug R into the first equation and solve for O.

to check yourself, i will hide the answer below
[hide][hide][hide] Olan is 48 yeas old and Rose is 45 years old[/hide][/hide][/hide]
Check your work for #1 @ zolojetto
And for those familiar with LaTeX sorry for not using that on the fractions, i am still not sure how to use it.

um how about you check your work, im right lol.  one half of 49 = 24.5 and one fourth of 46 = 11.5 plus 13 = 24.5 so sry but im right and ur not!!!

But there is one thing i will agree with you on

[quote="stevenmeow"]It seems like you just are too lazy to do your homework, or that you don't know how to do them.[/quote]

yes when five of almost the same problems are posted in a row you get suspicious and wonder if you are doing some one elses homework...not saying that you were just trying to get answers from us eeyahn23.

thank you guys so much...actually i had 25 problems...these are just the numbers that i can't figure out. i tried but every time i do the checking...it's just wrong.  the rest are ok.  actually it's not laziness, it's even worse, i've just given up on those numbers...i'm too sleepy but i still need to prepare for school.  anyway,  you both are heaven sent, specially for people like me that really takes up the whole night just to figure out math problems. hehehhe.  God bless!!!  may you continue helping other people.  :blush:

Well your very welcome and its releiving to know that i didnt just give away the answers to all of your homework problems.  :lol:

[quote="zolojetto"]um how about you check your work, im right lol.  one half of 49 = 24.5 and one fourth of 46 = 11.5 plus 13 = 24.5 so sry but im right and ur not!!!
[/quote]
[quote="eeyahn23"]  one-half of olan's age is thirteen years more than one-forth his sister's age one year ago.  find their ages now?
[/quote]
one half of 49 is supposed to be 13 more than one fourth of (46-1) not  one fourth of 46

so this should be one half of 48 is 13 more than one fourth of (45-1)

 [quote]so sry but im right and ur not!!![/quote]

sorry about the im right and ur not i have a reading problem thats why i failed mathocunts!!! ill try to delete that.


On the USAMO, or any olympiad for that matter, how are multiple solutions to the same problem generally treated?

That is to say, if I have two distinct approaches to a problem and I write a solution for each of them, then how would it be marked? If one of solutions is incorrect, how would I be penalized?

It means more work for the grader, which would probably make the grader unhappy. Why don't you just write one correct solution?

[quote="ComplexZeta"]Why don't you just write one correct solution?[/quote]

That's the thing. Under stressed conditions, it may not be easy to detect flaws.

Olympiads aren't like AMC, being able to write multiple solutions don't really give advantages to the contestants. So common sense tells me that the final score should be the maximum of the all the solutions given to that problem. I wonder if that's really the case .

Another scenario where this comes into play is if a contestant is so strong that he/she finishes all the problems and still have plenty of time left, so he/she decides to look for a more elegent solution or generalization.

[quote="ComplexZeta"]It means more work for the grader, which would probably make the grader unhappy.[/quote]

Not necessarily. If the grader hates grading solutions so much, then why would he/she join the jury in the first place? Besides, out of all likelyhood, most of what he/she would read would be half-nonsense.

I grade the Mandelbrot team round, which is obviously not quite olympiad level, but our philosophy is "If you cross something out, and the resulting proof is correct, it doesn't matter what the content of the crossed-out stuff was."

[quote]"If you cross something out, and the resulting proof is correct, it doesn't matter what the content of the crossed-out stuff was."[/quote]

So, even if you cross out the first proof, but the second proof is correct, you'll still give it full marks?

Right -- if a team were to submit two proofs, one of which was correct and one of which was horrendously wrong, they would get full credit (assuming everything was legible and so on).

Now, it happens that on Mandelbrot there's rarely enough time for doing anything like that, so it usually just amounts to ignoring a single line or statement or something.  But in theory, yes.  And the same thing should be true of Olympiads, which are generally graded on a rubric.

Do you think the same philosophy would apply to grading USAMO?

Well, if you're planning planning to write more than one solution to a problem on the USAMO, I would really suggest that you ask an AMC official beforehand to make absolutely sure, since I don't think there's a well-known policy on this subject and I personally wouldn't risk getting a 0 on a problem if I were not absolutely sure that writing two solutions is acceptable.

This is not the same thing, but it is related:

On MOP tests for the lower group, you are given 5 questions, of which you choose three to do. At one point someone asked if they could submit solutions to four problems, and they were told not to do this, because the graders would just count their lowest three scores.

So if the policy is similar, they might just score whichever proof is worse.

First, some historical perspective:  In observing the USAMO, I don't recall seeing a situation where a participant wrote two distinct correct solutions to a problem.  I also don't recall a situation where a participant wrote a wrong solution, kept it on the paper and also wrote another correct solution.   (Although both have probably happened sometime!) I have seen some papers where a wrong solution was started, even finished, then clearly crossed out, and a correct solution substituted.

Let's take those cases in reverse order.  The judges are good at reading papers, they've seen a lot.  And they will clearly recognize where a solution has been crossed out and will generally ignore anything crossed out (even if it is correct!).  The leftover, uncrossed-out material will be carefully read, considered and graded accordingly.

In the second situation, I am not sure what the judges would do, it's a hypothetical case, as I said.   I suspect that the paper would lose points, since the writer did not recognize an incorrect argument.  I don't think such a  paper would get 0, since the correct answer would be acknowledged as worthy of some partial credit.

The first case is even more hypothetical.  Finding a complete correct solution to a (USA)MO level problem is challenging enough, finding two in the limited time available stretches credulity.  But if someone offered two correct, complete solutions s/he would get the full 7 points credit, no more, except perhaps the sincere admiration of the judges.

My advice:  concentrate on getting one good correct solution or proof, then move on to another problem.

Having said that, let me offer a little advice about writing your solution:
(1) Work it out fully on scratch paper, then transcribe a neat, organized, legible, logical solution or proof on the solution sheets.
(2) Use a black pen, or #2 lead pencil, and stay within the borders.  Otherwise the judges may not be able to read what has been faxed.  If it can't be read, it won't be graded.
(3) Make the solution or proof easy to read, easy to follow, and complete, but don't write a book!  Irritating the reader is not a good strategy for gaining points!  A messy, tangled proof is irritating to read.  A proof in which every 1+1 = 2 step is agonizingly belabored is equally irritating.  Clarity counts!
(4) I'm sure there is addtional good advice, but I will leave that to Mr. Rusczyk and Mr. Crawford in the Math Jam.

Steve Dunbar
AMC Director

Mr. Dunbar,

Would it be possible to sort of establish an official standard for grading papers in the second situation (when a student writes both a correct and an incorrect solution to a problem)? I ask this because the standard could really influence how a student would act during the USAMO. For example, say I came up with two solutions to a geometry problem, one using vectors and the other using inversion. However, being a bit shaky in both techniques, I am not quite sure whether either of them are completely correct. Which solution should I write down, or should I write down both?


[quote="zfzf21"]Since about a week or two ago, when I try to copy posts with LaTeX from the forum to a Microsoft Word document, the LaTeX images come out as blank boxes. I have Microsoft Word 2000 and 97 on two different computers and it doesn't work for either, and this did not occur before two weeks ago. Does anyone know why this would happen? Thank you![/quote]I have no idea. There might be many issues, but mainly it is your machine at fault here.

Wild guess here - have you changed any security settings on your computer so that it wouldn't allow you to import remote images into a Word document?

[quote="nr1337"]The same strange thing is happenning to me.  It used to come out perfectly, but then a while ago, copying images just stopped working.  I think it may be something on the server.[/quote]FYI I just copy-pasted something with formulas from the forum and it went just fine.

In general as long as you can see the images in your browser, it has [u]absolutely[/u] nothing to do with the server ;). It is your client at fault here, so you must investigate your own configuration  :?

Maybe there was a change in the LATeX itself? Dunno, as I don't really use it...

[quote="zfzf21"]Since about a week or two ago, when I try to copy posts with LaTeX from the forum to a Microsoft Word document, the LaTeX images come out as blank boxes. I have Microsoft Word 2000 and 97 on two different computers and it doesn't work for either, and this did not occur before two weeks ago. Does anyone know why this would happen? Thank you![/quote]

two questions: Did you disconnect from the internet before you copied? If not, in word, go to tools>macros>security, and adjust it to medium.

also, try it on firefox, after all it is 4mb and takes only 20 secs on a dsl connection

ONe more thing, if you use xp, roll back your machine by 2 weeks. start>programs>accessories>system tools>system restore, and restore it to the day it was working.

i am having this problem too :? its weird


Given is a bilinear mapping $\phi : V_1 \times V_2 \rightarrow W$.
Show that $Im(\phi)$ needn't to be a vector space.
(Only an example)

Consider $f: %Error. "mathdd" is a bad command.
{R}^{2} \times %Error. "mathdd" is a bad command.
{R}^{2} \rightarrow %Error. "mathdd" is a bad command.
{R}^{2} \times %Error. "mathdd" is a bad command.
{R}^{2}$, $f((x,y),(a,b)) = ((xa,yb),(xb,ya))$. Then $f$ is bilinear, $((0,1),(0,0)),((1,0),(0,0)) \in Im(f)$ but $((1,1),(0,0)) \notin Im(f)$, whence $Im(f)$ is not a vector space.


number theory
greatest common divisor
modular arithmetic
LaTeX
number theory unsolved

show that if $ (a,b)\equal{}1$ and $ p$ is an odd prime then

$ (a\plus{}b, (a^p\plus{}b^p)/(a\plus{}b) )\equal{}1 or p$

Note that $ \gcd( a^m \minus{} b^m , a^n \minus{} b^n ) \equal{} a^{\gcd(m,n)} \minus{} 1$. Hence, $ \gcd(a \minus{} (\minus{}b) , a^p \minus{} (\minus{}b)^p ) \equal{} a \minus{} (\minus{}b)$. Since $ p$ is odd, we deduce $ \gcd(a \plus{} b , a^p \plus{} b^p ) \equal{} a \plus{} b$. Then it's clear that your conclusion follows.

as i know $ a^m\minus{}1, a^n\minus{}1)\equal{} a^gcd(m,n) \minus{}1$...Note that .i cant understand what you used in your solution...can you describe??

reduce the second number mod a+b and u will be able to prove it. Im unable to write the solution coz i dont know how to insert math symbols. sorry

Dear Ishfaq, I won't help you with this. Because some time ago I sent you a solution to a similar problem and this one can be solved using the same idea. If you read my solution carefully you should have been able to solve it.

Even though, here are some hints: Let $ \left(a + b,\frac {a^p + b^p}{a + b}\right) = g$. Then $ a\equiv - b\pmod g$, and $ a^{p - 1} - a^{p - 2}b + \cdots + b^{p - 1}\equiv 0\pmod g$. Try substituting the first congruence in the second one and see what happens.

[quote="lajanugen"]Im unable to write the solution coz i dont know how to insert math symbols. sorry[/quote]

To write in $ \text{\LaTeX}$, just put your symbols within two dollar signs. For example
[code]$(a^{m}-1,a^{n}-1)=a^{\gcd(m,n)}-1$[/code]
will give you: $ (a^{m} - 1,a^{n} - 1) = a^{\gcd(m,n)} - 1$. To learn more about $ \text{\LaTeX}$ codes, see [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:About]here[/url].


linear algebra
matrix
linear algebra unsolved

Solve in $M_2(R)$

$A^2+B^2=\left(\begin{array}{cc}2&0\\-8&2\end{array}\right)$

There are many possible solutions. Here are two possibilities:

$I^2+\begin{bmatrix} 1 & 0 \\ -4 & 1 \end{bmatrix}^2$

$\begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}^2+\begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}^2$

To you want to try to characterize all of them?


AMC AIME MATHCOUNTS USA(J)MO USAMO algebra polynomial

Hello,

Does anyone want to do this anymore? This one will be math. If you want to be the 1 or join the mob, post or PM me.

(I think the last one failed.)

I will join only if there is no trivia and only math.

[quote="lotsofmath"]There will be no trivia.[/quote]

OMGOMGOMG I AM IN

I'm in

The message is too small. Please make the message longer before submitting.

[quote="mathcountsfreak"]In.

Difficulty of math problems? Cause if you look in my sig...my skills have been declining recently (see my aime score :D).[/quote]

They will start out at MATHCOUNTS level and gradually get harder as the game progresses.

I'll join, but I'll die if it gets harder than basic high school.

Currently:

1: pythag011 ((s)he PM'd me)

Mob:

Brut3Forc3
cf249
1=2
shaggy75
mathcountsfreak
kimmystar94
hm29168

count me in if i can still join..:)... :lol:

You can still join. I haven't decided on the deadline to sign up yet.

dang, if both players get an integration problem correct it signals a requirement for evil problems

I did not use integration. I just noted that the answer has to be between 1 and \sqrt{2} :blush:

Let A be the integer such that 1+1/2+1/3+1/4+...+1/22+1/23=a/23!
Compute the remainder when a is divided by 13.

a.6
b.7
c.8

lol its been a while since i checked this topic, but i guess i happened to check on the right day :)

I am being very disgruntled that everyone (well, really, two people) is trouncing an entire arsenal of questions. Therefore I have submitted another problem, which will (hopefully) eliminate one of them.

[quote="Brut3Forc3"]Sorry, but I know pythag011, and he isn't liable to miss one of these... like, ever...[/quote]
Unless there is a calculus or hard USAMO level problems. But it multiple choice...

I sent in that problem.

My intended solution:

Since $ 1+\frac{1}{2}+\frac{1}{3}+\hdots+\frac{1}{22}+\frac{1}{23}=\frac{23!(1)+23!(\frac{1}{2})+23!(\frac{1}{3})+\hdots+23!(\frac{1}{23})}{23!}$.

Then, $ a=23!(1)+23!(\frac{1}{2})+\hdots+23!\frac{1}{23}$.

