There are two people A and B.  They can remove a coin at a time from a pile or split a pile into two non empty piles.  The last person to remove the coin loses.  The objective of the game is to lose.  The game starts with A1 piles of single coins and A2 piles of two coins...How would the person lose based on the sequence (A1,A2,A3...)?

I think this is the one about taking out a specific number that is bigger than the amount able to take out by one person but can be taken up if both people take the turn.

I think it is similar to this:(A very basic game)
A person can say 1,2 or 3numbers at a time.  Two people keep saying the numbers until the person to say 21 loses.  In this case, the second person always have to say up to the multiple of 4, which leads the person to victory.

So.. maybe there is a similar form of strategy to this game as well?

Amod lock this, the competition isnt over until 14th of october

Hey I read the downloads.  I can't find which question is the same is this can you tell me?

lol its number 11 person A=michael, person B=robin. Your friend just changed a bit so that other people wouldn't notice. Its basically the same idea :ninja:
If someone to you the losing solution, then you friend could just write the exact opposite of the losing solution :roll:  :wink:

Let:
$n:$ The number of the piles
$x:$ The total number of the coins
$m:$ The number of the single coins

If $m$ is even and $x-n$ is also even then we will say that we have a [u]W-state[/u].
Else we will say that we have an [u]L-state[/u].

The [b][u]objective[/u][/b] for each move is to reach a W-state.

[color=brown]We will show that, every W-state can lead only to L-states.
Also, in every L-state, there is (at least one) move such that we can go to a W-state.[/color]

From this point, when we say that $m$ changes, we will mean that its parity changes, that is, $m \mod 2$ changes.
When we say that $m$ keeps its value, we'll mean that $m \mod 2$ keeps its value
(for example, if the new value of $m$ is $m-2$, then we will say that $m$ doesn't change; it keeps its value)

[hide="Let us first see the possible moves"]
$\blacksquare$ Remove a coin from a pile with $k$ coins

For $x:$
The new value of $x$ is $x-1$. So $x$ [u]changes[/u]

For $n:$
If $k=1$ (if we remove from a pile with 1 coin) then, of course $n$ [u]changes[/u].
Else it doesn't.

If $k=1$, then $m$ becomes $m-1$, so $m$ [u]changes[/u]
If $k=2$, then $m$ becomes $m+1$, so $m$ [u]changes[/u]
In any other case $m$ doesn't change.

Summary:
[quote]$x-n$ keeps its value if $k=1$
$m$ keeps its value if $k>2$[/quote]

$\blacksquare$ Splitting
We split a pile of $k$ coins into two non-zero piles of $p$ and $q$ coins, such that $p+q=k$ (WLOG suppose that $p\leq q$)
One should notice that $k$ must be at least 2.

For $n:$
We loose a pile but we obtain two new piles.
So $n$ becomes $n+1 \Rightarrow n$ [u]changes[/u]

For $x:$
Of course we have the same number of total coins, so $x$ [u]doesn't change[/u]

For $m:$
We saw that $k\geq 2$. So we don't [u]loose[/u] any of the single coins by splitting.
We could only [u]obtain[/u] more single coins, depending on the values of $p,q$

If $1<p\leq q$, then we have the same number $m$ of single coins, that is, $m$ [u]doesn't change[/u]
If $1=p<q$, then we obtain 1 more single coin, so $m$ [u]changes[/u]
If $1=p=q$, then we obtain 2 more single coins, so $m$ [u]doesn't change[/u]

Summary for splitting:
[quote]$x-n$ changes
$m$ changes if [u]just one[/u] of the new piles is a single coin [/quote]
------------------------------------------------------------------------[/hide]
[hide="We will see that every W-state can lead only to L-states"]
Suppose we have a W-state, that is $m \equiv x-n \equiv 0 \mod 2$

We have two choices: split and remove

If we choose to split, $x-n$ changes, so we go to an L-state

If we choose to remove one coin from a pile with $k$ coins:
If $k>1$, as we saw, $x-n$ changes, so we go to an L-state
If $k=1$, then $x-n$ keeps its value, but $m$ changes. That is an L-state.

Conclusion
This means that, if we "give" to the opponent a W-state, then he can't give us a W-state.
He'll give to us an L-state.
Our goal is to convert an L-state to a W-state. Let's see how:
------------------------------------------------------------------------[/hide]
[hide="Now we will see that, if the state is L, then we can find a move, such that the result is a W-state"]
L-state means that at least one of $m,x-n$ is odd.

[color=brown]Note
If $x-n$ is odd, then there is at least one pile with $k>1.$
Indeed, if we have only single coins, then $x=n\Rightarrow x-n$ is even[/color]

We'll check 3 cases:

[hide="a) m odd and (x-n) even"]
Since $m$ is odd, there is at least one single coin.
We will remove this coin.
Remember the rules of removing:
[quote]$x-n$ keeps its value if $k=1$
$m$ keeps its value if $k>2$[/quote]
Here, $k=1$, so $x-n$ remains even and $m$ becomes even
So we have a W-state
------------------------------------------------------------------------[/hide]
[hide="b) m odd and (x-n) odd"]
We have the sub-cases:

$b.1)$ If there exists a pile with $k>2$ coins
We split this pile into two piles of $1$ and $k-1$ coins respectively.
Remember the rules of splitting:
[quote]$x-n$ changes
$m$ changes if [u]just one[/u] of the new piles is a single coin[/quote]

So $x-n$ changes and $m$ changes $\Rightarrow x-n \equiv m \equiv 0 \mod 2$

$b.2)$ If there is not a pile with $k>2$, then we can be sure that there exists at least one pile with $k=2$ coins (see note above)
We remove one coin from this pile
Remember the rules of removing:
[quote]$x-n$ keeps its value if $k=1$
$m$ keeps its value if $k>2$[/quote]

Here, it is $k=2$, so $x-n$ changes and $m$ changes $\Rightarrow x-n \equiv m \equiv 0 \mod 2$
------------------------------------------------------------------------[/hide]
[hide="c) m even and (x-n) odd"]
We have the sub-cases:

$c.1)$ If there is some pile with $k=2$ coins
Then we split this pile
Remember the rules of splitting:
[quote]$x-n$ changes
$m$ changes if [u]just one[/u] of the new piles is a single coin [/quote]

So $x-n$ changes and $m$ does not $\Rightarrow x-n \equiv m \equiv 0 \mod 2$

c.2) If there is not a pile with $k=2$ coins
Then we can be sure that there is some pile with $k>2$ coins (see note above)
We remove one coin from this pile

Remember the rules of removing:
[quote]$x-n$ keeps its value if $k=1$
$m$ keeps its value if $k>2$[/quote]

Here, it is $k>2$, so $x-n$ changes and $m$ doesn't $\Rightarrow x-n \equiv m \equiv 0 \mod 2$
------------------------------------------------------------------------[/hide]
----------------------------------------------------------------------------------------------------------
[/hide]

[size=134][color=blue]Conclusion[/color][/size]
So, we can always convert a W-state to an L-state

If the two players know this strategy, then the initial state will show who will be the winner.

If it is a W-state, then the first player can't convert it to a W-state, so he will loose
If it is an L-state, then the first player can convert it to a W-state, so he'll win

Nice proof pontios.
I came up with a similar solution.
I did
Let A=pile of coins
     B=number of piles
     0=even
If it is even(A)xodd(B)+odd(A)xeven(B)+even(A)xodd(B)+...
or      odd(A)xeven(B)+even(A)xodd(B)+odd(A)xeven(B)+...
During the person B's turn; It is always a win.
I have found this works up to any number of pile of coins up to 4.  The only catch was that If person A had 1odd left he would lose.  I was wondering would this strategy work?  And up to what extent?  Could someone contradict/point out/help me please?
Thanks.

PS:I have concluded that pontios's solution is the most correct so far.  However, is there any shorter solution?

Actually, when I posted my solution I couldn't simplify it more. And the proof was indeed hard. Thank you for revising this topic, because I had the chance to try it more. I wrote a script which calculates all the possible states starting from (1,2,3,4,5).
Finally, I found a simpler solution to this problem.

However, I can't understand some points in your post

[quote="max1546"]Let A=pile of coins
     B=number of piles
     0=even
[/quote]
what is A?

[quote="max1546"]If it is even(A)xodd(B)+odd(A)xeven(B)+even(A)xodd(B)+...
or      odd(A)xeven(B)+even(A)xodd(B)+odd(A)xeven(B)+...
[/quote]

what does this sentence mean?  :huh:

Sorry Pontios for being so unclear, I was rushing it.LOL
hmm I'll try to explain some of the things you asked.
Remember though, this MIGHT make you more confused.
So i'll do my best.

If there is 3coins stacked on top of each other, B=3, A=1.
If there are 6coins stacked so that it is 3coins per pile, it is B=3, A=2.

The other part was, I tried some trial and error and got the answer that if it is the person's turn and he starts off with;
1(A)x1(B)=Certain chance of winning.
1(A)x2(B)=Win
2(A)x1(B)=Lose
1(A)x3(B)=Win
1(A)x4(B)=Lose
1(A)x5(B)=Win
2(A)x2(B)=Win
And so on.I tried up to 4(A) and my sequence given worked.  I was trying to take in to account every single move either could make and if the person who starts has a chance of winning no matter what the other person does, he wins.  On the other hand, if the other person could win with at least one move no matter what the starting person does, I counted it as a loss.

Ahh, even I'm getting slightly confused.  I hope this explained slightly more.  Still I think your solution is the best so far.  If I have time, I'll try and improve my solution and post it again.  Hopefully it will be easier/clearer.

Thanks

Let's re-formulate the problem
[quote]There are two people A and B.  They can remove a coin at a time from a pile or split a pile into two non empty piles.
The last person to remove the coin [u][color=red]wins[/color][/u].

The game starts with
A1 piles of single coins
A2 piles of two coins...
A3 piles of three coins...
How would the person [u][color=red]win[/color][/u] based on the sequence (A1,A2,A3...)?
[/quote]

[hide="My new solution"]
Every state can be represented by the sequence $A_{1}A_{2}A_{3}A_{4}...$

Let $\boxed{x = A_{1}\mod 2}$ and $\boxed{y=(\sum{ A_{2k}})\mod 2}$

(we don't need the numbers A3,A5,A7,...)

Equivalently, we can say that we take the last binary digit of $A_{1}$ and $\sum{ A_{2k}}$

Next, we calculate the value $S=x \vee y$ from the folowing table:

$\begin{array}{|c|c|c|}\hline x & y & x \vee y \\ \hline 0 & 0 & 0 \\ \hline 0 & 1 & 1 \\ \hline 1 & 0 & 1 \\ \hline 1 & 1 & 1 \\ \hline \end{array}$ (Boolean OR)

And we are finished!

The state $(A_{1}A_{2}A_{3}A_{4}...)$is marked with the label $S$.
We will say that we have an [b]S[/b]-state

If you start playing from a [b]1[/b]-state, then you will [u]definitely[/u] win
If you start playing from a [b]0[/b]-state, then you will [u]definitely[/u] loose

So, after each move we should take care such that the result is a [b]0[/b]-state.

