induction
combinatorics unsolved
combinatorics

[i]Find all positive integer $ n$ such that :

  The set $ A \ \equal{} \ \{1 ;2 ; ....; 3n \}$ can be partitioned into $ n$ subsets $ A_1 ; A_2; ...; A_n$ with $ |A_1| \equal{} |A_2 | \equal{} ...\equal{} |A_n|\equal{}3$ and in each subset $ A_i ; i \equal{} 1 ; 2 ; ...; n$, there is a element which is sum of two other elements in $ A_i$ .[/i]

   [i]Sorry for my bad English :lol:  :lol:[/i]

[quote="nguyenvuthanhha"][i]Find all positive integer $ n$ such that :

  The set $ A \ \equal{} \ \{1 ;2 ; ....; 3n \}$ can be partitioned into $ n$ subsets $ A_1 ; A_2; ...; A_n$ with $ |A_1| \equal{} |A_2 | \equal{} ... \equal{} |A_n| \equal{} 3$ and in each subset $ A_i ; i \equal{} 1 ; 2 ; ...; n$, there is a element which is sum of two other elements in $ A_i$ .[/i]

   [i]Sorry for my bad English :lol:  :lol:[/i][/quote]

Perhaps it's a problem in China National Olympiad (I don't remember exactly the year  :mad:, but it's in 1990s)

[i]  Yes , this problem was chosen for a regional contests in Vietnam in 2009 . There is no contestant have point in that problem :lol: [/i]

I don't have anything remotely close to a rigorous proof but I've checked values for n=1,4, and 5 and have come to the conclusion that the solution is all n congruent $ 0$ or $ 1\mod 4$ the motivation for this is:
the sum of all the numbers in $ S$ is $ \dfrac{3n(3n\plus{}1)}{2}$ knowing this, we can conclude that the sum of the $ c$'s, where $ a_i\plus{}b_i\equal{}c_i$ for every $ a$, $ b$, and $ c$ in set $ A_i$, is $ \dfrac{3n(3n\plus{}1)}{4}$.
Now, we also know that $ 3n$ must be a $ c$ in some $ A_i$, by subtracting $ 3n$ from our sum, we find the sum of the rest of the $ c$'s equals $ \dfrac{9n(n\minus{}1)}{4}$, must be an integer so n is either $ 0$ or $ 1$ $ \mod 4$.  I know this argument is really weak so........ :|

That's a perfectly good argument that all such $ n$ must be 0 or 1 mod 4 (though the step of subtracting $ 3n$ from the sum is completely unnecessary) -- now we just need a construction to show that it's really possible for all those values.  I've tried playing around with building up from the $ n \equal{} 4$ case, but it hasn't led anywhere yet :(

I think I have a construction that works. Notice that if we put $ 1, 2, \ldots, n$ in different sets, we get that it's sufficient to find a string consisting of two of each of the numbers $ 1, \ldots, n$ with $ k\minus{}1$ numbers between the two appearances of $ k$. So for example, for $ n\equal{}4$ we'd use the string $ 11423243$ and for $ n\equal{}5$ we'd use the string $ 4115423253$.

So now suppose that we have the string for $ n\equal{}4k$. We're going to exploit the fact that it naturally splits into two parts, $ A_kB_k$, with $ A_1 \equal{} 11$ and $ B_1 \equal{} 423243$. Now, we're going to let $ A_{k\plus{}1} \equal{} B_k$ and $ B_{k\plus{}1} \equal{} (4k\plus{}4)(4k\plus{}2)(4k\plus{}3)(4k\plus{}1)A_k(4k\plus{}2)(4k\plus{}1)(4k\plus{}4)(4k\plus{}3)$. Then we obtain the desired string for multiples of $ 4$. For example, this gives the strings $ 4232438675116587$ for $ n\equal{}8$ and $ 8\ 6\ 7\ 5\ 1\ 1\ 6\ 5\ 8\ 7\ 12\ 10\ 11\ 9\ 4\ 2\ 3\ 2\ 4\ 3\ 10\ 9\ 12\ 11$ for $ n\equal{}12$.

Now we use the fact that $ B_k$ always starts with $ 4k$ to obtain the string for $ n\equal{}4k\plus{}1$. To do this, replace both instances with $ 4k$ with $ 4k\plus{}1$ and place a $ 4k$ between the first and second numbers. Then the string $ (4k)A_kB_k'$ will satisfy the desired conditions. For example, for $ k\equal{}1$ we have $ A_1 \equal{} 11$, $ B_1 \equal{} 423243$, so we have $ B_1' \equal{} 5423253$ and our string is $ 4115423253$.

It is easy to check that all of these constructions meet the necessary criteria.

[quote="Hamster1800"]I think I have a construction that works. Notice that if we put $ 1, 2, \ldots, n$ in different sets, we get that it's sufficient to find a string consisting of two of each of the numbers $ 1, \ldots, n$ with $ k \minus{} 1$ numbers between the two appearances of $ k$. So for example, for $ n \equal{} 4$ we'd use the string $ 11423243$ and for $ n \equal{} 5$ we'd use the string $ 4115423253$.

So now suppose that we have the string for $ n \equal{} 4k$. We're going to exploit the fact that it naturally splits into two parts, $ A_kB_k$, with $ A_1 \equal{} 11$ and $ B_1 \equal{} 423243$. Now, we're going to let $ A_{k \plus{} 1} \equal{} B_k$ and $ B_{k \plus{} 1} \equal{} (4k \plus{} 4)(4k \plus{} 2)(4k \plus{} 3)(4k \plus{} 1)A_k(4k \plus{} 2)(4k \plus{} 1)(4k \plus{} 4)(4k \plus{} 3)$. Then we obtain the desired string for multiples of $ 4$. For example, this gives the strings $ 4232438675116587$ for $ n \equal{} 8$ and $ 8\ 6\ 7\ 5\ 1\ 1\ 6\ 5\ 8\ 7\ 12\ 10\ 11\ 9\ 4\ 2\ 3\ 2\ 4\ 3\ 10\ 9\ 12\ 11$ for $ n \equal{} 12$.

Now we use the fact that $ B_k$ always starts with $ 4k$ to obtain the string for $ n \equal{} 4k \plus{} 1$. To do this, replace both instances with $ 4k$ with $ 4k \plus{} 1$ and place a $ 4k$ between the first and second numbers. Then the string $ (4k)A_kB_k'$ will satisfy the desired conditions. For example, for $ k \equal{} 1$ we have $ A_1 \equal{} 11$, $ B_1 \equal{} 423243$, so we have $ B_1' \equal{} 5423253$ and our string is $ 4115423253$.

It is easy to check that all of these constructions meet the necessary criteria.[/quote]

[i]Yes , after that lemma , we have a sequence $ a_1 a_2 ....a_{2n}$ , each number $ k \ \in \ \{1;2 ;...; n \}$ appear 2 times in that sequence with $ k \minus{} 1$ numbers between the two appearances of $ k$

  Call $ \sigma(k) ; \sigma'(k)$ be two positions of $ k$ in that sequence . Then the partition
$ \{ 1,\sigma(1) \plus{} n,\sigma'(1) \plus{} n \} , \{ 2,\sigma(2) \plus{} n,\sigma'(2) \plus{} n \} ; ...;\{ n,\sigma(n) \plus{} n,\sigma'(n) \plus{} n \}$ will satisfies all condition of the problem

 ( You can have $ n \equal{} 4k ; n \equal{} 4k \plus{} 1$ as sansuunoousama said )

  Very nice problem :lol: [/i]

[quote="Hamster1800"]   $ 8\ 6\ 7\ 5\ 1\ 1\ 6\ 5\ 8\ 7\ 12\ 10\ 11\ 9\ 4\ 2\ 3\ 2\ 4\ 3\ 10\ 9\ 12\ 11$ for $ n \equal{} 12$
[/quote]

[i] Seems to be vey nice , but it is a wrong construction

For example , in that string , there is $ \underline{7}$ numbers between two appearances of $ 9$[/i]

Yes, you are right. Perhaps we can modify the induction by going from $ 4k$ to $ 4k\plus{}1$ and then from $ 4k\plus{}1$ to $ 4k\plus{}4$, since the problem is that the spacing doesn't work out when adding four numbers, but rather when adding three numbers (as you can see when you have just $ 423243$). I'll try that tomorrow if nobody else gets to it first.

These sequences are known to exist (they are called Skolem Sequences and http://www.research.att.com/~njas/sequences/A014552 says that the number of them is positive for these cases), but I can't figure out a correct proof. Can anyone help?

I think that besides solving in that way , there is another proof because as Hamster1800 said : the sequence has been constructed only for small $ n$

   and that problem was created a long time ago . :lol:

You might want to check [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1054571#1054571]this[/url] out. The things the poster proves in the solution could be helpful in the general case.


An equilateral triangle has a perimeter of 24 cm. How many square centimeters are in the area of the triangle formed by connecting the midpoints of the sides of the original triangle? Express your answer in simplest radical form.

Side is 8, so midline is 4. Area: $ \frac{4^2\sqrt3}{4}\equal{}4\sqrt3$


inequalities
quadratics
algebra
algebra unsolved

This is part of my homework exercise.  I am unable to understand how it is done:

For what values of k does the system have no solution:

(2x - 1)(x + 3) = y
-5x^2 + 17x - k = y

Mostly I do not know where to start.
I believe it is a question of inequalities but i'm not sure.
If anyone could point me in the right direction.

[hide="Hint"]First, eliminate $y$ from both equations and set them equal.
$2x^2+5x-3=-5x^2+17x-k$

$\Longleftrightarrow 7x^2-12x+(k-3)=0$

Now, we want to find the value(s) of $k$ such that the quadratic equation $7x^2-12x+(k-3)=0$ has no solutions. So...
[hide="2nd hint"]Use discriminant ;) [/hide][/hide]

Thank you, that explanation was great.

I  used the Discriminant to solve for k, ended up with 57/7.
I tried to check 57/7 and it fails....

The Discriminant
D = 0 -- two equal, real roots
D > 0 -- two real roots
D < 0 -- two complex roots

What em I missing here?
I really don't think it is possible to use the discriminant to solve for k.  In that case then:
D = 57/7
Which means there are two real roots.

Still not sure how to get the values of the k...
Unless it is k > 57/7


No, this post isn't the least bit sarcastic. I've been to MC Nats, I know how extremely tough it is to do well under such extreme pressure. Especially this year, with high expectations. Beating 52 other teams is no small feat. I know you guys probably feel like you should've done better, but the intangible stuff you've taken out of Mathcounts hopefully means much more than the tangible should-haves.

Anyhow, I don't know individual rankings yet except for Justin, but good job guys!

wow lol lyndon dam i thought he was going to win seriously.
oh well atleast he has a chance to go to blue mop...
hmm yea 5th i know fanatic probably made this so the indiana team wouldnt feel disappointed, but i dont ever remembering indiana getting worst than 3rd in like 5 years or so...

Well... stupid mistake on #1 on the team round.  #9: Justin accidentally thought 8*4=12... But we can't complain because he was the only one who could do it quickly enough. 1 more team question would've gotten us 2nd place (Texas had perfect on the team round).

Rankings: Mark: 18th (beast)
Me: 52nd (ermmm. huge deprovement.)
Youkow: 76th..

is it true that you missed every other sprint question?

does ne one know how that one kid from virginia did. the one who used to live here. i believe that his aops name is unimpossible.

unimpossible got 33 = 30th place.

Ah. Mr. Frost is making fun of you again Lyndon.

yeah, dude......mathcrazed, is David Lu better than you at math? yeah, someone told me that if i say that, you'll get mad (and no it wasn't david)

btw: tan90=3/0, who r u?

[quote="frost13"]Congrats to Indiana, it really was fun hanging out with you guys pretty much the whole time.  Even Lyndon :lol:[/quote]

u sure?

i don't know why i'm posting on indiana forum...

[quote="cognos599"]yeah, dude......mathcrazed, is David Lu better than you at math? yeah, someone told me that if i say that, you'll get mad (and no it wasn't david)[/quote]

Um... was that person at Nats? And did his name begin with a U?

ummmmm lyndon gets mad if anyones better at him in math.......................................right?

[quote="unimpossible"][quote="tan90=3/0"]yes, how did you do?[/quote]

218th  :([/quote]
you liar i saw ur name on top 57 students

[quote="mathcrazed"]unimpossible got 33 = 30th place. [/quote]

WINK WINK :wink:

YAY WE CAUGHT THE LIAR UNIMPOSSIBLE :D  :D  :D

[quote="mathcrazed"]T.T... no I get sad xD... lol. david lu. what a beast[/quote]

Lol, david lu sucks at math.

[quote="cognos599"]alabama, massa, michigan, illinois, texas, and ohio have all invaded our forum.[/quote]

we should try to beat missouri.

EDIT: no wait wait, JAPAN HAS INVADED INDIANA

Missouri sucked this year, I blame it on my complete suckage of #10. Next year, when I inject sugar and caffeine into my veins immediately before th competition is when we will beast. Texas shall fall before us. Indiana too.

Ohio >>>> Indiana.

You guys have black highways

[quote="cognos599"]alabama, massa, michigan, illinois, texas, and ohio have all invaded our forum.[/quote]
 :D  :D  :D

O_0 but how do u know im from illinois?????? hmmmmm some1's being a suspicious stalker...>_>

[quote="cognos599"]how did you know that i was talking to you, when i said illinois..... but now that you said it, i know you are from illinois.[/quote]
lol :P  :P  :P


define $ f: \mathbb{N}\rightarrow \mathbb{N}^*$ such that $ f(m\plus{}n)\minus{}f(n)\minus{}f(m)$ is either $ 0$ or $ 1$ for all $ m,n$.
Given that $ f(2)\equal{}0, f(3)>0, f(9999)\equal{}3333$ , find $ f(2008)$.

$ \mathbb{N}^*$ is the natural numbers joined with zero.