Every term in this expression is divisible by $ 13$ except

$ 23!(\frac{1}{13})$. Consequently, the remainder when $ a$

is divided by $ 13$ is the remainder when $ \frac{23!}{13}$ is

divided by $ 1$. Since the remainders when $ 14, 15, 16, \hdots, 23$

are divided by $ 13$ are the same, respectively, as the remainders

when $ 1,2,3,\hdots,10$ are divided by $ 13$, the remianders $ R$

of the following are the same when divided by $ 13$:

$ (1 \cdot 2 \cdot 3 \cdot \hdots \cdot 12)(14 \cdot 15 \cdot 16
\cdot \hdots \cdot 23), (1 \cdot 2 \cdot 3 \cdot \hdots \cdot 9 \cdot 10)^2(11 \cdot 12), ((2 \cdot 7) \cdot (3 \cdot 9) \cdot (4 \cdot 10) \cdot (5 \cdot 8) \cdot 6)^2(11 \cdot 12)$.

Note that we have paired up numbers whose product is $ 1$ more

than a multiple of $ 13$. This means that the remainder when $ a$

is divided by $ 13$ is given by the remainder when $ 6^2 \cdot 11
\cdot 12$ is divided by $ 13$. Since $ 10 \cdot 11 \cdot 12=1320=13 \cdot 11+20$ gives the same remainder, we need

only look at the remainder when $ 20$ is divided by $ 13$.

Answer: $ \boxed{7}$.

[quote="BOGTRO"]
Since the remainders when $ 14, 15, 16, \hdots, 23$

are divided by $ 13$ are the same, respectively, as the remainders

when $ 1,2,3,\hdots,10$ are divided by $ 13$, the remianders $ R$

of the following are the same when divided by $ 13$:

$ (1 \cdot 2 \cdot 3 \cdot \hdots \cdot 12)(14 \cdot 15 \cdot 16 \cdot \hdots \cdot 23), (1 \cdot 2 \cdot 3 \cdot \hdots \cdot 9 \cdot 10)^2(11 \cdot 12), ((2 \cdot 7) \cdot (3 \cdot 9) \cdot (4 \cdot 10) \cdot (5 \cdot 8) \cdot 6)^2(11 \cdot 12)$.

Note that we have paired up numbers whose product is $ 1$ more

than a multiple of $ 13$. This means that the remainder when $ a$

is divided by $ 13$ is given by the remainder when $ 6^2 \cdot 11 \cdot 12$ is divided by $ 13$. Since $ 10 \cdot 11 \cdot 12 \equal{} 1320 \equal{} 13 \cdot 11 \plus{} 20$ gives the same remainder, we need

only look at the remainder when $ 20$ is divided by $ 13$.

Answer: $ \boxed{7}$.[/quote]

Note we could have also used Wilson's: $ 12!\equiv \minus{}1\text{ (mod 13)}$.

ahhhhhhhhhhhhhhhhhhhh! I forgot all about this thread!

Anyway,

In each iteration the Cantor Set is constructed by removing the middle third of each line segment. If the fractal dimension of the Cantor Set is x, which of the following is true?
a) 0.5<x<0.6
b) 0.6<x<0.7
c) 0.7<x<0.8

Due date Friday.

im already out...but isnt there a certain number of questions before the "1" just automatically wins...

[quote="lotsofmath"]No matter. pythag011 wins the game!!!!!!![/quote]Oh, I missed the problem. Sorry, I've been a bit busy. Is it too late to submit now? If it's too late, then what is the answer to the problem?


The shaded square has area $ \sqrt{2}^2\equal{}2$.  The total area of the large square is $ 4^2\equal{}16$.  The answer is 2/16, or $ \boxed{\frac{1}{8}}$.

If you rearrange the 1 by 1 right triangles in the grey square, you will get a 1 by 2 rectangle which has an area of 2 so the area os the grey square is $1*2 = 2$

The area of the large square is $4^2 = 16$

 $\frac{2}{16} = \boxed{\frac{1}{8}}$


The sum of the lengths of all the edges of a particular cube is 24 cm. When the cube is unfolded into a flat sheet (net) of six connected squares, each sharing at least one side with another square, what is the perimeter of this resulting polygon?

28 cm.

A cube has 12 edges, so 24/12 = 2 cm per edge. A diagram of the cube unfolded has 14 distinct edges, and so 28 cm worth of perimeter.


Consider the set $ T$ of equilateral triangles with side length $ 1$ all the vertices of which are among the vertices of a given convex $ n$-gon.

a) Prove that the set $ T$ contains at most $ \frac{2n}{3}$ triangles.

b) Prove that for every $ 0<a<\frac{2}{3}$ there exists a convex $ n$-gon such that the corresponding set $ T$ contains more then $ an$ triangles.


function
floor function
number theory unsolved
number theory

Let $n$ and $a_1,a_2,\dots,a_m$ be positive integers. Define the function $f$ as follows: for each
integer $k$ let $f(k)$ be the number of ordered $m-$tuples $(c_1,\dots,c_m)$ of integers such that $c_1+\cdots+c_m\equiv k\ (\text{mod }n)$ and $1\leqslant c_i\leqslant a_i$ for each $i$. Show that $f$ is constant if and only if $n$ divides at least one of $a_1,\dots,a_m$.

[hide="Here is my approach"]Let \[P(x)=(x+x^2+\cdots+x^{a_1})(x+\cdots+x^{a_2})\cdots(x+\cdots+x^{a_m})=x^m\frac{(x^{a_1}-1)(x^{a_2}-1)\cdots(x^{a_m}-1)}{(x-1)^m}\]

WLOG we may suppose $0\leqslant k<n$. Now we have
\[f(k)=\sum_{i=0}^{\left\lfloor\frac{a_1+a_2+\cdots+a_m-k}n\right\rfloor}(-1)^{a_1+\cdots+a_m-k-in}\sum_{\text{sym}}\alpha_1\alpha_2\cdots\alpha_{a_1+\cdots+a_m-k-in}\]

where $\alpha_1,\alpha_2,\dots,\alpha_{a_1+a_2+\cdots+a_m}$ are roots of $P(x)$, i.e. $a_1$th, $a_2$th, ..., $a_m$th roots of unity different from $1$ (also, $m$ of $\alpha$'s are equal to $0$).

I can't figure out what to do now, could somebody help?[/hide]

It is much more simple  :)
Take $ P(x) =
(x+x^2+\cdots+x^{a_1})(x+\cdots+x^{a_2}) \ldots (x+\cdots+x^{a_m}) $, as you wrote, and
we can write $ P(x) = \sum_{j=1}^{N} c_{j} x^{j} $, where $ c_{j} $ are integers and
we know that for every $ k $, $ \sum_{ j \equiv k ( mod n) } c_{j} = C = const $, therefore
if we denote by $ \zeta = e^{ \frac{2 \pi i}{n} } $ the $ n $-th root of unity, then
we see that $ P( \zeta ) = C ( \zeta + \zeta^{2} + \ldots + \zeta^{n} ) = 0 $,
so for some $ a = a_{j} $ we have that $  \zeta + \zeta^{2} + \ldots + \zeta^{a} = 0 $.
From this we see that $ 1 + \zeta + \zeta^{2} + \ldots + \zeta^{a-1} =
\frac{\zeta^{a} - 1 }{\zeta -1} = 0 $, so $ \zeta^{a} = 1 $, so $ n | a = a_{j} $.

Well, that's me - doing everything the harder way :) Thank you.


calculus
derivative
function
real analysis
real analysis unsolved

1) Find all $C^{2}$ function f(x,y) such that:
$\frac{d^{2}f}{dxdy}=0$ for all x,y
2) Find all $C^{2}$ function f(x,y) such that:
$\frac{d^{2}f}{dx^{2}}=\frac{d^{2}f}{dy^{2}}$
Can anyone solve it? :D

If $f_{xy}=0$, then $f_{x}$ does not depend on $y$, so...

The change of variables $x=u+v$, $y=u-v$ reduces 2) to 1).

It is a beautiful solution :D . Thank you very much :lol:


geometry
circumcircle
trapezoid
perpendicular bisector
geometry proposed

Let $ ABC$ be an acute triangle, $ CC_1$ its bisector, $ O$ its circumcenter. The perpendicular from $ C$ to $ AB$ meets line $ OC_1$ in a point lying on the circumcircle of $ AOB$. Determine angle $ C$.

Let $ \{X\} \equal{} OC_1\cap h_c$, where $ h_c$ is the altitude corresponding to the side $ AB$.
Moreover, let $ K$ be the intersection point of the line $ CC_1$ with the circumcenter of $ \triangle{ABC}$. Consider the inversion $ I$ of pole $ O$ and power $ CO^2$.
Then $ I(C_1) \equal{} X$, thus $ COKX$ is a cyclic trapezoid, which implies that $ AB$ is the perpendicular bisector of $ CX$.
Since $ AOBX$ is cyclic we infer that
\[ 180^{\cdot} \equal{} \angle{AOB} \plus{} \angle{AXB} \equal{} 2\angle{C} \plus{} \angle{C},
\]
hence $ \angle{C} \equal{} 60^{\cdot}$.

Sorry for bumping this post up like 10 years, but can someone explain how  $ COKX$ is a cyclic trapezoid in bravado's solution.


For a poitive integer $ n$, denote $ S(n)$ be the sum of digits of $ n$. Prove that there exists 2008 integrs $ a_1, a_2,\cdots, a_{2008}$ satisfying $ a_i\ne a_j$ for $ i\ne j$ and $ a_i \plus{} S(a_i) \equal{} const$ for every $ i \equal{} 0,1,\cdots,2008$

Obviosly $ 91\plus{}S(91)\equal{}101\equal{}100\plus{}S(100)$. Let $ a\equal{}91,b\equal{}100$
Therefore we can chose $ a_i\equal{}\sum_{k\equal{}0}^{m\minus{}1} c_k1000^k$, Obviosly $ S(a_i)\plus{}a_i\equal{}101\frac{1000^m\minus{}1}{999}\equal{}const$ for all $ 2^m$ cases chose $ c_k\equal{}a$ or $ c_k\equal{}b$ or chose $ a_i$.

[quote="Rust"]Obviosly $ 91 \plus{} S(91) \equal{} 101 \equal{} 100 \plus{} S(100)$. Let $ a \equal{} 91,b \equal{} 100$
Therefore we can chose $ a_i \equal{} \sum_{k \equal{} 0}^{m \minus{} 1} c_k1000^k$, Obviosly $ S(a_i) \plus{} a_i \equal{} 101\frac {1000^m \minus{} 1}{999} \equal{} const$ for all $ 2^m$ cases chose $ c_k \equal{} a$ or $ c_k \equal{} b$ or chose $ a_i$.[/quote]
Can you explain in more details? What is relationship between m and i? How is c_k denoted? Thanksso much.

No problem. It is very hard and I think its proof is not simple. I still hope a solution for this.  :)  thanks for your interests.

Let $ a_i\equal{}10^ka\minus{}b$, then $ T(a_i)\equal{}a_i\plus{}S(a_i)\equal{}10^ka\plus{}9k\plus{}S(a)\minus{}b\minus{}S(b)$ and $ T(10^ka\plus{}c)\equal{}10^ka\plus{}S(a)\plus{}c\plus{}S(c)$.
It give solutions. $ T(91)\equal{}101\equal{}T(100)$. Let $ a_{10}\equal{}91$, then $ a_{11}\equal{}101$
If $ T(a_i)\equal{}T(b_i),a_i,b_i<10^k$ we consider numbers $ a_{i\plus{}1,j}\equal{}10^ka\minus{}a_{i,j},a_{i\plus{}1,i\plus{}1}\equal{}10^ka\plus{}c$, suth that
$ T(a_{i\plus{}1,j}\equal{}10^ka\plus{}9k\plus{}S(a)\minus{}T(a_{ij}),T(a_{i\plus{}1,i\plus{}1})\equal{}10^ka\plus{}S(a)\plus{}T(c)$.
Obviosly $ T(a_{i\plus{}1,j}\equal{}T(a_{i\plus{}1,i\plus{}1})$ if $ 9k\equal{}T(a_{i,j})\plus{}T(c)$. We can find c, suth that $ 9|T(a_{i,j})\plus{}T(c)$ and find $ k\equal{}\frac{T(a_{i,j}\plus{}T(c)}{9}$ and continue procecc.
$ a_{00}\equal{}9\to 9k\equal{}18\plus{}T(c)\to c\equal{}0,k\equal{}2\to a_{10}\equal{}10^2\minus{}9,a_{11}\equal{}10^2\plus{}0\equal{}100.$
$ T(a_{10})\equal{}T(a_{11})\equal{}101\to 9k\equal{}101\plus{}T(c)\to c\equal{}17,k\equal{}14\to a_{20}\equal{}99999999999909,a_{21}\equal{}99999999999899,a_{22}\equal{}100000000000017$ e.t.c.


counting
distinguishability
combinatorics solved
combinatorics

Let S = {1, 2, ... , 10}. Find the number of unordered pairs A, B, where A and B are disjoint non-empty subsets of S.

If Card(S)=n , the required number is (1+3^n)/2 - 2^n

Ok, let's give full solution to that old problem before classify it into the 'solved section' :
Let S be a set such that |S| = n.

First, we choose any subset A from S distinct from  \emptyset  and S (to be able to choose B =/=  \emptyset ). Let k = |A|. Then there are C(n,k) such choices for A.
For any of thses choices, we have to select B =/=  \emptyset from the set S' = S-A where |S'|= n-k. Thus there are 2 n-k -1 possible choices for B.
Note that following the reasoning each pair {A,B} appears exactly 2 times.
Then, the required number is f(n) = (1/2)sum(k=1 to n-1) C(n,k)(2 n-k  - 1).

Using that C(n,k) = C(n, n-k) that sum(k=0 to n) 2 k = 3 n  and sum(k = 0 to n) C(n,k) = 2 n , it easily leads to Diogene's answer : f(n) = (1 + 3 n )/2 - 2 n .

In particular, returning to the initial question f(10) = 28501.

Pierre.