[color=darkred]Note
 - The state $(A_{1}A_{2}A_{3}A_{4}...)=(1000...)$ is clearly a [b]1[/b]-state (if you start from this state, you'll win)
 - The state $(A_{1}A_{2}A_{3}A_{4}...)=(0000...)$ can be defined as a [b]0[/b]-state[/color][/hide]

[hide="Proof"]
Again the proof is a litle complicated but no so much as the previous
[hide="opponent s moves"]
The opponent plays always starting from a [b]0[/b]-state, that is, $(x,y)=(0,0)$

That is, he has
 - an even number of single coins, and
 - an even number of even piles

[i][u]Even pile[/u]: a pile consisting of 2 or 4 or 6, ... coins[/i]

Can he make a move to go to a [b]0[/b]-state?
The answer is [b][u]no[/u][/b]

Let's see the possible opponent's moves:

$\hline$
[color=green][size=150]Removing[/size]
Remove 1 single coin . Bad move! $x$ becomes $1$
Remove 1 coin from a pile with $2$ coins. Bad move!  $x,y$ become $1$
Remove 1 coin from a pile with $m\geq 3$ coins. Then we have:
$A_{m}$ decreases by $1$
$A_{m-1}$ increases by $1$

Some of $m,m-1$ is even, so $y$ becomes $1$[/color]
$\hline$

[color=brown][size=150]Splitting[/size]
Split the pile $A_{m+n}$ into two nonzero piles $A_{m},A_{n}$ (suppose $m\geq n$)

If $m=n=1$ then he obtains two single coins ($x$ doesn't change) but he destroys one double-coin pile (so $y$ becomes 1)

If $m>n=1$ then he obtains 1 single coin ($x$ becomes 1)

If $n\geq 2:$

If $m+n$ was odd, then [u]exactly one[/u] of $m,n$ is even (Bad move; $y$ becomes 1)

If $m+n$ was even:

Either $m,n$ are both even
Or they are both odd

At either case he destroys the y-balance[/color]

So the opponent can't turn the game to [b]0[/b]-state
$\hline$[/hide]
[hide="Your moves"]You start from a [b]1[/b]-state. Let's see if there is a way to turn the game to a [b]0[/b]-state

In the following, $m$ will represent a natural number $m \in \{0,1,2,...\}$
There are three cases such that a state is a [b]1[/b]-state:

$(x,y) = (1,0)$
$(x,y) = (0,1)$
$(x,y) = (1,1)$

[hide="(1,0)"]There is an odd number of single coins. You want to make it even.
[color=red][b]Solution:[/b] Just remove a single coin[/color] ($x$ becomes 0)
$\hline$[/hide][hide="(0,1)"]
You have an odd number of even piles.
$\blacktriangleright$ If there is a pile with $m+3$ coins
[color=red][b]Solution: [/b] remove a coin from it.[/color]
You obtain a pile with $m+2$ coins, but you loose the initial pile with $m+3$ coins
Exactly one of $m+3$ and $m+2$ is even, so you adjust the y-balance ($y$ becomes 0)

$\blacktriangleright$ If there is not a pile with $m+3$ coins. Then you have only double coins (don't care about the single coins)
[color=red][b]Solution: [/b]Split one double-coin pile into two single coins.[/color] So you adjust the y-balance
$\hline$[/hide][hide="(1,1)"]
You have an odd number of single coins and an odd number of even piles

$\blacktriangleright$ If there is a pile with $m+3$ coins, then
[color=red][b]Solution: [/b]split 1 coin[/color]
You obtain a single coin so $x$ becomes 0
You obtain a pile with $m+2$ coins, but you loose the initial pile with $m+3$ coins. For the same reasons as above, you adjust the value of $y$

$\blacktriangleright$ If there is not a pile with $m+3$ coins. Then you have only double coins (and some odd single coins)
[color=red][b]Solution: [/b]Just remove one coin from a double-coin pile.[/color]
The double coin becomes single.
So you obtain one single and loose one double, as a result both the values of $x$ and $y$ are fixed
$\hline$[/hide]

You can see the diagram:

$(x,y)=\{\begin{array}{{l}(1,0) \Rightarrow  \text{Remove a single coin}} \\ \\ (0,1) \Rightarrow  \exists m>0 : A_{m+3}>0 \{\begin{array}{l}\exists \Rightarrow  \text{Remove a coin from }A_{m+3}\\ \not\exists\Rightarrow  \text{Split a coin from }A_{2}\end{array} \\ \\ (1,1) \Rightarrow  \exists m>0 : A_{m+3}>0 \{\begin{array}{l}\exists \Rightarrow  \text{Split a coin from }A_{m+3}\\ \not\exists\Rightarrow  \text{Remove a coin from }A_{2}\end{array} \end{array}$

[/hide][/hide]


The AoPSWiki Words of the Week for Nov 8-14 are [[Asymptote]] (the graph concept, not the vector graphics language used with LaTeX) and [[AMC historical results]].

Shouldn't each diff be posted by the admins?

[quote="1=2"]By "AMC Historical Results", do you mean perfect scores on AMC 8, or perfect scores on the AMC 12, or participating in IMO when you were 10 years old?[/quote]The community should decide these issues on the article's discussion page.  I started off the discussion but the page should be whatever people think is most relevant.


I know I can solve all of them and I'm more than happy to explain (I write my explanation on computers) to you. I might be able to PM you...

i lived in North Carolina too!

In Charlotte, Mecklenburg district


hi guys, i'm gr 12 this year, so everything for math is over for me, but i just want to know how u guys handle this so i can kick myself for this.

ok, i love math, i've done all of the waterloo books and more
when i was preparing for comc this year, every contest i can get around 65, but actual thing i got 52
same for euclid 2003, every practice i can get 80+ but i got 55 in 2003
this year, i think i can only get 7*
both from being too nervous
how do u guys get rid of nervousness so that u can do ur best?
plus what materials do u guys work on?

i know i can do a lot better, because last year, i got 32 on hypatia, which is a 83rd in canada
i know it is not very good compare to some of u guys, but it's good for me
btw, i got 111 on fermat because of nervous, and retarded mistakes because i was expecting 130+ because that's what i've been getting on my practice contests

long story short, i got my butt wooped because i was nervous, any ideas how to get past this? (other than, do more questions)

um, comments like yours don't help, bugzpodder.

he wants someone to give advice. you shouldn't have said anything if you didn't have any.

sorry i was trying to say everyone is nervous... it didnt came out right

It gets better after you participate in more contests (I don't mean just practice, rather actually sitting in the test room).

At first, I can barely handle the stress of a 30 min contest. After a while, you just get used to the idea of math competitions. Now, sitting 4.5h for say USAMO isn't a problem for me.

As well, after you do more problems (especially the proving ones), you get a sense of what solutions are right and what are not.

Then again, of the two COMCs that I've done, I never quite figured out where I lost the marks. The Olympiads are definitely better, at least for me. For both USAMO and APMO, I got marks divisible by 7 :).

Lastly, one of my philosophies for math contests is that, one's performance on a contest is no greater than 33% of his/her actual ability.

Remember it is just you and the problem. It is an icky little thing, but just slay it, take your time as you do with the practices. If you are getting panicky, do something else for a little while. I sometimes go for a walk to the toilet and back. Or eat something. Or take a few deep breaths. Feel your mind with thoughts of the problem, and not with thoughts like "I must solve this to get a silver".... concentrate, basically.


A certain coin is weighted such that the chance of flipping heads is $ \frac{1}{3}$ and the chance of flipping tails is $ \frac{2}{3}$.  Suppose that we win $ \$3$ if we flip a heads on a coin toss, but lose $ \$2$ if we flip tails.  What is the expected value, in dollars, of our winnings after one flip?  Express your answer as a common fraction.

By definition, this is $ 3\cdot \frac{1}{3}\plus{}(\minus{}2)\cdot \frac{2}{3} \equal{}\boxed{\frac{\minus{}1}{3}}$


analytic geometry
inequalities
calculus
derivative
function

If $a,b,c \in [0,\infty),bc\ne 0 ,$ show that \[1 \le \frac{a+b+c}{\sqrt{a^{2}+b^{2}+c^{2}}}\le \sqrt{3}\; .\; \; \; \; (\mbox{corrected !})\] [size=75][b]Try to generalize these inequalities.[/b][/size]

I think it generalized should be something like this:
$1\leq\frac{a_{1}+a_{2}+...+a_{n}}{\sqrt{a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}}}\leq\sqrt n$
Given that they are all >=0 and at least one is non-zero.

To prove this:
First, note that scaling all the variables by some factor k leads to the same equation since the k cancels in the numerator and denominator.  This means that we can restrict our interval to some finite interval, because anything outside that can be achieved by scaling.  Take all the variables to  be in the interval [0,1].

Now I want to say that you only need to consider the endpoints, that is each variable=0 or 1, but I'm not sure why that would be allowed.  If that works, then you can see that the minimum is 1 and the maximum is $\sqrt n$.

Another way to prove the minimum:
Take all the variables to be coordinates in n-dimensional space.  Then the minimum path from the origin to that point is the denominator.  The numerator is an alternative path.  That means the numerator>=denominator, and equality gives when it =1.

For [b]Wumbate[/b]'s generalization, RMS-AM gives the RHS: \[\frac{a_{1}+\ldots+a_{n}}{n}\le\sqrt{\frac{a_{1}^{2}+\ldots+a_{n}^{2}}{n}},\]  \[\frac{a_{1}+\ldots+a_{n}}{\sqrt{a_{1}^{2}+\ldots+a_{n}^{2}}}\le\sqrt{n}.\] Since all the $a_{i}$ are non-negative, then \[a_{1}^{2}+\ldots+a_{n}^{2}\le (a_{1}+\ldots+a_{n})^{2}.\] Hence, \[1\le\frac{a_{1}+\ldots+a_{n}}{\sqrt{a_{1}^{2}+\ldots+a_{n}^{2}}},\] giving the LHS.

Actually, the inequality cannot be strengthened past $1 \le \frac{a_{1}+a_{2}+...+a_{n}}{\sqrt{a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}}}\le \sqrt{n}$, because setting $a_{1}=a_{2}=...=a_{n-1}=0$ yields a value of 1 for the expression.  It is easy to prove the RHS using Cauchy, and the LHS follows as long as the terms are nonnegative.

[quote="K81o7"]Actually, the inequality cannot be strengthened past $1 \le \frac{a_{1}+a_{2}+...+a_{n}}{\sqrt{a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}}}\le \sqrt{n}$, because setting $a_{1}=a_{2}=...=a_{n-1}=0$ yields a value of 1 for the expression.  ........[/quote]
Thanks for interest,

Even if $bc=0$, you can let $a=0$, and take the limit of the expression as $b \to 0$ (with $c \neq 0$).  That value is the same: ie, by continuity, you can always make the expression as close to 1 as you want it to be.

[quote]... following problem arise: [b]prove or disprove that [/b]......  FALSE ![/quote]
Sorry, thanks for interest,Alex

Even if b=c>0, then you have this:
$\frac{a+2b}{\sqrt{a^{2}+2b^{2}}}$
Now take a->infinity and/or b->0 and this tends to 1.  So it would still have 1 as a lower bound.

For the general inequality, I still want to know if there is a way to reduce the "checking" to values of 0 or 1.

I think it would be nice if you could reduce all the variables to only 0 or 1 - it's easy, it happens to work, but I'm not sure with the reasoning.

If you take a partial derivative of the function with respect to one of the variables, and set that to 0, you will find the max and mins.  Now if you assume all the other variables are 0 or 1, then the only interesting value of that variable is 0 or 1.

I'm not too good with multivariable stuff, so if someone else could explain, that would be nice.


When are the student solutions for round 4 going to be posted?  I'm curious to see some of the commended techniques for solving this round's problems.


linear algebra
matrix
vector
probability and stats

hi I 'm looking for informations about this kind of regression.

We try to estimate a symetric $ m$x$ m$ matrix, $ A$ for the following regression model :

\[ y_ty_t^T\minus{}I_m\equal{}A(x_{t}x_{t}^T\minus{}I_m)A\plus{}E_t\]

with $ y_t,x_t$ two observations vectors of dimension $ m$, and $ E_t$ is a symetric matrix representing model errors.

To estimate this, $ A$ is calcutlated by minimising the the average Fröbenius norm of $ E_t$ matrices over the set of $ A$ in the space of symetric matrices.

This method is apparently known as a "non linear least square" method. I have tried to find some information about this kind of regression and estimation method in this particuliar context but i couldn't find anything. Basically I have found this strange regression in a statistical article where it is presented as "classic" but as I don't have a very large culture in statistics I didn't got much success on my googling research.

If anyone could lead me to some references or web links I would appreciate

bye

If it was presented in a research article then your best bet is to look at the references. I don't mean to state the obvious but if it is listed as "classic" then several other papers in the references should have information pertaining to it.


calculus
derivative
algebra
polynomial
induction
real analysis
real analysis solved

i dont know whether the problem is simple or not.It looks very harmless but I cant solve it.
Find the nth derivative of e^(-x 2 )

Looks like I am getting stupid day by day.
Please hurry I want the solution desperately.

I can feel that there will be some reduction or recursive type of formula but I want the result to involve only n,e,x.

Just heard about them.I will surely be glad if u tell me more about that.
I know how to solve recursions & generating functions.
Thats it.

[quote="the game"]Sorry but I didnt understand what Jayscholar has tried to say
Is it really that simple![/quote]
No, it isn't that simple. Jaysscholar is simply wrong, which is what jmerry was saying in #5.

[quote="jmerry"]Product rule...[/quote]
which states that $ (fg)' = f'g+fg'$

Certainly I know product rule.How will the knowledge of product rule help the solution of Hermite polynomials.I presume it will be solved by generating functions.

Thought it to be a lot easier than that.At least the recursion relation looked simple.Thanks any way.


number theory
greatest common divisor
modular arithmetic
relatively prime
Divisibility Theory

If $n$ is a natural number, prove that the number $(n+1)(n+2)\cdots(n+10)$ is not a perfect square.

The largest power of $ 2$ that divides $ \frac{(n\plus{}10)!}{n!}$ is $ 32$, so it cannot be a perfect square.

That doesn't quite work ($ n \equal{} 54$, for example).