[hide="solution"]I transform the recursion into $ f(m+n)=f(m)+f(n)+s$ for $ s\in{0,1}$. Looking at $ f(1+1)$, we quickly see that $ f(1)=0$, and then we find $ f(3)=1$. Looking at both $ f(1+3)$ and $ f(2+2)$, we see $ f(4)=1$. Using the first three values of $ f$ and the recursion, we quickly derive the following:

$ f(n+1)=f(n)$ or $ f(n+1)=f(n)+1$
If $ f(n-1)\nef(n-2), f(n)=f(n-1)$.
If $ f(n-1)=f(n-2)=f(n-3), f(n)=f(n-1)+1$
If $ f(n-1)=f(n-2)\nef(n-3), f(n)=f(n-1)$ or $ f(n)=f(n-1)+1$

In the latter case, if $ f(n)=f(n-1)$ always, then $ f(n)=\lfloor\frac{n}{3}\rfloor$, $ f(2008)=669$, and $ f(9999)=3333$. Every instance where, in the latter case, $ f(n)=f(n-1)$ translates the remainder of the function one to the left. This can be done once or twice in order to maintain $ f(9999)=3333$. If done twice, then $ f(2008)=670$. So $ f(2008)$ is either $ 669$ or $ 670$.[/hide]

[quote="Darmani"][hide="solution"]I transform the recursion into $ f(m + n) = f(m) + f(n) + s$ for $ s\in{0,1}$. Looking at $ f(1 + 1)$, we quickly see that $ f(1) = 0$, and then we find $ f(3) = 1$. Looking at both $ f(1 + 3)$ and $ f(2 + 2)$, we see $ f(4) = 1$. Using the first three values of $ f$ and the recursion, we quickly derive the following:

$ f(n + 1) = f(n)$ or $ f(n + 1) = f(n) + 1$
If $ f(n - 1)\nef(n - 2), f(n) = f(n - 1)$.
If $ f(n - 1) = f(n - 2) = f(n - 3), f(n) = f(n - 1) + 1$
If $ f(n - 1) = f(n - 2)\nef(n - 3), f(n) = f(n - 1)$ or $ f(n) = f(n - 1) + 1$

In the latter case, if $ f(n) = f(n - 1)$ always, then $ f(n) = \lfloor\frac {n}{3}\rfloor$, $ f(2008) = 669$, and $ f(9999) = 3333$. Every instance where, in the latter case, $ f(n) = f(n - 1)$ translates the remainder of the function one to the left. This can be done once or twice in order to maintain $ f(9999) = 3333$. If done twice, then $ f(2008) = 670$. So $ f(2008)$ is either $ 669$ or $ 670$.[/hide][/quote]

You wrote:
$ f(n + 1) = f(n)$ or $ f(n + 1) = f(n) + 1$
If $ f(n - 1)\nef(n - 2), f(n) = f(n - 1)$.
If $ f(n - 1) = f(n - 2) = f(n - 3), f(n) = f(n - 1) + 1$
If $ f(n - 1) = f(n - 2)\nef(n - 3), f(n) = f(n - 1)$ or $ f(n) = f(n - 1) + 1$

it is actually,
$ f(n + 1) = f(n)$ or $ f(n + 1) = f(n) + 1$
If $ f(n - 1)\not =f(n - 2), f(n) = f(n - 1)$.
If $ f(n - 1) = f(n - 2) = f(n - 3), f(n) = f(n - 1) + 1$
If $ f(n - 1) = f(n - 2)\not =f(n - 3), f(n) = f(n - 1)$ or $ f(n) = f(n - 1) + 1$
right?

Exactly, thanks. I was pretty sure the command for $ \not \equal{}$ was \ne, but I was clearly wrong.

Thanks ffao for catching that; I figured something was wrong when I got two answers.

$ \ne$ is \ne. The problem with your code was that you attached the "f" (from f(n-3)) to it. (\neq also works.)


Find $f: Z\to Z$ sastifying $f(f(x))=1-x^{4}$

yes!! your solution is the same as mine
an easier problem but we can see the way to have a solution for this problem
http://www.mathlinks.ro/Forum/viewtopic.php?t=110387

anyone have the solution for general problem! it will be an interesting solution! and it's hard to have it  :roll:


Help!!!

I tried to typeset a pdf in TeXnic, and I couldn't, because at the end of the page, my sentence couldn't fit, and I don't know how to start a new page.  It accounted for six errors.  So, how do you start a new page?

Help!!!   :huh:

LaTeX will normally insert page breaks automatically.

You're going to have to provide more specific details in order for anybody to be able to assist you.

To start a new page, insert the \pagebreak command.  So like:

[code]blahblahblahblah

\pagebreak

blahblahblahblah[/code]

I don't know if that's what you are asking, but it's worth a shot.


Prove for every natural n (n > 2) :
\[\begin{array}{l}\\ \sum_{i = 1}^{n-1}{i\left( \begin{array}{l}n \\ i+1 \\ \end{array}\right)(n-1)^{n-i-1}}= (n-1)^{n}\\ \end{array}\]


Doug can paint a room in $ 5$ hours. Dave can paint the same room in $ 7$ hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let $ t$ be the total time, in hours, required for them to complete the job working together, including lunch. Which of the following equations is satisfied by $ t$?

$ \textbf{(A)}\ \left(\frac{1}{5}\plus{}\frac{1}{7}\right)(t\plus{}1)\equal{}1 \qquad
\textbf{(B)}\ \left(\frac{1}{5}\plus{}\frac{1}{7}\right)t\plus{}1\equal{}1 \qquad
\textbf{(C)}\ \left(\frac{1}{5}\plus{}\frac{1}{7}\right)t\equal{}1 \\
\textbf{(D)}\ \left(\frac{1}{5}\plus{}\frac{1}{7}\right)(t\minus{}1)\equal{}1 \qquad
\textbf{(E)}\ (5\plus{}7)t\equal{}1$

[hide]
Each hour they work, they can paint $ \frac{1}{5}\plus{}\frac{1}{7}$ of the room.  They work for $ t\minus{}1$ hours.  So the correct equation is $ D$.
[/hide]

[hide]WLOG, let the total work be 1. Then the rates of Doug and Dave are $ \frac {1}{5}$ and $ \frac {1}{7}$, respectively. Since they took an hour off, we have the equation

$ \left(\frac {1}{5} \plus{} \frac {1}{7}\right)(t \minus{} 1) \equal{} 1 \Leftrightarrow \boxed{D}$, as desired.[/hide]

[hide=solution]Each hour Doug paints $\cfrac{1}{5}$ of the room and Dave $\cfrac{1}{7}$ of the room. Combined, they paint $\cfrac{1}{5}+\dfrac{1}{7}$ of the room. They work for $t-1$ hours, as they do not work during lunch. Therefore the correct answer is \[\boxed{\textbf{(D)} ~\left(\cfrac{1}{5}+\dfrac{1}{7}\right)\left(t-1\right)=1}\]


modular arithmetic
pigeonhole principle
geometry
rectangle

1) Show that for every n positive there exist n integers consecutives so that no one is prime or power of prime.

2) Given a 4*n chessboard show that there is no n for which a knight can touch every square only one time.

3) Given wichever set X of ten numbers in (10;99) it is possible to choose two sub-sets from X so that their intersection is 0 e they have the same sum.

[Sorry for my terrible english, but I'll improve it, promised)

1) Is a classical consequence of the chinese remainder theorem.

2) Color the $4 \times n$ as a classical checkoard. And consider the set $E$ of the two exterior rows, and the set $I$ of the two interior rows, to prove that if a circuit does exists then the unit squares which belong to one of this set all have the same color, which is clearly absurd.

3) smells like a classical pigeon-hole problem.

Pierre.

Is there anyone who can formalize  the solutions 1 and 3, plz?

For the first problem, you can take, for example, some sequence of distinct primes $p_1,p_1',p_2,p_2',\ldots,p_n,p_n'$. By the Chinese Remainder Theorem, thee is a number $k$ s.t. $k\equiv -i\pmod{p_i\cdot p_i'},\ \forall i\in\{1,2,\ldots,n\}$. This simply means that $k+i$ is divisible by two distinct primes $p_i,p_i'$ (for all $i$ from $1$ to $n$), so it's not a prime power.

I think Pierre was referring to something similar to this.

For 3.
Consider all 1023 distinct non-empty subsets of $X$. Sum of the elements in each of these subsets is in $[10,99\cdot 10]$ - less than 1023 different values, so due to Pigeonhole principle we conclude that two distinct subsets have equal sums. Excluding common elements from them we obtain two disjoint subsets of $X$ with the same sum.

Grobber : Not exactly  ;)  I tought about $n  = p_i - i \mod [p_i^2]$ for sufficientely large $n$.

Pierre.

Myth : That's exactly what I thought.  :P

Pierre.

about you previous message, Pierre:

It is strange, because my idea was exactly the same as grobber's one :)
 :lol:  :lol:  :lol:

That's why I said "something [b]similar[/b] to this" :). I knew Pierre had something tricky up his sleeve :D.

1) Which mxn rectangles are covered by L-tetrominoes?
2) Show that (1+2+3+...+n) divides (1^k + 2^k + 3^k + ... + n^k) for odd k.

Do you like Freddie Mercury, Simo_the_Wolf?

[quote]My soul is painted like the wings on butterflies,
fairytales of yesterday will grow but never die,
I can fly, my friends!!

-F.Mercury-[/quote]

The Show Must Go On?

Ok guys, you're so "fast and furious"! :)  I like the math of domino so ...

For wich n an nxn grid can be tiled by 3x1 tiles sparing only three vertices ?

I think (not sure) he means this: you have an $n\times n$ board. For which $n$ can you cover the entire board except for three corners (???) with $3\times 1$ rectangles?

Ops, ya , grobber's version is right and i'm definitively a wash-out at speaking english :blush: . I beg your pardon ;)

Right, but prove it :P so that everybody can learn such a beutiful math-mith  branch :D .

$n$ is necessarily a multiple of $3$.

Assume we want to leave the corners $(1,n),(n,1),(n,n)$ uncovered. Color the first line $1,2,3,1,2,3,\ldots$, the second line $2,3,1,2,3,1,\ldots$, the third line $3,1,2,3,1,2,\ldots$, the fourth line $1,2,3,1,2,3,\ldots$ and so on (I'm sure you can see the pattern). Each $3\times 1$ "domino" covers a $1$, a $2$ and a $3$. There are equally many $1$'s, $2$'s and $3$'s on the board, so the three uncovered corners must have $1,2,3$ written in them. However, both $(1,n)$ and $(n,1)$ have a $3$, so we have a contradiction.

There are N (with N>2 and odd) adjacent squares along a single line ( so Nx1), infinite 2 x 1 dimini and two players A end B. A puts down his domino for first. Who win?

Who lose in your game? The player, which cannot perform his turn. Am I right?

BTW, you should make the separate topic for this problem ;)

In your question about the 4 by n chessboard you say that it is always impossible for a knight to reach each square. I think it is possible when n=3. I think that the problem with your proof is that you can move from the interior to the interior. Before he moves from the inside to the inside he only visits squares of one color in E and the other color in I. After the knight visits the first 2n squares, he must move from the inside to the inside so that he can start visiting squares of the opposite colors.


Bob and Joe are playing a game where Bob roles three dice and Joe roles one die. If Joe's die is higher or equal to Bob's highest die, Joe wins. What is the probability of Joe winning? ;)

[quote]Bob and Joe are playing a game where Bob rolls three dice and Joe rolls one die.
If Joe's die is higher or equal to Bob's highest die, Joe wins.
What is the probability of Joe winning?[/quote][hide]
There are $6^4 = 1296$ ways that the four dice can turn up.

If Joe rolls a 6, Bob can roll any number on all three dice.
There are $6^3 = 216$ ways this can happen.

If Joe rolls a 5, Bob must roll 5 or less on all three dice.
There are $5^3 = 125$ ways this can happen.

If Joe rolls a 4, Bob must roll 4 or less on all three dice.
There are $4^3 = 64$ ways this can happen.

If Joe rolls a 3, Bob must roll 3 or less on all three dice.
There are $3^3 = 27$ ways this can happen.

If Joe rolls a 2, Bob must roll 2 or less on all three dice.
There are $2^3 = 8$ ways this can happen.

If Joe rolls a 1, Bob must roll 1 on all three dice.
There is $1^3 = 1$ ways this can happen.

Hence, Joe will win in $216 + 125 + 64 + 27 + 8 + 1 = 441$ of the outcomes.

Therefore: $P(\text{Joe wins}) \;= \;\frac{441}{1296} \;= \;\frac{49}{144}$[/hide]

[quote]Bob and Joe are playing a game where Bob roles three dice and Joe roles one die. If Joe's die is higher or equal to Bob's highest die, Joe wins. After rolling, Joe and Bob notice that all 4 die have different numbers showing. What is the probability that Joe has won?[/quote]
[hide="I probably got this wrong"]
If Joe gets a 6, Bob can roll 5 x 4 x 3, or 60 ways
If Joe gets a 5, Bob can roll 4 x 3 x 2 ways, or 24 ways
If Joe gets a 4, Bob can roll 3 x 2 x 1 ways, or 6 ways
If Joe gets a 3 or lower, Bob automatically wins

90/(6x5x4x3) = 90/360 = [b]1/4[/b]
[/hide]

Good.  Can you find a way to do it more quickly?

Any one played here a game called "Risk".. If yes, you would know such proablities by heart... :)

[quote="JBL"]Treething: you're still dividing by 216.  But there aren't 216 actual possibilities any more.

Also: can you find a way to do it more quickly?[/quote]

No I'm not, silly :P .


What is the sum of all two-digit positive integers whose squares end with the digits 01?

We must find all two digit numbers $ 10a \plus{} b$ such that $ 100a^2 \plus{} 20ab \plus{} b^2\equiv20ab \plus{} b^2\equiv1\pmod{100}$.
Since $ b \equal{} 1,9$ our two cases are:
i. $ b \equal{} 1$

$ 20a \plus{} 1\equiv1\pmod{100}\implies20a\equiv0\pmod{100}\implies a\equiv0\pmod{5}$.