You could also solve it this way: Suppose you have a set S, where |S|=n as stated already. Now prepare 3  indistinguishable bins for our elements labeled 'A', 'B' and 'C'.  The number of ways to place the n elements in the bins is of course, 3^n. This isn't the answer however because:
 1) This counts the (2^n - 1) + (2^n - 1) = 2^(n+1) -2 arrangements with either A or B empty.
 2) This also  counts the case where both A and B are empty, and
 3) It counts each valid A:B 'pair' exactly twice (since they are supposed to be indistinguishable- i.e. {1,2,3}:{4,5} is the same as {4,5}:{1,2,3}).
 To correct for these shortcomings we have:
 (3^n - (2^(n+1) -2) - 1)/2 = (3^n +1)/2   - 2^n

...or would you rather tackle it by solving the non-homogenious recurrence relation?
 a_n = 3*a_(n-1) + 2^(n-1) - 1
 I believe this would be harder however but I would be interested in seeing if someone solved it this way.

Let b_n=a_n+2 n-1/2.
Notice that b_(n+1)=3*b_n.

Or you cold notice that (3 n+1)/2-2 n checks the recurrence you (JJJ) wrote and check for the first term of the sequence.


Compute $x$ if $\text{Arctan }x+\text{Arctan }1=2\left(\text{Arctan }x-\text{Arctan }\frac{1}{3}\right)$

this was posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30264]here[/url], so im not sure if we should post answers... :?

edit: day 7 has ended, so i will post my answer here

[hide]let $y=\arctan{x}, z=\arctan{\frac{1}{3}}$
$y+45=2(y-z)$
$45+2z=y$
draw triangle for $z$
$\sin{z}=\frac{1}{\sqrt{10}}, \cos{z}=\frac{3}{\sqrt{10}}$
$\sin{2z}=2\sin{z}\cos{z}=\frac{3}{5}$
so it forms a 3-4-5
$x=\tan{y}=\tan{(45+2z)}=\frac{\tan{45}+\tan{2z}}{1-\tan{45}\tan{2z}}=\frac{1+\frac{3}{4}}{1-(1)(\frac{3}{4})}=\boxed{7}$[/hide]


How do I type Chinese in the TeXnicCentre? Do I have to download a package?

P.S. My version of TeXnicCentre is stange. It is called: Version 1 Beta 1.61 (Fawks). Is there a way for me to update it?

The (slightly) newer version is 1 beta 7.01; go to http://www.texniccenter.org to download it.

I don't know if it has Chinese character support.

Thanks for the link of the new version!

I updated my TeXnicCentre and installed a CJK package from the MiKTeX Package manager, but I don't know how to use it......


[u][b]Problem 1 :[/b][/u]

In a square $ABCD$ let $F$ be the midpoint of $\left[ CD\right] $ and let $E$ be a point on $\left[ AB\right] $ such that $AE>EB$ . the parallel with $\left( DE\right) $ passing by $F$ meets the segment $\left[ BC\right] $ at $H$.
Prove that the line $\left( EH\right) $ is tangent to the circle circumscribed with $ABCD$.

[u][b]Problem 2 :[/b][/u]

Find all the positive integers $x,y,z$ satisfiing : $x^{2}+y^{2}+z^{2}=2xyz$

[u][b]Problem 3 :[/b][/u]

consider $n$ point $A_{1},A_{2}...A_{n}$ on a cercle.
How many ways are there if we want to color these points by $p$ colors, so that each two neighbors points are colored with two different colors ?

[u][b]Problem 4 :[/b][/u]

consider the set : $A=\left\{ 1,2,3,...,2046\right\} $
and $S$ a subset of $A$ formed of 21 elements.
Prove that there exist three element $a,b,c$ from $S$ verifiing : $bc<2a^{2}<4bc$

Thanks [b] erdos [/b] for the problems :)

Here are the links to the problems:

Problem 1: http://www.mathlinks.ro/Forum/viewtopic.php?p=184289#p184289
Problem 2: http://www.mathlinks.ro/Forum/viewtopic.php?p=183940&highlight=#183940
Problem 3: http://www.mathlinks.ro/Forum/viewtopic.php?p=183923&highlight=#183923
Problem 4: http://www.mathlinks.ro/Forum/viewtopic.php?p=183928&highlight=#183928

how many time you have for this problems?

[quote="zhaoli"]Would you please tell which year did these problems were? Are they hold in 2005?[/quote]
These olympiads took place on March 11 , 2005
And the time allowed was 3h.

How do you think: were problems quite hard?

yeeeh a little bit,  :lol:
especially the two last ones, I didn't solve them at all.  :(
But it's not important , the important tests are on next month , the moroccan team selection tests for IMO 2005, I hope I will pass  :lol:


induction algebra polynomial Gauss quadratics calculus integration

Let x be a positive real number and let k be a positive integer. Assume that $x^k+\frac{1}{x^k}$and$ x^{k+1}+\frac{1}{x^{k+1}}$ are both rational numbers. Prove that $x+\frac{1}{x}$, is also a rational number.

For each integer $nb$, let $S_n = x^n + x^{-n}$. Then $S_{-n} = S_n$
Suppose that $S_k$ and $S_{k+1}$ are rational numbers of some positive integer $k$.
Note that the result is obvious if $x=1$, then we may assume that $x \neq 1$.

It is easy to verify that, for each integers $n,p$, we have
$S_nS_p = S_{n+p} + S_{n-p}$ (1)
And, in particular : $S_n^2 = S_{2n} + 2$. (1')

It follows that $S_{n+1}-S_n = S_n(S_1-2) + S_n-S_{n-1}$, then by induction we deduce that the sequence $\{ S_n \}$ is increasing (that's where we use that $x \neq 1$).

If $k =1$, then an easy induction using (1) proves that $S_n$ is rational for all $n$.
Now assume that $k > 1$.
Another easy induction using (1) and (1') gives that $S_{nk}$ and $S_{n(k+1)}$ are rational for all $n$.
It follows that both $S = S_1+S_{2k+1}=S_kS_{k+1}$ and $P = S_1S_{2k+1} = S_{2k} + S_{2(k+1)}$ are rational numbers.
Thus, $S_1$ is a root of $X^2 - SX + P$. (2)

In another hand, both $S' = S_1 + S_{2k^2-1} = S_{2k^2}S_{2(k^2-1)}$ and $P'= S_1S_{2k^2-1} = S_{2k^2} + S_{2(k^2-1)}$ are rational numbers too.
Thus $S_1$ is a root of $X^2-S'X+P'$. (3)
Moreover, since $ \{ S_n \}$ is increasing, it follows that $S' = S_1 + S_{2k^2-1} > S_1 + S_{2k+1} = S$.
Thus, from (2) and (3), $S_1 = \frac {P'-P} {S'-S}$ is rational.

Note that, it is now easy to prove by induction that all $S_n$ are rational numbers.
Moreover, it is not difficult to adapt the above proof if we replace 'rational' by 'integer'.

Pierre.

At our last IMO training, in Bremen, one of the twelve matboj problems was this problem, but for $k$, $k+1$ replaced by arbitrary coprime integers and "rational" replaced by "integer" in the problem. I was the only one who solved it(my first proof used galois theory, later I found a nice argument not using galois theory... you consider the set of common zeroes of the two polynomials $x^{n}+x^{-n}-a$ and $x^{m}+x^{-m}-b$ for some integers $a,b$ and conclude, using that $m,n$ are coprime, that there are most 2 common zeroes. therefore their common divisor which has to have integers coefficients, is of degree at most 2. thus it must be of the form $t(x+x^{-1}+s$ with integers $t,s$; by using gauss lemma, we then conclude $t=1$ and we are done).

Peter, can you show your proof using Galois theory?

Here's another proof for the problem mentioned by Peter:

$x^{m}$ and $x^{n}$ are algebraic integers in the quadratic fields $\mathbb Q\left(\sqrt d\right)$ and $\mathbb Q\left(\sqrt{d'}\right)$ respectively for some square-free integers $d,d'$. Because $x^{mn}$ belongs to both of these fields and is not an integer (we disregard the trivial case $x=1$), we have $d'=d$, i.e. $x^{m},x^{n}\in\mathbb Q\left(\sqrt d\right)$.

Choose integers $p,q$ such that $pm+qn=1$. We then have $x=x^{pm+qn}=(x^{m})^{p}\cdot(x^{n})^{q}\in\mathbb Q\left(\sqrt d\right)$, so the minimal polynomial of $x$ has degree two (and integral coefficients). Also notice that the free term of this minimal polynomial is either $1$ or $-1$. If it is the former, we are done: $x+x^{-1}\in\mathbb Z$. If it is the latter, then $x-x^{-1}\in\mathbb Z$, so if $m$ is odd, for instance (at least one of $m,n$ is odd), then $x^{m}-x^{-m}\in\mathbb Z$. This means that $2x^{m},2x^{-m}\in\mathbb Z$, which is absurd.

[quote="grobber"]
Choose integers $p,q$ such that $pm+qn=1$. We then have $x=x^{pm+qn}=(x^{m})^{p}\cdot(x^{n})^{q}\in\mathbb Q\left(\sqrt d\right)$, so the minimal polynomial of $x$ has degree two (and integral coefficients). Also notice that the free term of this minimal polynomial is either $1$ or $-1$. If it is the former, we are done: $x+x^{-1}\in\mathbb Z$. If it is the latter, then $x-x^{-1}\in\mathbb Z$, so if $m$ is odd, for instance (at least one of $m,n$ is odd), then $x^{m}-x^{-m}\in\mathbb Z$. This means that $2x^{m},2x^{-m}\in\mathbb Z$, which is absurd.[/quote]
I don't know what is the meaning of the free term of this minimal polynomial.Could somebody explain a liitle more? :blush:


I used a TI-89 heavily for one of my solutions.  Can I still get 5/5 for it, even though I won't get commended?

I m not a grader and this is my first year doing USAMTS but I think that the more you use the calculator the less probability that you get a high score on that problem.  They are probably looking for solutions that don't need a calculator for anything but maybe checking some arithmetic.  This is what I found on the USAMTS portion of this site:

[quote]More specifically, full credit will be given only for complete, well presented solutions; answers alone will be rewarded more frugally.[/quote]

I would consider a solution that relies on the calculator comparable to just an answer alone (less extreme however).

[quote="dts"]I think you'll still get a 5.  If my memory serves, they actually let you write computer programs that scan the problem for possible answers and you can still get a 5 (but I'm not sure, and even if I am right, you are still supposed to submit a description of your program and explain why it will get all possible solutions...).[/quote]

This is basically right - you have to be able to convince the grader that your use of (calculator/computer) generates all solutions.  If you do so, you should be fine.


Fie $ A_1A_2A_3...A_n$, $ n\geq3$ un poligon regulat,A,B,C trei varfuri consecutive si M un punct pe cercul circumscris astfel incat dreapta AC sa separe punctele B si M.Sa se arate ca are loc relatia:$ MA\plus{}MC\equal{}2MBcos\pi/n$

daca nu posteaza nimeni in seara asta solutia voi reveni maine cu prezentarea ei..mi e cam somn :P

chiar nimeni nu a putut posta o solutie? problema nu e grea...voi posta eu solutia...:D cam scolareasca dar merge.Fie $ z_A\equal{}1$, $ z_B\equal{}\epsilon$ , $ z_C\equal{}\epsilon^2$, unde $ \epsilon\equal{}cos \frac{2\pi}{n}\plus{}isin \frac{2\pi}{n}$ iar afixul lui M il vom nota $ z_M\equal{}cost\plus{}isint$.
\[ MA\equal{}|z_M\minus{}1|\equal{} \sqrt{(cost\minus{}1)^2\equal{}sin^2t}\equal{} \sqrt{2\minus{}2cost}\equal{}2sin\frac{t}{2}\]
\[ MB\equal{}2sin(\frac{t}{2}\minus{} \frac{\pi}{n})\]
\[ MC\equal{}2sin(\frac{t}{2}\minus{} \frac{2 \pi}{n})\]
utilizand aceste rezultate egalitatea din enunt devine
\[ 2sin \frac{t}{2}\plus{}2sin( \frac{t}{2}\minus{} \frac{2\pi}{n})\equal{}4sin( \frac{t}{2}\minus{} \frac{\pi}{n})cos \frac{\pi}{n}\]Se transforma in produs membrul stang si inegalitatea este evidenta.

sau putem direct, mult mai direct cu Ptolemeu


Let $ a,b,c,d$ be positive numbers such that: $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}4$ .Prove that:
$ a^4\plus{}b^4\plus{}c^4\plus{}d^4 \ge \frac{8}{3}\plus{}\frac{8}{3}.\frac{a^3\plus{}b^3\plus{}c^3\plus{}d^3}{a^3\plus{}b^3\plus{}c^3\plus{}d^3\plus{}4abcd}$

[quote="quykhtn-qa1"]Let $ a,b,c,d$ be positive numbers such that: $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 4$ .Prove that:
$ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \ge \frac {8}{3} \plus{} \frac {8}{3}.\frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3}{a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \plus{} 4abcd}$[/quote]
IT's easy because $ a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \plus{} 4abcd \leq 8$! :P


"Dihydrogen Monoxide (DHMO) is a colorless and odorless chemical compound, also referred to by some as Dihydrogen Oxide, Hydrogen Hydroxide, Hydronium Hydroxide, or simply Hydric acid. Its basis is the highly reactive hydroxyl radical, a species shown to mutate DNA, denature proteins, disrupt cell membranes, and chemically alter critical neurotransmitters. The atomic components of DHMO are found in a number of caustic, explosive and poisonous compounds such as Sulfuric Acid, Nitroglycerine and Ethyl Alcohol."

Please read about the adverse effects of dihydrogen monoxide: http://www.dhmo.org/facts.html

You do know that we are over 66% Dihydrogen Monoxide?
And that with less than 30% DHMO in our bodies we are dead?

herefishfishy has already showed me this. :D

I'm writing a paper on this right now. Thank you for the statistics! :)

Did you know that a good percentage of the US congress voted for banning Dihydrogen Monoxide when some ingenious guy submitted a proposal to ban it? :roll:

[quote="alkjash"][quote="stevenmeow"]oh snap i drank some dihidrogen monixide recent;y... hopefully i dont experience any adverse effects[/quote]

Experts say that humans are genetically addicted to DHMO. Withdrawal effects include loss of consciousness and certain death.[/quote]

oh yeah effects should start kicking in after a few days

anyway... what was the percentage?