Suppose the product is a perfect square. Among the ten terms in the product:

- five are even
- at most two are odd and divisible by 3
- at most one is odd and divisible by 5
- at most one is odd and divisible by 7

So at least one term is not divisible by any of 2,3,5,7. This term is relatively prime with the others since any common divisor of two terms is at most 9. So this term is a perfect square, and at least $ 11^{2}$.

Now note that this is the only square in our product, otherwise we would have a square greater than or equal to $ 11^{2}$ differing from another square by at most 9.

In particular, there must be exactly one term not divisble by any of 2,3,5,7. So we must have equality in all of our counts, and no odd term can be divisible by 15, 21, or 35.

Consider the odd terms divisible by 3. Neither are divisible by 2,5,7, and the gcd between either of these terms and any another one is at most 9, therefore a power of 3.

So we can factor out the largest power of 3 from each of these terms. Each quotient that results is relatively prime to all other terms and are both squares. So it follows that each odd term divisible by 3 is either a square or three times a square.

We know from above that neither can be squares, so both are three times a square. But this also produces two squares that are too close (in particular, they differ by 2). This is a contradiction, so the product cannot be a square.

We can prove it by using problem A9 in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=150377]here[/url].

From ten consecutive numbers we have one which is relatively prime to others. Remove this number from the list we have nine numbers which we can chose five from them which are consecutive. For five consecutive we will prove that there's one prime to others else. Indeed, there are at most two ones which is divisible by $ 3$, and if there're two ones then one is odd and one is even. In case we remove one which is even, so one left there're at most two even numbers from $ 4$ number left or $ 3$ for the left case. Remove them we have one odd left or two odd but in which  there's one is not divisible by $ 3$. So there's one number which is odd, not divisible by $ 3$. This number from list $ 5$ consecutive numbers is relatively prime to others else. Remove it we get four numbers which always find two are consecutive, and then of cause relative each other. So we could find three numbers which is relatively to any $ 9$ numbers else. Now note that if $ a$ is square then $ a\plus{}1,a\plus{}2$ are not and $ a\minus{}1,a\minus{}2$ are not of $ a>2$. So in ten consecutive numbers it could not contains more than two squares. This proves the problem.

[quote="pluricomplex"]
So we could find three numbers which is relatively to any $ 9$ numbers else. [/quote]

this isn't true. for example, take n=209, then only 211 is relatively prime to other nine numbers because 210 212 214 216 218 are even numbers;213 216 219 are multiple of 3; 210 and 215 are multiple of 5; 210 and 217 are multiple of 7

the mistake is that you may choose a number in the five consecutive numbers which is relatively prime to other $ 4$ numbers, but it maybe not relatively prime to other $ 9$ numbers.

There was such a problem at Latvian Open Olympiad in Mathematics:

It is given that $ n$ is a natural even number. Let $ R \equal{} n\left(n\plus{}1\right)\left(n\plus{}2\right)\left(n\plus{}3\right)$.
[list][i]a[/i]) Can $ R$ be a square of a natural number?[/list][list][i]b[/i]) Can $ R$ be a qube of a natural number?[/list]

I think all your solutions don't work, why?
We note that $n \equiv 1 \pmod{8},n \equiv 1 \pmod{9},n \equiv 1 \pmod{25}$
$n \equiv 43 \pmod{49},n \equiv 111 \pmod{11^2}$,
By Chinese remainder theorem, there always exists $n$ satisfies all conditions above, then $11^2.2^8.3^4.7^2.5^2||(n+1)(n+2)...(n+10)$ so $mod(11,2,3,7,5)$ don't work!, In that cases, we only can consider the modular of prime number which is large enough but It's impossible

[quote="mszew"][quote="Peter"]If $ n$ is a natural number, prove that the number $ (n \plus{} 1)(n \plus{} 2)\cdots(n \plus{} 10)$ is not a perfect square.[/quote]

There is a much stronger result proven by Erdos, "The product of consecutive integers is never a power" you can check it at:
[url]http://www2.renyi.hu/~p_erdos/1975-46.pdf[/url][/quote]

The URL cannot use anymore.
I found a new one [url]http://www.renyi.hu/~p_erdos/1975-46.pdf[/url]

just consider divisibility by all primes less then $10$ and you'll face contradiction


[color=purple]Trang chủ của ML có phần Problem of day anh Tuân có thể cho em biết điều kiện để bài toán của mình được chọn ko ạ  :)
   Thanks sếp  :D [/color]

Tôi cũng muốn hỏi vấn đề này. Nhưng tôi nghĩ muốn đưa bài lên là do các master làm.


Is it possible to find three conics in the plane such that any straight line in the plane intersects at least two of the conics and through any point of the plane pass tangents to at least two of them?


topology
function
real analysis
real analysis solved

could prove that $\mathbb{R}^{\mathbb{R}}$ is seprable or disprove it .it is not that hard that $\mathbb{R}^{P(P(\mathbb{R}))}$ is not seprable with product topology.

Lemma: The cardinality of a separable space is at most that of the real numbers.
Proof: Take a countable dense subset. Every element of the space can be expressed as the limit of a sequence in that set. There are as many sequences as there are real numbers.

Now, $\mathbb{R}^{\mathbb{R}}$ has larger cardinality than $\mathbb{R}$, so it cannot be separable.

[quote="jmerry"] Every element of the space can be expressed as the limit of a sequence in that set. [/quote]
I would really love to see your proof of that, jmerry. It is not obvious at all if the base of topology at a point is not countable...

My previous post assumed a metric topology.

Yeah, I thought so too. The problem is that $\mathbb R^{\mathbb R}$ with the product topology is not metrizable...

Yes, I think it's not even first-countable. :?

yes not being first countable is almost obvious you can take these points :all with finite zero and the other index is 1 then the point that has all zero in all index is limit point of these sequence but you can not find the sequense  such that its limit is all zero in all index so it is not first countable .

If the product topology is weak i.e the basis is of the form $\Pi{ \{U_{\alpha}\}}_{\alpha \in \mathbb{R}}$ where all but finite $U_{\alpha}$ are the whole $\mathbb{R}$ and the others are open sets in $\mathbb{R}$, then I think we can show that the space is seperable as follow :
 Let $I$ be the set of all horizontal intervals in $\mathbb{R}^2$ whose endpoints have rational coordinates. This set is countable. Let $Q$ be the set of function $f$ ( and there fore $Q$ is a subset of $\mathbb{R}^{\mathbb{R}}$) whose graph deleted the points $(x,f(x)), f(x)=0$ is consisted of finite number of intervals in $I$. $Q$ is also countable. We will prove that $Q$ is densed in $\mathbb{R}^{\mathbb{R}}$.
 Now let $\Pi {\{{U_{\alpha}\}}_{\alpha \in \mathbb{R}}}$ be an open set in the space as above.
Call those open sets $U_{\alpha}$ which are not the whole $\mathbb{R}$, $U_1,\ldots ,U_n$ and their indices are $\alpha_1 < \ldots <\alpha_n$.
Let $p_1,\ldots ,p_n$ be rational points in $U_1,\ldots ,U_n$ respectively. Choose rationals $s_i,t_i$ such that $s_1 < \alpha_1 < t_1 < \alpha_2 < \ldots < \alpha_n < t_n$
Consider $f \in Q$ the function that $f(s_i,t_i) = p_i$ and $0$ otherwise. It is easy to see that $f \in \Pi {\{{U_{\alpha}\}}_{\alpha \in \mathbb{R}}}$ because $f(\alpha_i) = p _i\in U_i$ and for other values $\alpha$ we have $ f(\alpha) \in U_{\alpha}$ because $U_{\alpha} = \mathbb{R}$

That looks like it works. This is still a monster.

The theorem proved here: if $A$ is separable and $B$ has a Hausdorff topology with a countable base, $A^B$ is separable in the product topology. This probably restricts the cardinality of $B$ to no more than that of the real numbers; I don't believe there are any separable (and Hausdorff) topological spaces larger in cardinality than $\mathbb{R}^\mathbb{R}.$

[quote="jmerry"]I don't believe there are any separable (and Hausdorff) topological spaces larger in cardinality than $\mathbb{R}^\mathbb{R}.$[/quote]
That's right and easy to prove. Let $X$ be a Hausdorff space and let $D$ be a countable dense subset of $X$. For each point $x\in X$ consider the set of all subsets of $D$ of the form $U\cap D$ where $U$ is a neighborhood of $x$ in the topology of $X$. It is easy to check that the corresponding families of subsets are different for different $x$. Thus the cardinality of $X$ does not exceed that of $P(P(D))$, which is the cardinality of $\mathbb R^{\mathbb R}$.


is anyone else taking idea math? which class?
I'm taking "Math Circles"


hi all, pls help mi with this... :)

why does the cis isomer has lower melting point but higher boiling point, solubility and dipole moment than the trans isomer?

thx lots

Welcome to this forum.

The molecules of a cis isomer, as they are less simetric relative to the molecules of the trans isomer, don't pack so readily in an ordered crystaline structure, and so their melting points are, in general, higher.

In the cis isomer, the dipole moments of the two cis bonds add, whereas in trans isomer they cancel each other (this is specially true when the groups are the same). So, cis isomer has a greater dipole moment, i.e., is more polar. This higher polarity os cis isomer molecules explain why they have higher boiling points (stronger intermolecular forces) and better solubilities (stronger interactions with solvent).


find all [b]x,y[/b] and [b]z[/b] in positive integer:
[b]1!+2!+...+x!=y z. [/b]

I think you mean $z \geq 2$ otherwise there are many solutions  :D
Assume that $ z \geq 2$ and $x >1$ (we clearly have that $(1,1,z)$ is a solution).
Let $S(x)= 1! + \cdots + x!$

- If $z = 2$ : Then for $i \geq 5$, we have $i! = 0 \mod [10]$. Thus, if $x \geq 4$, we have $S(x) = S(4) = 3 \mod [10]$. But no square has unit digit $3$, so there is no solution in that case.
Direct checking leads to the only solution $(3,3,2)$.

- If $z \geq 3$ : As above, we see that for $x \geq 2$ we have $S(x) = 0 \mod [3]$, so that $3$ has to divide $y$. But, since $ z \geq 3$, it follows that $S(x) = 0 \mod [27]$. It is easy to verify that this cannot happen when $x \geq 8$.
Direct checking shows that there is no solution for $x \leq 7$, and we are done.

Therefore, the solutions are $(x,y,z) = (1,1,z)$ and $(3,3,2)$.

Pierre.


geometry
inequalities
trigonometry
inequalities unsolved

[quote="Sung-yoon Kim"][quote="arqady"]Prove that for all acute triangle holds
$\sqrt3R\geq\frac{a^{2}+b^{2}+c^{2}}{a+b+c}.$[/quote]

I think it is not true.. Try $A \rightarrow \frac{\pi}{2}$ and $B=C$.[/quote]
If $a=\sqrt2$ and $b=c=1$ then $R=\frac{1}{\sqrt2}$ and
$\sqrt3R\geq\frac{a^{2}+b^{2}+c^{2}}{a+b+c}\Leftrightarrow\sqrt{\frac{3}{2}}\geq\frac{4}{2+\sqrt2},$ which true.

What about $A \to 0$, $B=C$? Then $b=c$ and $a \to 0$, but $R \to \frac{b}{2}$, right? (I suck at geometry so I hope I haven't misunderstood notation.) Then your expression approaches $\frac{\sqrt{3}}{2}> 1$, which false.

[color=darkblue][b]Remark.[/b] Suppose the isosceles triangle $ABC$ ($b=c$). If $a\searrow 0$, then $b=c\nearrow 2R$.
Thus, $\frac{a^{2}+b^{2}+c^{2}}{a+b+c}=\frac{a^{2}+2b^{2}}{a+2b}\rightarrow b\rightarrow 2R>R\sqrt 3\ !$ For example, the triangle $(1,4,4)\ .$

[b]Proposed problem.[/b] Let $ABC$ be a triangle for which there is $0<2r<x$ so that $a=x-r$, $b=x$, $c=x+r\ .$ Then exists the relation $\frac{a^{2}+b^{2}+c^{2}}{a+b+c}\le R\sqrt 3\ .$
[b]Proof.[/b] $p=\frac{3x}{2}\ ,\ p-a=\frac{x}{2}+r\ ,\ p-b=\frac{x}{2}\ ,\ p-c=\frac{x}{2}-r\ .$ Thus, $S\equiv [ABC]=\frac{x}{4}\sqrt{3(x^{2}-4r^{2})}$ and $R\sqrt 3=\frac{x^{2}-r^{2}}{\sqrt{x^{2}-4r^{2}}}\ .$ Therefore,
$(1)\ \ \frac{a^{2}+b^{2}+c^{2}}{a+b+c}\le R\sqrt 3\Longrightarrow \frac{3x^{2}+2r^{2}}{3x}\le \frac{x^{2}-r^{2}}{\sqrt{x^{2}-4r^{2}}}$  $\Longleftrightarrow$ $(3x^{2}+2r^{2})^{2}(x^{2}-4r^{2})\le 9x^{2}(x^{2}-r^{2})^{2}\ .$ Denote $\frac{x^{2}}{r^{2}}=t>4\ .$
The last inequality becomes :
$(3t+2)^{2}(t-4)\le 9t(t-1)^{2}\Longleftrightarrow$ $6t^{2}+53t+16\ge 0$, what is truly.
[b]Remark.[/b] If $2r<x<4r$, then $C>90^{\circ}$ (obtuse $\triangle$ : 2,3,4) ; if $x=4r$, then $C=90^{\circ}$ (right $\triangle$ : 3,4,5) ; if $x>4r$, then $C<90^{\circ}$ (acute $\triangle$ : 4,5,6 ).