This means that $ 51$ is the only possible answer for this case.

ii. $ b \equal{} 9$

$ 180a \plus{} 81\equiv1\pmod{100}\implies80a\equiv \minus{} 80\pmod{100}$

$ \implies a\equiv \minus{} 1\equiv4\pmod5$.

This means that $ 49,99$ are the only possibilities.

$ 49 \plus{} 51 \plus{} 99 \equal{} \boxed{199}$.


[quote="eddie77"]Prove true for all $x\in\Re$ in the following equation -
\[ x^6-6x+5\geq 0  \][/quote]

$\Longleftrightarrow (x-1)^2((x^2+2)(x+1)^2+3)\geq 0$

which is obvious

Just found an simple prove by AM-GM

$x^6+1+1+1+1+1\geq 6\sqrt[6]{x^6}=6|x|\geq 6x$

hence $x^6-6x+5\geq 0$


MATHCOUNTS
number theory
least common multiple
greatest common divisor
prime factorization

The least common multiple of 54 and some natural number N is 432.

The greatest common divisor of 54 and N is 6.

What is N?

48. Oops, forgot the solution  :mrgreen:

432/54 is 8. The prime factorization of 432 is 3 :^3: (2 :^4: ), which is equal to 9(6)(8). 9(6) is 54, so N must have 8/2 :^3:  in it. If the gcf is 6, the least N can be is 48, which satisfies all the requirements.

s0mp,

What can you say about the GCD and LCM of any two numbers?

Correct me if I'm wrong, but by the looks of it, the gcd is the product of the common prime factors of the two numbers, with the condition that you choose the lowest power possible of each factor.

And for the lcm, its just the product of the highest power of each factor of both numbers, but if there is a 2 :^3: in one and 2 :^2: in the other, you choose the 2 :^3:.....i think.

Just to clarify this as-clear-as-mud post, an example is 8(2 :^3:) and 76(19*2 :^2:) . The lcm is 19*2 :^3:, and the gcd would be 2 :^2:. :?

And since you choose the higher power for LCM and lower for GCD, what do you get when you multiply the LCM and GCD together?

Ahh.. its the product of the two numbers in question
 :D I see what that implies...that because gcd(lcm)=N1(N2), as this relates to your challenge question, 6(432)=54N, so all you have to do is divide by 54 and you get your answer...i get it

using this formula thingie i got from my broski a few years ago the the
gcd x lcm=xy
so the gcd is 6 and the lcm is 432 and one of the numbers is 54 so plugging that into the equation

6 x 432=(54)(y)
2592=(54)(y)
divide both sides by 54 using simple algebra to get
48=y
vala
thats a nifty little thing that one should rememorize (yes i spelled that on purpose)
so remember gcd x lcm=xy
okay then. bring on the next problem. nuff said

that too. i figured out why and such. but yeah. oh nm. nuff said

[quote="MCrawford"]Noooooooo!

Don't memorize!  Understand why it's true (read above).

Understanding is far more flexible and powerful than memorization.[/quote]

I think that's true, although I learned it the hard way.

of course one should always know how. sorry about that

Chinaboiiiiiiiiii,
                                   what did you learn the hard way?

I spent almost 2 weeks on one subject (my mom was teaching me), cause I tried to memorize the formula, but then the questions were always changing from this to that.

Yeah. Basically thats the difference between 'school math' and 'problem solving math'. At school you memorise the formulas and know when to use them, its all about whether you can use them or not. In problem solving its all about thinking about <how> to do the problem, you <aren't> told how. Once you are told how to do it the problems are (almost) always simple. Even international mathematical olympiad problems. Memorizing formulas never really helps (apart from things like double angle formulas and stuff like that.)

im sorry. if anything i hate the math school teaches cause they barely get halfway through the book ranting about some topic that has nothing to do to school. at least thats how it is in oklahoma. probably why oklahoma never does well in mathcounts. lool. its just like in geometry. if you understand the proofs, its pretty easy from there knowing the theorems and stuff.


calculus
integration
trigonometry
function
limit
calculus computations

Study the convergence of the integral
\[ \int_1^{ \plus{} \infty } {\sin \left( {1 \plus{} x \plus{} x^2  \plus{} ... \plus{} x^{2007} } \right)dx}.
\]
I will post my solution at weekend. Have fun  :roll: .

It converges. (Conditionally, of course.) Just do an integration by parts.

$ \int_1^{\infty}\sin(1 + x + \cdots + x^{2007})\,dx =$

$ \left.\frac { - \cos(1 + x + \cdots + x^{2007})}{1 + 2x + \cdots + 2007x^{2006}}\right|_1^{\infty} - \int_1^{\infty}\frac {(2 + 3\cdot2x + \cdots + 2007\cdot2006x^{2005}) \cos(1 + x + \cdots + x^{2007})}{(1 + 2x + \cdots + 2007x^{2006})^2}\,dx$

That second integral converges absolutely by comparison with a constant times $ x^{ - 2007}.$

Note that these denominators have no zeros for $ x>0.$ We need that.

I suggest to use Abel test or Dirichlet test.

You can study a more general problem:

Let $ f$ be a differentiable function, monotonic increasing on the inverval $ (1,\infty)$ such that $ \lim\limits_{x\rightarrow \infty}f(x)=\infty$.

Study the convergence of the following integral:

$ \int\limits_{0}^{\infty}\sin(f(x))dx$.

Remark: When $ a=0$ and $ f(x)=x^{2}$ we get the famous Fresnel integral $ \int\limits_{0}^{\infty}\sin(x^{2})dx$.

Copying the proof I gave above for the more general question will leave us with

$ \left.\frac { \minus{} \cos(f(x))}{f'(x)}\right|_0^{\infty} \minus{} \int_0^{\infty}\frac {f''(x)\cos(f(x))}{(f'(x))^2}$

In order to make this work, we would need to know something about the rate of growth not just of $ f$ but also of $ f'.$

You are right Kent.

Here is the general case:

The integrals $ \int\limits_{a}^{\infty}\cos f(x)dx$ and $ \int\limits_{a}^{\infty}\sin f(x)dx$ converge if $ f'$ is monotonically increasing and $ \lim\limits_{x\rightarrow \infty}f'(x)\equal{}\infty$.


A regular pyramid with a square base has edges of length 6 inches and height 4 inches. What is the ratio of the number of cubic inches in the volume of the pyramid to the number of square inches in its surface area?

Volume ($ \frac {1}{3} B*h$)= $ \frac {1}{3}(6^2)(4) \equal{} 48$.
Surface area ($ B \plus{} 4 lateral sides' area$*)= $ 36 \plus{} 4(5) \equal{} 56$.

$ \frac {48}{56} \equal{} \fbox{6/7}$.

*We find the slant height for the height of the triangle with the Pythagorean formula: half an edge of base ($ \frac {1}{2}(6)$) is $ 3$. Height is $ 4$. $ 3^2 \plus{} 4^2 \equal{} 5^2$.

EDITE: oh...whoops. :oops: Thanks for catching it!

[quote="r15s11z55y89w21"]Volume ($ \frac {1}{3} B*h$)= $ \frac {1}{3}(6^2)(4) \equal{} 48$. [/quote]

Correct.

[hide="However..."]

SA = $ 36 \plus{} 6 \times 5 \times 4 \div 2$ or $ 96$

Thus, our answer is $ \boxed{\frac12}$.[/hide]


There are 6 doors, each one having either 1 prize behind it or 1 goat.  There are a total of 4 goats and 2 prizes.  By the way, a goat is different from a prize.  The doors are closed and you cannot see what is behind them.  You pick two of the six doors and they are fixed.  I open two other doors and reveal two goats behind them.

Question:  If you are required to switch BOTH of your picked doors together in one move (and one move only), in favor of the two remaining doors to try to increase your chances of having both prizes, would you switch your two doors?

The probability you got both 0 prizes first off (4/6)(3/5)=(2/5)
Therefore, after you swictch your doors, the probability you get the 2 is 3/5, much great than the 1/15 of you getting both of them off the mark...


calculus
MATHCOUNTS
USAMTS
AMC 8
AMC 10
AMC

Has anyone here ever started their own math team? My school doesn't have a math team, and I have been trying to convince other students to do the AMC-12 exam, which is already extremely difficult. Furthermore, the math department leader at my school has been extremely discouraging (i.e. repeatedly telling me that no one will want to do it) and she has put the burden of paying all costs on any students who want to join. I have tried my best to ask everyone I can, but most people just don't want to. This isn't surprising since most of them associate math contest with math class -- > homework --> boring! Any tips?

Go for math teachers -- if the head of the department is being difficult, first try and line up a teacher to help you with it.  You probably won't be able to get anything big, but if you can establish something with 5 or 6 people, that would be a good start.  Get permission to put up posters, if your school lets you do this.  Try something with an interesting problem or mathematical fact on it to try and get interested people to come to a meeting.  Those would be my first suggestions.  If any of that works, go talk to the department head again and show her what you've got.

Go for it!  I started a math team at my school this year, and while we haven't done extremely well on, say, mandelbrot, it's a start.

And don't worry if your math team turns out to be really small. Our math team has about 25 people per grade, but there should only be about 5 people in my class. You can go faster without having to worry about the others who are behind (most of the people in my class just want to pad their GPA). Also, if you want, you could probably compete in math tournaments at local schools by yourself. Usually they aren't very expensive, and I did this one time in sixth grade when we were just starting a math team. I also agree with Joel. Show your math teacher that you are very interested in math by asking them questions or showing them interesting problems you figured out. Our math team was not student-started, but most of the people see that math is fun and cool once the "math team season" :D starts (which is about mid January here).
Good luck!

[quote="mathnovice"]Has anyone here ever started their own math team? . . . . Any tips?[/quote]

I was the first coach for a new math team, beginning two years ago. A parent in our homeschooling group had a son in the UMTYMP

http://www.itcep.umn.edu/umtymp/

calculus class here in Minnesota, and the son had heard about math competitions from classmates. She wanted her son to have a chance to join competitions, and I was willing to give a whirl at being coach if my son (then fourth-grade age) could tag along. We first called friends and friends of friends to scratch up (barely) a six-person team for Minnesota Junior High Mathematics League

http://members.aol.com/mathleague/

with the families splitting the cost of registering one team. That was a pretty good experience. Some of the kids on my team were quite good, although not experienced in competition problems, but it was challenging for everybody. The MN Junior High Mathematics League founder recommends that everyone sign up for AMC 8 and for MATHCOUNTS, but we didn't that year--all this stuff was too new for us. At the end of the year, one of the parents gave me a thank-you gift: the book  [url=http://www.amazon.com/exec/obidos/tg/detail/-/0425179834/learninfreed]Math Coach: A Parent's Guide to Helping Children Succeed in Math[/url] by Wayne Wickelgren, which has a lot of good tips about forming a team. I suppose this site should link to that book.

Last year we did MN Junior High Math League again, with some changes in personnel, and we had five kids try the AMC 8. I was afraid our group of "ringers" (all recruited from a gifted student homeschooling group) would be accused of cheating, because they were in a good position to score well, but we had good proctoring from a public school in my district, and when I got the official score sheets I smiled, because each test-taker made a different pattern of mistakes, so it was pretty clear they weren't cheating off of one another. One guy in our group tied the top score in MN last year (23), and three of the five were Honor Roll, with one of the two others being Merit Roll. AGAIN we didn't try MATHCOUNTS--my daughter was just being born then, and I was too tired to take on another project. But later in the school year another parent in our group registered a team for Math Masters of Minnesota

http://www.mathmastersmn.org/

and we entered sixth-grade and fifth-grade teams in that contest, with the fifth-grade team doing better, because it PREPARED after the sixth-grade team went in too overconfident and got creamed. (The two meets were about a month apart.)

This year we are in the MN Junior High Math League, where it all began, fielding a short-handed  :cry: team that scores very well in the team part of the competition. I have just registered a team for MATHCOUNTS--but I don't know who all four of my team members will be. We just did the AMC 8, and I hope to join another homeschooling parent for her AMC 10 administration. And I HOPE we can field a Math Masters of Minnesota team (or two) again this year.

Next year I'd like to get some of us into USAMTS, Mandelbrot, and other contests.

Hope this helps!


algebra
polynomial
algorithm
induction
algebra proposed

Let $P(x)$ be a polynomial with degree $ n\in{\mathbb{N}}$. If $ P(0),\ P(1),\ \ldots ,\ P(n)$ are integers, then prove that $ P(k)$ is integer for any integer $k$.

An algorithm for constructing the values of $ P$:
Let $ \delta P(k)\equal{}P(k\plus{}1)\minus{}P(k)$, $ \delta^2 P(k)\equal{}\delta P(k\plus{}1)\minus{}\delta P(k)$, and so on. For $ 0\le i\le n$ and $ 0\le j\le n\minus{}i$, $ \delta^i P(j)$ is an integer; it comes from repeatedly taking differences of integers.

Now, set $ \delta^n P(k)\equal{}\delta^n P(0)$ for all $ k$; $ \delta^n P$ must be constant since $ P$ is a polynomial of degree $ \le n$. We can then build back down the difference table: $ \delta^{n\minus{}1}P(\minus{}1)\equal{}\delta^{n\minus{}1}P(0)\minus{}\delta^n P(\minus{}1),\dots,P(\minus{}1)\equal{}P(0)\minus{}\delta P(\minus{}1)$ and similarly $ \delta^{n\minus{}1}P(2)\equal{}\delta^{n\minus{}1}P(1)\plus{}\delta^n P(1),\dots,P(n\plus{}1)\equal{}P(n)\plus{}\delta P(n)$. We repeat indefinitely to get $ P$ at all positive and negative integers. Every step of the algorithm uses only addition and subtraction, so all of the answers are integers.