[quote="dragon96"]Dihydrogen=2(H)=h^2
Monoxide=1(O)=o^1
dihydrogen monoxide=$ H^2O$

I like that trick.[/quote]

It's actually $ H_2O$...

It is illegal to ban dihydrogen monoxide. Doing so is equally as immoral as exiling an entire population for no reason.

Yeah we're just a bunch of AoPS clowns. Dont worry about us.

Ban it!  :mad:  We cannot not allow the use of such a dangerous poison! I tried washing a stain of DHMO away with water, but it didn't work!

This is terrible! OUr government has failed us yet again by allowing us to use this chemical.

[quote="AIME15"][b]WE ARE ADDICTED BECAUSE OUR PARENTS ARE [/b][/quote]
Actually, addiction is not inherited. It is a self-gained habit. I will research about why animals and humans need it.

[quote="1=2"][b]KILL THE PARENTS!!!!!!!!![/b][/quote]
Darn you!

Mathelete, swearing is bad.  I suggest you stop.

By the way, please don't @#&% 1=2.  At least while I'm here.  *moves away*

[quote="AIME15"][b]WE ARE ADDICTED BECAUSE OUR PARENTS ARE [/b][/quote]

Just as I stated earlier. Thus, if we can make a BIG effort and make our children not drink it before they reproduce, it will produce a generation of non-addicts! Problem is, I don't think our children can survive that long.

[quote="mewto55555"]Down with DHMO! Up with HOH![/quote]
That is a contradiction. DHMO=H+HO=H+OH=HOH

PS:
Down with 1=2! Up with AOPS!

I would say...ban it in the States.  They don't know how to use it right and it threatens the environment.  But keep it everywhere else.

*feels like mod*
This is becoming spammy. Please turn discussion back :).

[quote="mathelete"][quote="zephyredx"]No! Stop this slaughter! Killing the parents won't save their children! It's the children's children who need to be stopped from drinkind DHMO![/quote]
If they stop, they will die.[/quote]

If they don't, many more will die.

[quote="mewto55555"]Down with DHMO! Up with HOH![/quote]

hurrah1  somebody who shares me thots!

O.K. now I have a few thoughts:

1. mathlete is bringing a flame war against me into this topic.  Keep that in PMs and blogs.
2. mathlete is killing this topic by stressing the fact that DHMO=HOH and that we are all idiots for still discussing this. He can keep those mean and nasty thoughts to himself.

Now lets stay on topic.

1=2, you typoed.

[quote="1=2"]
2. mathlete is killing this topic by stressing the fact that DHMO=HOH and that we are all idiots for still discussing this.  He can keep those mean and nasty thoughts to himself.
[/quote]

ITS NOT A FACT!!! ItS A STUPID OPINION.

THBAPSO 1=2.  TY.


suppose $ A,B \subset N$ and we have $ |B|,|A|\geq2$.
$ R(x),Q(x) \in Z[x]$and both of them are injective. we define$ M$as follow:
$ M$={$ p(prime)|R(a_{i}) \plus{} Q(b_{j}) |a_{i} \in A,b_{j} \in B$}.
prove that if $ |M| < \infty$ we have both$ |A| < \infty$ and $ |B| < \infty$.


geometry
3D geometry
algebra
polynomial
algebra proposed

The first part of the following result of Davenport is probably known to many of the forum members; the second part is much more interesting, though.

Problem. (i) [Easy.] If $f,g \in \mathbb{C}[x]$ are co-prime polynomials with $\deg f = 2k, \deg g = 3k$, then
\[
\deg (f^3-g^2) \geq k+1.
\]
(ii) [Incredibly difficult.] Prove that the bound is sharp: for each $k$, there exist co-prime $f_k,g_k$ of respective degrees $2k,3k$ such that $\deg (f_k^3-g_k^2) = k+1$.

Corollary of Part 2. For each $\varepsilon, \delta > 0$, there exist co-prime integers $x,y \in \mathbb{N}$ such that
\[
y^3 < x^2 < y^3 + \delta y^{1/2 + \varepsilon}.
\]

The well-known Hall conjecture asserts that the exponent $1/2$ cannot be improved.

Just a question : You mention Zvonkin. Which one? I know a Dimitri Zvonkine who is a great young mathematician. But, the one you are talking about is maybe his father...

Pierre.

Actually, I just saw the name of A. Zvonkin mentioned in the text (but not the references) of an article of Stepan Orevkov. According to the article, A. Zvonkin obtained a topological proof of the second part of the problem (first solved by Stothers), but did not publish it because he thought it was identical to Zannier's.

--Vesselin

[quote="vess"]Problem. (i) [Easy.] If $f,g \in \mathbb{C}[x]$ are co-prime polynomials with $\deg f = 2k, \deg g = 3k$, then
\[
\deg (f^3-g^2) \geq k+1.
\]
[/quote]
I don't think it is so easy. I know proof using Mason's theorem.

Yes, it is an immediate corollary of Mason's theorem. Well, I agree that, for those who are not familiar with the abc theorem for polynomials, part (i) is probably very difficult as well. I wanted to emphasize, however, that (somewhat surprisingly, perhaps) the construction of a pair of polynomials for which the result is optimal is much more difficult than the result itself.

--Vesselin

[quote="vess"]I wanted to emphasize, however, that (somewhat surprisingly, perhaps) the construction of a pair of polynomials for which the result is optimal is much more difficult than the result itself.[/quote]
I hope you will tell us about it. ;)


Fibonacci numbers: $ F_1\equal{}F_2\equal{}1$, $ F_{n\plus{}2}\equal{}F_{n\plus{}1}\plus{}F_n$

Pell numbers: $ P_1\equal{}1$, $ P_2\equal{}2$, $ P_{n\plus{}2}\equal{}2P_{n\plus{}1}\plus{}P_n$

Find all Fibonacci numbers being Pell numbers as well.

I don't know if the solution is known or not so I put it here...

I was able to prove that the intersection of the Fibonacci and Pell numbers consists only of the following four numbers: $ 0, 1, 2,$ and $ 5$.
However, my proof is rather complicated to post it here - I'd better put it at arXiv soon.

I've eventually prepared the paper and put it in arxiv:
http://arxiv.org/abs/1002.1679

In particular, Theorem 8 addresses the problem of finding numbers that are both Fibonacci and Pell.


geometry
3D geometry
modular arithmetic
algebra
binomial theorem

1)Prove that any odd number raised to an even number is of the form $8r+1$

2)Prove that the cube of a number when divided by 7 leaves remainder 0,1 or 6.

1)[hide="my method for #1"]So let the thing be $(2n+1)^{2k}$
$=(4n^2+4n+1)^k$
Now,$4n^2+4n$ is always divisible by 8(Right?).
let $4n^2+4n$=$a$
continuing...
$=(a+1)^k$
So now use binomial theorem and expand it to get blah+blah+....+$1$ where all the blah blah things are divisible by 8(Since $a$ is divisible by 8).
$=\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{8r+1}}}}}}$[/hide]

[hide="1"] $(2a+1)^{2b}-1=((2a+1)^b+1)((2a+1)^b-1)$. Both are even, and one of two consecutive even integers is always divisible by 4. So there is a factor divisible by 4 and another even factor. So the product is divisible by 8.[/hide]

[hide="2"] If $7|a$, then $7|a^3$. Otherwise, $7|(a^6-1)=(a^3+1)(a^3-1)$ by Fermat's Little Theorem. Then 7 divides either $a^3-1$ or $a^3+1$.[/hide]

[quote="dondigo"]An easy way of solving such type of problems is by using congruences:

First problem:
$k\equiv{1}\pmod{2}$
So $k\equiv$ 1 or 3 or 5 or 7 $\pmod{8}$

In the first case we've got thesis. In the second we have $3^2\equiv{1}\pmod{8}$. Similarly in other cases.
My reasoning uses the fact that if:
$a\equiv{b}\pmod{m}$, then $a^k\equiv{b^k}\pmod{m}$ and transitivity

The second problem:
$a\equiv{0}\pmod{7}\Rightarrow a^3\equiv{0}\pmod{7}$
$a\equiv{1 }\pmod{7}\Rightarrow a^3\equiv{ 1}\pmod{7}$
$a\equiv{2 }\pmod{7}\Rightarrow a^3\equiv{ 1}\pmod{7}$
$a\equiv{ 3}\pmod{7}\Rightarrow a^3\equiv{ 6}\pmod{7}$
$a\equiv{ 4}\pmod{7}\Rightarrow a^3\equiv{1 }\pmod{7}$
$a\equiv{ 5}\pmod{7}\Rightarrow a^3\equiv{ 6}\pmod{7}$
$a\equiv{ 6}\pmod{7}\Rightarrow a^3\equiv{6 }\pmod{7}$[/quote]

how do you get $k\equiv$ 1 or 3 or 5 or 7 $\pmod{8}$ from $k\equiv{1}\pmod{2}$??

Because if $k \equiv 2x \mod 8$ then $k \not\equiv 1 \mod 2$.

[hide="the elementary ways - middle school thinking"]
1. Let the number be $2k+1$, where $k$ is an integer.

Then $(2k+1)^2=4k^2+4k+1$. Consider the $4k^2+4k$ term. It can be factored into $4(k(k+1))$, and since $k(k+1)$ is always even, $4k(k+1)$ will be divisible by 8. Thus $4k^2+4k+1$ is in the form 8r+1.

2. We take all the possible remainders when divided by 7 (0, 1, 2, ..., 6) and cube them. We have 0, 1, 8, 27, 64, 125, 216, which leaves remainders of 0, 1, 1, 6, 1, 6, and 6 when divided by 7. Thus only remainders of 0, 1 and 6 are possible when a cube of a positive integer (is this what you meant?) is divided by 7. [/hide]


calculus
integration
function
real analysis
real analysis unsolved

Construct a sequence $ ( f_{n})_{n\geq 1}$ of continuous functions  from  [0,1]  to [0, 1] such that $ \int f_{n}$ $ \to\0$ 0 but the sequence $ ( f_{n}(x)) _{n\geq 1}$ divergent for each x in [0, 1]

Thanks you Prof. Merryfield for the links


:wink:
I- if a , b , c >0 and  abc = 1000 .
   find Min (1+a)(1+b)(1+c)
II- If : tana+tanb+tana.tanb=1  find a+b=?

for problem 1,we will use AM-GM inequality.We know that (1+a)>=2a^1/2
so we get (1+a)(1+b)(1+c)>=8.(abc)^1/2 =8.10.10^1/2=80.10^1/2
  so we get the min. of the given.

2.we know that tan(a + b)=(tan a +tan b)/(1-tan a.tan b)
here putting tan a + tan b= 1-tan a.tan b[ given that tan a + tan b + tan a.tan b=1]
we get tan (a+b) = 1 so we get a +b= nr + 45 degree[ I can't find pi in the key board here r = pi]


15 students join a summer course. Every day, 3 students are on duty after school to clean the classroom. After the course, it was found that every pair of students have been on duty together exactly once. How many days does the course last for?

[hide]

The number of pairs of students is $\binom{15}{2}=105$.  Each day 3 pairs are used up.  The course lasts $\frac{105}{3}=35$ days.

[/hide]

[quote="E^(pi*i)=-1"][hide]

The number of pairs of students is $\binom{15}{2}=120$.  Each day 3 pairs are used up.  The course lasts $\frac{120}{3}=40$ days.

[/hide][/quote]

$\binom{15}{2} \neq 120$.  $\binom{16}{2}$ does.

[quote="Yooper"]For N=6 students {A,B,C,D,E,F}, the same method gives (6*5/2)/3 = 5 days.

Challenge:  Prove that the correct answer is "Impossible"; no such scheduling scheme exists.
(or prove that I am wrong :roll: )

BTW: An equivalent geometry problem is to first draw a 'complete graph' of a regular hexagon (that is, 6 points and all 15 connecting line segments).  Then, can you draw 5 triangles that use each of the segments exactly once?[/quote]

For N students, you know that each student has been paired exactly once with each of the N - 1 other students, and because the student is paired with 2 people that the student has never been with before each time the student is on duty, the number of days the student is on duty is $\frac{N - 1}{2}$, which is impossible if N is even, so it is impossible for N = 6, but is possible for N = 15.

[hide]I think it's also imposslible for $n=15$[/hide]


function
integration
probability and stats

Let X1,...Xn, form a random sample of size n from a continuous distribution for which the density function is F;
Define Y1=min{X1,...,Xn} & Yn=max{X1,...Xn}
I'm trying to find E(F(Y1)) & E(F(Yn)).

start with n=2,

$ X_1, X_2$ with density $ f_1, f_2$ then $ Z\equal{}X_1\minus{}X_2$ has density $ \int_{\minus{}\infty}^\infty f_1(x\plus{}y)f_2(y)dy\equal{}[f_1(x) * f_2(\minus{}x)]$

then $ E[Y_1]\equal{}E[\min(X_1,X_2)]\equal{}E[X_1\cdot 1_{X_1<X_2}\plus{}X_2\cdot 1_{X_1\ge X_2}]$

$ \equal{}E[X_1\cdot 1_{Z<0}]\plus{}E[X_2\cdot 1_{Z\ge 0}]\equal{}E[X_1]\plus{}E[(X_2\minus{}X_1 )\cdot 1_{Z\ge 0}]$

$ \equal{}E[X_1]\minus{}E[Z \cdot 1_{Z\ge 0}]\equal{}\int_{\minus{}\infty}^\infty xf_1(x)dx \minus{}\int_0^\infty x[f_1(x) * f_2(\minus{}x)]dx$

$ \equal{}\int_{\minus{}\infty}^\infty x (f_1 \minus{} [f_1(x) * f_2(\minus{}x)]1_{x\ge 0}) dx$

so the density of $ Y_1$ is $ f_1 \minus{} [f_1(x) * f_2(\minus{}x)]1_{x\ge 0}$

and if we have n i.i.d's, we can use this formula n-1 times to get something not very nice looking, but calculable i guess.
you would get a similar mess with $ Y_n$, but maybe this isnt what you wanted..