[b]Another counter example :[/b] $(1,3,\sqrt{10})$ - right triangle.

[b]Question.[/b] Exist $0<\epsilon ,\delta <20^{\circ}$ so that the obtuse triangle $(\cos \epsilon\ ,\ \cos (\epsilon+\delta )\ ,\ \sin \delta )$ with $R=1$ doesn't verify the relation $(1)$ ?? Indeed, for $0<\epsilon ,\delta <20^{\circ}$ we have :
$\cos \epsilon >\cos (\epsilon+\delta)\ ,\ \cos \epsilon >\sin\delta\ ;$ $\cos\epsilon <\cos (\epsilon+\delta)+\sin\delta\ ;$ $\cos^{2}\epsilon>\cos^{2}(\epsilon+\delta)+\sin^{2}\delta\ .$  [/color]

ARQADY, see the my last message : I added a question with which I tried to find usually a counter exemple - obtuse triangle !


inequalities
rearrangement inequality
inequalities proposed

[quote="Parkdoosung"]Let $a+b+c+d=4$ and $a,b,c,d\geq0$.
Prove that $a^{2}bc+b^{2}cd+c^{2}da+d^{2}ab\leq4$[/quote]
It is necessary to prove, that $(a+b+c+d)^{4}\geq64(a^{2}bc+b^{2}cd+c^{2}da+d^{2}ab).$
If $a=\min\{a,b,c,d\}$ and $b=a+x,$ $c=a+y,$ $d=a+z$ then it is killing. But it is very ugly.

Are there other inequalities where we can apply this :?:

[quote="Sung-yoon Kim"] I found that $a^{2}bcd+b^{2}cde+c^{2}dea+d^{2}eab+e^{2}abc \leq 5(\frac{a+b+c+d+e}{5})^{5}$ can be solved by the same idea. [/quote]
I see only that if $a+b+c+d+e=5$ and $a,b,c,d,e$ are non-negative then $a^{2}bcd+b^{2}cde+c^{2}dea+d^{2}eab+e^{2}abc+3abcde\leq8.$

[quote="Sung-yoon Kim"]I found that $a^{2}bcd+b^{2}cde+c^{2}dea+d^{2}eab+e^{2}abc \leq 5(\frac{a+b+c+d+e}{5})^{5}$ can be solved by the same idea.[/quote]
Now I see it!  :lol:

[quote="Sung-yoon Kim"][quote="Parkdoosung"]Let $a+b+c+d=4$ and $a,b,c,d\geq0$.
Prove that $a^{2}bc+b^{2}cd+c^{2}da+d^{2}ab\leq4$[/quote]

Let ${p,q,r,s}={a,b,c,d}$ and $p \geq q \geq r \geq s$. Then by rearrangement inequality,
$a^{2}bc+b^{2}cd+c^{2}da+d^{2}ab=a(abc)+b(bcd)+c(cda)+d(dab)$
$\leq p(pqr)+q(pqs)+r(prs)+s(qrs)=(pq+rs)(pr+qs)$
$\leq (\frac{pq+rs+pr+qs}{2})^{2}=\frac{1}{4}((p+s)(q+r))^{2}$
$\leq \frac{1}{4}((\frac{p+q+r+s}{2})^{2})^{2}$
$=4$.

Equality holds iff q=r=1 and p+s=2. So equality holds if two of them are equal to 1. Applying this, we can get the equality conditions $(a,b,c,d)=(1,1,1,1),(2,1,1,0)$ or any cyclic forms.

Nice problem![/quote]

I think this ineq is cyclic, it isn't symmetric sothat we can't assume that $a\ge b\ge c\ge d$.  :maybe:

[quote="M4RI0"]We are not assuming that  :wink: .[/quote]
Oh, how stupid I am. I read too carelessly. Thank you very much

[quote="Sung-yoon Kim"]And by this idea, we can solve that
$a^{2}b+b^{2}c+c^{2}a \leq \frac{4}{27}$ if $a+b+c=1$
which was from Canada.

If $\{ p,q,r \}= \{ a,b,c \}$, $p \geq q \geq r$, then as $pq \geq pr \geq qr$,
$a^{2}b+b^{2}c+c^{2}a =a(ab)+b(bc)+c(ca) \leq p(pq)+q(pr)+r(qr)$
$=q(p^{2}+pr+r^{2}) \leq q(p+r)^{2}= \frac{1}{2}(2q)(p+r)(p+r)$
$\leq \frac{1}{2}(\frac{(2q)+(p+r)+(p+r)}{3})^{3}$
$=\frac{1}{2}(\frac{2}{3})^{3}=\frac{4}{27}$.[/quote]

Edited: Sorry I don't really understand the solution but now I do.

Hmm.. that's different with assuming without loss of generality

Hahaha I think I am influenced by the post that tells it is cyclic and can't think clearly  :D

[quote="BoesFX"]
But, this is the incomplete solution.[/quote]
Why? Thank you.

[quote="Sung-yoon Kim"]And by this idea, we can solve that
$a^{2}b+b^{2}c+c^{2}a \leq \frac{4}{27}$ if $a+b+c=1$
which was from Canada.

If $\{ p,q,r \}= \{ a,b,c \}$, $p \geq q \geq r$, then as $pq \geq pr \geq qr$,
$a^{2}b+b^{2}c+c^{2}a =a(ab)+b(bc)+c(ca) \leq p(pq)+q(pr)+r(qr)$
$=q(p^{2}+pr+r^{2}) \leq q(p+r)^{2}= \frac{1}{2}(2q)(p+r)(p+r)$
$\leq \frac{1}{2}(\frac{(2q)+(p+r)+(p+r)}{3})^{3}$
$=\frac{1}{2}(\frac{2}{3})^{3}=\frac{4}{27}$.[/quote]
Why can we assume $p\geq q\geq r$?

Is $p,q,r$ a permutation of $a,b,c$?

[quote="hitek"]
Why can we assume $p\geq q\geq r$?

Is $p,q,r$ a permutation of $a,b,c$?[/quote]
Yes!
[quote="Sung-yoon Kim"]

If $\{ p,q,r \}= \{ a,b,c \}$, $p \geq q \geq r$, then...[/quote] :wink:

Suppose that  $ (x;y;z;t)$ is a permutation of  $ a;b;c;d$ such that $ x\geq y\geq z\geq t$
=>$ xyz\geq xyt\geq xzt\geq yzt$
and $ x.xyz \plus{} y.xyt \plus{} z.xzt \plus{} t.yzt\geq a^{2}bc \plus{} b^{2}cd \plus{} c^{2}da \plus{} d^{2}ab$
Then
$ x.xyz \plus{} y.xyt \plus{} z.xzt \plus{} t.yzt \equal{} (xy \plus{} zt)(xz \plus{} yt)\leq\frac {1}{4}(xz \plus{} xy \plus{} yt \plus{} zt)^{2}\leq 4$
Done!

I mean, there's an identical solution above, chien than.

And we prove by the same method:

$ \sum \frac {a_k}{k^2} \geq \sum \frac {1}{k}$

For $ a_k$ positive integers and distinct (ai#aj for all i,j)
Wich is from IMO 78
 :)  :wink:

[quote="Sung-yoon Kim"]And by this idea, we can solve that
$ a^{2}b \plus{} b^{2}c \plus{} c^{2}a \leq \frac {4}{27}$ if $ a \plus{} b \plus{} c \equal{} 1$
which was from Canada.

If $ \{ p,q,r \} \equal{} \{ a,b,c \}$, $ p \geq q \geq r$, then as $ pq \geq pr \geq qr$,
$ a^{2}b \plus{} b^{2}c \plus{} c^{2}a \equal{} a(ab) \plus{} b(bc) \plus{} c(ca) \leq p(pq) \plus{} q(pr) \plus{} r(qr)$
$ \equal{} q(p^{2} \plus{} pr \plus{} r^{2}) \leq q(p \plus{} r)^{2} \equal{} \frac {1}{2}(2q)(p \plus{} r)(p \plus{} r)$
$ \leq \frac {1}{2}(\frac {(2q) \plus{} (p \plus{} r) \plus{} (p \plus{} r)}{3})^{3}$
$ \equal{} \frac {1}{2}(\frac {2}{3})^{3} \equal{} \frac {4}{27}$.[/quote]

$ a,b,c$ must me positive, I think

[quote="Sung-yoon Kim"][quote="Parkdoosung"]Let $ a \plus{} b \plus{} c \plus{} d \equal{} 4$ and $ a,b,c,d\geq0$.
Prove that $ a^{2}bc \plus{} b^{2}cd \plus{} c^{2}da \plus{} d^{2}ab\leq4$[/quote]

Let $ {p,q,r,s} \equal{} {a,b,c,d}$ and $ p \geq q \geq r \geq s$. Then by rearrangement inequality,
$ a^{2}bc \plus{} b^{2}cd \plus{} c^{2}da \plus{} d^{2}ab \equal{} a(abc) \plus{} b(bcd) \plus{} c(cda) \plus{} d(dab)$
$ \leq p(pqr) \plus{} q(pqs) \plus{} r(prs) \plus{} s(qrs) \equal{} (pq \plus{} rs)(pr \plus{} qs)$
$ \leq (\frac {pq \plus{} rs \plus{} pr \plus{} qs}{2})^{2} \equal{} \frac {1}{4}((p \plus{} s)(q \plus{} r))^{2}$
$ \leq \frac {1}{4}((\frac {p \plus{} q \plus{} r \plus{} s}{2})^{2})^{2}$
$ \equal{} 4$.

Equality holds iff q=r=1 and p+s=2. So equality holds if two of them are equal to 1. Applying this, we can get the equality conditions $ (a,b,c,d) \equal{} (1,1,1,1),(2,1,1,0)$ or any cyclic forms.

Nice problem![/quote]

I don't see the point for introducing $ p, q, r, s$ to replace $ a, b, c, d$.
Also, how do one prove that

$ a(abc) \plus{} b(bcd) \plus{} c(cda) \plus{} d(dab)$
$ \leq p(pqr) \plus{} q(pqs) \plus{} r(prs) \plus{} s(qrs)$?

[quote="Sung-yoon Kim"][quote="Parkdoosung"]Let $ a \plus{} b \plus{} c \plus{} d \equal{} 4$ and $ a,b,c,d\geq0$.
Prove that $ a^{2}bc \plus{} b^{2}cd \plus{} c^{2}da \plus{} d^{2}ab\leq4$[/quote]

[b]Let $ {p,q,r,s} \equal{} {a,b,c,d}$ and $ p \geq q \geq r \geq s$.[/b] Then by rearrangement inequality,
$ a^{2}bc \plus{} b^{2}cd \plus{} c^{2}da \plus{} d^{2}ab \equal{} a(abc) \plus{} b(bcd) \plus{} c(cda) \plus{} d(dab)$
$ \leq p(pqr) \plus{} q(pqs) \plus{} r(prs) \plus{} s(qrs) \equal{} (pq \plus{} rs)(pr \plus{} qs)$
$ \leq (\frac {pq \plus{} rs \plus{} pr \plus{} qs}{2})^{2} \equal{} \frac {1}{4}((p \plus{} s)(q \plus{} r))^{2}$
$ \leq \frac {1}{4}((\frac {p \plus{} q \plus{} r \plus{} s}{2})^{2})^{2}$
$ \equal{} 4$.

Equality holds iff q=r=1 and p+s=2. So equality holds if two of them are equal to 1. Applying this, we can get the equality conditions $ (a,b,c,d) \equal{} (1,1,1,1),(2,1,1,0)$ or any cyclic forms.

Nice problem![/quote]
I don't know why we can assume this.