My solution is almost same as yours except for my using induction, jmerry. :)

[quote="kunny"]Let$ P(x)$ be polynomial with degree $ n\in{\mathbb{N}}$. If $ P(0),\ P(1),\ \cdots P(n)$ are integers, then prove that $ P(k)$ is integer for any integers $ k$.[/quote]

If we write $ P(x)\equal{}a_0\plus{}\sum_{i\equal{}1}^na_i\frac{\prod_{k\equal{}0}^{i\minus{}1}(x\minus{}k)}{i!}$

we have $ a_0\equal{}P(0)$ and, for $ p\in[1,n]$, $ P(p)\equal{}a_0\plus{}\sum_{i\equal{}1}^{p}a_i\binom{p}{i}$, and so $ a_1\equal{}P(1)\minus{}P(0)$ and $ a_p\equal{}P(p)\minus{}a_0\minus{}\sum_{i\equal{}1}^{p\minus{}1}a_i\binom{p}{i}$ Fpr, $ p>1$

And so all $ a_p$ are integers.

Now, for any $ m>n$ : $ P(m)\equal{}a_0\plus{}\sum_{i\equal{}1}^na_i\frac{\prod_{k\equal{}0}^{i\minus{}1}(m\minus{}k)}{i!}$ $ \equal{}a_0\plus{}\sum_{i\equal{}1}^na_i\binom{m}{i}$ and is an integer.

Now, for any $ m<0$ : $ P(m)\equal{}a_0\plus{}\sum_{i\equal{}1}^na_i\frac{\prod_{k\equal{}0}^{i\minus{}1}(m\minus{}k)}{i!}$ $ \equal{}a_0\plus{}\sum_{i\equal{}1}^na_i(\minus{}1)^{i}\binom{i\minus{}1\minus{}m}{i}$ and is an integer.

And so $ P(n)$ is an integer fo any integer $ n$.


limit
logarithms
real analysis
real analysis unsolved

Let $(a_{n})_{n \in N^{*}}$ be a sequence, strictly decreasing with $a_{n} >0$ $\forall n  \in N^{*}$ with $\displaystyle \lim_{n\to \infty} a_{n}=0$. Let $k \in N^{*}$, $k \neq 1$ and $\beta_{n,k}$ the solutions of the ecuation: $( \log_{k+1+a_{n}} x) (\ln (k)) =\ln(x-1)$. Compute $L= \displaystyle \lim_{k\to \infty} \displaystyle \lim_{n\to \infty} \beta_{n,k}$.

Do you really think there aren't other more interesting problems than those from ADMITERE ASE? I hope you won't post also those from POLITEHNICA, BAC, ...

I don't see why you always pick on me!  :?  :?

Anyway, I can post whatever I like! I guess everybody should send you private messeges to ask you which problem to post and which not!

Whatever......harazi!

I'm verry sorry that you interpreted this in this wrong way. I didn't say that you aren't allowed to post anything you want. I do not understand why consider that what I said is something bad related to your person. Please... No, I just said that some problems really aren't interesting and I don't think this site is their place. But if I'm not allowed even to say this, Ok, I apologise for having a point of view. And yes, I always pick on you, I want to see you dead  :D . Come on, Lagrangia... you know, we have some ages now, we don't play anymore woth toys... And I beg everyone not to send me so many private messages, since my inbox is already full and I don't want to empty it all the time. So, everyone, please nice problems. But anyway, you can post whatever you want ( maybe this is the only good part in what you have said) . It's not my site, I'm not the judge here. I just have some observations, which seem to be missunderstood.

Well I am sorry, but this is the proper section to post this Harazi, so Lagrangia is right.

Maybe a different section will be created in the future, containing all "beautifull" problems, but for now we just have to bear with each other.

Just keep in mind that not all people have the same background/math levels, and the same preferences, so while you dislike this problems, others might enjoy it.

My dear people, but I have just expressed a simple opinion. Did I interdict to anyone posting problems? I don't remember to have done this.

Yes I know Harazi, and I understand your point of view, but I also understand Lagrangia's, that's why I was trying to create a middle path :)

Stay calm, I have nothing with Lagrangia , though he thinks the opposite. Ok, I apologise and promisse I will never say how ugly are those problems. I give my word that I will never say something about any useless problem posted. So be it!
 Ps: I really son't have anything with Lagrangia.


logarithms
trigonometry
MATHCOUNTS
AoPS Books

i was wondering what books are good to look at if u want to do well at mathcounts.

The AoPS books are super! Mostly the first one though; you don't usually use logarithms or advanced trigonometry at MC.

Also, Mathematical Circles is more advanced. You might want to get it after you've done AoPS Vol. 1.

Old Mathcounts tests help. I think the Mathcounts Foundation publishes some of the old tests in book form.

Billy

Math books, of course!!! :lol:

I would also recommend the Art and Craft of Problem Solving by Paul Zeitz.

i dont think strict textbook learning is going to help because mathcounts requires some outside of the box reading

And LOTS AND LOTS OF PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE PRACTICE. You won't believe how many stupid mistakes I make every time. There's a Sticky in this forum about "Mathcounts resources".

mathcounts.saab.org is what I've used.


You mean, what's USAMTS?
[url]http://www.usamts.org/[/url]


Let $f$ a degree $n$ polynomial with integer coefficients. Prove that there is $\xi\in I$ such that $f(\xi)\in\mathbb{Z}$. ;)

$I$ is the set of irational numbers.

Either I don't understand either there is something missing  :|

If $f(x) = x$ such $\xi$ can't exist.


Euler
geometry
circumcircle
perpendicular bisector
geometry unsolved

if the euler line passes through any vertex then the triangle is either right angled or isosceles or both-prove it!

Suppose that the Euler line of $ \triangle ABC$ passes through $ A$, and let $ O$ and $ H$ respectively denote the circumcenter and the orthocenter of $ \triangle ABC$. If $ H\neq A$ then we must have $ AO\perp BC$. Hence $ A$ lies on the perpendicular bisector of $ BC$ and therefore the triangle is isosceles. If $ H\equal{}A$ then obviously the triangle is right (also not necessarily scalene) and the conclusion follows.


Μια απλή αλλά όχι όμορφη λύση είναι η παρακάτω:

Η ανισότητα γράφεται $\frac{a}{b}+\frac{b}{\frac{1}{ab}}+\frac{1}{a}\geq a+b+1 \Rightarrow (b^3-b+1) a^2 - (b^2+b) a + b \geq 0$

που είναι τριώνυμο ως προς $a$    :D   με διακρίνουσα $D=-b(b-1)^{2}(3b+4) \leq 0$

και $(b^3-b+1)=(b+1)(b-1)^{2}+b^2 >0$ .     :)

Για $D=0$ δηλαδή για $b=1$ παίρνουμε $a=1$ και $c=1$ άρα έχουμε ισότητα για $a=b=c=1$.

Θα προσπαθήσω για κάτι καλύτερο  :)

Η άσκηση είναι πραγματικά παράξενη ,άρα θα βάλλω και μια παράξενη λύση . :D

Από την ανισότητα Cauchy έχουμε ότι $\frac{a}{b}+\frac{b}{c}+\frac{1}{a}\geq\frac{(a+b+1)^2}{ab+bc+a}$

άρα αρκεί νδο   $\frac{(a+b+1)^2}{ab+bc+a}\geq a+b+1$  ή ισοδύναμα

$b+abc\geq b(a+c)$  ή ισοδύναμα $(1-a)(1-c)\geq 0$  το οποίο ισχύει !!!!!  αφού το

είπε και  ένας από τους  καλύτερους στο σάιτ  στις  ανισότητες   Vasc  εδώ

http://www.mathlinks.ro/Forum/viewtopic.php?p=455275#p455275

 :D

Μια άλλη λύση είναι η εξής . Τα φέρνουμε όλα στο πρώτο μέλος και θεωρούμε συνάρτηση  $f(a,b,c)$  και χρησιμοποιούμε τη μέθοδο mixing variables .

Αποδεικνύουμε ότι  $f(a,b,c)\geq f(t,t,c)$  όπου  $t=\sqrt{ab}$

Στη συνέχεια αρκεί νδο  $f(t,t,c)\geq 0$     η οποία είναι ισοδύναμη με

$\frac{\sqrt{ab}}{c}+\frac{1}{\sqrt{ab}}-2\sqrt{ab}\geq 0$

ή ισοδύναμα $\frac{1}{c\sqrt c}+\sqrt{c}-\frac{2}{\sqrt{c}}\geq 0$

ή $(c-1)^2\geq 0$  που ισχύει

@Ramanajuan: πολύ ωραία η λύση σου νομίζω ότι είναι η πιο φυσιολογική

[quote="silouan"]
$\frac{(a+b+1)^2}{ab+bc+a}\geq a+b+1$
[/quote]
Σιλουανέ όσον αφορά την πρώτη λύση σου,νομίζω ότι η παραπάνω ανισότητα  δεν ισχύει για $a=2,b=1,c=1/2$ .
Ο Vasc αναφερόταν μάλλον σε συμμετρικές ανισότητες,όπου μπορούμε να βρούμε πάντα δυο μεταβλητές από τα $a,b,c$ ώστε
$(1-a)(1-b)\geq 0$  ή $(1-a)(1-c)\geq 0$ ή $(1-c)(1-b)\geq 0$ ,αφού το γινόμενο τον αριθμών στα αριστερά μέλη είναι πάντα θετικό.
Ξανα δέσ'το!

Στέλιος... :!:

Μάλλον δεν είμαι καλά σήμερα έχεις δίκιο .Απλά παρασύρθηκα

[quote="Ramanujan"]Προσπάθησα να βρώ μια ακόμη πιο φυσιολογική λυση θέτοντας $a=\frac{x}{y} ,b =\frac{y}{z} ,c= \frac{z}{x}$

για να αποφύγω τη συνθήκη αλλα δεν τα κατάφερα. Όταν βρώ χρόνο θα την ξανακοιτάξω εκτός αν με προλάβεις.

 :)[/quote]

σε αυτο στηριζεται η λυση μου, μακροσκελης και παραξενη..... ;)

λοιπον, ας παρουσιασω εν συντομια τη λυση μου.....

συμφωνα με τα προηγουμενα, παρατηρουμε οτι οι τριαδες $(\frac{1}{y^2},\frac{1}{z^2},\frac{1}{xz})$ και $(xz,xy,yz)$ ειναι ομοιοτροπα διατεταγμενες.
Μετα τσεμπισεβ και λιγες πραξεις....

 :)

[quote="socrates"]
Μετα τσεμπισεβ και λιγες πραξεις....
 :)[/quote]

Θα ήθελα να τις δω αν έχεις χρόνο  :D   :)

[quote="Ramanujan"]
Θα ήθελα να τις δω αν έχεις χρόνο  :D   :)[/quote]

οκ!  :)

Με εφαρμογη της ανισοτητας τσεμπισεφ λαμβανουμε

$\text{LHS}\geq \frac{1}{3} \cdot (\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{xz})(xy+yz+zx)$

oποτε αρκει νδο $(\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{xz})(xy+yz+zx)\geq 3\frac{x}{y}+3\frac{y}{z}+3$

Αναπτυσσοντας και ομαδοποιωντας έχουμε

$(\frac{xz}{y^2}+\frac{x}{y}+\frac{x}{z})+(\frac{xy}{z^2}+\frac{y}{z}+\frac{y}{x})+(\frac{z}{y}+\frac{y}{z}) \geq 3\frac{x}{y}+3\frac{y}{z}+2$

που ισχυει ως αθροισμα των ανισοτητων

$\frac{xz}{y^2}+\frac{x}{y}+\frac{x}{z}\geq 3\frac{x}{y}$

$\frac{xy}{z^2}+\frac{y}{z}+\frac{y}{x}\geq 3\frac{y}{z}$

$\frac{z}{y}+\frac{y}{z}\geq 2$

που προκυπτουν απο την ανισοταυτοτητα του Cauchy (AM-GM)!

Eλπιζω να ήμουν κατανοητος και κυριως σωστος στους συλλογισμους και τις πραξεις!
Ελεγξτε τη λυση για επιβεβαιωση της ορθοτητας!

καλο πασχα!
 :lol:

Μπράβο  Socrates πολύ όμορφη λύση  :)


Donna and Rod bought a new van at an "end of the year sale" for "dealer cost plus $ 5\%$." If they paid $ \$17,325.00$, what was the dealer cost in dollars?


geometry
trigonometry
inequalities
cyclic quadrilateral
geometry proposed

In a cyclic quadrilateral $ ABCD, a,b,c,d$ are its side lengths, $ Q$ its area, and $ R$ its circumradius. Prove that:

$ R^2\equal{}\frac{(ab\plus{}cd)(ac\plus{}bd)(ad\plus{}bc)}{16Q^2}$.

Deduce that $ R \ge \frac{(abcd)^{\frac{3}{4}}}{Q\sqrt{2}}$ with equality if and only if $ ABCD$ is a square.

$ Q\equal{}area(ABD)\plus{}area(BCD)\equal{}\frac{ad\sin A}{2}\plus{}\frac{bc\sin C}{2}\equal{}$

$ \equal{}\frac{(ad\plus{}bc)\sin A}{2}\equal{}\frac{(ad\plus{}bc)BD}{4R}\Rightarrow R\equal{}\frac{BD(ad\plus{}bc)}{4Q}$

$ Q\equal{}area(ABC)\plus{}area(ADC)\equal{}\frac{(ab\plus{}cd)\sin B}{2}\equal{}\frac{(ab\plus{}cd)AC}{4R}\Rightarrow$
$ \Rightarrow R\equal{}\frac{AC(ab\plus{}cd)}{4Q}$

Then $ R^2\equal{}\frac{AC\cdot BD(ad\plus{}bc)(ab\plus{}cd)}{16Q^2}$

But ABCD is cyclic and $ AC\cdot BD\equal{}ac\plus{}bd$

So $ R^2\equal{}\frac{(ad\plus{}bc)(ab\plus{}cd)(ac\plus{}bd)}{16Q^2}$

To prove the inequality apply AM-GM inequality.

The equality holds if $ ab\equal{}cd, \ ac\equal{}bd, \ ad\equal{}bc\Rightarrow a\equal{}b\equal{}c\equal{}d$


group theory
abstract algebra
superior algebra
superior algebra unsolved

1. Let $G$ be a finite group such that $[G,G]=G$. Prove that if $K$ is a cyclic, normal  subgroup of $G$ then $K$ is a subgroup of $Z(G)$.