Very nice fact: if $ X$ has continuous d.f. $ F$, then $ F(X)$ is uniformly distributed on $ [0,1]$.

Thus $ F(Y_1) \equal{} \min\{U_1,\dots,U_n\}$, where $ U_i$ are iid uniform on $ [0,1]$ and the same holds for $ Y_n$ with $ \max$.

Hence, $ E[F(Y_1)] \equal{} 1/(n \plus{} 1)$, $ E[F(Y_n)] \equal{} n/(n \plus{} 1)$.


limit
real analysis
real analysis theorems

(1) State the definition of the sum of the infinite series which converges.

(2) If $ \sum_{n \equal{} 1}^{\infty} a_n \equal{} 1,\ \lim_{n\to\infty} na_n \equal{} 0$, then prove that  $ \sum_{n \equal{} 1}^{\infty} n(a_n \minus{} a_{n \plus{} 1}) \equal{} 1.$ on basis the definition in (1).

By writing $ \sum_{n\equal{}1}^N n(a_n\minus{}a_{n\plus{}1})\equal{}\sum_{n\equal{}1}^N (na_n\minus{}(n\plus{}1)a_{n\plus{}1})\plus{}\sum_{n\equal{}1}^N a_{n\plus{}1}$ the conclusion follows by telescoping and  $ \lim_{n\to\infty} na_n \equal{} 0$

$ \sum_{n \equal{} 1}^{N}a_{n \plus{} 1} \equal{} \sum_{n \equal{} 1}^{N \plus{} 1}a_n \minus{} a_1 \to 1 \minus{} a_1$ according to (1).

By the way you may "hide" $ \lim na_n\equal{}0$ by requiring the monotonicity of $ a_n.$


ratio
geometry
incenter
angle bisector
geometry unsolved

In triangle $ ABC$, let $ AB\equal{}c,BC\equal{}a,CA\equal{}b$. Let $ I$ be the incenter, and let the incircle meet $ BC$ at $ P$. Let $ X\equal{}IC\cap AP$.

Find $ \frac{IX}{IC}$ in terms of $ a,b,c$.

Let $ CI \cap AB \equal{} Q$
By Menelaus, PB/PC * CX/XQ * QA / AB = 1

Now QA/QB = b/(a+b) and PB/PC = (a+c-b) / (a+b-c)

BY angle bisector thm., CI/ IQ = AC/AQ = BC/BQ = (AC+BC)/(AB) = a+b / c

Let CX / XQ = p, CI / IQ = q
then (p-q)/(p(q+1)) = IX / IC

So: (p-q)/p = 1 - q/p = 1 - q * QA/QB * PB/PC = 1 - b(a+c-b) / c(a+b-c) = (ca - ab +b^2-c^2) / c(a+b-c) = (b-c)(b+c-a)/c(a+b-c)

Thus \[ \frac{IX}{IC} \equal{} \frac{(b\minus{}c)(b\plus{}c\minus{}a)}{ (a\plus{}b\plus{}c)(a\plus{}b\minus{}c) }\]
if i didn't make computation mistakes


Using three straight lines, what is the maximum number of regions into which the enclosed region can be divided?  Attached is the image.

you just want each line to intersect all other lines and the figure 4 times.


logarithms
integration
function
real analysis
real analysis unsolved

Prove

$ \sum_{k\equal{}1}^\infty \frac{1}{k^3 \, 2^k}\equal{}\frac78 \zeta(3)\minus{}\frac{\pi^2}{12} \ln 2\plus{}\frac16 \ln^3 2$

Maybe it's  hint.
I will speak generally .
suppose we need ti find serie :
$ P(a,z) : \equal{} \sum_{k\geq 1} \frac {1}{k^{z}a^{k}}$
if $ a \equal{} 1$ then we san say that it's in exact Riemann zeta function.

Therefore we suppose that $ a > 1$

note that :
\[ \frac {1}{k^{z}a^{k}} \equal{} \frac {1}{\Gamma(z)}\int_{0}^{\infty}x^{z \minus{} 1}e^{ \minus{} x(k )}a^{ \minus{} k }
\]
(for example consider $ I(\alpha) \equal{} \int_{0}^{\infty}e^{ \minus{} x\alpha}dx)$ and take $ k$ -times derivetive of $ I(\alpha)$)
Ok.
then we see:
\[ P(a,z) : \equal{} \sum_{k\geq 1} \frac {1}{k^{z}a^{k}} \equal{} \frac {1}{\Gamma(z)} \sum_{k \equal{} 0}^{\infty} x^{z \minus{} 1}(e^{x}a)^{ \minus{} (k \plus{} 1)}dx \equal{} \frac {1}{\Gamma(z)}\int_{0}^{\infty} \frac {x^{z \minus{} 1}}{e^{x}a \minus{} 1}dx
\]
by substitution: $ e^{x}a \equal{} e^{y}$ We have :
\[ P(a,z) \equal{} \frac {1}{\Gamma(z)} \int_{\ln a}^{\infty} \frac {(y \minus{} \ln a)^{z \minus{} 1}}{e^{y} \minus{} 1}dy
\]
now if we suppose that $ z \in \mathbb{N}$ then we have :
\[ P(a,z) \equal{} \frac {1}{\Gamma(z)} \int_{\ln a}^{\infty} \frac {\sum_{k \equal{} 0}^{z \minus{} 1}\binom{z \minus{} 1}{k}y^{z \minus{} 1 \minus{} k}(\ln a )^{k}(\minus{}1)^{k}}{e^{y} \minus{} 1}dy
\]
the last term can be calculated exactli ( $ \int \frac {1}{e^{t} \minus{} 1}$)
now the question is how to calculate
$ \int_{\ln a}^{\infty} \frac {y^{n}}{e^{y} \minus{} 1}dy$ when $ n \geq 1$
we can write:

$ \int_{\ln a}^{\infty} \equal{} \int_{0}^{\infty} \minus{} \int_{0}^{\ln a}$  and by this way get zeta function again but now I have no idea how to calculate $ \int_{0}^{\ln a} \frac {y^{n}}{e^{y} \minus{} 1}dy$ later I will think about it .

Now I've found a solution.

We need the polylogarithm functions

$ \text{Li}_2(x)\equal{}\sum_{k\equal{}1}^\infty \frac{x^k}{k^2}$ and $ \text{Li}_3(x)\equal{}\sum_{k\equal{}1}^\infty \frac{x^k}{k^3}$ , whereat $ \text{Li}_3'(x)\equal{}\frac{\text{Li}_2(x)}{x}$.

And we need the identities

$ \text{(I1)} \quad \text{Li}_2(x)\plus{}\text{Li}_2(1\minus{}x)\equal{}\frac{\pi^2}{6}\minus{}\ln(x)\, \ln(1\minus{}x)$

$ \text{(I2)} \quad \text{Li}_2\left(1\minus{}\frac{1}{x}\right)\plus{}\text{Li}_2(1\minus{}x)\equal{}\minus{}\frac12 \ln^2 x$

already shown [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=916534#916534]here[/url].

$ \left[\text{Li}_3\left(1\minus{}\frac{1}{x}\right)\right]'\equal{}\frac{\text{Li}_2\left(1\minus{}\frac{1}{x}\right)}{1\minus{}\frac{1}{x}}\, \frac{1}{x^2}\equal{}
\frac{\minus{}\text{Li}_2\left(1\minus{}\frac{1}{x}\right)}{x\, (1\minus{}x)}$

Now use $ \text{(I2)} \quad \minus{}\text{Li}_2\left(1\minus{}\frac{1}{x}\right)\equal{}\frac12 \ln^2 x\plus{}\text{Li}_2(1\minus{}x)$ and write $ \frac{1}{x\, (1\minus{}x)}$ as $ \frac{1}{x}\plus{}\frac{1}{1\minus{}x}$ .

$ \left[\text{Li}_3\left(1\minus{}\frac{1}{x}\right)\right]'\equal{}\underbrace{\frac{\frac12 \ln^2 x}{x}}_{\left[\frac16 \ln^3 x\right]'}\plus{}\frac{\frac12 \ln^2 x}{1\minus{}x}\plus{}\frac{\text{Li}_2(1\minus{}x)}{x}\plus{}\underbrace{\frac{\text{Li}_2(1\minus{}x)}{1\minus{}x}}_{\big[\minus{}\text{Li}_3(1\minus{}x)\big]'}$ .

Therefore

$ \left[\text{Li}_3(1\minus{}x)\plus{}\text{Li}_3\left(1\minus{}\frac{1}{x}\right)\minus{}\frac16 \ln^3 x\right]'$$ \equal{}\frac{\frac12 \ln^2 x}{1\minus{}x}\plus{}\frac{\frac{\pi^2}{6}\minus{}\ln(x)\,\ln(1\minus{}x)\minus{}\text{Li}_2(x)}{x}$ .

And so $ \left[\text{Li}_3(x)\plus{}\text{Li}_3(1\minus{}x)\plus{}\text{Li}_3\left(1\minus{}\frac{1}{x}\right)\minus{}\frac16 \ln^3 x\minus{}\frac{\pi^2}{6}\ln x\right]'$

$ \equal{}\frac{\frac12 \ln^2 x}{1\minus{}x}\minus{}\frac{\ln(x)\, \ln(1\minus{}x)}{x}\equal{}\minus{}\left[\frac12 \, \ln^2(x)\, \ln(1\minus{}x)\right]'$ .

Hence

$ \text{Li}_3(x)\plus{}\text{Li}_3(1\minus{}x)\plus{}\text{Li}_3\left(1\minus{}\frac{1}{x}\right)\equal{}\frac16 \ln^3 x\plus{}\frac{\pi^2}{6}\ln x\minus{}\frac12 \ln^2 (x)\, \ln(1\minus{}x)\plus{}C$ .

When we choose $ x\equal{}1$ we see that the constant $ C$ must be $ \text{Li}_3(1)\equal{}\zeta(3)$ .

Now let's choose $ x\equal{}\frac12$ :

$ 2\,\text{Li}_3\left(\frac12\right)\plus{}\underbrace{\text{Li}_3(\minus{}1)}_{\minus{}\frac34 \zeta(3)}\equal{}\minus{}\frac16 \ln^3 2\minus{}\frac{\pi^2}{6} \ln 2\plus{}\frac12 \ln^3 2\plus{}\zeta(3)$

and this is $ \zeta(3)\minus{}\frac{\pi^2}{6}\ln 2\plus{}\frac13 \ln^3 2$ .

So $ \text{Li}_3\left(\frac12\right)\equal{}\frac78\zeta(3)\minus{}\frac{\pi^2}{12}\ln 2\plus{}\frac16\ln^3 2$ .

Does anyone know what sum(1/(k^4*2^k), k = 1 .. infinity) or sum(1/(k^5*2^k), k = 1 .. infinity) equals?


Can I get a second opinion on these answers:

The 3 problems are here: http://img125.imageshack.us/img125/2599/3questions3ts.jpg

and I got

1). C

2). D

3). A

Thanks

For the $x$-intercept is 5, the answer is $C).$
For the vertical asymptote, the answer is $D).$
For $x$-axis is its horizontal asymptote, the answer is $C).$


calculus
trigonometry
calculus computations

In my calculus AB course, I keep seeing that the average value of the sine from 0 to pi is always 2/pi, but 2/pi is the answer to Buffon's needle problem. Is this a coincidence?

edited

http://en.wikipedia.org/wiki/Buffon's_needle
See for yourself.

[url=http://en.wikipedia.org/wiki/Buffon%27s_needle]Fixed link[/url]. Apostrophes break links if they're in the link text.


given $ A\equal{}[a_{ij}]_{4\times 4}$ satifying:
$ a_{ij}\equal{}0$ if i=j ,
and $ a_{ij}\equal{}\plus{}\minus{}1$ if $ i\neq j$
prove that we always have $ detA \neq 0$

Counterexample: Let $ A\equal{}\begin{pmatrix}
0 & 1 & \minus{}1\\
1 & 0 & 1\\
1 & 1 & 0
\end{pmatrix}$

The size matters; the original problem specifies $ 4\times 4$, and that matters.

Work in the field with two elements. For the $ n\times n$ matrix $ J$ of all ones, we have $ J^2\equal{}nJ$, and $ (J\minus{}I)(\frac1{n\minus{}1}J\minus{}I)\equal{}\frac{n}{n\minus{}1}J\minus{}J\minus{}\frac1{n\minus{}1}J\plus{}I\equal{}I$. When $ n\equal{}4$, $ \frac13\equal{}1$, and $ J\minus{}I$ is its own inverse. Since it's invertible mod 2, its determinant is odd when brought back to the rationals.

Yeah, I was wrong. I didn't see the size of that matrix.
But Jmerry, could you please explain this part of your solution?
[quote="jmerry"]...Since it's invertible mod 2, its determinant is odd when brought back to the rationals.[/quote]

The determinant of a matrix $ M$ as described in the original problem is some integer.  The determinant of $ J$ is the determinant of $ M \bmod 2$ (because we have just reduced every entry $ \bmod 2$ and done all our computations $ \bmod 2$).  Because this determinant is $ 1$, we know that $ \det M$ is odd and hence not equal to zero.


problema 1

Sa se determine functiile continue f:R->R pentru care exista funcii injective F:R->R,astfel incat:

f(F(x+y))=F(f(x)+f(y))           Sursa:concursul interjudetean Traian Lalescu 2001

Cate metode de rezolvare stiti?

Fie
$\ x+y=z+t \Longrightarrow F(x+y)=F(z+t) \Longrightarrow f(F(x+y))=f(F(z+t)) \Longrightarrow F(f(x)+f(y))=F(f(z)+f(t))$
Pt ca $\ F$ este injectiva rezulta ca $\ f(x)+f(y)=f(z)+f(t)$
Punem $\ z=0$ rezulta $\ f(x)+f(y)=f(x+y)+f(0)$
Fie $\ g(x)=f(x)-f(0)$  $\ g$ este continua $\ g(0)=0$ si verifica ecuatia lui Cauchy
$\ g(x+y)=g(x)+g(y)$ . Fiind continua rezulta ca $\ g(x)=cx$ si de aici rezulta $\ f(x)=ax+b$.