[quote="Sung-yoon Kim"]Well, I found that $a^{2}bcd+b^{2}cde+c^{2}dea+d^{2}eab+e^{2}abc \leq 5(\frac{a+b+c+d+e}{5})^{5}$ can be solved by the same idea. I tried to generalize for the number of variables, but failed. But I got that if $2n$ satisfies, then $2n+1$ satisfies, so the remaining cases are even numbers bigger than or equal to 6.[/quote]
Let $\{a,b,c,d,e\}=\{p,q,r,s,t\}$.  WLOG $p\le q\le r\le s\le t$ and similarly we have $pqrs\le prst\le pqst\le pqrt\le qrst$.  Therefore, by rearrangement:

 \begin{eqnarray} &&\text{LHS}=a(abcd)+b(bcde)+c(cdea)+d(deab) \\
&\le& p(pqrs)+q(prst)+r(pqst)+s(pqrt)+t(qrst) \\
&=& qrs(p^2+t^2+3pt)=qrs[(p+t)^2+pt] \\
&\le& \frac{(2q)(2r)(2s)(p+t)(p+t)}{8}+pqrst \\
 &\le& \frac18(\frac{2q+2r+2s+p+t+p+t}{5})^5+(\frac{p+q+r+s+t}{5})^5 \\
&=& 5(\frac{a+b+c+d+e}{5})^5 \end{eqnarray}

[quote="Binomial-theorem"]Let $\{a,b,c,d,e\}=\{p,q,r,s,t\}$.  WLOG $p\le q\le r\le s\le t$ and similarly we have $pqrs\le prst\le pqst\le pqrt\le qrst$. [...][/quote]
It seems to be false  :maybe:

[quote="shuchang"][quote="Sung-yoon Kim"]And by this idea, we can solve that
$a^{2}b+b^{2}c+c^{2}a \leq \frac{4}{27}$ if $a+b+c=1$
which was from Canada.

If $\{ p,q,r \}= \{ a,b,c \}$, $p \geq q \geq r$, then as $pq \geq pr \geq qr$,
$a^{2}b+b^{2}c+c^{2}a =a(ab)+b(bc)+c(ca) \leq p(pq)+q(pr)+r(qr)$
$=q(p^{2}+pr+r^{2}) \leq q(p+r)^{2}= \frac{1}{2}(2q)(p+r)(p+r)$
$\leq \frac{1}{2}(\frac{(2q)+(p+r)+(p+r)}{3})^{3}$
$=\frac{1}{2}(\frac{2}{3})^{3}=\frac{4}{27}$.[/quote]

wo also have$a^{2}b+b^{2}c+c^{2}a+abc \leq \frac{4}{27}$ if$a+b+c=1$[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=534056&p=3541033#p3541033
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=768&sid=c3c5b1d6f4ddcaf0a1d3e963d9763e32#p768
https://artofproblemsolving.com/community/c6h1136308p5308238

[quote=Sung-yoon Kim][quote="Parkdoosung"]Let $a+b+c+d=4$ and $a,b,c,d\geq0$.
Prove that $a^{2}bc+b^{2}cd+c^{2}da+d^{2}ab\leq4$[/quote]

Let ${p,q,r,s}={a,b,c,d}$ and $p \geq q \geq r \geq s$. Then by rearrangement inequality,
$a^{2}bc+b^{2}cd+c^{2}da+d^{2}ab=a(abc)+b(bcd)+c(cda)+d(dab)$
$\leq p(pqr)+q(pqs)+r(prs)+s(qrs)=(pq+rs)(pr+qs)$
$\leq (\frac{pq+rs+pr+qs}{2})^{2}=\frac{1}{4}((p+s)(q+r))^{2}$
$\leq \frac{1}{4}((\frac{p+q+r+s}{2})^{2})^{2}$
$=4$.

Equality holds iff q=r=1 and p+s=2. So equality holds if two of them are equal to 1. Applying this, we can get the equality conditions $(a,b,c,d)=(1,1,1,1),(2,1,1,0)$ or any cyclic forms.

Nice problem![/quote]
classical and nice

[quote=Binomial-theorem]
Let $\{a,b,c,d,e\}=\{p,q,r,s,t\}$.  WLOG $p\le q\le r\le s\le t$ and similarly we have $pqrs\le prst\le pqst\le pqrt\le qrst$. [/quote]

This is false.  The inequality chain should be $pqrs\le pqrt\le pqst\le prst\le qrst$.

[quote=Parkdoosung]Let $a+b+c+d=4$ and $a,b,c,d\geq0$.
Prove that $a^{2}bc+b^{2}cd+c^{2}da+d^{2}ab\leq4$[/quote]

a^2bc<((2a+b+c)/4)^4=((S+x)/4)^4=f(x)
S=a+b+c+d
X=a-d  Y=b-a  Z=c-b  T=d-c
(f(x)+f(y)+f(z)+f(t))/4<f((x+y+z+t)/4)=((a+b+c+d)/4)^4
solved with yensen inequality


Anyone have any recommendations on which AP exams are better to take than if one would take the course at a community college, assuming both are accepted for credit? I would prefer to take everything I'm taking in highschool AP, but I'm not sure how easy it would be to self-study, say, Psychology or APUSH.

to rephrase, can anyone recommend good self-study APs (I already took Stats this year, that seems to be the one everyone thinks is a joke and should only be taken self-study :P)? I'm concerned that I'd be missing something if I didn't take certain courses with a group (Lit, Psychology, things like that) - should that be a big concern?

also, if there are any college people who can recommend some good generic/transferable college courses to take that aren't APs (speech communication? Intro to Philosopy? Technical Writing?) I'd appreciate it :)

Also, if anyone has any comments about transferability of APs and/or college credits taken as a highschooler, that'd be great too :) thanks in advance.

Any AP can be self-studied and any self-study will give you more than a high school class typically will.  And faster.

If you know how to write you'll be fine.  If you can grade your own papers, do so.  Or, if you have any teacher-buddies, ask one of them.

This is a very complicated subject. First of all, because different schools accept different AP exams, and do so in different ways from each other. And a lot of top Universities will accept AP credits for certain courses, but won't accept credit for those same courses taken at other Universities (without taking the AP exam).

[Incidentally, the AP course system was designed to keep students from going to college or taking college courses too early. In the olden days it was very typcal for bright students to graduate from high school really early and go to college].

But nowadays not accepting credit from other Universities is for both philosophical and practical (i.e. money) reasons. Philosophically, it's because if they give you a degree, they expect the majority of the classes taken for that degree to be from their University. But mostly, it's because they want to get as much money from you as possible, and they have a lucrative system established with the College Board who administers the exams.

If you really want to take a class at another Univeristy and get college credit for it at another University, you could always take the corresponding AP exam (assuming there is one, and assuming the exam is accepted at the other University).

All that being said, there's nothing really advantageous in my opinion about accumulating credits to graduate early from college (except for money conciderations). College is a great place to learn as much as possible - it's not something to just get through for the sake of getting through. Even if you plan on going on to graduate school, rushing through college isn't necessarily going to put you in the best position to be successful there. The gap between high school and graduate school is very large, and you need as much knowledge as you can get before going there. Most of the people that I known who've been successful in grad school have had at least an equivalent of 4 years of undergrad. (some maybe took classes in the summer to graduate early).

That being said, AP's can be very good for fulfilling prerequisites once you get to college, and increasingly that's how top Universities are using them - sort of as placement exams.

To be quite frank I don't take AP's just for the hope of getting a college credit.  I personally do it for the sake of learning.  My view might change in college about whether or not I should get a college credit or not but right now it does not influence my schedule.

[quote="applepuppy"]:/ I guess I forgot this when I was making my post - essays: am I missing out that much if I have no one to grade my essays and I have to grade them myselves? or are the college board's practice free response guidlines sufficient? I think that was my main concern, especially for the AP englang and lit tests.[/quote]

I'm pretty sure that most teachers don't grade their students' essays according to the College Board guidelines - at least, in most classes (English may be an exception.)  My biology teacher gave us some free-response questions to do, and then gave us the answer key, but she didn't take the time to grade them herself.  So there was no advantage of taking a class (as opposed to self-studying) in that respect.

o.O this is all very intimidating.

[quote="gauss202"]But nowadays not accepting credit from other Universities is for both philosophical and practical (i.e. money) reasons. Philosophically, it's because if they give you a degree, they expect the majority of the classes taken for that degree to be from their University. But mostly, it's because they want to get as much money from you as possible, and they have a lucrative system established with the College Board who administers the exams.[/quote]

i see... I think i'm just really confused now o.O maybe instead of University, I should've said "Community College", would that change anything? the couple of colleges I looked at accept AP and College Credits and offer sophomore standing if you want it and have enough credits, is there some general category for places that won't do that? (Ivy league, tiny liberal arts, etc. - you mentioned "top" universities (i assumed you meant undergraduate), is that by prestige or quality?)

[quote="gauss202"]All that being said, there's nothing really advantageous in my opinion about accumulating credits to graduate early from college (except for money conciderations). College is a great place to learn as much as possible - it's not something to just get through for the sake of getting through. Even if you plan on going on to graduate school, rushing through college isn't necessarily going to put you in the best position to be successful there. The gap between high school and graduate school is very large, and you need as much knowledge as you can get before going there. Most of the people that I known who've been successful in grad school have had at least an equivalent of 4 years of undergrad. (some maybe took classes in the summer to graduate early). [/quote]

okay, so what you're saying is there's nothing advantageous to graduating early, in your opinion, right? I can't believe that there's nothing advantageous about taking college courses as opposed to taking the same course at a public high school...

[quote="mathfanatic"]I'm pretty sure that most teachers don't grade their students' essays according to the College Board guidelines - at least, in most classes (English may be an exception.) My biology teacher gave us some free-response questions to do, and then gave us the answer key, but she didn't take the time to grade them herself. So there was no advantage of taking a class (as opposed to self-studying) in that respect.[/quote]

that's comforting... I think... (wait wait, she gave YOU the answer keys to grade YOURSELF?? what did she do exactly in the way of teaching?? that [i]is[/i] comforting then...)

thanks all, I appreciate the comments... I don't really know any, uh, AOPS-like people that I can talk to in person, so I think this helps me a lot.

[quote="applepuppy"]
okay, so what you're saying is there's nothing advantageous to graduating early, in your opinion, right? I can't believe that there's nothing advantageous about taking college courses as opposed to taking the same course at a public high school... .[/quote]

O think what he meant was that there are advantages of course but the disadvantages that can affect you in the long run level the advatages out.  I love learning so I think that I have no need to graduate from it early(I'm just posting my feelings).

[quote="joml88"]
O think what he meant was that there are advantages of course but the disadvantages that can affect you in the long run level the advatages out.  I love learning so I think that I have no need to graduate from it early(I'm just posting my feelings).[/quote]

advantages to taking courses... check....

disadvantages in the long run? huh? you mean, by skipping the freshman year? I really don't understand...  especially if that's what one wants to do...

I like learning too :P i just think I'd like working and accomplishing things better than staying in an undergraduate school an extra year.

another question; how do you, (personally, not, "how does one...") study for the APs? not review, what do you use to teach yourself? will any used college textbook work?

Sorry, my post before was kind of vague. I was trying to address too many questions at once. Here, let me break it down a little bit.

- Studying for and taking AP Exams: A plus. It's always good to learn new things, and passing lots of AP exams can never hurt. I was just saying that for people who are taking numbers of AP exams in the "single digits" it's okay. You're still be in good shape for college.

- Taking college courses while in high school: A plus. Again, it never hurts to learn new things. If you take a class at a local college for which there's a corresponding AP exam (e.g. Chemistry or Calculus), you should probably take the AP exam at the end of the year.

- Taking lots of AP exams to try to graduate from college early: This is a "so-so" to me. The main advantages are for people planning on going on to professional schools like Med School, Law School, or Business School. I think you can get a solid enough foundation after 3 years for most of those programs, and the money you save on u-grad you can use for professional school. The other reason it could be an advantage is to students who want to travel abroad for a year or a semester. Most of the students that I know who had enough AP credits for a year of exemption used it for this purpose.

- Disadvantages to graduating early from college are: for one, you're missing out on a great time in your life that you won't be able to get back. The purpose of college is more than just getting the degree - it should be a learning and personal growth experience. Now I'm not saying you should spend like 6 years in college - but 3 might not be enough for most people. One option for people who have completed all of their u-grad requirements at some schools is to use their 4th year to work on a Masters Degree. Some really good schools like the U. of Chicago offer a 4 year Masters Degree program for students with enough advanced standing in their subject area.