2. Prove that having a finite group $G$ and a normal subgroup $N$ of $G$ the implication is true :
if there is exactly one subgroup $D$ of $G$ such that $ND=G$ and $D \cap N =1$ then
$N$ is a simple group.

I'm sure it's simple but I haven't found any simple approach, any ideas ?

[quote="Rafal"]1. Let $G$ be a finite group such that $[G,G]=G$. Prove that if $K$ is a cyclic, normal  subgroup of $G$ then $K$ is a subgroup of $Z(G)$.
[/quote]

Write $K=<a>$. It is enough to show that $a \in Z(G)$. Take $u \in G$. Then $u = xyx^{-1}y^{-1}$ for some $x,y \in G$. Now:

$ua=au \Leftrightarrow$
 $xyx^{-1}y^{-1}a=axyx^{-1}y^{-1} \Leftrightarrow$
$y^{-1}x^{-1}axy = x^{-1}y{-1}ayx$.

But both sides belong to $K$ (since $K$ is normal in $G$) and they have same order, whence they are equal, and the initial equality holds.

[quote="ZetaX"][quote="julien_santini"]But both sides belong to $K$ (since $K$ is normal in $G$) and they have same order, whence they are equal, and the initial equality holds.[/quote]
Why are they equal then¿[/quote]

They are not necessarily equal, I went too fast  :oops: .

Well eventually all I could get is the case where $|K|$ has a $p$-Sylow $S$ of order $|p|$, where $p$ is the smallest prime dividing $G$. In this case $G$ acts on $S$ by conjugation (since $K$ is normal, and every conjugate of an element of $K$ of order $p$ has order $p$, whence $S$ is normal in $G$ since $K$ is abelian), meaning we have an homomorphism $f: G \rightarrow S(K)$, and since $|f(G)|=1$ or $p$, then $f(G)$ is abelian, meaning $f$ contains $G=[G: G]$ in its kernel, which gives the result.
Note that I did not use the cyclicity of $K$ for this weaker result (just its commutativity).

I did not find anything more in the general case; any idea ?

[quote="Rafal"]2. Prove that having a finite group $G$ and a normal subgroup $N$ of $G$ the implication is true :
if there is exactly one subgroup $D$ of $G$ such that $ND=G$ and $D \cap N =1$ then
$N$ is a simple group.[/quote]

As I understand the problem, it is not true. Take $n$ to be any group of odd order, and $G=N\times\mathbb Z_2$. Any group $D$ satisfying $ND=G,\ N\cap D=1$ must have order $2$, and there's only one element of order $2$ in $G$, so $D$ is unique. However, $N$ can be taken to be non-simple. WHat am I missing here? :?

your explenation seems and even is correct, I just translated it precisely from an exercise book and probably there was somewhere a typo.


Indiana University-Purdue University Indianapolis most likely

[quote="s0mp"]mathfanatic, what does iupui stand for? sounds familiar.[/quote]

I've heard of the name from college basketball, but yep, Syntax is right, I think.

yep, they're in the mid-continental league (meaning the crap of the crap teams). the only way a team from mid-con makes it to march madness is by winning the conference tourny, usually its valporaiso though.

I as well got 1000.  No complex mathematical proof though but probably the same thinking, his was just done mathematically.  First I simplified the top and bottom of the various expressions in each parenthesis.  I had (12)(30)(56)(90)/(6)(20)(42)(72).  I noticed several patterns emerge that proved useless.  I then simplified and multiplied together to get 5.  I started to think that the number of terms on the top or bottom + 1 = solution.  I decided to test that by calculating the next two terms and my assertion held true.  I divided 1998 by 2 and then added 1 and got 1000.


What is the product of the first 20 non-negative even numbers.

Please show all of your work. ;)

[hide]0 is the first non-negative even number, so the answer is 0.[/hide]

I don't know, but here is my guess.  Even means not odd, and 1,3,5... and -1,-3,-5... are odd.  Since 0 is not on that list, 0 is even.  :?

An integer is even if it is divisible by two (in other words, an integer $n$ is even iff $n \equiv 0 \bmod 2$).  $\frac{0}{2}$ leaves remainder $0$ upon division by two, so it is divisible by two and it is the first non-negative even integer.

However, if the question were modified to ask for positive integers instead, then the answer is $40!!$ (where $n!! = n(n-2)(n-4)...(4)(2)$ is the multifactorial) $= 2^{20} 20! \approx 2.55108... \times 10^{24}$.

Robinhe: Odd numbers are 1,3,5....etch and -1,-3,-5...etc. Even numbers are 2,4,6,8...etc and -2,-4,-6...... zero isn't any of those numbers so..............

t0rajir0u: As you have said, an even number gives a remainder of 0 when divided by 2. Yet zero gives a remainder of 0 when divided by 3,1,5,7.....etc......so it might be odd......but it isn't......

This leaves us where we started..............I still am not sure why 0 is even and not odd or undefined.

[quote="dogseatcheese"]Yet zero gives a remainder of 0 when divided by 3,1,5,7.....etc......so it might be odd......but it isn't......[/quote]

$6$ also leaves remainder 0 when divided by 3.  Is it odd?   :D

An odd number is defined as one that is not divisible by two, that is, an integer $n$ is odd iff $n \equiv 1 \bmod 2$.

I agree with t0rajir0u:

If a number is even, then $n\equiv0\pmod{2}$.

If a number is odd, then $n\equiv1\pmod{2}$.

0 clearly is not 1 (mod 2), and it is evenly divisible by 2, so it is EVEN.

hm....

when I first learned about integers, one of my textbooks said 0 is a neutral, number still debated wheather it's even or odd...

so I think you really have to watch the wordings of this problem

No, I don't think so.  Any integer is either $\equiv 0 \bmod 2$ or $\equiv 1 \bmod 2$, and those definitions correspond exactly to the notions of evenness and oddness for natural numbers, so it's a perfectly logical extension to all integers, which includes zero.

[quote="dogseatcheese"]Yes....that is the correct answer....But why is 0 even? What is the definition of even? I don't know it.........[/quote]
A number $n$ is even iff $n\equiv 0\pmod 2$.
A number $m$ is odd iff $m\equiv 1\pmod 2$.

I hope that helps. :)

Masoud Zargar

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

as a little extra reference.

[hide]$0$[/hide]

the link is use full...(236 factorial) ;)

What are the quotient and the rest for the following divisions:
$0.\ \ \boxed {0\ : \ 7}\ ;\ \boxed {0\ : \ (-7)}\ ;\ \boxed {3\ : \ 7}\ ;\ \boxed {3\ : \ (-7)}\ ;\ \boxed {(-3)\ : \ 7}\ ;\ \boxed {(-3)\ : \ (-7)}\ ;$
$1.\ \ \boxed {109\ : \ 7}\ ;$
$2.\ \ \boxed {109\ : \ (-7)}\ ;$
$3.\ \ \boxed {(-109)\ : \ 7}\ ;$
$4.\ \ \boxed {(-109)\ : \ (-7)}\ .$
Conclude a general rule for the division $\boxed {\ a\ : \ b\ }$, where $a\in Z\,\ b\in Z^*.$

[u]Remark.[/u] For any $a\in Z,\ b\in Z^*$ there are and are unique the[b] quotient [/b]$q\in Z$ and the [b]rest[/b] $r\in N$ so that $0\le r<|b|$ and $\boxed {\ a=q\cdot b+r\ }\ .$

[b]Prove that[/b] $q=(sgn\ b)\cdot \left [\frac{a}{|b|}\right]$ and $r=|b|\cdot \left\{\frac{a}{|b|}\right\}$.
Denote $x\in R,\ x=[x]+\{x\},\ [x]\le x<[x]+1,\ 0\le \{x\}<1,\ x=|x|\cdot sgn\ x.$


geometry
incenter
circumcircle
search
geometric transformation
homothety
geometry unsolved

M,N,P are midpoint of BC,CA,AB respectively.F is the Feuerbach point, MF,NF,PF cut ⊙I (incircle) at R,S,T, respectively. I,G,O are incenter, centroid,circumcenter respectively.prove:
(1)if NF>MF , NF>PF, then MF+PF=NF
(2) AR , BS ,CT , OI , FG concurrent at Z, the interior similar center of ⊙O and ⊙I

What is Feurbach point ? Is it center of Feurbach circle ?

Dear "thaithuan",
tha Feuerbach's point is the point of tangancy of the incircle with the nine points circle of a triangle.
I found after my research the first complete synthetic proof of this result (see my website).
Sincerely
Jean-Louis

Dear plane geometry,

your first problem is quite well-known. See http://www.mathlinks.ro/viewtopic.php?search_id=1436568649&t=24959 for Darij's solution.
I think the second problem would be better if you didn't give away that $ Z$ is $ X_{55}$.

[b]Proof:[/b]
Because $ F_e$ is the external center of similitude of $ (I)$ and $ (W)$, we have that $ IR$ and $ WM$ are parallel. A homothecy with center $ H$ and factor $ 2$ maps $ W$ to $ O$ and $ M$ to the antipode of $ A$ in $ (O)$. Hence, $ IR \parallel WM \parallel OA$, which proves that $ AR$ passes through $ X_{55}$, the internal center of similitude of $ (O)$ and $ (I)$. Similarly, we prove that $ BS,CT$ pass through $ X_{55}$.

Now, we only need to prove that $ F_eG$ passes through $ X_{55}$. Because $ H$ and $ G$ divide $ W$ and $ O$ harmonically, we have that $ G$ is the internal center of similitude of $ (W)$ and $ (O)$. By Monge-d'Alembert, $ FG$ passes through $ X_{55}$.

Thank you, dear Jan，It is rather nice！
This problem belongs to a friend of mine,who is very strong in Geometry


function
analytic geometry
graphing lines
slope
symmetry

A certain function $f$ has the properties that $f(3x)=3f(x)$ for all positive real values of $x$, and that $f(x)=1-\mid x-2 \mid$ for $1\leq x \leq 3$. Find the smallest $x$ for which $f(x)=f(2001)$.

[hide="Solution"] The graph of $f(x)$ just consists of copies of the path $(1, 0) \to (2, 1) \to (3, 0)$ "expanded" by $3$ every time $x$ jumps to the next power of $3$, so you just need to figure out the distance $f(2001)$ is from the closest power of $3$ (which is $2187$), and since the graph's slope doesn't change this means that $f(2001) = f(2187)+186 = 186$.

Now, between $3^{k}$ and $3^{k+1}$ the function reaches a maximum of $3^{k}$, and the smallest value of $k$ for which $3^{k}> 186$ is $k = 5$, so somewhere between $3^{k}= 243$ and $3^{k+1}= 729$ we will have another $186$.  By symmetry, this occurs at $243+186 = \boxed{ 429 }$.

I think.  It's late, and when it's late I mess up calculations.  :([/hide]

p4: First note that $f(2001) = 3^6f\left(\frac{2001}{3^6}\right) = 186$, since we can extend our given equation to $f(x)=3^k f\left(\frac{x}{3^k}\right)$. Next, note that the range of $1-|x-2|$ is $[0,1]$, so we want to be sure our value of $k$ gives a value for $\frac{186}{3^k}$ in the range $[0,1]$. The only working value is $k = 5$, giving $243 f\left(\frac{x}{243}\right) = 186$. We now require $f\left(\frac{x}{243}\right) = \frac{62}{81}$, and we can solve for $x$ to get $x = \boxed{429}, 543$.

I got 186 but forgot to look at the range of $f(x)$ :noo


Ned and Nellie were playing on the elevator in their 12-story apartment complex one evening.  They entered the elevator on the first floor, and decided to reach their eleventh-floor apartment by continually repeating the following sequence of steps.  Ned pushed the button of the floor that was three floors above where they were currently, and upon arriving, Nellie then pressed the floor of the nearest even floor below where Ned's button led them.  Assuming no one interrupted their game, how many times did they have to press a button before they could get out on the eleventh floor?

[hide]9. I did find a pattern, but there aren't like a hundred floors, so there's no point using it (btw, though, it is: After you get to the fourth floor after four button presses, every four button presses from then on will make you go higher by 4 floors). So I basically kept track of what floor I was on until I got to 11. [/hide]

[hide="My solution"]
There are 2 ways to solve this problem:
[list]1) [b]Just write out the numbers.[/b] There aren't a lot, making this the faster method:
$1\rightarrow 4\rightarrow 2\rightarrow 5\rightarrow 4\rightarrow 7\rightarrow 6\rightarrow 9\rightarrow 8\rightarrow 11$ : $10-1$ (first # doesn't count) $=9$.

2) [b]Find a pattern.[/b] In this particular problem, it would take a bit longer, but it is better to use this method if you know that there are too many moves to count or if you don't know how many moves there are. Here, after reaching $4$, every two-move sequence brings you up 2 levels:
$1\rightarrow 4\rightarrow 2\rightarrow 5\rightarrow 4$ : $4$ moves + $8-4=4$ moves (to go from $4$ to $8$) + $1$ move from $8$ to $11$ : $9$ moves total.
[/list][/hide]

EDIT: Long post, made several typos

[hide]Nine times.  I didn't need to use a pattern though I did find one.  1+3=4 and 4-2=2 Every sequence of two button pushes after that was always the next even number because even+odd=odd and odd-1=even.  In this case, our odd increasement is 3 so 3-1=2.[/hide]


inequalities
function
inequalities solved

Hint :
consider f(x)=x/(1+x 2 ) . Prove:
 If (x+y)(xy-3) \leq 0 then  f(x)+f(y) \leq 2f((x+y)/2)

How can  you  use  Jensen?? This  function  has not  f''>0  or  f''<0  at  the  interval  [-3/4,+00)...

We have $(4a + 3)(3a - 1)^2 \geq 0$ (and the analogous inequalities).
Hence $36a^3 + 3a^2 - 14a + 3 \geq 0$ so \[36(a^3 + a) + 3(a^2 + 1) \geq 50a.\]
Dividing by $a^2 + 1$ we get \[36a + 3 \geq \frac{50a}{a^2 + 1}\] and the two analogous inequalities.
Adding these we get \[\sum \frac{a}{a^2 + 1} \leq \frac{9 + 36 \sum a}{50} = \frac{9}{10}.\]


Here's a problem my uncle gave me in an email.  I'm stumped on this.  I'd appreciate a solution for this as good as possible.