[quote="siriusra"]sa-mi ziceti daca-s simple astea sa nu mai postez de genu asta....sau daca le stiti.... cu ceva idei puteti sa raspundeti,indicatii...  :)[/quote]Se posteaza cate [b]o singura[/b] problema per topic.

nu mi s-a parut ca ar fi un topic de o problema de asta....plus ca le-am gasit cam in acelasi loc si mai ales nu as fi vrut sa aglomerez subforumul ...


What was the hardest AIME question ever in your opinion (mine the 2001 aime1 #15(octohedron))

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=394736]1994 #11[/url]

Not the hardest, but it's easily my least favorite  :|

Can someone explain to me #15(its unsolved)
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=384211&sid=f2d5a5a58f5436c4a70ca27d5bab10e9#p384211


I see that in many solving of the problems you suppose that a+b+c=1 and then use it to prove a problem. For example in the imo01 inequality problem the solving is by supposing that a+b+c=1. When can I suppose that and when I cant??
I think its when if the ineq is true for a,b,c and the ineq is true for ka,kb,kc as well

Yeah, you can add condition like that when the inequality is homogenous (every term has the same degree).


integration
limit
calculus
trigonometry
calculus computations

[quote="Virgil Nicula"][color=darkblue]Denote $a_{n}=\int_{\frac{1}{n+1}}^{\frac{1}{n}}\arctan (nx)\ dx$ , ${b_{n}=\int_{\frac{1}{n+1}}^{\frac{1}{n}}\arcsin (nx)\ dx}$ and $c_{n}=\int_{\frac{1}{n+1}}^{\frac{1}{n}}\frac{\arctan(nx)}{\arcsin (nx)}\ dx\ .$

Prove (without the [u]Leibniz-Newton's theorem[/u]) that :

$1.\blacktriangleright\ \lim_{n\to\infty}a_{n}=\lim_{n\to\infty}b_{n}=\lim_{n\to\infty}c_{n}=0\ .$

$2.\blacktriangleright\ \lim_{n\to\infty}n^{2}\cdot c_{n}=\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=\frac{1}{2}\ .$[/color][/quote]
Indeed, very good exercises by [b]Virgil Nicula[/b]. Thanks! :)
[b]
Solution (1)[/b]: Using the first mean value theorem for integration, we have
$a_{n}= \tan^{-1}(nx_{1}) \int_{\frac{1}{n+1}}^{\frac{1}{n}}\, dx = \tan^{-1}(nx_{1}) \left( \frac{1}{n}-\frac{1}{n+1}\right)$, where $x_{1}\in (\frac{1}{n+1}, \frac{1}{n})$.

Note that $\frac{1}{n+1}< x_{1}< \frac{1}{n}\Rightarrow \frac{n}{n+1}<nx_{1}< \frac{n}{n}= 1.$ So, by the squeezing theorem, $\lim_{n\rightarrow \infty}nx_{1}= 1.$

$\therefore \lim_{n\rightarrow\infty}a_{n}= \tan^{-1}(1) (0-0) = 0.$

Similarly, $b_{n}= \sin^{-1}(nx_{1})\int_{\frac{1}{n+1}}^{\frac{1}{n}}\, dx = \sin^{-1}(nx_{1}) \left( \frac{1}{n}-\frac{1}{n+1}\right)$, where $x_{1}\in (\frac{1}{n+1}, \frac{1}{n})$.

Reasoning the same way as above, we again have
$\lim_{n\rightarrow\infty}b_{n}= \sin^{-1}(1) (0-0) = 0.$

The argument is the same for evaluating $\lim_{n\rightarrow \infty}c_{n}$, which is also equal to $0.$
[b]

Solution to (2)[/b]: Note that $c_{n}= \frac{\tan^{-1}(nx_{1}) }{\sin^{-1}(nx_{1})}\left( \frac{1}{n}-\frac{1}{n+1}\right)$, where $x_{1}\in (\frac{1}{n+1}, \frac{1}{n})$.

$\therefore n^{2}\cdot c_{n}= \frac{n^{2}}{n(n+1)}\frac{\tan^{-1}(nx_{1})}{\sin^{-1}(nx_{1})}$

$\Rightarrow \lim_{n\rightarrow \infty}n^{2}\cdot c_{n}= \frac{\frac{\pi}{4}}{\frac{\pi}{2}}= \frac{1}{2}.$

Finally, $\frac{a_{n}}{b_{n}}= \frac{\tan^{-1}(nx_{1})}{\sin^{-1}(nx_{2})}$, where $x_{1}, x_{2}\in (\frac{1}{n+1}, \frac{1}{n}).$

$\Rightarrow \lim_{n \rightarrow \infty}\frac{a_{n}}{b_{n}}= \frac{\frac{\pi}{4}}{\frac{\pi}{2}}= \frac{1}{2}.$


The techniques for solving linear equations and linear inequalities are similar, yet different. Explain and give an example of both a linear equation and a linear inequality that demonstrates this difference.

Don't double-post.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159438


Consider $ p$ as a odd prime number.
Calculate how many ways we can choose $ p$ numbers from $ \{1,2,..,2p\}$ that sum of those numbers is divisible by p.  :roll:


Find the maximum k such that,
a^4 + b^4 + c^4 + k(a^2bc + ab^2c + abc^2) ≥ (k+1)(a^2b^2 + b^2c^2 + c^2a^2)
(a, b, c, k is positive real number.)

[quote="Hong Quy"]a=b=1  and c->0 .
 hence k=1.[/quote]
I think, $ k\leq1.$ It remains to prove that $ \sum_{cyc}(a^4\minus{}2a^2b^2\plus{}a^2bc)\geq0.$
But $ \sum_{cyc}(a^4\minus{}2a^2b^2\plus{}a^2bc)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^4\minus{}a^3b\minus{}a^3c\plus{}a^2bc)\plus{}\sum_{cyc}(a\minus{}b)^2ab\geq0,$ which true by Schur.


inequalities
Sequence
algebra
bounded
IMO Shortlist

Let $a_1,a_2,\ldots$ be an infinite sequence of real numbers, for which there exists a real number $c$ with $0\leq a_i\leq c$ for all $i$, such that \[\left\lvert a_i-a_j \right\rvert\geq \frac{1}{i+j} \quad \text{for all }i,\ j \text{ with } i \neq j. \] Prove that $c\geq1$.

Let $k$ be an arbitrary natural number.

Let $\{m_1,m_2,\ldots{},m_k\}$ be a permutation of $\{1,2,\ldots{},k\}$ such that $a_{m_1} < a_{m_2} < \cdots{} < a_{m_k}$.

Note that we can never have equality since $|a_{m_i} - a_{m_{i+1}}| \ge \frac{1}{m_i+m_{i+1}}$.

Let $\overline{a_ia_j} = |a_i-a_j|$.

By looking at the $a_i$ as a set of intervals on $[0,c]$, it makes sense that $\overline{a_{m_1}a_{m_k}} = \sum \limits_{i=1}^{k-1} \overline{a_{m_i}a_{m_{i+1}}}$.

$\overline{a_{m_i}a_{m_k}} \ge \sum\limits_{i=1}^{k-1} \frac{1}{m_i+m_{i+1}}$.

By the Arithmetic Mean Harmonic Mean Inequality,

$\frac{(a_1+a_2) + (a_2+a_3) + \ldots{} + (m_{k-1}+m_k)}{k-1} \ge \frac{k-1}{\frac{1}{m_1+m_2} + \ldots{} + \frac{1}{m_{k-1}+m_k}}$.

$(m_1+2m_2+\ldots{}+2m_{k-1}+2m_k)\left(\frac{1}{m_1+m_2} + \ldots{} + \frac{1}{m_{k-1}+m_k}\right) \ge (k-1)^2$.

$(\overline{a_{m_1}a_{m_k}})(m_1+2m_2+\ldots{}+2m_{k-1}+m_k) \ge (k-1)^2$.

The right term of the left-hand side is less than $2(m_1+m_2+\ldots{}+m_k)$:

$2\overline{a_{m_1}a_{m_k}}(m_1+m_2+\ldots{}+m_k) > (k-1)^2$

Since $\{m_1,m_2,\ldots{},m_k\}$ is a permutation of $\{1,2,\ldots{},k\}$,

$2\overline{a_{m_1}a_{m_k}} \cdot \frac{k(k+1)}{2} > (k-1)^2$.

$\overline{a_{m_1}a_{m_k}} > \frac{(k-1)^2}{k(k+1)} = \frac{k-1}{k} \cdot \frac{k-1}{k+1} > \left(\frac{k-1}{k+1}\right)^2 = \left(1-\frac{2}{k+1}\right)^2$.

If $\overline{a_{m_1}a_{m_k}} < 1$ for all $k \in \mathbb N$, we can easily find a $k$ such that $\left(1-\frac{2}{k+1}\right)^2 > \overline{a_{m_1}a_{m_k}}$, causing a contradiction.

So $\overline{a_{m_1}a_{m_k}} \ge 1$ for some integers $m_1$, $m_k$.

$|a_{m_1}-a_{m_k}| \ge 1$.

Since both terms are positive, it is clear that at least one of them is greater than or equal to $1$.

So $c \ge 1$, as desired.

I think this is a faster finish - it's basically the same solution.

Let the first $ n$ terms be $ a_{1},a_{2},\ldots,a_{n}$, and let $ b_{1},b_{2},\ldots,b_{n}$ be the permutation of them such that $ b_{1}<b_{2}<\cdots<b_{n}$. Then, we need $ b_{n}\minus{}b_{1}\le c$. But we have

$ b_{n}\minus{}b_{1}$;
$ \equal{}(b_{n}\minus{}b_{n\minus{}1})\plus{}(b_{n\minus{}1}\minus{}b_{n\minus{}2})\plus{}\cdots\plus{}(b_{2}\minus{}b_{1})$;
$ \ge\dfrac{1}{b_{n}\plus{}b_{n\minus{}1}}\plus{}\dfrac{1}{b_{n\minus{}1}\plus{}b_{n\minus{}2}}\plus{}\cdots\plus{}\dfrac{1}{b_{2}\plus{}b_{1}}$.

By AM-HM, this is greater than

$ \dfrac{(n\minus{}1)^{2}}{(b_{1}\plus{}b_{2})\plus{}(b_{2}\plus{}b_{3})\plus{}\cdots\plus{}(b_{n\minus{}1}\plus{}b_{n})}$;
$ \equal{}\dfrac{(n\minus{}1)^{2}}{2(b_{1}\plus{}b_{2}\plus{}\cdots\plus{}b_{n})\minus{}(b_{1}\plus{}b_{2})}$;
$ \ge\dfrac{(n\minus{}1)^{2}}{n(n\plus{}1)\minus{}3}$.

Note that $ \lim_{n\rightarrow\infty}\dfrac{n^{2}\minus{}2n\plus{}1}{n^{2}\plus{}n\minus{}3}\equal{}1$. Thus, if $ c<1$, we can find sufficiently large $ n$ such that $ b_{n}\minus{}b_{1}>c$ (I guess by Epsilon-Delta?), contradiction. So $ c\ge1$.

Clarification on the above: $ b_{1},b_{2},\ldots,b_{n}$ should be a permutation of $ 1,2,\ldots,n$ such that $ a_{b_{1}}<a_{b_{2}}<\cdots<a_{b_{n}}$, and $ a_{b_{n}}\minus{}a_{b_{1}}\le c$. And the first two lines of equalities should read $ a_{b_{n}}$, etc.

[quote="orl"]Let $ a_1,a_2,\ldots$ be an infinite sequence of real numbers, for which there exists a real number $ c$ with $ 0\leq a_i\leq c$ for all $ i$, such that
\[ \left|\,a_i \minus{} a_j\,\right|\geq{1\over i \plus{} j}{\rm \forall}i,j \quad \textnormal{with} \quad i\ne j.\]
Prove that $ c\geq1$.[/quote]
Nice problem  :lol:
Anyone try solve this problem? Let $ c\geq 1$ such that there exists an infinite sequence $ a_1,a_2,\ldots$ of real numbers, such that $\lim_{n\to \infty} x_n \equal{} c$, and for all $ i\neq j$ to have $|\,a_i \minus{} a_j|\geq{1\over i \plus{} j}$. Find all such $ c$.

Consider $a_1,a_2,\cdots a_n$ and let $b_1,b_2,\cdots b_n$ be this sequence in sorted order, and suppose $b_{k}=a_{i_k}$. Note that
\begin{align*}
c&\ge b_n-b_1\\
&=\sum_{k=1}^{n-1} |b_{k+1}-b_{k}|\\
&\ge \sum_{k=1}^{n-1} \frac{1}{i_k+i_{k+1}}\\
&\ge \frac{(n-1)^2}{i_1+2i_2+\cdots + 2i_{n-1}+i_{n}}\\
&\ge \frac{(n-1)^2}{1+2+2(3+\cdots +n)}\\
&=\frac{(n-1)^2}{n^2+n-3}
\end{align*}
so as $n$ approaches infinity, we recover $c\ge 1$.