- Taking and AP class in high school vs. Studying on your own vs. Taking the same class at a local college: It all depends. It depends on  the quality of the high school teacher, the quality of the class, how many other obligations you have, and the quality of the teacher and class at the college. In general, I think it's always better to have at least a mentor (if not a teacher) to help you learn a subject. You might be able to study a subject well enough to pass the AP Exam, but that doesn't really mean you understand all of the intricacies of the subject. There's no a priori advantage to one option over the other though. The key is just to learn a subject as well as you can. I knew plenty of people who had 5's on the BC exam who got smoked in college math and science classes because they didn't really have a good [i]understanding[/i] of calculus.

- So my last point, passing lots of AP exams vs. learning a few subject well: needless to say, learn the few subjects well.

Thank you gauss202... I understand everything now :P

[quote="gauss202"]
- Disadvantages to graduating early from college are: for one, you're missing out on a great time in your life that you won't be able to get back. The purpose of college is more than just getting the degree - it should be a learning and personal growth experience. Now I'm not saying you should spend like 6 years in college - but 3 might not be enough for most people. One option for people who have completed all of their u-grad requirements at some schools is to use their 4th year to work on a Masters Degree. Some really good schools like the U. of Chicago offer a 4 year Masters Degree program for students with enough advanced standing in their subject area.
[/quote]

I think that's what I'm planning to do, get my Master's in the same time it takes everyone else to get a Bachelor's, or at least take one year less time.

[quote="gauss202"]
- Taking and AP class in high school vs. Studying on your own vs. Taking the same class at a local college: It all depends. It depends on  the quality of the high school teacher, the quality of the class, how many other obligations you have, and the quality of the teacher and class at the college. In general, I think it's always better to have at least a mentor (if not a teacher) to help you learn a subject. You might be able to study a subject well enough to pass the AP Exam, but that doesn't really mean you understand all of the intricacies of the subject. There's no a priori advantage to one option over the other though. The key is just to learn a subject as well as you can. I knew plenty of people who had 5's on the BC exam who got smoked in college math and science classes because they didn't really have a good [i]understanding[/i] of calculus.
[/quote]

I see exactly what you mean... I'll definitely remember that and everything else, thanks a lot, this has been a big help :)


Anyone applying to SUMaC from AoPS this year? :D

SUMaC? never hear dof  it. please explain. I might be interested in the future.

actually, I have been accepted to the SUI... run by the same ppl, but different dates. I am consider not going if i get into HCCSiM or MathCamp. BUT after searching for SUMaC, it sounds okay.

Of the 35 people who attended SUMaC 2001 (I was among them), 7 made USAMO 2002, so the students who attend are pretty good.

I have heard good things about SUMaC from alumni on a
Brand X discussion forum. It's on our short list of summer programs for the future.

i applied.

i got waitlisted last year, but i really hope i make it.

I was head counselor at SUMAC one year [actually, the same year that Simon attended].

I think it should be pointed out that there are a number of strong students who attend SUMAC, but also a wide range of other skill levels as well.  Also, even though there are some rather strong students, the material being covered is not particularly formidable [compared to some other camps I have worked at], so you should not feel intimidated by the earlier remark.

[quote="chenwb"]Another thing--the SUMaC website said that it was especially considerate of those students who did not have many opportunities at their school. This definitely does not describe me. Is it still worth applying? It sounds like a great program  :)[/quote]

That all depends on how much time you have on your hands.  I would say that, when in doubt, apply.

I applied. However, my transcript will probably be mailed next week (since my school is on spring break). Is this ok?

email the director. addy is on the website.

chenwb, there were two students from Stuyvesant in my year. Their opportunities would have been roughly equal to yours, and they got in, so it's definitely possible.

Hi Rudel. It's nice to see you on this forum.

does anyone have any idea of how many people admitted in previous years solved all the problems.  they weren't very  hard.

See you all there this summer.  Just got the email today.

I guess I'm going too.

What program is everyone in?

I'm doing Program II apparently.

Lu


I created a new environment called 'example' that I will use in my documents to give an example.  However, I don't know how to insert numbering - that is to say, I want the document to look like:

Example 1: blah blah

....

Example 2: blah blah

....

Example 3: blah blah blah

I suspect that this involves creating a counter of some sort, and then creating a reference to that counter and adding one after each example use.  However, I have no idea how to do this, and can't find it in lshort.pdf.

Alternatively, I could number the examples by section number, such as:

Example 1.2.1: blah blah

....

Example 4.2.3: blah blah

I'm sure there's an easy way to say 'insert the current section number here', but again, I don't know it.  Please help!

Thanks.

I'll answer my own question in case anyone else has a similar query.  Basically, use the \newcounter{counter-name} command to initialize a new counter (default value is 0).  Then, use the \stepcounter{counter-name} command to increase the counter by 1.  Finally, use a command like \arabic{counter-name} to display the counter's current value.  (There are other commands that \arabic{} that make the number look slightly different, but this is the most common one.)

Counters are useful in creating new environments or commands.  I used it to create an environment called [i]definition[/i] (among others), which took the format

[quote]Definition 1: median.

The median is the middle number of a set of numbers arranged in numerical order.  If the number of values in a set is even, then the median is the sum of the two middle values, divided by 2.[/quote]

This would require a fair bit of formatting, and I'd have to remember what definition number I was on.  Instead, I created an environment that did everything (including indexing) for me, so that the only input required was:

\begin{definition}{median}
The median is the middle number of a set of numbers arranged in numerical order.  If the number of values in a set is even, then the median is the sum of the two middle values, divided by 2.
\end{definition}

I have used this in an (incomplete) book-like thing on statistics, found on the LaTeX section of the AoPS website.  (http://www.artofproblemsolving.com/LaTeX/AoPS_L_PictExam.php)


1) Compute the following sums:
a)sum(k=0 to n) 1/[(n-k)!*(n+k)!]=?
b)sum (k=m to n) [r^k* A(m k) * C( k n)], 1<=m<=n

where r is from C and A(m k) = m!/(m-k)! and C(k n) = n!/[k!*(n-k)!]

bye

I got the first one. Let S be the sum. Then consider n!*S. If n is even and thus equal to 2k, we have (2k)!*S=(2k choose k)+(2k choose k+1)+...+(2k choose 2k)=2^(n-1)+1/2*(n choose n/2). If n is odd, we don't have the (2k choose k) term, so then n!*S=2^(n-1). Thus

S=

2^(n-1)/n! if n is odd
2^(n-1)/n!+(n choose n/2)/(2*n!) if n is even.


geometry
geometric transformation
rotation
complex numbers

hey evry1
i dont actually want a solution to this one but i just want to ask something
"Go from the pine-tree to the birch, turn to the right, and walk the same distance. Go from the pine-tree to the oak-tree, turn to the left, and walk the same distance. Dig up in the middle." You come to the island with this guidance from the pirate Rogers. You see the oak-tree and the birch but no pine-tree. Is it possible to find the treasure? "
if ne1 has seen this problem b4,can he xplain wot "Dig up in the middle"
means (basically my question is "in the middle of wot"? is it the line joining the pine and the oak or sumthing else?
cheers

My guess is that the problem shoud sound like this:

"Go from the pine-tree to the birch, turn to the right, and walk the same distance.
YOU GET POINT A.
Go from the pine-tree to the oak-tree, turn to the left, and walk the same distance.
YOU GET POINT B.
 Dig up in the middle   OF AB."
You come to the island with this guidance from the pirate Rogers. You see the oak-tree and the birch but no pine-tree. Is it possible to find the treasure? "

Pirates sail to a deserted island to bury their treasure, and they decide to place it in a spot that they can find later without using a map.  They mark three trees A,B, and C, and then from B they pace off the distance to A, turn left 90 degrees and pace off the same distance, at which point they place a stake.  Then they go back to B and pace off the distance to C, turn
right 90 degrees, and place another stake at this location.  They
bury the treasure mid-way between the two stakes.  Then they remove
the stakes and sail away.  Unfortunately, when they return to dig up
the treasure they discover that a storm has washes away the "B" tree.
The problem is to locate the treasure.

this is the corrected problem.and now since i've got the solution,this can be moved to the propsed prob section

This is the perfect example for the application of complex numbers in geometry. Let a, b, c be the complex nuymbers denoting pts A, B, C, d is the complex number denoting the first post and e for the second post. We can quickly see (rotation) that e=c+i(b-c) and d=a-i(b-a). The treasure is at the middle of DE and this middleis represented by (e+d)/2. We can see that e+d=a+c +i(a-c), so it does not depend on b, so we don't need point B.

It can also be done by considering rotations of angles 45, -45.


this inequaliti easy but interesting:
let a,b,c be positive numbers. prove that:
∑a/(9ab+(a+b+c)^2)≥ 1/(2(a+b+c)).

Sorry because I can't type the LATEX.PLEASE help me. :(

[quote="arqady"][quote="quykhtn-qa1"]this inequaliti easy but interesting:
let a,b,c be positive numbers. prove that:
∑a/(9ab+(a+b+c)^2)≥ 1/(2(a+b+c)).
  [/quote]
Let $ a,b,c$ be positive numbers. prove that:
$ \sum_{cyc}\frac {a}{9ab \plus{} (a \plus{} b \plus{} c)^2} \geq\frac {1}{2(a \plus{} b \plus{} c)}.$[/quote]
Thank you arqady!

[quote="arqady"]This is obviously true after expanding. :wink:[/quote].
YES.CERTAINLY. :)

[quote="nguoivn"]It's Vasc's one.
Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 3.$
Prove that:
$ \sum\ \frac {a}{ab \plus{} 1} \geq\ \frac {3}{2}$
The solution by Am-Gm is very well-known  :maybe:
For the stronger and also nice:
$ \sum\ \frac {a}{ab \plus{} 2} \geq\ 1$
(with the same condition)
We can prove it by pqr after expand.[/quote]
you can prove thís inequaliti 2 ? :)

Sorry, my friend. Indeed, ineq (2) is not trues (example a=2.2;b=0.8;c=0)
I haven't checked it carrefully (of course, if it's true, we can prove it easily by pqr but not so nice). Sorry again  :oops:


real analysis
topology
logarithms
real analysis unsolved

I assume the reader already knows something about the Cantor set. If you don't, look [url=http://en.wikipedia.org/wiki/Cantor_set]on Wikipedia[/url].

First, let me construct the set.

In the construction of the Cantor set, in each iteration we delete the middle third of every interval. Let $C_{n}$ denote the set we would obtain by doing the same construction, but deleting the middle $\frac{1}{n}$ of every interval on each iteration. (Thus $C_{3}$ is the Cantor set.)

Now consider
$\mathcal{C}= \bigcup_{n=2}^\infty C_{n}$

Here are some interesting things I thought might be worth thinking about:

* Is $\mathcal{C}$ closed? (I didn't try to show it but I think not)

* Is $\mathcal{C}$ everywhere dense? (I didn't try to show it but I think so)

* $\mathcal{C}$ has Lebesgue measure 0. However, the Hausdorff dimension of $C_{n}$ is
$\frac{\log(2)}{\log\left(\frac{2n}{n-1}\right)}$
so I think $\mathcal{C}$ has Hausdorff dimension 1. (If I'm wrong, please correct me and explain why!) So it's a relatively simple example of a set whose (Hausdorff) measure is 0 or $\infty$ in every dimension.
(See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=107277]here[/url] about Hausdorff measure.)

The complement of $C_{n}$ in $[0,1]$ does not contain any open intervals of length more than $1/n$. Therefore, the complement of $C$ does not contain nonempty open intervals. This answers the first two questions.
As for the third one, $A\subset B$ implies $\dim A\le \dim B$ by the definition of Hausdorff dimension. Therefore, $\dim C\ge \dim C_{n}$ for all $n$, hence $\dim C=1$.

Another way to modify the standard Cantor set is to remove segments of different relative length at each iteration. For example, take out the $1/(n+2)$ middle part on the $n$th step, $n\ge 1$. This gives another set of measure zero (why?) and dimension $1$, and this one is compact.


The following question sounbs familiar :

Consider the $n \times n$ plane grid formed with the points $(a,b)$ where $a,b$ are integer and $1 \leq a,b \leq n$.

What is the maximal number $M(n)$ of points you can select from this grid so that no three are collinear (in any direction) ?

Clearly, the pigeon-hole principle gives $M(n) \leq 2n$.

With some constructions, by hands, it might be $M(n) = 2n-1$ for $n \geq 5$.

Does anybody have some results about that?

The question arised in a non-math jouranl (namely, 'Le Monde') where it was asked only for $M(8)$ but the given construction of a set with $16$ points was wrong....

Pierre.

I answer to myself....

http://mathworld.wolfram.com/IntegerLattice.html

Pierre.


calculus
integration
function
calculus computations

Hi all,

Could anyone please post the integrals of the six basic trigonometric functions, as well as those of their inverses?