[hide="The problem"]I am really stumped with this.  I have researched it (online and elsewhere) and have asked for expertise from two math teacher friends, but they were equally confused.

When dividing items among three people, what is the fairest way to take turns picking the items each person wants?  This came up recently because the three kids in my family were recently dividing up the estate items of a maiden aunt who'd died.  We shuffled together three pieces of paper (printed with the numbers 1, 2 and 3) and then each person blindly picked one of the pieces of paper.  I thought we'd go #1, #2, #3, and then the same thing over and over again.  But my brother said, no, the fairest way to do it is that the order reverses so that the "rounds" of picking items go like this:

1, 2, 3
3, 2, 1
1, 2, 3
3, 2, 1
etc.

It seems to me that while this is pretty fair to #1 and #3, it's less so to #2.  But I can't find any mathematical/probability stuff on this to prove what I suspect is true.  Is there some mathematically correct way to ensure fair and equal turn-taking in a situation like this?

As it turns out, I was #2 and got everything I wanted and then some--so this isn't a sour grapes issue.  But I got to thinking about it purely from a math standpoint and wondering what is the simplest, fairest, most equal system.  Thanks for your input.[/hide]

See http://en.wikipedia.org/wiki/Fair_division and http://mathworld.wolfram.com/CakeCutting.html .

The discrete problem is harder than the continuous problem, but incidentally a friend of mine at RSI (edit: that's the poster below!) was doing research on this subject.  The three-person problem has an explicit solution, but I don't remember it.

There are a number of algorithms that can be used here to try to allocate the items fairly, but there is no "best" way, as long as the items are indivisible. (Imagine if all three people wanted one of the items a lot and didn't care for the other two. There is quite literally no way to resolve this issue.)

On the other hand, if you introduce a few other allowable steps, it gets more fun. For example, if all three individuals have a lot of cash, they can bid on the items to divide them among each other, by assigning a value to each item. You would want to use the Knaster procedure, which works something like this (I will use just a single item in this example--it can easily be extended to more by simply repeating the procedure):

Let all three individuals "bid" on the item to be given out. In other words, have them each individually write down what they believe the item to be worth. Then the bids are revealed and the person who placed the highest bid receives the item. He keeps the item and pays 2/3 the value of the item. From that amount, each of the other two individuals then receives 1/3 the value of the item (as they recorded it). Since each of the other two individuals' bids were less than the bid of the person who received the highest bid, there will be money left over. This money is distributed equally among all three individuals.

Here is a more concrete example. Qiaochu, Sam, and Winston have come into possession of a vintage edition of Harry Potter, a V for Vendetta DVD, and a cane. They want to divide the three items between each other.

So each person bids secretly on each item. Suppose the bids are as follows:

Qiaochu:
Harry Potter: $210 V for Vendetta:$150
Cane: $60 Sam: Harry Potter:$90
V for Vendetta: $60 Cane:$90

Winston:
Harry Potter: $240 V for Vendetta:$120
Cane: $75 Note that Qiaochu wins V for Vendetta, Sam wins the cane, and Winston wins Harry Potter. Now, we calculate how much money each person pays or receives (remember that if a person receives an item, they pay 2/3 of what he or she believes its value is, and if a person does not receive an item, they receive 1/3 of what he or she believes its value is): Qiaochu: 1/3($210) + V for Vendetta - 2/3($150) + 1/3($60) = V for Vendetta and pays $10 Sam: 1/3($90) + 1/3($60) + Cane - 2/3($90) = Cane and pays $10 Winston: Harry Potter - 2/3($240) + 1/3($120) + 1/3($75) = Harry Potter and pays $95 Now this means that there is 10 + 10 + 95 =$115 floating around, and it is divided equally between Qiaochu, Sam, and Winston.

To see why this algorithm works, consider the players and what they receive with respect to a  single item. Either a) they receive the item and pay 2/3 of its perceived worth, or b) they do not receive the item but they receive 1/3 of its perceived worth. Either way, they should be satisfied. Also note that there is money left over that is divided equally, so each person will actually receive a bit more than 1/3 the value of the item.

[quote="winluo2"]Qiaochu:
Harry Potter: 210 dollars
V for Vendetta: 150 dollars
Cane: 60 dollars

Sam:
Harry Potter: 90 dollars
V for Vendetta: 60 dollars
Cane: 90 dollars

Winston:
Harry Potter: 240
V for Vendetta: 120 dollars
Cane: 75 dollars

Note that Qiaochu wins V for Vendetta, Sam wins the cane, and Winston wins Harry Potter. Now, we calculate how much money each person pays or receives (remember that if a person receives an item, they pay 2/3 of what he or she believes its value is, and if a person does not receive an item, they receive 1/3 of what he or she believes its value is):

Qiaochu: 1/3(210) + V for Vendetta - 2/3(150) + 1/3(60) = V for Vendetta and pays 10 dollars

Sam: 1/3(90) + 1/3(60) + Cane - 2/3(90) = Cane and pays 10 dollars

Winston: Harry Potter - 2/3(240) + 1/3(120) + 1/3(75) = Harry Potter and pays 95 dollars

Now this means that there is 10 + 10 + 95 = 115 dollars floating around, and it is divided equally between Qiaochu, Sam, and Winston.[/quote]

Dollar signs.  Tsk tsk.  Also, I hate you.  :P


superior algebra
superior algebra unsolved

1. Let G be a group. Prove that if there are 3 distinct elements x, y, z in G s.t. x^2=y^2=z^2=e, e is the identity, then G is NOT a cyclic group.

2. Prove that if k>=3 then (Z_(2^k))* is NOT a cyclic group.

1.: more general, the equation $x^{n}=e$ has at most $n$ solutions in a cyclic group. We even have the converse:
If the equation $x^{n}=e$ has always at most $n$ solution in the finite group $G$, then $G$ is cyclic.

2. Posted before the last days, it's not hard to find it.


ratio
trigonometry
geometry proposed
geometry

[color=blue]Let $A,C,B,D$ four collinear points (in this order) so that $\frac{CA}{CB}=\frac{DA}{DB}\ .$ Denote the middlepoints $M\ ,$ $N$ of the segments $[AB]\ ,$ $[CD]$ and the point $P$ for which $PA=PD\ ,$ $PA\perp PD\ .$ Prove that $m(\widehat {MPN})=45^{\circ}\ .$

See the post http://www.mathlinks.ro/Forum/viewtopic.php?t=85445 [/color]

WLOG, A, C, B, D follow on the line in this order. Denote AB = c, BC = a, CA = b. Obviously, b > a. Point C cuts the segment AB internally in the ratio $\frac b a$, point D cuts the segment AB externally in the ratio $\frac b a.$

$AC = \frac{bc}{b + a},\ \ \ AD = \frac{bc}{b - a}$

$CD = AD - AC = \frac{bc}{b - a} + \frac{bc}{b + a} = \frac{2abc}{b^2 - c^2}$

Let Q be the foot of a normal from P to the given line. Since PA = PD, Q is the midpoint of AD.

$PQ = \frac{AD}{2} = \frac{bc}{2(b - a)}$

$MQ = \frac{AD}{2} - \frac{AB}{2} = \frac{bc}{2(b - a)} - \frac c 2 = \frac{ca}{2(b - a)}$

$NQ = \frac{AD}{2} - \frac{CD}{2} = \frac{bc}{2(b - a)} - \frac{abc}{b^2 - c^2} = \frac{bc}{2(b + a)}$

$\tan \widehat{MPQ} = \frac{MQ}{PQ} = \frac a b$

$\tan \widehat{NPQ} = \frac{NQ}{PQ} = \frac{b - a}{b + a}$

$\tan \widehat{MPN} = \frac{\tan \widehat{MPQ} + \tan \widehat{NPQ}}{1 - \tan \widehat{MPQ} \tan \widehat{NPQ}} = \frac{\frac a b + \frac{b - a}{b + a}}{1 - \frac a b \cdot \frac{b - a}{b + a}} =$

$= \frac{a(b + a) + (b - a)b}{b(b + a) - a(b - a)} = \frac{a^2 + b^2}{b^2 + a^2} = 1$

$\angle MPN = 45^\circ$


http://www.youtube.com/watch?v=uSQnK5FcKas

Fortunately enough, there are quite some people with digital cameras or camcorders and conscience nowadays.

[color=green]Moderator says: there is [i]never[/i] a good reason to block-quote the entire post preceding yours.[/color]

One still shot from a TV station does not a conspiracy make.

[color=green]Moderator says: there is [i]never[/i] a good reason to block-quote the entire post preceding yours.[/color]

What would you like to say after most of these "innocent" media confessed their falsifications, openly or covertly?

I understand that it is very unpleasant to be proved as a liar, so the fact that some media like CNN confessed but began new falsification on the same topic is also expectable and understandable.

Do not be so CNN!

Obviously, the media interprets all events through the lens that best serves its own interests (in this case, by espousing anti-Chinese rhetoric).  The events in Tibet are no exception to this rule.  However, there is no denying that China has ruthlessly occupied Tibet for decades and responded in a disproportionate and abhorrent manner to civil unrest in Tibet.

[quote="Begoner"]However, there is no denying that China has ruthlessly occupied Tibet for decades[/quote]

If the word "occupied" is to be used, then you should replace "decades" with "more than one thousand years"

[quote="Begoner"]responded in a disproportionate and abhorrent manner to civil unrest in Tibet.[/quote]

How do you know this? Via western liars like CNN, BBC et al et al?


trigonometry
function
calculus
calculus computations

Consider a sequence defined with:

$ \sin(\phi_n \minus{} \phi_{n \plus{} 1}) \equal{} \Delta \sin(\phi_n)$

where:

$ n \in \mathbb{N}_0$

$ 0 < \Delta < 1$

$ \phi_0 \equal{} \sqrt {1 \minus{} \Delta^2}$

Try to calculate:

$ P \equal{} \frac {\Delta}{2} \sum^\infty_{n \equal{} 0} \sin(\phi_n)$

The problem came from a situation where you have an infinite number of sticks of height $ 1$ positioned vertically on the $ x$ axis and apart from each other by $ \Delta$ (so the first one is at $ x \equal{} 0$, the second at $ x \equal{} \Delta$, the third at $ x \equal{} 2\Delta$ and so on). The one at position $ x \equal{} 0$ is fixed, while others are pushed (actually, the one at position $ x \equal{} \infty$ is pushed to the left, and they knock each other out - like domino) to lean one on the other. Now, the question is, what is the surface that is created under these sticks. I believe that the sum above describes it. Do you think there is another way to calculate this? An easier one, perhaps? Or can you do this one? Also, I know it's much, but what about the series. Can we get an explicit solution (my opinion - not a chance)? Or a good approximation at least? I tried with defining a new sequence by $ \sin(\phi_n) \equal{} f_n$ but it didn't get me far. Any ideas are welcome!!

I realize that this does not directly address your question, but I'll throw it in anyway.

Suppose we have a function $ f(x)$ such that the tangent line at any point $ (x,y)\equal{}(t,f(t))$ intersects the $ x$ axis at the point $ (x,y)\equal{}(t\plus{}\Delta, 0)$. This is fundamentally similar to your setup, except that the tangents are taken continuously and independently instead of sequentially.

We want the points $ (t,f(t))$ and $ (t\plus{}\Delta,0)$ to solve the equation $ y\equal{}mx\plus{}b$. Plugging in these two pairs gives that the tangent line is $ y \equal{} \minus{} \frac{f(t)}{\Delta}x \plus{} \frac{(t\plus{}d) f(t)}{\Delta}$. Thus, $ f'(t) \equal{} \minus{} \frac{f(t)}{\Delta} \implies f(t) \equal{} C e^{\minus{} \frac{t}{\Delta}}$

Thus, the function that satisfies this continuous condition is $ f(x) \equal{} Ce^{\minus{} \frac{x}{\Delta}} \qquad C \in \mathbb{R}^{\geq 0}$

While your condition is not the same, and the joint points on the curve are taken at at non-constant invervals, you can see that the end behavior of your function will resemble exponential decay.

That being said, I think you'd be better off allowing the sequence of points to be $ (a_n, b_n)$, where $ a_n$ and $ b_n$ are interlaced sequences, by finding the intersection between lines and circles. Then you can find a relationship between $ a$ and $ b$, so that it suffices to look at only one sequence, and each item will be defined in terms of the previous. You can then see if you can find the formula of a curve that fits those points.

More precisely, you can show that $ \varphi_n\equal{}A(n)(1\minus{}\Delta)^n$ where $ A(n)$ tends to a finite positive limit $ A$ as $ n\to\infty$. I doubt we can explicitly find $ A$ in terms of $ \Delta$ though.


function
rate problems
calculus
calculus computations

Could someone please walk me through these problems including the answers! Thanks

Fin the average rate of change in the function -4x^2-4x-5 over the interval [-4,-4+h].

Fin the average rate of change in the function x^3+7x over the interval [-2,-2+h].

Is this right?(f[-2+h]-f[-2])/(-2+h-(-2)
=(f[-2+h]-f[-2])/h

That's the definition (in this case, for the second problem). Now calculate and simplify.

[quote="knowltng"]Could someone actually do the problems below for me?[/quote]
Well, no. That's not what we do around here. We're willing to give you hints and suggestions, and if you'll post your own work, we'll comment on it and correct it. But you need to bring your own efforts into this.

I am  very sorry and I understand I have done the work but did not want to take the time to type it all in here. Here is my work for the problems above could you tell me if they are right. If they are wrong can you correct them and walk me to the right answer?
Fin the average rate of change in the function x^3+7x over the interval [-2,-2+h].

f(-2+h)=(-2+h)^3-14+7h

f(-2)=-22

f(-2+h)-f(-2)=(-2+h)^3+8+7*h, which simplifies to
19*h-6*h^2+h^3.

When you divide this by h, you get 19-6h+h^2 is this the answer?