[hide=Solution]
Fix $n\geq 2$ and let $f(k)$ be a bijective function mapping $\{1, 2, \cdots, n\}$ to $\{1, 2, \cdots, n\}$ such that $a_{f(1)}<a_{f(2)}<\cdots<a_{f(n)}$. From the condition, we have: \begin{align*}a_{f(2)}-a_{f(1)}\geq \frac{1}{f(2)+f(1)} \\ a_{f(3)}-a_{f(2)}\geq \frac{1}{f(3)+f(2)} \\ \dots \\ a_{f(n)}-a_{f(n-1)}\geq \frac{1}{f(n)+f(n-1)}.\end{align*} Adding these inequalities and applying Titu's lemma gives \begin{align*}c\geq a_{f(n)}-a_{f(1)} =\left(a_{f(n)}-a_{f(n-1)}\right)+\left(a_{f(n-1)}-a_{f(n-2)}\right)+\dots +\left(a_{f(3)}-a_{f(2)}\right)+\left(a_{f(2)}-a_{f(1)}\right) \\ \geq \frac{1}{f(1)+f(2)}+\frac{1}{f(2)+f(3)}+\dots+\frac{1}{f(n-1)+f(n)}\\ \geq \frac{(n-1)^2}{f(1)+f(n)+2(f(2)+f(3)+\dots+f(n-1))}\\ \geq \frac{(n-1)^2}{2(f(1)+f(2)+\dots+f(n))} =\frac{(n-1)^2}{n(n+1)}.\end{align*} The inequality $c\geq \frac{(n-1)^2}{n(n+1)}$ implies that $n^2(c-1)+n(c+2)-1\geq 0$, which for large enough $n$ forces $c\geq 1$, as desired..
[/hide]

[quote=K.titu]I have a question:
What the limit of $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n-1}$ wher approaches infinity?[/quote]

$ln(2)$

[hide=Solution]
Let $\sigma\in S_n$ such that $a_{\sigma(1)}\le a_{\sigma(2)}\le\cdots\le a_{\sigma(n)}$. We have that
\[c\ge a_{\sigma(n)}-a_{\sigma(1)} = |a_{\sigma(2)}-a_{\sigma(1)}| + |a_{\sigma(2)}-a_{\sigma(1)}| +\cdots+|a_{\sigma(n)}-a_{\sigma(n-1)}|\ge\frac{1}{\sigma(1)+\sigma(2)}+\cdots+\frac{1}{\sigma(n-1)+\sigma(n)}.\]
But note that by AM-HM,
\[\frac{n}{\frac{1}{\sigma(1)+\sigma(2)}+\cdots+\frac{1}{\sigma(n-1)+\sigma(n)}}\le \frac{2(\sigma(1)+\cdots+\sigma(n))-\sigma(1)-\sigma(n)}{n}\le\frac{n^2+n-1-2}{n},\]
so
\[c\ge\frac{1}{\sigma(1)+\sigma(2)}+\cdots+\frac{1}{\sigma(n-1)+\sigma(n)}\ge \frac{n^2}{n^2+n-3}.\]
Thus, $c\ge 1-\epsilon$ for arbitrarily small positive $\epsilon$, so $c\ge 1$, as desired.
[/hide]

Since $a_i \neq a_j$, for each natural $n$ there exist a permutation $(t_i)_{1 \le i \le n}$ of $(i)_{1 \le i \le n}$ for which $a_{t_{k+1}}-a_{t_k} \ge \frac{1}{t_k+t_{k+1}}, \forall 1 \le  k \le n-1.$ Then $$c \ge a_{t_n}-a_{t_1} \ge \sum_{k=1}^{n-1} \frac{1}{t_k+t_{k+1}} \ge \frac{n^2}{2 \sum t_i - t_1 - t_n} > \frac{n^2}{n(n+1)} = \frac{n}{n+1}$$ for all naturals $n$, then clearly $c \ge 1$.

[quote=aops29]This is much different from the other solutions, so I don't know if it's correct...
[hide=Solution]
Suppose $c<1$. It suffices to show that for some $i,j > 1$ the given inequality fails to hold.

For all natural numbers $n$, there exists an infinite subsequence $\{a_{i_1}, a_{i_2}, ...\}$ of $\{a_1, a_2,...\}$ such that for any positive integers $j_1 \not = j_2$ we have $|a_{i_{j_1}} - a_{i_{j_2}}| < 1/n$. This is true by the pigeonhole principle.

As $n$ tends to infinity, clearly there are distinct integers $j_1, j_2$ such that $i_{j_1} + i_{j_2} < n$. However, then we have $$|a_{i_{j_1}} - a_{i_{j_2}}| < 1/n < 1/(i_{j_1} + i_{j_2})$$, which is a contradiction. Hence $c>=1$ and we are done.
[/hide][/quote]

The infinite subsequence could start at $a_{n+1}$ for all $n.$ The line "clearly there are distinct integers..." does not hold because of this.

Order $\{a_i\}_i$ as $0\leq a_{\pi(1)}\leq a_{\pi(2)}\leq \cdots \leq a_{\pi(n+1)}$ for some $n$. Then,
\[a_{\pi(n+1)} \geq \sum_{i=1}^{n} \frac1{\pi(i) + \pi(i+1)  }\]
Which by Jensen's on the convexity of $\frac1x$, is
\[\geq n \cdot \frac{1}{\frac1{n} \sum_{i=1}^{n} \pi(i) + \pi(i+1)}= \frac{n^2}{2\cdot (1+2+\cdots n) - \pi(1)-\pi(n+1)} \]
\[\geq \frac{n^2}{n(n+1)} = \frac{n}{n+1}\]
which as $n\to \infty$ converges to 1. Thus, if $c<1$, then selecting $n$ such that $\frac{n}{n+1} > c$ would yield a contradiction on some inequality for $i,j\leq n$, and we're done. $\blacksquare$.

Addendum: Construction. I believe that the equality case is
\[a_1, a_n, a_{3}, a_{n-2}, a_5 ,\cdots \]

I don't know if my method is correct but I gave it a try.

We have the sequence a_1; a_2; a_3; ......., suppose we rearrange it to b_1; b_2; b_3;..... such that b_1>b_2>b_3>......

We know that |a_i-a_j|>1/(i+j) so we can also say that |b_i-b_j|>1/(i+j). If we let j>i we know that b_i>b_j so that gives b_i-b_j>1/(i+j), which gives b_i>1/(i+j) + b_j. Setting j=i+1 results in b_i>1/(2i+1) + b_(i+1). Filling in i=i+1 and j=i+2 gives b_(i+1)>1/(2i+3) + b_(i+2) so b_i>1/(2i+1) + b_(i+1)> 1/(2i+1) + 1/(2i+3) + b_(i+2).

Doing this a few times shows us that:
b_i> 1/(2i+1) + 1/(2i+3) + 1/(2i+5) + ...... + 1/(2i+1+2n) + b_(i+n+1)
> 1/(2i+1) + 1/(2i+3) + 1/(2i+5) + ...... + 1/(2i+1+2n). If we plug in i=1 and n=6 we get that b_1 > 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/13 + 1/15 > 1. If c<1 we get that b_1 is less than 1 which is less than b_1, a contradiction. Therefore c>1
Q.E.D.

We do the dumbest possible thing. Assume, for the sake of contradiction, that there exists a valid ordering such that $c < 1$. Take some integer $k$, and say that the values $a_1,a_2,\ldots,a_k$ sorted are $a_{\sigma(1)}, a_{\sigma(2)}, \ldots, a_{\sigma(k)}$; i.e., $\sigma$ is a bijection from $[n] \rightarrow [n]$ and $a_{\sigma(1)} < a_{\sigma(2)} < \cdots < a_{\sigma(k)}$. Then, write: \[ c \geq \sum\limits_{i=1}^{k-1} \frac{1}{a_{\sigma(i)} + a_{\sigma(i+1)}} \] by applying the given equation to consecutive terms, and then by C-S we have  \[ \left( \sum\limits_{i=1}^{k-1} \frac{1}{a_{\sigma(i)} + a_{\sigma(i+1)}}\right) \left(\sum\limits_{i=1}^{k-1} a_{\sigma(i)} + a_{\sigma(i+1)} \right) \geq \left(\sum\limits_{i=1}^{k-1} 1\right)^2 \] and note that the right hand side is equal to $(k-1)^2$. Furthermore, we have: \[ \sum\limits_{i=1}^{k-1} (a_{\sigma(i)} + a_{\sigma(i+1)}) < 2\sum\limits_{i=1}^{k} a_i = k(k+1) \] and therefore $c \geq \frac{(k-1)^2}{k(k+1)}$. However, the right hand side approaches $1$ as $k$ goes to infinity, so at some point the $\frac{(k-1)^2}{k(k+1)}$ will be larger than $c$, contradiction.

Note that something stronger can be claimed. For every infinite sequence $\big( a_k\big)_{k=1}^{\infty}$ with $0\le a_k\le 1$ there exists sufficiently large $N$ and $i,j; N\le i<j\le 2N$ such that
$$|a_i-a_j|>\frac{1}{2(j-i)}.$$
This is stronger since
$$\frac{1}{2(j-i)} \ge \frac{1}{2N}\ge \frac{1}{i+j}$$
when $N\le i<j\le 2N$.

One can see how it is done in this [url=https://artofproblemsolving.com/community/c6h254140p1389559]Miklos Schweitzer, 1980 problem[/url].

Suppose FTSOC that $c<1$. Then we can pick arbitrarily large $n$ such that $0\le a_i\le c$ for all $1\le i\le n$. Consider a nonincreasing sequence of numbers $a_{\sigma(1)},a_{\sigma(2)},\dots,a_{\sigma(n)}$. Note that
\begin{align*}
a_{\sigma(1)}-a_{\sigma(n)} &= (a_{\sigma(1)}-a_{\sigma(2)})+(a_{\sigma(2)}-a_{\sigma(3)})+\dots+(a_{\sigma(n-1)}-a_{\sigma(n)}) \\
&\ge \frac{1}{\sigma(1)+\sigma(2)}+\frac{1}{\sigma(2)+\sigma(3)}+\dots+\frac{1}{\sigma(n-1)+\sigma(n)} \\
&\ge \frac{(n-1)^2}{n(n+1)-\sigma(1)-\sigma(n)} \\
&\ge \frac{n^2-2n+1}{n^2+n-3} \\
&= 1-\frac{3n-4}{n^2+n-3}
\end{align*}
so that taking $n$ to be sufficiently large, we obtain a contradiction. $\blacksquare$

Let $n$ be large and $\sigma$ be the permutation of $(1,2,\dots,n)$ for which $a_{\sigma(1)}<a_{\sigma(2)}<a_{\sigma(3)}<\dots<a_{\sigma(n)}.$
Note that \[c\ge a_{\sigma(n)}-a{\sigma(1)}=\sum_{k=1}^{n-1}{a_{\sigma(k+1)}-a{\sigma(k)}}\ge\sum_{k=1}^{n-1}{\frac{1}{\sigma(k+1)+\sigma(k)}}\]
By AM-HM, \[c\ge \frac{(n-1)^2}{n(n+1)-\sigma(1)-\sigma(n)}\ge \frac{(n-1)^2}{n(n+1)-3}.\] Since the RHS approaches $1$ as $n$ approaches infinity, $c\ge 1.$

[quote=mihaig]Great problem[/quote]

I concur

can we construct such a sequence for $c=1$?


Would you recommend Ti-84 or Ti-Nspire for math class and competitions.

what forum does a question like this belong?

I have both of them, and I think both have their strengths.  The N-Spire has slightly more features, a bigger screen, and easily acessible letters (unfortunately they're raised over the normal keys, slightly annoying), but is hard to get used to from the 84's, and has a smaller keyboard, most functions must be accessed from a gigantic library.  Using it may be difficult as many teachers have never even seen it before.  It does come with a keyboard to make it basically at TI-84, but the keyboard is different.

The TI-84.  Though it's a couple years old, it's a good calculator and a standard for a lot of math classes, at my school all classrooms use TI-84's, someone with an 89 or an N-spire or another brand of calculator would have a harder time, with instruction done for the 84.

For competition math, I think the one that is easier to use is the best, with most math competition phasing out calculators, it really doesn't matter as much.  I don't think one is better than the other for this.

I also have all of them, and I completely agree with Quickster. I would not need a CAS for math competitions like NYSML, ARML, or the iTest, which are the only high-school tests that I know allow the use of calculators. The only middle-school test that I can see using a calculator is MathCounts.


Find all $ f,g: \mathbb{Z}\rightarrow\mathbb{Z}$ such that for all integers $ a,b$,

$ f(a\plus{}b)\equal{}f(a)g(b)\plus{}g(a)f(b)$ and $ g(a\plus{}b)\equal{}g(a)g(b)\minus{}f(a)f(b)$

My solution is too lengthy and stupid so I believe that there must be much better methods.
[hide]Substitute a=b=0 into two equations and then solve them
   we get f(0)=0 and g(0)=0 ......Case 1or
             f(0)=0 and g(0)=1 ......Case 2   as f(0) and g(0) are integers
Substitute b=-a into two equations, we get
[b]For Case 1[/b],
$  \{\begin{array}{c} \displaystyle f(a)g( - a) + g(a)f( - a) = 0 ........3 \\
 \\
g(a)g( - a) - f(a)f( - a) = 0........4 \end{array} $
$ 3^2 + 4^2$, we get sum of the squares of f(a)g(-a), g(a)f(-a), g(a)g(-a) and f(a)f(-a) is equal to zero
      So $  \{\begin{array}{c} \displaystyle f(a) = 0 or g( - a) = 0 ......5 \\
 \\
g(a) = 0 or f( - a) = 0 ......6 \end{array} $
for any a,
Let a=1, if f(1)=g(1)=0, we easily get f(x)=g(x)=0 for all integers x by substituting b=1.It's easy to check that f(x)=g(x)=0 satisfies the condition.
If f(1)=f(-1)=0 ,then by 4 we get either g(1)=0 or g(-1)=0, which returns to our last claim.The case that g(1)=g(-1) are similar.
[b]For Case 2[/b], Using substitution b=-a again we get g(a)g(-a)-f(a)f(-a)=1 ......7 ,which means gcd(g(a),f(a))=1 and gcd(g(-a),f(-a))=1
,Using this and 3, we get │f(a)│=│f(-a)│=0 and │g(a)│=│g(-a)│=1
or │f(a)│=│f(-a)│=1 and │g(a)│=│g(-a)│=0.Then
    [b]Case 2 i)[/b] f(1)=f(-1)=0 then │g(1)│=│g(-1)│=1
     Substitute b=1 we get f(a+1)=f(1)g(a) and g(a+1)=g(1)g(a)
     As g(1)=1 or -1 in this case, we get f(a)=0 and g(a)=1 or
f(a)=0 and g(a)=(-1)^a
            [b] 2 ii)[/b] g(1)=g(-1)=0 then │g(a)│=│g(-a)│=1 and g(a) is not equal to g(-a)
,by similar substitution and a few steps we get f(a)=(-f(1))^a

,By substituing a= 1 , ... we get f(1)= -1 ,then f(-1)=1^(-1)=1=f(1), a contradiction.