Thanks.

Come on! Just type "trigonometric integral" in Google or use Mathworld's Integrator. Well
$\int \text{sin }x \text{ d}x=-\text{ cos }x+C$
$\int \text{cos }x \text{ d}x=\text{ sin }x+C$
$\int \text{tan }x \text{ d}x=-\text{ ln }|\text{cos }x|+C$
$\int \text{cot }x \text{ d}x=\text{ ln }|\text{sin }x|+C$
$\int \text{sec }x \text{ d}x=\text{ ln }|\text{ sec }x+\text{ tan }x|+C$
$\int \text{csc }x \text{ d}x=\text{ ln }|\text{ csc }x-\text{ cot }x|+C$

$\int \text{arc sin }x \text{ d}x=x\text{ arc sin }x+\sqrt{1-x^{2}}+C$
$\int \text{arc cos }x \text{ d}x=x\text{ arc cos }x-\sqrt{1-x^{2}}+C$
$\int \text{arc tan }x \text{ d}x=x\text{ arc tan }x-1/2\text{ ln }(x^{2}+1)+C$
$\int \text{arc cot }x \text{ d}x=x\text{ arc cot }x+1/2\text{ ln }(x^{2}+1)+C$
$\int \text{arc sec }x \text{ d}x=x\text{ arc sec }x-\text{ ln }x\left(\sqrt{\frac{x^{2}-1}{x^{2}}}+1\right )+C$
$\int \text{arc csc }x \text{ d}x=x\text{ arc csc }x+\text{ln }\left (x\left (\frac{\sqrt{x^{2}-1}}{x^{2}}+1\right ) \right )+C$


$ sin^3x\equal{}\frac 14 (3sinx \minus{}sin 3x)$.
Therefore
\[ \sum_{n\equal{}1}^{N}3^nsin^3\frac{\pi}{3^n}\equal{}\frac 14 \sum_{n\equal{}1}^{N}3^{n\plus{}1}sin\frac{\pi}{3^n}\minus{}3^nsin\frac{\pi}{3^{n\minus{}1}}\equal{}\frac{3\pi}{4}\frac{3^N}{\pi}sin\frac{\pi}{3^N}.\]
Therefore $ \lim_{N\to \infty}S(N)\equal{}\frac{3\pi}{4}.$


What is the volume, in cubic inches, of a right circular cone of altitude 6 inches and a base diameter of 8 inches? Express your answer in terms of $ \pi$.

be careful, it's a cone, so you have to divide by 3...

so $ \frac{1}{3}\cdot\pi\cdot r^2\cdot h\equal{}\frac{1}{3}\cdot\pi\cdot 4^2\cdot 6\equal{}\boxed{32\pi}$

Yes, area of a cone:

1/3( ╥•radius^2•height)


x,y,z are positive real numbers whose sum is 1. Prove:
$ \sum_{cyc}{\frac{3x^{2}}{1+x^{2}}}\geq\sum_{cyc}{\frac{x}{1+x^{2}}}$

It can be solved with the rearrange inequality:
$ \sum_{cyc}\frac{x}{1+x^{2}}=\sum_{cyc}\frac{x^{2}+xy+xz}{1+x^{2}}$
So the initial inequiality is equivalent to:
$ \sum_{cyc}\frac{2x^{2}}{1+x^{2}}\geq \sum_{cyc}\frac{xy+xz}{1+x^{2}}$
If $ 1 \geq x \geq y \geq z \geq 0$
we have $ \frac{x}{1+x^{2}}\geq \frac{y}{1+y^{2}}\geq \frac{z}{1+z^{2}}\geq 0$
So, by rearrange:
$ x\left(\frac{x}{1+x^{2}}\right)+x\left(\frac{x}{1+x^{2}}\right)+y\left(\frac{y}{1+y^{2}}\right)+y\left(\frac{y}{1+y^{2}}\right)+z\left(\frac{z}{1+z^{2}}\right)+z\left(\frac{z}{1+z^{2}}\right)$
Is bigger than:
$ y\left(\frac{x}{1+x^{2}}\right)+z\left(\frac{x}{1+x^{2}}\right)+x\left(\frac{y}{1+y^{2}}\right)+z\left(\frac{y}{1+y^{2}}\right)+x\left(\frac{z}{1+z^{2}}\right)+y\left(\frac{z}{1+z^{2}}\right)$
QED


Mortimer gave some pencils to each of his three sisters. He gave half of his pencils plus one more to Gwendolyn. He then gave half of what he had left plus two more pencils to Guinevere. Finally, he gave half of what he had left plus two more pencils to Matilda. He then had no pencils left. How many pencils did he originally have?

If Matilda recieved half of what was left and 2 more to get all of Mortimer's pencils, the "what was left" must be 4. We continue working backwards. To reach 4, he gave 8 pencils to Guinevere, for a total of 12. The last step gives 14 to Gwendolyn, for a grand total of $ \boxed{26}$ pencils.


algebra
polynomial
linear algebra
matrix
similar
linear algebra unsolved

Let A, B be square matrices over $ \mathbb{C}$ of the same size, and suppose that for all $ t\in \mathbb{C}$, A is similiar to A+tB.

Does it follows that B=0?

Thanks!

$ P(t) \equal{} \text{det}(\mathbf{A} \plus{} t \mathbf{B})$ is a polynomial of degree (at most) $ n$, but since similarity transformations preserve determinants it must be a constant polynomial, so yes, I believe so.

All that proves is that $ B$ is nilpotent, I think.  If you look at the case $ n \equal{} 2$, you will find that this argument is not quite enough.

Hey, our avatars are characters drawn by the same artist. :)

[quote="LSH"]Let A, B be square matrices over $ \mathbb{C}$ of the same size, and suppose that for all $ t\in \mathbb{C}$, A is similiar to A+tB.

Does it follows that B=0?[/quote]

The answer is no, for $ n>2$.  Take $ A\equal{}\left(\begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & \minus{}1 \\ 0 & 1 & 0\end{array}\right), B\equal{}\left(\begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{array}\right)$.  Then $ A\plus{}tB\equal{}\left(\begin{array}{ccc} 0 & t\plus{}1 & 1 \\ 1 & 0 & t\minus{}1 \\ 0 & 1 & 0\end{array}\right)$, and you can check that the characteristic polynomial of this matrix is $ x^3\minus{}1$ for any $ t$, which uniquely determines its similarity equivalence class. (since this polynomial has three unique roots, there is a basis of linearly independent eigenvectors)

In general, probably all we can show is that $ B$ is nilpotent, which I think we can do as follows:

Let $ a_k(M)$ be the coefficient of $ x^k$ in the characteristic polynomial of $ M$.  Then $ a_k: M_n \minus{}> \mathbb{C}$ is a continuous function satisfying $ a_k(\alpha M) \equal{} \alpha ^{n\minus{}k} a_k(M)$ for all $ \alpha \in \mathbb{C}$.  Suppose that we had $ a_k(B)\neq 0$ for some $ k<n$.  Since similarity preserves the characteristic polynomial, we have:

$ a_k(A)\equal{}\lim_{t\minus{}>\infty} {a_k(A\plus{}tB)}\equal{}\lim_{t\minus{}>\infty} {t^{n\minus{}k} a_k(t^{\minus{}1}A\plus{}B)}\equal{}a_k(B) \lim_{t\minus{}>\infty} {t^{n\minus{}k}}\equal{}\pm \infty$.  So we must have $ a_k(B)\equal{}0$ for all $ k<n$, hence the characteristic polynomial of $ B$ is $ x^n$.  In other words, $ B$ is nilpotent.

Hi LydianRain,
the charac. polynomial of your $ A\plus{}tB$ is $ \minus{}x^3\plus{}2xt\plus{}1$.
As you wrote $ B$ is nilpotent. But it isn't sufficient. For example:
 if $ A\equal{}I$ then $ B\equal{}0$.
If $ A$ has n distinct eigenvalues, then if $ B$ commutes with $ A$ then $ B$ works; if $ B$ doesn't commute with $ A$ then I don't know.

EDIT: Sorry, I was wrong [i]again[/i] when I posted this.  Here is a correct example:

$ A \equal{} \left(\begin{array}{ccc} 0 & 0 & 1 \\
1 & 0 & 0 \\
0 & \minus{} 1 & 0\end{array}\right), B \equal{} \left(\begin{array}{ccc} 0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0\end{array}\right)$

Then the characteristic polynomial of $ A \plus{} tB$ is always $ \minus{} x^3 \minus{} 1$, which has three distinct roots, hence the similarity equivalence class is uniquely determined.

...and actually it's much simpler than this.  Take $ A \equal{} \left(\begin{array}{cc} 1 & 0 \\
0 & \minus{}1\end{array}\right), B \equal{} \left(\begin{array}{cc} 0 & 1 \\
0 & 0\end{array}\right)$.

I'm going to venture a guess that for any $ A$ that is not a scalar multiple of the identity, we can find some $ B$ with this property.

Assume that $ A$ has $ n$ distinct eigenvalues.
 In my last post I wrote that if $ AB\equal{}BA$ then $ B$ works; but this is irrelevant because there don't exist any such nilpotent matrix (except $ B\equal{}0$). We may assume that $ A$ is diagonal.
Fact: in the case $ n\equal{}2$, $ A\equal{}diag(u,v)$ then the solutions are: $ B\equal{}\begin{pmatrix}0&a\\0&0\end{pmatrix}$ or $ B\equal{}\begin{pmatrix}0&0\\a&0\end{pmatrix}$.
in the case $ n\equal{}3$, $ A\equal{}diag(u,v,w)$ then the NSC over $ B$ are: $ B$ is nilpotent, has a $ 0$-diagonal and satisfies the supplementary condition: $ ub_{23}b_{32}\plus{}vb_{31}b_{13}\plus{}wb_{12}b_{21}\equal{}0$.
   Thus I conjecture this result:
If $ A$ is diagonal with $ n$ distinct eigenvalues then a convenient matrix $ B$ has a $ 0$-diagonal.

That quite easy to prove, as the characteristic polynomial of any upper triangular matrix is determined by its diagonal elements.  Hence any upper triangular $ B$ with zero diagonal will work.

No; I wanted t say:
 Thus I conjecture this result:
If A is diagonal with n distinct eigenvalues then ANY convenient matrix B has a 0-diagonal.

If I have not made a mistake, we have:

[color=blue][b]Theorem 1.[/b] Let $ K$ be an infinite field, let $ n$ be a positive integer, and let $ A$ and $ B$ be two matrices from $ \text{M}_n\left(K\right)$ such that $ A \equal{} \text{diag}\left(a_1,a_2,...,a_n\right)$ for some pairwise distinct elements $ a_1$, $ a_2$, ..., $ a_n$ from $ K$. If the matrix $ A$ is similar to the matrix $ A \plus{} tB$ for every $ t\in K$, then all diagonal entries of $ B$ are $ 0$'s.[/color]

Here is a [i]sketch of a proof[/i]:

We denote the diagonal entries of $ B$ by $ b_1$, $ b_2$, ..., $ b_n$.

For every $ t\in K$, the matrices $ A$ and $ A \plus{} tB$ have the same characteristic polynomial (since they are similar). Thus, $ \det\left(A \minus{} uI_n\right) \equal{} \det\left(A \plus{} tB \minus{} uI_n\right)$ for every $ t\in K$ and every $ u\in K$. In other words, $ \det\left(A \minus{} uI_n\right) \equal{} \det\left(\left(A \minus{} uI_n\right) \plus{} tB\right)$ for every $ t\in K$ and every $ u\in K$.
Thus, the value of $ \det\left(\left(A \minus{} uI_n\right) \plus{} tB\right)$ does not depend on the value of $ t$. Hence, $ \det\left(\left(A \minus{} uI_n\right) \plus{} tB\right)$, seen as a polynomial of $ t$ (while $ u$ is a constant), is constant, and therefore, its coefficient before $ t$ is equal to $ 0$ for every $ u$ (here we use that $ K$ is infinite). But the coefficient of the polynomial $ \det\left(\left(A \minus{} uI_n\right) \plus{} tB\right)$ before $ t$ is equal to the sum of the determinants of all $ \left(n \minus{} 1\right)\times\left(n \minus{} 1\right)$ minors of the matrix $ A \minus{} uI_n$ multiplied with the determinants of the corresponding "opposite" $ 1\times 1$ minors of the matrix $ B$. Now, $ A \minus{} uI_n \equal{} \text{diag}\left(a_1 \minus{} u,a_2 \minus{} u,...,a_n \minus{} u\right)$, so that the only $ \left(n \minus{} 1\right)\times\left(n \minus{} 1\right)$ minors of the matrix $ A \minus{} uI_n$ which have nonzero determinant are the $ n$ "diagonal" minors (i. e. the minors whose rows and columns have the same indices), and their determinants are equal to $ \left(a_1 \minus{} u\right)\left(a_2 \minus{} u\right)...\widehat{\left(a_i \minus{} u\right)}...\left(a_n \minus{} u\right)$ for $ i\in\left\{1,2,...,n\right\}$ (here, $ x_1x_2...\widehat{x_i}...x_n$ means $ \prod_{1\leq j\leq n;\ j\neq i}x_j$); the corresponding "opposite" $ 1\times 1$ minors of the matrix $ B$ are $ b_i$, respectively. Hence, the coefficient of the polynomial $ \det\left(\left(A \minus{} uI_n\right) \plus{} tB\right)$ before $ t$ equals $ \sum_{i \equal{} 1}^n\left(a_1 \minus{} u\right)\left(a_2 \minus{} u\right)...\widehat{\left(a_i \minus{} u\right)}...\left(a_n \minus{} u\right)\cdot b_i$. Thus, $ \sum_{i \equal{} 1}^n\left(a_1 \minus{} u\right)\left(a_2 \minus{} u\right)...\widehat{\left(a_i \minus{} u\right)}...\left(a_n \minus{} u\right)\cdot b_i \equal{} 0$. This is now a polynomial identity in $ u$, and since it holds for all $ u\in K$ (and $ K$ is infinite), we conclude that all coefficients of this polynomial identity are $ 0$.