#2
Fin the average rate of change in the function -4x^2-4x-5 over the interval [-4,-4+h].
F(-4+h)=(-4+h)^2-16-5
F(-4)=-37
F(-4+h)-F(-4)=(-4+h)^2+7*h At this point I am so lost with this problem I know I have to simplify but could someone finish the problem for me. For some reason i just can't come up with the answer

#1: You're right.
#2: You haven't calculated the values at $ -4+h$ or at $ -4$ correctly, and the constant terms won't go away when you subtract unless you expand out the square first. The value at $ -4$ is actually $ -64+16-5=-53$.


Is there a way to number the pages so that the first page is not numbered and the second page starts off as page 1?
Thanks.

On the first page put
[code]\thispagestyle{empty}
\setcounter{page}{0}[/code]

I would like to adjust my page numbering. What I want is a little bit more difficult:

The page numbering should look like:

\Roman{chapter}-\arabic{page}

I have tried to do use the following code:

[code]

\let\stdchapter\chapter

\renewcommand{\chapter}[1]{\stdchapter{#1}\addjust}

\newcommand{\addjust}{\setcounter{page}{1}}

\renewcommand{\thepage}{\Roman{chapter}-\arabic{page}}
[/code]

This isn't the way to go because you get in trouble with tableofcontents and bibliography

Does anyone know the answer?

[quote="Sirinial"]This isn't the way to go because you get in trouble with tableofcontents and bibliography[/quote]
I'm not sure what you mean. Provided you put the \addjust macro before you use it, rather than after as you had it, then the page numbers of \tableofcontents will be correct. Not sure about the relevance of the bibliography.

However, why not use a package ready made for this [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=chappg]chappg[/url]?


1. Prove that if $n$ is odd, then $\frac{2}{n}$ can be written as the sum of two distinct unit fractions.

2. Prove that any unit fraction $\frac{1}{a}$ can be written as the sum of two distinct smaller unit fractions.


topology
superior algebra
superior algebra unsolved

Suppose M and N are two orientable manifolds without boundary. Show that M x N is also orientable.

I'm not sure about this, but if we let $ M*$ an $ N*$ be the orientable double covers, we can use the fact that the product of these covering spaces is (i think) the orientable double cover for $ M\times N$ (certainly it covers it and is 2-1, but i'm not sure if it is the cover we want to look at). it is connected if and only if $ M\times N$ is not orientable, but because neither $ M*$ nor $ N*$ were connected to begin with, we conclude that $ M\times N$ is orientable.


geometry
circumcircle
trapezoid
AMC
USA(J)MO
USAMO
perpendicular bisector

Let $ ABC$ be a non-isosceles triangle. Let $ M,N,P$ denote the midpoints of $ BC,CA,AB$ respectively. $ D$ and $ E$ are two points on $ AM$ such that $ DA \equal{} DB$ and $ EA \equal{} EC$, $ BD$ meets $ CE$ at $ F$. Prove that $ A,N,F,P$ lie in a circle.

The A-median $ AM$ cuts the triangle circumcircle $ (O)$ again at $ M_o.$ $ BD, CE$ cut $ (O)$ again at $ D_o, E_o.$ Since $ PD, NE$ are perpendicular bisectors of $ AB, CA,$ $ D_oM_o \parallel AB$ and $ E_oM_o \parallel CA.$ In addition, $ \angle BE_oM_o \equal{} \angle BAM_o$ and $ \angle CD_oM_o \equal{} \angle CAM_o.$ It follows that $ BE_o \parallel CD_o$ are both parallel to the A-symmedian $ AK.$ $ F \equiv BD_o \cap CE_o$ is the diagonal intersection of the isosceles trapezoid $ BE_oD_oC.$ Using central projection, project the circumcircle $ (O)$ to a circle $ (O')$ with center $ M.$ The projected $ \triangle A'B'C'$ becomes right at $ A'$ and the A-symmedian $ AK$ of the original triangle is projected to its A'-altitude $ A'K'$. Since $ BE_o \parallel AK \parallel CD_o$ concur at $ Q$ at infinity, it follows that $ B'E_o', C'D_o'$ meet at $ Q' \in A'K'.$ Since $ B'C'$ is a diameter of the projected circle $ (O'),$ the angles $ \angle B'E_o'C', \angle C'D_o'B'$ are right and $ F' \in A'K'$ is the orthocenter of the $ \triangle Q'B'C'.$ As a result, the original point $ F \in AK.$ But the common perpendicular bisector $ OF$ of $ BE_o, CD_o$ is perpendicular to $ AK \equiv AF,$ hence the angle $ \angle AFO$ is right. Consequently, $ F$ is on the circle $ \odot(APON)$ with diameter $ AO.$

[quote="mr.danh"]Let $ ABC$ be a non-isosceles triangle. Let $ M,N,P$ denote the midpoints of $ BC,CA,AB$ respectively. $ D$ and $ E$ are two points on $ AM$ such that $ DA = DB$ and $ EA = EC$, $ BD$ meets $ CE$ at $ F$. Prove that $ A,N,F,P$ lie in a circle.[/quote]
An ugly solution :
Let $ AM$ cut $ (ABC)$ at X .
Then we have :
${ (YO,YD) = (Y0,YA) (\mod \pi})$
But because $ A,B$ is symmetric through $ OD$ so
$ (YO,YA)\equiv (BO,BD) (\mod \pi)$
So $ (YO,YD)\equiv (BO,BD) (\mod \pi)$
It follows that $ B,O,X,D$ is cyclic .
Similar $ C,O,E,X$ is cyclic.
So $ (CO,CF)\equiv (BO,BF) (\mod \pi)$
$ \Rightarrow F \in (BOC)$
$ (FO,FD) = (FE,FO)$ (1)
Easy to check that $ (FA,FO,FD,FE) = - 1$(2)
(1)(2) $ \Rightarrow FO$ and $ FA$ perpendicular .
So $ F\in (ANP)$

[url=http://www.mathlinks.ro/viewtopic.php?p=1116181#p1116181]USAMO 2008 Problem 2[/url]

We can solve this problem by the following method:
Let O be the intersection of PD and NE, hence O is the center of the circumcircle (ABC).
When P and O is inside triangle ABC:
We have $ \hat{BFC}\equal{}\hat{BDM}\plus{}\hat{CEM}\equal{}2\hat{BAM}\plus{}2\hat{CAM}\equal{}2\hat{BAC}\equal{}\hat{BOC}$
$ \equal{}> B,F,O,C$ are lie in a circle, hence $ \hat{DFO}\equal{}\hat{OCB}\equal{}\hat{OBC}\equal{}\hat{EFO}$ => FO is bisector of angle DFE.
$ \frac{DF}{DB}.\frac{MB}{MC}.\frac{EC}{EF}\equal{}1 \equal{}> \frac{FD}{FE}\equal{}\frac{BD}{CE}\equal{}\frac{AD}{AE}$
=> FA is the external angle bisector os angle DFE. Hence angle AFO is right.
So A,N,F,P are lie a circle, whose diameter is AO.

Let’s call M’ the common point of AM and (ABC), different of A, B’ and C’ the intersection of BD and CE with (ABC). Then let’s denote K, L, R and S the projections of B and C on AM, and of A on BD and CE respectively. We claim that BC=B’C’ and that A belongs to the angle bisector of $ \angle B'FC'$.
Since AD=BD we get $ \angle BAM'\equal{}\angle ABB'$ , while AE=CE gives $ \angle CAM'\equal{} \angle ACC'$, so $ \angle BAC \equal{} \angle B'M'C'$, with BC=B’C’ and BCB’C’ isosceles trapezoid.
Then it is easy to see $ \triangle ABR \equal{} \triangle BAK$, with AR=BK, similarly getting AS=CL, but BK=CL, M being the middle of BC, hence the distances from A to FB’ and FC’ are equal, and AF is actually the angle bisector of $ \angle B'FC'$, as claimed.
But from the isosceles trapezoid is follows that the angle bisector of the angle made by its diagonals is parallel to its parallel sides, BC’ and CB’, from here it follows that, if AF cuts again (ABC) at A”, F is the middle of AA”.
If AA’ is a diameter of the circle (ABC), $ A'A"\perp AA"$, hence $ OF\perp AF$, and F belongs to the circle with OA diameter, which passes through N and P.

Best regards,
sunken rock


3^x+4^x=5^x is the problem. he told us that x=0,2 are answers, but there are more or something.

I think $x=2$, because equation equivalent to $f(x)=(\frac{3}{5})^{x}+(\frac{4}{5})^{x}=1$, note $f(2)=1$ and $f(x)$ is monotone on $\mathbb{R}$.

@integral90: I removed your identical post from the Algebra subforum. Please, don't double post in the future.


inequalities geometry rectangle perimeter

Prove that if $ ar\minus{}2bq\plus{}pc\equal{}0$ and $ ac\minus{}b^2>0,$ then, $ pr\minus{}q^2\leq 0.$

This question seems correct intuitively, but here is a half-proof..

We have $ ar\plus{}pc \equal{} bq \plus{}bq$ and $ ac>b^2$, we want to show that $ pr<q^2$

Consider the counter argument
If $ pr>q^2 \longrightarrow acpr>b^2q^2 \longrightarrow (ar)(pr)>(bq)(bq)$

So we can reduce the equation and prove an equivalent one;
 If $ ab\equal{}c^2$ proove: $ a\plus{}b\ge c\plus{}c$

This is true because we know for a rectangle with given perimeter, its area maximised when it is a square.

Therefore it follows that if $ pr>q^2$ then $ ar\plus{}pc > 2bq$, but we need equality so $ pr<q^2$... :huh:

Not sure if it's correct.

[list]$ ar\plus{}pc\equal{}2bq\Longrightarrow ar^2c\plus{}prc^2\equal{}2bqrc$

$ ac\minus{}b^2>0\Longrightarrow ac>b^2$.[/list]
Hence,

[list]$ r^2ac\plus{}prc^2>b^2r^2\plus{}prc^2\Longrightarrow 2bqrc > b^2r^2\plus{}prc^2\geq 2brc\sqrt{pr}\text{ (AM\minus{}GM)}$[/list]
Therefore

[list]$ q^2\geq pr.$[/list]
But i think the statement should inform that $ p,r>0.$

$ 4b^2q^2\equal{} (ar \plus{} pc)^2 \ge 4arpc$, from which the solution follows immediately.


Let   $ x,y>0$ and $ x\plus{}y\equal{}1$ prove that:
\[ \frac{x}{\sqrt{1\minus{}x}}\plus{}\frac{y}{\sqrt{1\minus{}y}}\geq\sqrt2\]

[color=red][Moderator edit: LaTeX formula edited.][/color]

tabe f(x)=x/√(1-x)  mohadab ast pas ba jensen darim f(x)+f(y)≥2f((x+y)/2)

ke masale hale :wink:

dast karie hokm:
$ \frac {x}{\sqrt {1 \minus{} x}} \plus{} \frac {y}{\sqrt {1 \minus{} y}}\geq\sqrt2$
$ \frac x{\sqrt y} \plus{} \frac y{\sqrt x}\geq\sqrt2$
  [ham makhraj va tavan 2]
$ \frac {x^3 \plus{} y^3}{xy}\geq2$
Hokme jadid:
$ x^3 \plus{} y^3\geq2xy$
Va chon $ x,y > 0$  va $ x^2 \plus{} y^2\geq2xy\Rightarrow$
$ x^3 \plus{} y^3\geq2xy$

Mn 3 ruz tamam say mikardam in soale mozheko ba vasete ha hal konam!mokham dasht mipokid be khoda. Fek konam ino dadashe 5om ebtedayi mam hal kone.albate man hanuz haal nakardam tabe bekhunam vaghti ketabesho baz mikonam sar dard migiram!vali 2 sal vaght daram…!
Nokteye akhlaghi:hamishe ye rah hast ke rahat tar az unie ke behesh fekr mikonim ( :D

Saay konid LaTeX yad begirid. Az hamun link Latex balaye saretun ettelaat khubi be dast miyarid.