So the solutions are (1) f(x)=g(x)=0 (2) f(x)=0 and g(x)=1 (3) f(x)=0 and g(x)=(-1)^x[/hide]


Hello AoPSers. Game 2 is over, and game 1 is coming to an end, so I have decided to create game 3. AoPS Kingdom. In this game everybody that signs up is a villager that lives in the AoPS Kingdom. I am currently the king, however I am getting old, and I am soon to die. So one of the villagers has to replace me. So I have decided to have all the people that are willing to be king/queen come to the arena. At the arena you guys will compete in challenges where people will lose, or where people will earn points that will help later in the game. When there are only six people left in the game, then I will get rid of the person that has the least points. The amount of points that each person has will be known by them only (and me.) Then we will be left with 5 competitors. We will have a battle between them. The person that has the fifth most points will compete in a challenge against the person with the fourth most points. The winner will remain, but the loser will leave. The person who won this challenge will face off against the person that got the most points in another challenge. The other two will be competing in the same challenge. The winners of these challenges will try to convince me to name them king/queen. Then we will start another game in this thread, and the current king/queen can host it, but if he/she chooses not to, then the runner up can host it, but if he/she doesn't want to, then the previous host can host it, but if they don't want to then the runner up for that game will host, and so on.

Here are the only rules that you must follow if you are the host of this competition;
1. If you are host, then you can not people join during the game.
2. If you are host, then you can not replace people that quit or are inactive. They are simply taken out of the game.
3. No bringing people back into the game.
4. When we reach the top six the game must end in the way I described in which people will recieve rankings based on their number of points and 6th autimatically loses, 5th goes against 4th in a challenge. The winner of this challenge goes against 1st, while 2nd and 3rd go against each other, and the two winners of these two challenges will try to convince the current host to pick them to be the new host.

Sign-ups begin now;

bleh. i'll play...even though i don't really get it.

i guess i join...

what are the challenges?

I guess I'm in even though I don't really get it... :)

/in. I don't really get it, but I think I understand what the game is about.

read the rules, for a start.  :P

Can we start with 12 people?????

Hey, like abcak said, I posted 11 hours after lotsofmath! I think I should have 14 points now?

Oh I'm sorry, but this is the new current score;

The Current Score;
mathfreak123 - 14 pts.
eggylv999 - 7 pts.
yaofan - 6 pts.
abcak - 5 pts.
lotsofmath - 3 pts.
bluecarneal - 1 pt.
jonathanchou711 - 0 pts.

Bump.  But I don't have to participate in the challenge right?

[size=150][b]Elimination Challenge 2: Convince your Competition[/b][/size]

Nominees for Elimination;

jonathanchou711
bluecarneal
lotsofmath

One of these three will be leaving the competition unfortunately. For five days the thread will be their's. They can post whatever they want to defend themselves, and nobody else can post in the thread. When the five days is over, anybody can talk in the thread. They will discuss which of the three nominees will be eliminated. The nominees are free to answer questions directed at them during this challenge. When a majority of the players voting have voted for the same person, that person will be packing their bags, and going home. The other two live to see another day. This is sort of why alliances may come into play. All votes/unvotes must be bolded. Let the challenge begin.

Okay, you three only have about two days left.

[quote="lotsofmath"]I'm the only one to be in this thread in three days.[/quote]

[quote="bluecarneal"]Ack! really busy...

Guys!.... VOTE FOR ME I am the best candidate b/c I m probably the only one who posted to defend himself.  And... I like math :rotfl:[/quote]

ummm.......

[quote="bluecarneal"]But that doesn't mean he did what the challenge said and defended himself.......[/quote]

you were too late.

[quote="game.guy"]Is this dead?[/quote]
well you really ought to be more active  :wink:

I don't think that anybody is interested in this game anymore. Tbh, I'm not either.


Given triangle $ ABC$,$ H$ is orthocenter of triangle $ ABC$ and the circumcircle of triangle $ ABC$ has center $ O$.Let $ G$,$ A_1$,$ B_1$,$ C_1$are centroid of triangle $ ABC$,$ HBC$,$ HCA$,$ HAB$ respectively.
a)Prove that :
$ \overrightarrow {OA_1 } \plus{} \overrightarrow {OB_1 } \plus{} \overrightarrow {OC_1} \equal{} 5\overrightarrow {OG}$

b)If triangle ABC is right triangle($ \widehat{BAC} \equal{} 90^\circ$).Prove that :
$ \sin B.\sin C \equal{} \frac {1}{2}\left( {\frac {r}{R}} \right)^2 \plus{} \frac {r}{R}$

I think a question (a) is incorrect.
what do you think ?


IN THE NAME OF GOD

If you begin with a one digit integer, multiply by 3, add 8, divide by 2 and subtract 6, you will get the integer back.
Find the number.

THANKS
SALAH

Hi salahcool.

$ \frac{3n\plus{}8}{2} \minus{}6 \equal{} n$

$ 3n\plus{}8 \minus{}12 \equal{} 2n$

$ n\equal{}4$.

thank you very much AndrewTom
this is true solution
salah
 :rotfl:


Hello all, I have come up with a good idea of a team tournament. You can get with 2 other people you know and make a team of 3. If we can get 8 teams we can have a single elim tournament. The format would be you face each member of the opposing team 2 times so you play a total of 6 matches 3, each of your teammates will do the same, after all matches are completed the team with the most matches won wins and will move on to the next round, this process will continue until the finals when you play each opponent 3 times. Please reply in this thread with your 3 team members. The seedings will be random

ya.
uh.....I think this thread is dead now??
If so, please lock

look at the date: May...

if a thread is this old, just ignore it


geometry
3D geometry
modular arithmetic
Diophantine Equations
pen

What is the smallest positive integer $t$ such that there exist integers $x_{1},x_{2}, \cdots, x_{t}$ with \[{x_{1}}^{3}+{x_{2}}^{3}+\cdots+{x_{t}}^{3}=2002^{2002}\;\;?\]

[b]Answer:[/b] $ t \equal{} 4$
[b]Solution:[/b] Consider the equation modulo $ 9$. Note that the cube of every integer is $ \minus{}1,0,$ or $ 1\pmod 9$. However, $ 2002^{2002}\equiv 4^{4}\pmod 9\equiv 256\pmod 9\equiv 4\pmod 9$. So, $ x_{1}^{3}\plus{}x_{2}^{3}\plus{}...\plus{}x_{t}^{3}\equiv 4\pmod 9$ but as $ |x_{i}\pmod 9|\leq 1$ we must have $ t\geq 4$.

Now, if we take $ x_{1}\equal{} x_{2}\equal{} 10*2002^{667}, x_{3}\equal{} x_{4}\equal{} 2002^{667}$ then $ x_{1}^{3}\plus{}x_{2}^{3}\plus{}x_{3}^{3}\plus{}x_{4}^{3}\equal{} 2002^{667*3}(10^{3}\plus{}10^{3}\plus{}1^{1}\plus{}1^{3}) \equal{} 2002^{2001}*(2002) \equal{} 2002^{2002}$
So, for $ t \equal{} 4$ there exist the required integers $ x_{1}, x_{2},...,x_{t}$ and as $ t\geq 4$ then $ 4$ is the smallest positive integer satisfying the required conditions.

[hide=hints]Find the remainder when L.H.S is divided by 9


Can someone explain to me why though change in free energy of a weak acid dissociation is positive, but the reaction still occurs?

The closest thing I could find to an explanation is a picture on this page: http://staff.um.edu.mt/jgri1/teaching/che2372/notes/05/03/03/equil_applications.html#p2_b (scroll down a little)

But that potential energy graph seems to suggest change in free energy is negative.... is it that the graph is almost right or am i just way off..

thx!

The graph is saying that the free energy decreases for a small amount of dissociation, but quickly reaches a minimum.  After that, it doesn't dissociate any further.  That's why weak acids dissociate only slightly - strong acids, as shown, have a minimum much further along and dissociate to a large extent before reaching it.

[quote="adchia"]Can someone explain to me why though change in free energy of a weak acid dissociation is positive, but the reaction still occurs?
[/quote]

This is what I think. I am just wondering if you are confusing Free Energy with "Standard" Free Energy.

Standard Free Energy can be calculated via $ \Delta G^{\circ} \equal{} \minus{}RT \ln K$. Since the $ K$ for most weak acid is a lot smaller than $ 1$ thus $ \Delta G^{\circ} >0$.

However, under non-standard conditions (as it is usually the case) we have $ \Delta G \equal{} \Delta G^{\circ} \plus{} RT \ln Q$ or equivalently $ \Delta G \equal{} \minus{}RT \ln \frac{K}{Q}$. If $ K>Q$ then the process is spontaneous as the Free Energy had be smaller than $ 0$.

Consider then a simple example: $ HA \rightleftharpoons H^{\plus{}} \plus{} A^{\minus{}}$. If we start off with a large amount of $ HA$ and negligible amount of the ions then the value of $ Q$ is going to be a lot smaller than $ K$ resulting in a spontaneous dissociation.

ye hah banished that makes soo much sense...

as clarification, $ \Delta G^\circ$ calculates the free energy difference if the reaction went from beginning to completion right? (so THAT would never happen). And it is difficult to find how much free energy there actually IS at equilibrium right? (hopefully I'm not confusing even more things)

Also, im wondering whether the graph of weak acid looks more like this: (instead of their graph, where the change is negative)

No, delta G calculated the free energy when all aqueous reactants and products are present at 1 M concentration, and all gases are present at 1 atm pressure.

That is the standard free energy change $ \Delta G^{\circ}$. I am not sure what adchia means by free energy at equilibrium. You can't really measure free energy, only change in free energy. If that's what adchia means then the $ \Delta G \equal{} 0$ at equilibrium.

I think what's trying to be said is that the weak acid equilibrium is so overwhelmed by the fact that there are no H+ or A- ions currently present that it dissociates, until it reaches equilibrium, that is.


[quote="pleurestique"][hide]by induction, it's easy to show that $\sum_{k=1}^{n} \arctan\frac{1}{2k^2} = \arctan \frac{n}{n+1}$. then $\sum_{k=1}^{\infty} \arctan\frac{1}{2k^2} = \lim_{n\rightarrow \infty} \arctan \frac{n}{n+1} = \frac{\pi}{4}$[/hide][/quote]

[hide="question"]
How did you come up with $\sum_{k=1}^{n} \arctan\frac{1}{2k^2} = \arctan \frac{n}{n+1}$?  I realize it's true, but did you memorize this or derive it somehow?[/hide]

well, i tried small cases; n = 1;n = 2; and noticed a pattern...then proved it was true by induction

Here's another interesting series:

Compute $\sum_{k=0}^{n}\arctan\left(\frac{1}{k^{2}+k+1}\right)$.

[hide="Hint"]$\arctan\alpha+\arctan\beta=\arctan\left(\frac{\alpha+\beta}{1-\alpha\beta}\right)$[/hide]

You don't need to use that formula. Just be all like $\tan \left( \arctan a + \arctan b\right) = \frac{a+b}{1-ab}$.

Yea except in the first one there are an infinite number of terms so I don't see how the tangent addition formula for two angles will help us very much.

Not the first one.

$\frac{1}{k^{2}+k+1}=\frac{\left(k+1\right)-k}{1+k\left(k+1\right)}$

Maybe you guys have another, better way to do this.  From the above, do you see why I said to use the addition formulas?


Counting! :-D

Have you tried a smaller number of tiles? Say...six tiles 1,1,2,2,3,3

:-)

I have an inelegant solution. I hope someone can post a more elegant one.[hide]The first one you pick can be anything(1), the second one you pick must be either same or different(1/11 or 10/11). If the first two are same, then you could pick is anything again(1), if the first two are different, we must have the third one to match with one of them, or 2/10.......

That gave us a pattern.

if n=3,5,7,9,11 then when we have n tiles left, the prob. of picking is either 1/n, or (n-1)/n because that's when you have one tile in your hand.

  if n=4,6,8,10, fraction in front is 1/(n+1), then prob. of picking is 1 because you have no tile in your hand,

  if n=4,6,8,10, fraction infront is n/(n+1), then prob.of picking is 2/n because you now have two different tiles in your hand.

Even denominator(n) cancel out either way

so it simplifies into a/11*b/9*c/7*d/5*e/3, where a,b,c,d,e can be 1 or 2. total prob.= sum of all possible a/11*b/9*c/7*d/5*e/3

(1+5C1(2)+5C2(4)+5C3(8)+5C4(16)+5C5(32))/(11*9*7*5*3)=(3^5)/11*9*7*5*3= 243/11*9*7*5*3=9/385, which is the answer

[hide][/hide][/hide]

Fierytycoon wrote:A bag contains 12 tiles marked 1, 1, 2, 2, ... , 6, 6. A player draws tiles one at a time at random and holds them. If he draws a tile matching a tile he already holds, then he discards both. The game ends if he holds three unmatched tiles or if the bag is emptied. Find the probability that the bag is emptied.

I think this solution is pretty elegant:

[hide]In the first three draws, the probability of him "failing" 10*8/(11*10) = 8/11. So the probability of him being able to discard is (1-8/11). After this set of three, he is left with one tile, so for the next two tiles, we use the same method, getting the probability (1-8/11)(1-6/9), so the final probability is (1-8/11)(1-6/9)(1-4/7)(1-2/5) = (3/11)(1/3)(3/7)(3/5) = 9/385.[/hide]

[hide]Quote:After this set of three, he is left with one tile, so for the next two tiles, we use the same method, getting the probability (1-8/11)(1-6/9)

But you said that we are left with one tile, so how can we apply the same method? For the initial case, you had to choose 3 tiles, but after that, you only have to choose 2 tiles (since you have one left over).[/hide]

[quote="Fierytycoon"]
But you said that we are left with one tile, so how can we apply the same method? For the initial case, you had to choose 3 tiles, but after that, you only have to choose 2 tiles (since you have one left over).[/quote]

For the initial case, I'm working with three tiles, given any first tile (doesn't matter what it is), i.e. let the first tile be 1. Therefore, w/e extra tile pops out of my first three is the "first tile" in the next three, which are the third, fourth, and fifth draws.