This means that for every $ k\in\left\{1,2,...,n\right\}$, we have $ \sum_{i \equal{} 1}^n\sigma_k\left(a_1,a_2,...,\widehat{a_i},...,a_n\right)\cdot b_i \equal{} 0$ (here, $ \sigma_k$ is the $ k$-th elementary symmetric polynomial, and $ a_1,a_2,...,\widehat{a_k},...,a_n$ is an abbreviation for "all numbers $ a_1$, $ a_2$, ..., $ a_n$ except of $ a_k$"). This is a system of $ n$ equations in $ n$ unknowns $ b_1$, $ b_2$, ..., $ b_n$. Remembering that we want to show that these $ n$ unknowns $ b_1$, $ b_2$, ..., $ b_n$ are all equal to $ 0$ (because these unknowns are the diagonal entries of $ B$), we see that it remains to prove that these $ n$ equations are all linearly independent (because $ n$ linearly independent linear equations in $ n$ unknowns have only one solution - and obviously, $ \left(b_1,b_2,...,b_n\right) \equal{} 0$ is a solution). This comes down to showing that the $ n\times n$ matrix $ \left(\sigma_k\left(a_1,a_2,...,\widehat{a_i},...,a_n\right)\right)_{1\leq k\leq n;\ 1\leq i\leq n}$ (this is the matrix of the coefficients of our $ n$ linear equations) has nonzero determinant. But it is easy to see that the determinant of this matrix is equal to the Vandermonde determinant $ \prod_{1\leq i < j\leq n}\left(a_i \minus{} a_j\right)$, maybe up to sign; and the Vandermonde determinant is nonzero because $ a_1$, $ a_2$, ..., $ a_n$ are all pairwise distinct.

I hope this was halfways understandable. Honestly, I don't like this proof enough to write it down properly...

  darij

I like that proof.

Took me a moment to see that that determinant has to be the Vandermonde determinant, but of course if you set any two variables equal...

Hi Darij,
your proof is essentially correct.
Here however two remarks:
1) The calculated coefficient is before $ t$ and not before $ t^{n-1}$.
2) The end of your proof is too complicate: you have
$ \sum_{i = 1}^n(a_1 - u)(a_2 - u)...\widehat{(a_i - u)}...(a_n - u)\cdot b_i$. Take $ u=a_1$; then $ (a_2 - a_1)\cdots{(a_n-a_1)}b_1=0$ and $ b_1=0$ and so on.

Thanks, I have corrected my post according to remark 1). As for 2), that's a really good point. We could probably even simplify the whole argument by considering $ \det\left(\left(A \minus{} a_iI_n\right) \plus{} tB\right)$ rather than $ \det\left(\left(A \minus{} uI_n\right) \plus{} tB\right)$. The matrix $ A \minus{} a_iI_n$ has only one $ \left(n \minus{} 1\right)\times\left(n \minus{} 1\right)$ minor with nonzero determinant...

  darij


A crop of potatoes yields an average of 16 metric tons of proteins persquare km of planted are, while corn yields 24 metric ton per square km.... in what proportion must potatoes be planted in order to yield 21 tons of protein persquare km from the combined crops.?    :?:  :?:

hey can someone explain this formula?? :maybe:

The formula 16x+24y/x+y=21 is just taking the protein from each (16 from potatoes and 24 from corn) and multiplying it by the number of metric tons over the tons to give you 21 metric tons of protein in a square km.  The x+y erase the metric tons, so you just get 21 instead of 21 metric tons in which you would have to create another variable.  So you solve the formula like thus:
16x+24y/x+y=21
          *x+y=*x+y\
16x+24y=21x+21y
-16x-21y=-16x-21y
3y=5x
and there's your proportion.


Can someone explain what they are? I have been seeing them over and over again but don't know how to do them. A few problems would be nice, too.

Recall that the coefficients of a polynomial and the symmetric sums of the roots of that polynomial are related.  That is, if you have $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ then we have that $-\dfrac{a_{n-1}}{a_n}$ is the sum of the roots, $\dfrac{a_{n-2}}{a_n}$ is the sum of the roots taken two at a time and so on.

Newton's sums use these facts, and give a way to compute the sum of the $k$th power of the roots of a polynomial, i.e. $s_k=r_1^k+r_2^k+\cdots+r_n^k.$

Newton basically says that

\[ a_ns_j+a_{n-1}s_{j-1}+\cdots+ja_{n-j}=0. \]

Sorry for the terse explanation.  Ask if you have questions.
[i]
Edited for corrections.[/i]

That formula should actually be $a_ns_j+a_{n-1}s_{j-1}+\cdots+ja_{n-j}s_0=0$.

Actually, I think the last term should just be $ja_{n-j}.$

Well $s_0=1$ anyways, so they are identical.

edit>Nevermind, no it isn't. it should be just $ja_{n-j}$.

Hmm,

\[ s_0=r_1^0+r_2^0+\cdots+r_n^0=\underbrace{1+1+\cdots+1}_{n \ \text{ones}}=n. \]

Or am I missing something?

<edit: ok, I see your edit ; ) >


How many $ 8''\times8''$ square tiles are needed to tile a $ 6'\times8'$ rectangular floor?

A 6' by 8' rectangle has an area of $ (6 \times 12) \times (8 \times 12) \equal{} 6912$ square inches. The tiles each have an area of $ 64$ square inches. Thus, there are $ 6912 \div 64 \equal{} \boxed{108}$ square tiles needed.

If 6912/64=108 is too time consuming (it is, considering that this is countdown):

6'=72", 72/8=9
8'=96", 96/8=12
9*12=108


I recently had this problem suggested by a person.It seems interesting and definitely may have been a problem given somewhere already.
The Fibinocci series is well known. On similar lines, can one characterize any recursive series for k sums i.e.,
The nth term equal the sum of the  previous k terms. The nth term and the sum of n terms are to be found out. I know a good book on Recursive Series by Yasuhara. Such a problem is briefly discussed there.
Sathej

what was really your question?
i think for $k$ terms it is surely like fibonacci  numbers.


trigonometry
induction
calculus
calculus computations

$a\otimes b=\frac{a+b}{1+ab}\\ (..(1\otimes 2)\otimes 3)...)\otimes n=?$

edited ;)

[quote="PFloyd"]$a\otimes b=\frac{a+b}{1+ab}\\ (..(a\otimes b)\otimes c)...)\otimes n=?$[/quote]
I don't understand. What do you want?

$a\otimes b = \tanh(\tanh^{-1}a+\tanh^{-1}b)$
$a\otimes b \otimes c = \tanh(\tanh^{-1}a+\tanh^{-1}b+\tanh^{-1}c)$
$a\otimes b \otimes c \otimes d = \tanh(\tanh^{-1}a+\tanh^{-1}b+\tanh^{-1}c+\tanh^{-1}d)$
$\cdots$

[quote="N.T.TUAN"]What is $\tanh{\alpha}$? I know $\tan{\alpha}$, then $\tan{(a+b)}=\frac{\tan a+\tan b}{1-\tan{a}\cdot\tan{b}}$.[/quote]

It's the hyperbolic tangent.  I still don't understand the question.  Are the ...s going from negative infinity to infinity?

No, I think PFloyd's question is from $1$ to $n$ :)

Let $a = 2 \otimes \cdots \otimes n$. Then

$1 \otimes \cdots \otimes n = 1 \otimes a = \frac{1+a}{1+1\cdot a}= 1$

brilliant :) thanks

and what about $2\otimes 3\otimes...\otimes n$?

$2\otimes 3\otimes\cdots\otimes n= \frac{\binom{n+1}{2}+(-1)^{n}}{\binom{n+1}{2}-(-1)^{n}}$. Prove this by induction- I found it by calculating a few values and noticing the pattern.


Find all integers $ n > 1$ such that any prime divisor of $ n^6\minus{}1$ is a
divisor of $ (n^3\minus{}1)(n^2\minus{}1).$

Baltic way.

Ευχαριστω για την αφιερωση ρε Δημητρη.
Βαζω μια προχειρη λυση.(πες μου αν χανει καπου ή αν κατι γραφω γρηγορα-σημερα εχω την αποφοιτηση και τρεχω  , soz  :wink:  )
ειναι $ (n^6 \minus{} 1) \equal{} (n^3 \minus{} 1)(n \plus{} 1)(n^2 \minus{} n \plus{} 1)$
Θεωρω τυχαιο πρωτο διαρετη, $ p$, του $ (n^2 \minus{} n \plus{} 1)$ τοτε πρεπει να διαρει και το $ (n \minus{} 1)(n \plus{} 1)(n^3 \minus{} 1)$
- Αν διαρει το $ n^3 \minus{} 1$ ευκολα $ p \equal{} 2$, ατοπο αφου $ n^2 \minus{} n \plus{} 1$περιττο παντα..
-Aν διαιρει το $ n \minus{} 1$ τοτε $ p|1$και ατοπο
- Αν διαρει το $ n \plus{} 1$ τοτε $ p|n \minus{} 2$ αρα $ p \equal{} 3$
Συνεπως ειναι $ n^2 \minus{} n \plus{} 1 \equal{} 3^x$
ομως αν την θεωρησουμε σαν δευτεροβαθμια ως προς n πρεπει να εχει διακρινουσα τελειο τετραγωνο,
Θεωρω $ x > 1$ τοτε,
$ 1 \minus{} 4(1 \minus{} 3^x) \equal{} m^2 \leftrightarrow 3 \minus{} 4*3^x \equal{} m^2 \equal{} > 3|m \equal{} > 1 \minus{} 4*3^{x \minus{} 1} \equal{} 3z^2 \equal{} > 3|1$ ατοπο..
(το ατοπο προκυπτει επισης με περιπτωσεις και modulo9)
αρα $ x \equal{} 1$, $ n(n \minus{} 1) \equal{} 2 \equal{} > n \equal{} 2$
με επαληθευση λοιπον μονη λυση η $ n \equal{} 2$. :)


geometry
incenter
trigonometry
geometry solved

Does  anyone  has  a prove  for  this one ?

I  is  incenter , H  is orthocenter  of  a  triangle  ABC . Then [tex]  IH^2 = 2r^2 - 4R^2 \cos A \cos B \cos C  [/tex]

Thanks

See [url=http://groups.yahoo.com/group/Hyacinthos/messages/6147?viscount=41&expand=1]Hyacinthos messages #6147 and #6195[/url] for a trigonometric proof by Nikolaos Dergiades. This is the nicest proof I know for this identity.

  Darij


[quote="kihe_freety5"]x,y,z≥0 ; x²+y²+z²=1;
prove that:

b) $ \frac {}$sqrt{xyz}${ } \sqrt {3(x + y + z)}$ ≥ $ \frac {1}{x}$+$ \frac {1}{y}$+$ \frac {1}{z}$+xyz;
c)1+$ \sqrt {}$frac{y}{xz}${ }$≥xy+yz[/quote]

[b]retype the original[/b] :)

$ x,y,z\ge0$ ; $ x^2+y^2+z^2=1$
prove that:

b) $ \frac{\sqrt{xyz}}{\sqrt {3(x + y + z)}} \ge \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + xyz$
c) $ 1+\sqrt{\frac{y}{xz}} \ge xy+yz$