[quote="chakavak"]dast karie hokm:
x/√(1-x)+y/√(1-y)≥√2
x/√y+y/√x≥√2
  [ham makhraj va tavan 2]
(x^3+y^3)/xy≥2
Hokme jadid:
x^3+y^3≥2xy
Va chon x,y>0  & x^2+y^2≥2xy  =>
x^3+y^3≥2xy

Mn 3 ruz tamam say mikardam in soale mozheko ba vasete ha hal konam!mokham dasht mipokid be khoda. Fek konam ino dadashe 5om ebtedayi mam hal kone.albate man hanuz haal nakardam tabe bekhunam vaghti ketabesho baz mikonam sar dard migiram!vali 2 sal vaght daram…!
Nokteye akhlaghi:hamishe ye rah hast ke rahat tar az unie ke behesh fekr mikonim (kalemate ghassar!jelde 2) :D[/quote]

[quote="chakavak"]dast karie hokm:
$ \frac {x}{\sqrt {1 \minus{} x}} \plus{} \frac {y}{\sqrt {1 \minus{} y}}\geq\sqrt2$
$ \frac x{\sqrt y} \plus{} \frac y{\sqrt x}\geq\sqrt2$
  [ham makhraj va tavan 2]
$ \frac {x^3 \plus{} y^3}{xy}\geq2$
Hokme jadid:
$ x^3 \plus{} y^3\geq2xy$
Va chon $ x,y > 0$  va $ x^2 \plus{} y^2\geq2xy\Rightarrow$
$ x^3 \plus{} y^3\geq2xy$
[/quote]
$ \frac {x^3 \plus{} y^3}{xy}\geq2$ in namosavi ghalate!mesale naghzesham $ x \equal{} y \equal{} \frac {1}{2}$
age makhraj moshtarak begirim va be tavane 2 beresunim in namosavi bedast nemiad!in be dast miad:
$ \frac {x^3 \plus{} y^3 \plus{} 2xy\sqrt {xy}}{xy}\geq{2}$ ke ba tavajoh be inke $ x \plus{} y \equal{} 1$ moadel ast ba:
$ {(x \plus{} y)((x \plus{} y)^2 \minus{} 3xy) \plus{} 2xy\sqrt {xy}\geq{2xy}}\Leftrightarrow{1 \minus{} 5xy \plus{} 2xy\sqrt {xy}\geq0}$ke agar gharar dahim $ t \equal{} \sqrt {xy}$ kafi ast sabet konim:
$ 2t^3 \minus{} 5t^2 \plus{} 1\geq0$
$ \Leftrightarrow{2t^3 \minus{} t^2 \minus{} (4t^2 \minus{} 1)} \equal{} (2t \minus{} 1)(t^2 \minus{} 2t \minus{} 1) \equal{} (1 \minus{} 2t)((1 \minus{} t)(1 \plus{} t) \plus{} 2t)\geq0$
ke ba tavajoh be inke:$ 0\leq\sqrt{xy}\equal{}t\leq\frac{x\plus{}y}{2}\equal{}\frac {1}{2}$
namosaviye akhar niz badihi ast.

are ALI666 manham hamino bedast avordam vali hoseleye neveshtanesho inja nadashtam ke didam jensen kootahtare va oono neveshtam :)

va dar rabete ba chakavack bayad goft har masaleie arzeshe khase khodesho dare va nabayad daste kam gereftesh

kheyli az hamatun mamnunam.ba m bahador movafegham kheyli zud ghezavat kardam :blush: eshtebaham maskhare bud vali ba hamunam hal mishe tozihesh bashe vase bad :!: age in dafe ham ba word type konam omid hatami davam mikone :( .pas bye ta vaghti ke latex yad begiram :lol:  :P  :D

rahe halaye digeie vase in soal ke taze post shode :)

http://www.mathlinks.ro/viewtopic.php?p=1135885&sid=5478c9d083fd7cff746cc3ec28207242#1135885

be nazare man ke asanam soale asooni nis:

[url]http://www.mathlinks.ro/viewtopic.php?search_id=1160077791&t=160324[/url]

Manam ye rah hal daram ke bad nist :
farz konid $ x \leq y$ . (bedoone vared shodane eshkal be kolliat) dar in soorat agar $ x \equal{} \sin^{2}{a}$ va $ y \equal{} \cos^{2}{a}$ . mishe farz kard $ a$ too bazeye $ [0,\frac {\pi}{4}]$ hastesh . (ba hamoon farze $ x\leq y$)
hala hokm moadele ine ke neshoon bedim :
$ \sin^{3}{a} \plus{} \cos^{3}{a} \geq \sqrt2sin{a}cos{a}$
pas bayad neshoon bedim $ (\sin{a} \plus{} \cos{a})(1 \minus{} sin{a}cos{a})\geq \frac {\sqrt2}{2}\sin{2a}$
ama $ \sin{a} \plus{} cos{a} \equal{} \sqrt2\sin{(a \plus{} \frac {\pi}{4})}$ . vali $ a \plus{} \frac {\pi}{4}\geq 2a$ hast va darnatije :
$ \sin{(a \plus{} \frac {\pi}{4})}\geq \sin{2a}$
ham chenin $ 1\minus{}sin{a}cos{a}\geq \frac {1}{2}$ ke ba estefade az ina hokm bedast miad.

[quote="khashi70"]Manam ye rah hal daram ke bad nist :
farz konid $ x \leq y$ . (bedoone vared shodane eshkal be kolliat) dar in soorat agar $ x \equal{} \sin^{2}{a}$ va $ y \equal{} \cos^{2}{a}$ . mishe farz kard $ a$ too bazeye $ [0,\frac {\pi}{4}]$ hastesh . (ba hamoon farze $ x\leq y$)
hala hokm moadele ine ke neshoon bedim :
$ \sin^{3}{a} \plus{} \cos^{3}{a} \geq \sqrt2sin{a}cos{a}$
pas bayad neshoon bedim $ (\sin{a} \plus{} \cos{a})(1 \minus{} sin{a}cos{a})\geq \frac {\sqrt2}{2}\sin{2a}$
ama $ \sin{a} \plus{} cos{a} \equal{} \sqrt2\sin{(a \plus{} \frac {\pi}{4})}$ . vali $ a \plus{} \frac {\pi}{4}\geq 2a$ hast va darnatije :
$ \sin{(a \plus{} \frac {\pi}{4})}\geq \sin{2a}$
ham chenin $ 1 \minus{} sin{a}cos{a}\geq \frac {1}{2}$ ke ba estefade az ina hokm bedast miad.[/quote]
eyval...kheili rahe ghashangi bud...

ajab soali bude khabar nadashtim :lol:

rahe khodam az hame rahattare :wink: albatte age moshkeli nadashte bashe :rotfl:

$ 2xy\sqrt{xy}$be har hal mosbate pas chon $ {x^2}\plus{}{y^2}>2xy$

be har hal $ {x^3}\plus{}{y^3}\plus{}2xy\sqrt{xy}$ ham bozorgtar mosavie $ 2xy$ e.

chakavak hamoontori ke bacheha bala goftan to nemitooni az $ 2xy\sqrt{xy}$ sarfe nazar koni dar vaghe nemitooni begi $ x^{3}\plus{}y^{3}\geq 2xy$ chon in mesale naghz dare oonam $ x\equal{}y\equal{}\frac{1}{2}$ hast .

bayad sabet konim
\[ \frac x{\sqrt y} \plus{} \frac y{\sqrt x}\geq\sqrt2
\]

\[ \frac x{\sqrt y} \plus{} \frac y{\sqrt x} \equal{} \frac {x^2}{x\sqrt y} \plus{} \frac {y^2}{y\sqrt x}
\]
Lemma :
\[ \frac {a^2}{b} \plus{} \frac {c^2}{d}\geq\frac {(a \plus{} c)^2}{b \plus{} d}
\]
is true for all
\[ a,b,c,d\geq0
\]
ba tavajoh be Lem darim :
\[ \frac {x^2}{x\sqrt y} \plus{} \frac {y^2}{y\sqrt x}\geq\frac {(x \plus{} y)^2}{(x\sqrt y \plus{} y\sqrt x)} \equal{} \frac {1}{(x\sqrt y \plus{} y\sqrt x)}
\]
pas bayad sabet konim :
\[ \frac {1}{(x\sqrt y \plus{} y\sqrt x)}\geq\sqrt2
\]
Midanim Min in ebarat vaghti ast ke makhraj max bashad . midanim
AM-GM :
\[ {(x \plus{} y)}\geq{2}\sqrt xy
\]
==>
\[ \frac {1}{2}\geq\sqrt xy
\]
cooshy :
\[ {(x \plus{} y)}{(1 \plus{} 1)}\geq{(\sqrt x \plus{} \sqrt y)^2} \equal{} \equal{} > \sqrt2 \geq{(\sqrt x \plus{} \sqrt y)}
\]

\[ {(x\sqrt y \plus{} y\sqrt x)} \equal{} {\sqrt xy}{(\sqrt x \plus{} \sqrt y)}\leq \frac {1}{2} * \sqrt2 \equal{} \frac {\sqrt2}{2}
\]
==>
\[ \frac {1}{(x\sqrt y \plus{} y\sqrt x)}\geq\frac {1}{\frac {\sqrt2}{2}} \equal{} \sqrt2
\]
soale khoobi bood :)

age eshtebah nakonam male india 1992 bud soale rahatie rahe khodam be jensen hast vali yeki az bacheha ba miangin kheili ghasghang hal karde bud

in forum ham kam kam dare khub mishe


Given two concentric circles with radii $r$ and $R.$ Find the maximal number $n$ of circles touching both of these circles simultaneously and not intersecting each other. Show that this number $n$ lies between the numbers

$\begin{equation*}
\frac{3}{2}\cdot
\frac{\sqrt{R}+\sqrt{r}}{\sqrt{R}-\sqrt{r}}-1\text{\ \ \ \ \ \ \ \
\ \ and\ \ \ \ \ \ \ \ \ \ }\frac{63}{20}\cdot \frac{R+r}{R-r}.
\end{equation*}$

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)


Find an expression for 3/5 as a finite sum of distinct reciprocals of
positive integers. (For example: 2/7 = 1/7+1/8+1/56.)

Prove that any positive rational number can be so expressed.

Follow these steps to get your proof:

a) Prove that if $ 1/(n\plus{}1)<x<1/n$ for some $ n$, then the numerator in this sort of calculation must always decrease; conclude that $ x$ can be written as a finite sum of distinct numbers $ 1/k$.

b) Now prove the result for all $ x$ by using the divergence of $ \sum_{n\equal{}1}^{\infty}1/n$

[hide]Let the fraction be $ \frac {p_0}{q_0}$ and $ \frac {p_{n \plus{} 1}}{q_{n \plus{} 1}} \equal{} \frac {p_n}{q_n} \minus{} \frac {1}{k_n}$ where $ q_n \equal{} p_nk_n \minus{} r_n$ for some $ 0\leq r_n < p_n$. Taking $ q_{n\plus{}1}\equal{}q_nk_n$, we have $ p_{n \plus{} 1} \equal{} r_n$. Then $ \{p_n\}$ is nonnegative and decreasing. Therefore at some point $ p_n \equal{} 0$. Then $ \frac {p_0}{q_0} \equal{} \frac {1}{k_0} \plus{} ... \plus{} \frac {1}{k_{n \minus{} 1}}$ is the desired sum.
[/hide]


The mean of six numbers is 4 and the mean of four other numbers is 6. What is the mean of all ten numbers? Express your answer as a decimal rounded to the nearest tenth.

The sum of all the numbers is $ 2(6)(4)\equal{}48$, so the mean is $ \boxed{4.8}$.


calculus
integration
inequalities
limit
complex analysis

Using residue theorem compute:
$ \int_{\minus{}\infty}^{\plus{}\infty}{\frac{1}{x^6\minus{}x}}dx$

See diagram below.  Then consider the contour integral:

$ \mathop\oint\limits_{\text{Blue}}\frac {1}{z(z^5 - 1)}dz$

And consider the limit as the radius, $ \rho$, of the indentations around the two poles on the real axis goes to zero.  Since the poles there are simple, the integrals over these indented contours as the indentations go to zero are just $ - r_0\pi i$ and $ - r_1\pi i$ where $ r_0$ and $ r_1$ are the residues there.  Need to next evaluate the integral over the arc from $ 0$ to $ \pi$ as the radius, $ R$, goes to infinity.  That looks bounded by $ \frac {2\pi}{R^5 - 1}$ by the ML inequality so it goes to zero in the limit.  We're left then with:

$ \mathop\lim\limits_{\substack{\rho\to 0 \\ R\to\infty}}\left\{\mathop\oint\limits_{\text{Blue}} \frac {1}{z(z^5 - 1)}dz\right\} = \text{P.V.}\left\{\int_{ - \infty}^{\infty}\frac {dx} {x(x^5 - 1)}\right\} - r_0\pi i - r_1\pi i = 2\pi i(r_3 + r_4)$

The principal value looks like zero then to me.


trigonometry
analytic geometry
graphing lines
slope

There is a rectangular glass slab in the plane.It's refrative index $ \mu(x)$ varies with the distance from base $ x$ as
$ \boxed{\mu(x)\equal{}1\plus{}e^{4\cos^{2}x\plus{}19\sin x\plus{}x^{2}}}$
Find the slope to the equation of the path followed by light at a distance $ x$ from base .Assume the light entered the slab from air at an angle $ \alpha$ such that $ \sin\alpha\approx 1$.

[quote="Pardesi"]Find the slope to the equation of the path followed by light[/quote]

What do you mean by that?

well by that i meant if $ y\equal{}f(x)$ be the equation of the path followed by light assuming Newton's corposcular theory and that the size of objetcs corncerned is negligible as compared to the wavelength of light then find $ f'(x)$

[hide="hint"]for parallel glass slabs $ \mu\sin(i)$ is constant[/hide]

Very helpful hint. Is this approach correct?
$ i\equal{}\frac{\pi}{2}\minus{}\theta$ where $ \theta$ is the inclination.
Hence, $ \mu (x)\cos\theta$ is constant, or $ \mu (x)\sqrt{\frac{1}{1\plus{}f'(x)^{2}}}$ is constant.
Thus we have $ f'(x)\equal{}\frac{\mu (x)^{2}(1\plus{}\alpha)}{(1\plus{}e^{4})^{2}}$

yes this problem otherwise solved would be tidious . :)

(i) Is $ \;\alpha\;$ relative to x?

(ii) What could be the "otherwise solved would be tedious" way?

(iii) Isn't $ \; i\; \equal{}\;\theta,\;$ instead the complement?

[quote="Immanuel Bonfils"](i) Is $ \;\alpha\;$ relative to x?[/quote]
$ \alpha$ is the initial angle of incidence as stated in the original problem.

[quote="Immanuel Bonfils"](ii) What could be the "otherwise solved would be tedious" way?[/quote]
well u could do it like this
let at a particulat time $ t$ let the ray be at a distance $ x$ from the base then let the coordinates be $ (x,y)$ .now say sfter time $ dt$ the coordinates become $ (x\plus{}dx,y\plus{}dy)$.well now for a small length $ dy$ the thing behaves very nearly like rectangular slab.but we know for a rectangular slab displacement
$ d \equal{} t(1\minus{}\frac{\cos i}{\sqrt{\mu^{2}\minus{}\sin^{2}i}})\sin i$
using this we can arrive at a defferential equation
[quote="Immanuel Bonfils"](iii) Isn't $ \; i\; \equal{}\;\theta,\;$ instead the complement?[/quote]
no because $ \theta$ is the inclination with the $ X$ axis

We can derive this formula using Fermat Principle.

The Optical length is the lagrangian $ L\equal{}n(y(t)) \sqrt{y'(t)^2\plus{}1}$ that need to be minimized.

By Beltrami Identity.

$ L\minus{}y' \frac{\partial L}{\partial y'}\equal{}\frac{n(y)}{\sqrt{\left(y'\right)^2\plus{}1}}\equal{}c_1$

And we can formulate a more stronge law that can solve the problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=176819]There[/url]